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					<description><![CDATA[Click here to see more posts about NMR Only 15$ per sample for interpreting of your NMR spectrum Payment Upon Completion Send your results... Nuclear Magnetic Resonance (NMR) spectroscopy is an incredibly powerful tool for characterizing molecular structures. When submitting to the FDA or other regulatory agencies, full structural characterization by NMR provides crucial evidence [&#8230;]]]></description>
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<p>Nuclear Magnetic Resonance (NMR) spectroscopy is an incredibly powerful tool for characterizing molecular structures. When submitting to the FDA or other regulatory agencies, full structural characterization by NMR provides crucial evidence of compound identity. A combination of 1-dimensional and 2-dimensional NMR experiments are necessary for complete confidence in chemical structure.</p>



<span id="more-643"></span>



<p> This post will walk you through the steps to fully characterize a molecule by 1- and 2-dimensional NMR, including on how to perform NMR interpretation.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/Typical-Outline-of-NMR-Experiments-for-Structure-Elucidation.png" alt="Typical Outline Of NMR Experiments For Structure Elucidation" title="Typical Outline Of NMR Experiments For Structure Elucidation"/></figure>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/Thymidine-image.jpg" alt=""/></figure>



<h2 class="wp-block-heading" id="step-1-¹h-nmr">Step 1:&nbsp;¹H-NMR</h2>



<p>The first step in structural characterization is 1-dimensional proton ¹H-NMR. The chemical shift, multiplicity, coupling constants, and integration are all factors to consider when assigning protons. In this example, only three protons can be assigned by the proton spectrum alone: protons 3, 4, and 6.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/1H-edited2.jpg" alt=""/></figure>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Chemical Shift (ppm)</strong></td><td><strong>Multiplicity</strong></td><td><strong>Coupling Constant (Hz)</strong></td><td><strong>Integration</strong></td></tr><tr><td>11.256</td><td>s</td><td>&nbsp;–</td><td>1H</td></tr><tr><td>7.690</td><td>q</td><td>1.2</td><td>1H</td></tr><tr><td>6.163</td><td>t</td><td>6.8</td><td>1H</td></tr><tr><td>5.209</td><td>d</td><td>4.0</td><td>1H</td></tr><tr><td>4.999</td><td>t</td><td>5.2</td><td>1H</td></tr><tr><td>4.233</td><td>m</td><td>&nbsp;–</td><td>1H</td></tr><tr><td>3.754</td><td>q</td><td>3.7</td><td>1H</td></tr><tr><td>3.564</td><td>m</td><td>&nbsp;–</td><td>2H</td></tr><tr><td>2.068</td><td>m</td><td>&nbsp;–</td><td>2H</td></tr><tr><td>1.770</td><td>d</td><td>1.2</td><td>3H</td></tr></tbody></table></figure>



<p>To begin, let’s start with&nbsp;<strong>proton 3</strong>. Proton 3 is the only methyl group in the structure, and therefore must integrate to 3 protons. The only peak with an integration of 3 is the doublet at 1.770 ppm. The high field chemical shift supports this assignment. The peak is split into a doublet with a coupling constant of 1.2 Hz, reflecting the long-range coupling between protons 3 and 4, which also supports this assignment.</p>



<p>Protons that are coupled to each other should exhibit the same coupling constant. The long-range coupling constant observed for proton 3 (J=1.2 Hz, split into a doublet by proton 4) is reflected in the coupling constant for proton 4 (J=1.2 Hz, split into a quartet by proton 3). Therefore, the peak at 7.690 ppm must represent&nbsp;<strong>proton 4</strong>! The integration and chemical shift support the assignment, as proton 4 is the only aromatic proton in the structure.</p>



<p>There is only one singlet in the ¹H-NMR spectrum. The only proton that should show up as a singlet is&nbsp;<strong>proton 6</strong>, as it has no neighboring protons that would split the peak (the nearest proton is 5 bonds away!). The chemical shift of 11.256 ppm supports this assignment, as imide protons often show up far downfield. The peak also integrates to 1 proton, supporting the assignment.</p>



<p>The remaining protons are doublets, triplets, and multiplets that can be assigned by 2-dimensional COSY.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/Integration-Flowchart.png" alt=""/></figure>



<h2 class="wp-block-heading" id="step-2-¹h-¹h-cosy">Step 2:&nbsp;¹H-¹H COSY</h2>



<p>¹H-¹H Correlation Spectroscopy (COSY) shows the correlation between hydrogens which are coupled to each other in the ¹H NMR spectrum. The ¹H spectrum is plotted on both 2D axes. While 2-bond and 3-bond ¹H-¹H coupling is easily visible by COSY, long range coupling can also be observed with long acquisition times. The cross-peaks (not on the diagonal) that are symmetric to the diagonal show the COSY correlations. For example, protons 3 and 4 are coupled to each other, since they form a box pattern symmetric to the diagonal. This confirms assignments 3 and 4 made from the proton spectrum alone.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/Thymidine-COSY.png" alt=""/></figure>



<p><em><strong>Two types of COSY coupling:</strong>&nbsp;3-bond short range coupling between protons 7 and 8 (red) and 4-bond long range coupling between protons 3 and 4 (blue).</em></p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/cosy-zoom-edited.jpg" alt=""/></figure>



<p>My favorite way to analyze a COSY spectrum with many unassigned protons is to make a table of correlations, like the one seen here. Look at the table for any clear differences in correlation and begin there! In this example, all unassigned protons show one or two COSY correlations-except the proton at 4.233 ppm, which correlates to&nbsp;<em>three</em>other protons by COSY. The only proton expected to correlate with three nonequivalent protons is&nbsp;<strong>proton 9</strong>!</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Chemical Shift<br></strong><strong>(ppm)</strong></td><td><strong>COSY<br></strong><strong>correlations</strong></td><td><strong>Assignment</strong></td></tr><tr><td>11.256</td><td>none</td><td>6</td></tr><tr><td>7.690</td><td>4-3</td><td>4</td></tr><tr><td>6.163</td><td>one</td><td>?</td></tr><tr><td>5.209</td><td>one</td><td>?</td></tr><tr><td>4.999</td><td>one</td><td>?</td></tr><tr><td>4.233</td><td>three</td><td>?</td></tr><tr><td>3.754</td><td>two</td><td>?</td></tr><tr><td>3.564</td><td>two</td><td>?</td></tr><tr><td>2.068</td><td>two</td><td>?</td></tr><tr><td>1.770</td><td>3-4</td><td>3</td></tr></tbody></table></figure>



<p>Now that proton 9 has been assigned, the fun really begins. Thymidine’s structure suggests that proton 9 should couple protons 8, 10, and 11. Based on the COSY, proton 9 couples protons at 2.068 ppm (2H), 3.754 ppm (1H), and 5.209 ppm (1H). From this list, we can easily assign&nbsp;<strong>proton 8</strong>&nbsp;as the peak at 2.068 ppm based on its integration of 2 protons. To differentiate protons 10 and 11, take a look at our COSY table; 3.754 ppm shows two COSY correlations, while 5.209 ppm only shows one. Therefore, we can assign&nbsp;<strong>proton 10</strong>&nbsp;as 5.209 ppm and&nbsp;<strong>proton 11</strong>&nbsp;as 3.754 ppm.</p>



<p>Once proton 8 has been assigned, we can easily assign&nbsp;<strong>proton 7</strong>&nbsp;based on the remaining COSY correlation for proton 8. Proton 7’s peak at 6.163 ppm is split into a triplet by the two 8 protons, confirming the assignment.</p>



<p>All that remains are protons 12 and 13. We can assign&nbsp;<strong>proton 12</strong>&nbsp;(3.564 ppm) based on its integration of 2H and its COSY correlation to proton 11. The last remaining peak at 4.999 ppm must be&nbsp;<strong>proton 13</strong>; this is confirmed by COSY correlation with proton 12, triplet multiplicity based on splitting by proton 12, and integration of one proton.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/Thymidine-COSY-1H-Correlation-Flowchart.png" alt=""/></figure>



<p>Now we have a fully assigned ¹H-NMR spectrum! This spectrum will help us assign our carbons using HSQC and HMBC NMR spectroscopy.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/1h-side-black-edited-2-2.jpg" alt=""/></figure>



<h2 class="wp-block-heading" id="step-3-¹³c-nmr">Step 3: ¹³C-NMR</h2>



<p>Carbon NMR is a necessary step in full structural characterization. However, ¹³C-NMR alone does not provide enough information to assign the carbons in the molecule. The NMR spectrum below does confirm the number of carbons in the molecule; however, HSQC and HMBC (we will get to these soon!) are necessary to assign the carbons with confidence. Note that one of the carbons is hidden beneath the solvent signal but is clearly visible after zooming into that region.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/13c-with-zoom-edited.jpg" alt=""/></figure>



<h2 class="wp-block-heading" id="step-4-dept-45-90-and-135">Step 4: DEPT-45, 90, and 135</h2>



<p>Distortionless Enhancement of Polarization Transfer (DEPT) experiments help assign carbon peaks by determining the number of protons attached to each carbon. For very simple molecules, DEPT may be enough to partially or fully assign all carbons. In complex molecules, DEPT and HSQC together are useful for confirming both carbon and proton assignments. For example, the DEPT experiments below can only identify&nbsp;<strong>carbon 3</strong>-it is the only CH₃&nbsp;peak. I always go back and use DEPT to confirm the carbons I assigned by HSQC.</p>



<ul class="wp-block-list">
<li><strong>DEPT-45</strong>&nbsp;shows CH, CH₂, and CH₃&nbsp;carbons as positive peaks. Carbons with no protons are not visible.</li>



<li><strong>DEPT-90</strong>&nbsp;shows only CH peaks as positive peaks. Carbons with no protons, CH₂, and CH₃&nbsp;carbons are not visible.</li>



<li><strong>DEPT-135</strong>&nbsp;shows CH and CH₃&nbsp;carbons as positive peaks and CH₂&nbsp;carbons as negative peaks. Carbons with no protons are not visible.</li>
</ul>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/dept-overlay-no-labels-edited.jpg" alt=""/></figure>



<h2 class="wp-block-heading" id="step-5-¹h-¹³c-hsqc">Step 5: ¹H-¹³C HSQC</h2>



<p>¹H-¹³C Heteronuclear Single Quantum Coherence Spectroscopy (HSQC) shows which hydrogens are directly attached to which carbon atoms. The ¹H spectrum is shown on the horizontal axis and the ¹³C spectrum is shown on the vertical axis. The HSQC spectrum is most valuable when protons have already been assigned.</p>



<p>For example, HSQC shows a correlation between proton 4 and the carbon at 136.113 ppm; this carbon is now assigned as carbon 4.&nbsp;<strong>Carbons 3, 4, 7, 8, 9, 11, and 12</strong>&nbsp;are assigned by HSQC. Only 1-bond correlations are observed, so HSQC assignments are relatively straightforward. The DEPT experiments also confirm these assignments. HSQC is also useful in confirming proton assignments of nitrogen or oxygen-bound protons; they show no signal by HSQC. This further supports the assignments of protons 6, 10, and 13.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/HSQC-edited.jpg" alt=""/></figure>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/HSQC-Thymidine-Structure.png" alt=""/></figure>



<p><em>An example correlation between proton and carbon 4 is observed by HSQC.</em></p>



<h2 class="wp-block-heading" id="step-6-¹h-¹³c-hmbc">Step 6: ¹H-¹³C HMBC</h2>



<p>¹H-¹³C Heteronuclear Multiple Bond Correlation Spectroscopy (HMBC) shows the correlations between protons and carbons that are separated by multiple bonds. The ¹H spectrum is shown on the horizontal axis and the ¹³C spectrum is shown on the vertical axis. Correlated atoms are shown in blue and the connecting atoms are shown in red. Note that direct hydrogen-carbon bonds (1-bond correlations) are generally not seen. For example, hydrogen 4 shows correlations with carbons 1, 2, 3, 5, and 7, but not carbon 4.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/HMBC-Thymidine-Structure.png" alt=""/></figure>



<p><em>HMBC interactions between proton 4 and carbons 1, 2, 3, 5, and 7.</em></p>



<p>HMBC is incredibly useful for assigning carbons that have no protons attached. In this example, carbons 1, 2, and 5 have no protons attached.&nbsp;<strong>Carbon 1</strong>&nbsp;is assigned by HMBC interactions with protons 3, 4, and 6;&nbsp;<strong>carbon 2</strong>&nbsp;by interaction with protons 3, 4, 6, and 7; and&nbsp;<strong>carbon 5</strong>&nbsp;by interactions with protons 4 and 7 only. The chemical environment of carbon 5 suggests it would appear more downfield than carbon 1, which confirms these assignments.</p>



<figure class="wp-block-table"><table><tbody><tr><td>HMBC</td><td>Proton</td></tr><tr><td>Carbon</td><td>3</td><td>4</td><td>6</td><td>7</td></tr><tr><td>1</td><td>x</td><td>x</td><td>x</td><td></td></tr><tr><td>2</td><td>x</td><td>x</td><td>x</td><td>x</td></tr><tr><td>5</td><td></td><td>x</td><td></td><td>x</td></tr></tbody></table></figure>



<p>HMBC also confirms assignments that were based solely on the proton and COSY spectrum. For example, protons 10 and 13 are differentiated by HMBC; proton 10 is confirmed by interactions with&nbsp;<strong>carbons 8, 9, and 11</strong>, while proton 13 is confirmed by interactions with&nbsp;<strong>11 and 12</strong>. HMBC supports all proton and all carbon assignments, unambiguously confirming both the structure and analysis of thymidine.</p>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/HMBC-edited.jpg" alt=""/></figure>



<figure class="wp-block-image"><img decoding="async" src="https://emerypharma.com/wp-content/uploads/2018/03/hmbc-zoom-b-edited.jpg" alt=""/></figure>



<p>At Emery Pharma, we are experts in 1D and 2D NMR characterization and structure elucidation; in fact, 2D NMR projects are some of our favorites! We have supported numerous pharmaceutical companies in full NMR characterization for API submissions to regulatory agencies, as well as complete structure elucidation of impurities. We provide a fully annotated report with images similar to those seen here and support our results with high resolution mass spectrometry and elemental analysis.&nbsp;</p>



<p>Some nuclei rotate around their axis like electrons. In the presence of an external magnetic field, a rotating nucleus has only a small number of stable orientations. Nuclear magnetic resonance (NMR) occurs when a spinning core is excited from a lower energy orientation to a higher energy orientation in the presence of a magnetic field by absorbing enough electromagnetic radiation. Nuclear magnetic resonance spectroscopy involves measuring the amount of energy required to change spin nuclei from a stable orientation to a more unstable orientation in a magnetic field. Because spin-core nuclei change direction in a magnetic field at different frequencies, different frequencies of absorbing radiation are needed to change the orientation of spin-core nuclei. The frequency at which the absorption takes place is used for analysis and spectroscopy [1].</p>



<p>Nuclear magnetic resonance was first discovered independently in 1946 by Felix Bloch of Stanford University and Edward Parcel of Harvard University. They were able to show the absorption of electromagnetic radiation as a result of the transfer of the energy level of the nucleus in a strong magnetic field. The two physicists won the Nobel Prize in 1952 for their work. In the first five years after the discovery of the nuclear magnetic resonance method, chemists discovered that the molecular environment of objects affects the absorption of radiation by nuclei in the presence of a magnetic field, and this effect could be related to the structure of the molecule. Since then, the growth of magnetic resonance spectroscopy has been explosive and this method has had a significant effect on the development of organic chemistry, inorganic chemistry and biochemistry [2]. In 1999, a team of Canadian physicists developed a new method using the Beta Nuclear Magnetic Resonance Method, which is capable of demonstrating the magnetic and electrical properties of very thin layers and surfaces. BetaNMR methods are used in nanoscience. Be [3].</p>



<p>The magnitude of the spin angle motion in the nuclei is determined by the quantum number of the nucleus spin. Quantum number The core spin of any number can be integer or semi-integer. In 16 O and 12C non-spin nuclei, the quantum spin number of the nucleus is zero. Cores that are not spin and therefore do not have the magnitude of the spin angle motion can not be detected by NMR spectroscopy. Spin-core cores with spherical charge distribution have a spin quantum number of 1/2. Examples of these nuclei include 13C, 19F, 3H, 15N, 31P and 1H, which have a quantum number of 1/2 and a magnetic moment. In order for a nucleus in a magnetic field to absorb a large amount of electromagnetic radiation, it must have a high frequency in the sample and also have a relatively large magnetic moment (µ). Cores that have both properties in question include 1H, 19F, 21P. Most NMR measurements are usually performed for 1 h. Measurements of other nuclei are often performed using signal amplification methods to observe the spectrum. Usually, among the nuclei with low relative frequency that show the magnetic resonance of the nucleus, 12 C, 15N, 16O are the most important for chemists. The magnetic resonance method of the hydrogen nucleus (1H), which is used more than other nuclei, has a magnetic torque of about 79.2 برای. It will be magnetic. For other cores used for nuclear magnetic resonance spectroscopy, the magnetic torque for 21P, 19F 12C is 6873.2, 1305.1 and 0.7022, respectively [4]. In most cases, the sensitivity of non-proton core magnetic resonance devices, such as 12C, etc., is lower than that of HNMR. Also, in most compounds, the natural abundance of non-proton magnetic nuclei is significantly lower than that of protons. This factor causes the NMR spectra of non-proton nuclei to have a relatively low noise signal. The peaks of these spectra are small, and often the spectrum cannot be determined if the same device used for proton nucleus (PMR) NMR is used. Due to the low signal-to-noise ratio in these cases, most devices designed to record the NMR spectra of non-proton nuclei use multiple traverses with signal averaging techniques. The most common devices for spectral peak extraction use the Fourier transform. Fourier transformers are also used to prepare PMR spectra of dilute solutions and complex molecules, such as proteins, in which the amount of a particular proton in the molecule is small. The difference between PMR spectra and other NMR spectra is in the range of chemical displacement. The chemical displacement range for PMR is 10PPM in most cases. While for the 12C core the chemical displacement is up to about 200PPM, for the 19F and 21P spectra it is 300 and 400PPM, respectively. In NMR methods, the units used are usually time (seconds), angle (degrees or radians), temperature (Kelvin), magnetic field strength (Tesla, T), energy (joules), vibration (rpm) and power ( Watts) is. [5] Components of the NMR Device The important components of an NMR spectrometer are shown schematically in Figure (1). A brief description of each component is given below.</p>
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					<description><![CDATA[Click here to see more posts about NMR Only 15$ per sample for interpreting of your NMR spectrum Payment Upon Completion Send your results... Over the past fifty years nuclear magnetic resonance spectroscopy, commonly referred to as nmr, has become the preeminent technique for determining the structure of organic compounds. Of all the spectroscopic methods, [&#8230;]]]></description>
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<p>Over the past fifty years nuclear magnetic resonance spectroscopy, commonly referred to as nmr, has become the preeminent technique for determining the structure of organic compounds. Of all the spectroscopic methods, it is the only one for which a complete analysis and interpretation of the entire spectrum is normally expected. Although larger amounts of sample are needed than for mass spectroscopy, nmr is non-destructive, and with modern instruments good data may be obtained from samples weighing less than a milligram.&nbsp;<strong>To be successful in using nmr as an analytical tool, it is necessary to understand the physical principles on which the methods are based</strong>.</p>



<span id="more-640"></span>



<p>The nuclei of many elemental isotopes have a characteristic spin (<strong>I</strong>). Some nuclei have integral spins (e.g. I = 1, 2, 3 &#8230;.), some have fractional spins (e.g. I = 1/2, 3/2, 5/2 &#8230;.), and a few have no spin, I = 0 (e.g.&nbsp;<sup>12</sup>C,&nbsp;<sup>16</sup>O,&nbsp;<sup>32</sup>S, &#8230;.). Isotopes of particular interest and use to organic chemists are&nbsp;<sup>1</sup>H,&nbsp;<sup>13</sup>C,&nbsp;<sup>19</sup>F and&nbsp;<sup>31</sup>P, all of which have I = 1/2. Since the analysis of this spin state is fairly straightforward, our discussion of nmr will be limited to these and other I = 1/2 nuclei.</p>



<figure class="wp-block-table"><table><tbody><tr><th>For a table of nuclear spin characteristics&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr11" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<p><strong>The following features lead to the nmr phenomenon:</strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>1.</strong>&nbsp;A spinning charge generates a magnetic field, as shown by the animation on the right.<br>The resulting spin-magnet has a magnetic moment (<strong>μ</strong>) proportional to the spin.</td><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nucspin1.gif"></td></tr><tr><td><strong>2.</strong>&nbsp;In the presence of an external magnetic field (<strong>B<sub>0</sub></strong>), two spin states exist,&nbsp;<strong>+1/2</strong>&nbsp;and&nbsp;<strong>-1/2</strong>.<br>The magnetic moment of the lower energy +1/2 state is aligned with the external field, but that of the higher energy -1/2 spin state is opposed to the external field. Note that the arrow representing the external field points North.</td><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nucspin2.gif"></td></tr><tr><td><strong>3.</strong>&nbsp;The difference in energy between the two spin states is dependent on the external magnetic field strength, and is always very small. The following diagram illustrates that the two spin states have the same energy when the external field is zero, but diverge as the field increases. At a field equal to B<sub>x</sub>&nbsp;a formula for the energy difference is given (remember I = 1/2 and μ is the magnetic moment of the nucleus in the field).</td></tr><tr><th><a href="javascript:chg1();"></a></th></tr><tr><td>Strong magnetic fields are necessary for nmr spectroscopy. The international unit for magnetic flux is the tesla (<strong>T</strong>). The earth&#8217;s magnetic field is not constant, but is approximately 10<sup>-4</sup>&nbsp;T at ground level. Modern nmr spectrometers use powerful magnets having fields of 1 to 20 T. Even with these high fields, the energy difference between the two spin states is less than 0.1 cal/mole. To put this in perspective, recall that infrared transitions involve 1 to 10 kcal/mole and electronic transitions are nearly 100 time greater.<br>For nmr purposes, this small energy difference (ΔE) is usually given as a frequency in units of MHz (10<sup>6</sup>&nbsp;Hz), ranging from 20 to 900 Mz, depending on the magnetic field strength and the specific nucleus being studied. Irradiation of a sample with radio frequency (rf) energy corresponding exactly to the spin state separation of a specific set of nuclei will cause excitation of those nuclei in the +1/2 state to the higher -1/2 spin state. Note that this electromagnetic radiation falls in the&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/UV-Vis/spectrum.htm#uv2" target="_blank" rel="noopener">radio and television broadcast spectrum</a>. Nmr spectroscopy is therefore the energetically mildest probe used to examine the structure of molecules.&nbsp;<br>The nucleus of a hydrogen atom (the proton) has a magnetic moment μ = 2.7927, and has been studied more than any other nucleus.&nbsp;The previous diagram may be changed to display energy differences for the proton spin states (as frequencies) by mouse clicking anywhere within it.</td></tr><tr><td><strong>4.</strong>&nbsp;For spin 1/2 nuclei the energy difference between the two spin states at a given magnetic field strength will be proportional to their magnetic moments. For the four common nuclei noted above, the magnetic moments are:&nbsp;<sup>1</sup>H μ = 2.7927,&nbsp;<sup>19</sup>F μ = 2.6273,&nbsp;<sup>31</sup>P μ = 1.1305 &amp;&nbsp;<sup>13</sup>C μ = 0.7022. These moments are in nuclear magnetons, which are 5.05078•10<sup>-27</sup>&nbsp;JT<sup>-1</sup>. The following diagram gives the approximate frequencies that correspond to the spin state energy separations for each of these nuclei in an external magnetic field of 2.35 T. The formula in the colored box shows the direct correlation of frequency (energy difference) with magnetic moment (h = Planck&#8217;s constant = 6.626069•10<sup>-34</sup>&nbsp;Js).</td></tr><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nucfreq1.gif"></th></tr></tbody></table></figure>



<p><strong>&nbsp; &nbsp; &nbsp; 2. Proton NMR Spectroscopy</strong><br>This important and well-established application of nuclear magnetic resonance will serve to illustrate some of the novel aspects of this method. To begin with, the nmr spectrometer must be tuned to a specific nucleus, in this case the proton. The actual procedure for obtaining the spectrum varies, but the simplest is referred to as the&nbsp;<strong>continuous wave</strong>&nbsp;(CW) method. A typical CW-spectrometer is shown in the following diagram. A solution of the sample in a uniform 5 mm glass tube is oriented between the poles of a powerful magnet, and is spun to average any magnetic field variations, as well as tube imperfections. Radio frequency radiation of appropriate energy is broadcast into the sample from an antenna coil (colored red). A receiver coil surrounds the sample tube, and emission of absorbed rf energy is monitored by dedicated electronic devices and a computer. An nmr spectrum is acquired by varying or sweeping the magnetic field over a small range while observing the rf signal from the sample. An equally effective technique is to vary the frequency of the rf radiation while holding the external field constant.</p>



<figure class="wp-block-table"><table><tbody><tr><th>For a description of the pulse Fourier transform technique, preferred by most spectroscopists over the older CW method,&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#pulse" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/spctrmtr.gif" alt=""/></figure>



<p>As an example, consider a sample of water in a 2.3487 T external magnetic field, irradiated by 100 MHz radiation. If the magnetic field is smoothly increased to 2.3488 T, the hydrogen nuclei of the water molecules will at some point absorb rf energy and a resonance signal will appear. An animation showing this may be activated by clicking the&nbsp;<strong>Show Field Sweep</strong>&nbsp;button. The field sweep will be repeated three times, and the resulting resonance trace is colored red. For visibility, the water proton signal displayed in the animation is much broader than it would be in an actual experiment.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/sweep1.gif" alt=""/></figure>



<p>Since protons all have the same magnetic moment, we might expect all hydrogen atoms to give resonance signals at the same field / frequency values. Fortunately for chemistry applications, this is not true. By clicking the&nbsp;<strong>Show Different Protons</strong>&nbsp;button under the diagram, a number of representative proton signals will be displayed over the same magnetic field range. It is not possible, of course, to examine isolated protons in the spectrometer described above; but from independent measurement and calculation it has been determined that a naked proton would resonate at a lower field strength than the nuclei of covalently bonded hydrogens. With the exception of water, chloroform and sulfuric acid, which are examined as liquids, all the other compounds are measured as gases.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/shield1.gif" alt=""/></figure>



<p><strong>Why should the proton nuclei in different compounds behave differently in the nmr experiment ?</strong>&nbsp;<br>The answer to this question lies with the electron(s) surrounding the proton in covalent compounds and ions. Since electrons are charged particles, they move in response to the external magnetic field (B<sub>o</sub>) so as to generate a secondary field that opposes the much stronger applied field. This secondary field&nbsp;<strong>shields</strong>&nbsp;the nucleus from the applied field, so B<sub>o</sub>&nbsp;must be increased in order to achieve resonance (absorption of rf energy). As illustrated in the drawing on the right, B<sub>o</sub>&nbsp;must be increased to compensate for the induced shielding field. In the upper diagram, those compounds that give resonance signals at the higher field side of the diagram (CH<sub>4</sub>, HCl, HBr and HI) have proton nuclei that are more shielded than those on the lower field (left) side of the diagram.&nbsp;<br>The magnetic field range displayed in the above diagram is very small compared with the actual field strength (only about 0.0042%). It is customary to refer to small increments such as this in units of&nbsp;<strong>parts per million</strong>&nbsp;(ppm). The difference between 2.3487 T and 2.3488 T is therefore about 42 ppm. Instead of designating a range of nmr signals in terms of magnetic field differences (as above), it is more common to use a frequency scale, even though the spectrometer may operate by sweeping the magnetic field. Using this terminology, we would find that at 2.34 T the proton signals shown above extend over a 4,200 Hz range (for a 100 MHz rf frequency, 42 ppm is 4,200 Hz). Most organic compounds exhibit proton resonances that fall within a 12 ppm range (the shaded area), and it is therefore necessary to use very sensitive and precise spectrometers to resolve structurally distinct sets of hydrogen atoms within this narrow range.&nbsp;In this respect it might be noted that the detection of a part-per-million difference is equivalent to detecting a 1 millimeter difference in distances of 1 kilometer.</p>



<h4 class="wp-block-heading" id="chemical-shift">Chemical Shift</h4>



<p>Unlike infrared and uv-visible spectroscopy, where absorption peaks are uniquely located by a frequency or wavelength, the location of different nmr resonance signals is dependent on both the external magnetic field strength and the rf frequency. Since no two magnets will have exactly the same field, resonance frequencies will vary accordingly and an alternative method for characterizing and specifying the location of nmr signals is needed. This problem is illustrated by the eleven different compounds shown in the following diagram. Although the eleven resonance signals are distinct and well separated, an unambiguous numerical locator cannot be directly assigned to each.</p>



<figure class="wp-block-image"><a href="javascript:chg4();"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nmrtotl1.gif" alt=""/></a></figure>



<p>One method of solving this problem is to report the location of an nmr signal in a spectrum relative to a reference signal from a standard compound added to the sample. Such a reference standard should be chemically unreactive, and easily removed from the sample after the measurement. Also, it should give a single sharp nmr signal that does not interfere with the resonances normally observed for organic compounds.&nbsp;<strong>Tetramethylsilane</strong>, (CH<sub>3</sub>)<sub>4</sub>Si, usually referred to as&nbsp;<strong>TMS</strong>, meets all these characteristics, and has become the reference compound of choice for proton and carbon nmr.<br>Since the separation (or dispersion) of nmr signals is magnetic field dependent, one additional step must be taken in order to provide an unambiguous location unit.&nbsp;This is illustrated for the acetone, methylene chloride and benzene signals by clicking on the previous diagram. To correct these frequency differences for their field dependence, we divide them by the spectrometer frequency (100 or 500 MHz in the example),&nbsp;as shown in a new display by again clicking on the diagram. The resulting number would be very small, since we are dividing Hz by MHz, so it is multiplied by a million, as shown by the formula in the blue shaded box. Note that ν<sub>ref</sub>&nbsp;is the resonant frequency of the reference signal and ν<sub>samp</sub>&nbsp;is the frequency of the sample signal. This operation gives a locator number called the&nbsp;<strong>Chemical Shift</strong>, having units of parts-per-million (ppm), and designated by the symbol&nbsp;<strong>δ</strong>&nbsp;&nbsp;&nbsp;Chemical shifts for all the compounds in the original display will be presented by a third click on the diagram.</p>



<p>The compounds referred to above share two common characteristics:</p>



<p><strong>•&nbsp;</strong>The hydrogen atoms in a given molecule are all&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/suppmnt1.htm#nom1" target="_blank" rel="noopener">structurally equivalent</a>, averaged for fast conformational equilibria.&nbsp;<br><strong>•&nbsp;</strong>The compounds are all liquids, save for neopentane which boils at 9 °C and is a liquid in an ice bath.</p>



<p>The first feature assures that each compound gives a single sharp resonance signal. The second allows the pure (neat) substance to be poured into a sample tube and examined in a nmr spectrometer. In order to take the nmr spectra of a solid, it is usually necessary to dissolve it in a suitable solvent. Early studies used carbon tetrachloride for this purpose, since it has no hydrogen that could introduce an interfering signal. Unfortunately, CCl<sub>4</sub>&nbsp;is a poor solvent for many polar compounds and is also toxic. Deuterium labeled compounds, such as deuterium oxide (D<sub>2</sub>O), chloroform-d (DCCl<sub>3</sub>), benzene-d<sub>6</sub>(C<sub>6</sub>D<sub>6</sub>), acetone-d<sub>6</sub>&nbsp;(CD<sub>3</sub>COCD<sub>3</sub>) and DMSO-d<sub>6</sub>&nbsp;(CD<sub>3</sub>SOCD<sub>3</sub>) are now widely used as nmr solvents. Since the deuterium isotope of hydrogen has a different magnetic moment and spin, it is invisible in a spectrometer tuned to protons.</p>



<figure class="wp-block-table"><table><tbody><tr><th>For the properties of some common nmr solvents&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmrsol" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<p>From the previous discussion and examples we may deduce that one factor contributing to chemical shift differences in proton resonance is the&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/intro2.htm#strc3b" target="_blank" rel="noopener"><strong>inductive effect</strong></a>. If the electron density about a proton nucleus is relatively high, the induced field due to electron motions will be stronger than if the electron density is relatively low. The shielding effect in such high electron density cases will therefore be larger, and a higher external field (B<sub>o</sub>) will be needed for the rf energy to excite the nuclear spin. Since silicon is less electronegative than carbon, the electron density about the methyl hydrogens in tetramethylsilane is expected to be greater than the electron density about the methyl hydrogens in neopentane (2,2-dimethylpropane), and the characteristic resonance signal from the silane derivative does indeed lie at a higher magnetic field. Such nuclei are said to be&nbsp;<strong>shielded</strong>. Elements that are more electronegative than carbon should exert an opposite effect (reduce the electron density); and, as the data in the following tables show, methyl groups bonded to such elements display lower field signals (they are&nbsp;<strong>deshielded</strong>). The deshielding effect of electron withdrawing groups is roughly proportional to their electronegativity, as shown by the left table. Furthermore, if more than one such group is present, the deshielding is additive (table on the right), and proton resonance is shifted even further downfield.</p>



<figure class="wp-block-table"><table><tbody><tr><th>Proton Chemical Shifts of Methyl DerivativesCompound(CH<sub>3</sub>)<sub>4</sub>C(CH<sub>3</sub>)<sub>3</sub>N(CH<sub>3</sub>)<sub>2</sub>OCH<sub>3</sub>Fδ0.92.13.24.1Compound(CH<sub>3</sub>)<sub>4</sub>Si(CH<sub>3</sub>)<sub>3</sub>P(CH<sub>3</sub>)<sub>2</sub>SCH<sub>3</sub>Clδ0.00.92.13.0</th><th></th><th>Proton Chemical Shifts (ppm)Cpd. / Sub.X=ClX=BrX=IX=ORX=SR<strong>CH<sub>3</sub>X</strong>3.02.72.13.12.1<strong>CH<sub>2</sub>X<sub>2</sub></strong>5.35.03.94.43.7<strong>CHX<sub>3</sub></strong>7.36.84.95.0&nbsp;</th></tr></tbody></table></figure>



<p>The general distribution of proton chemical shifts associated with different functional groups is summarized in the following chart. Bear in mind that these ranges are approximate, and may not encompass all compounds of a given class. Note also that the ranges specified for OH and NH protons (colored orange) are wider than those for most CH protons. This is due to hydrogen bonding variations at different sample concentrations.</p>



<figure class="wp-block-table"><table><tbody><tr><th>Proton Chemical Shift Ranges*</th></tr><tr><th>Low Field<br>Region</th><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/hnmr1.gif"></td><th>High Field<br>Region</th></tr><tr><td></td><td>&nbsp;&nbsp;<strong>*</strong>&nbsp;For samples in CDCl<sub>3</sub>&nbsp;solution. The δ scale is relative to TMS at δ = 0.</td><td></td></tr></tbody></table></figure>



<p>To make use of a calculator that predicts aliphatic proton chemical shifts&nbsp;<a href="http://www.colby.edu/chemistry/NMR/H1pred.html" target="_blank" rel="noopener">Click Here</a>. This application was developed at Colby College.</p>



<h4 class="wp-block-heading" id="signal-strength">Signal Strength</h4>



<p>The magnitude or intensity of nmr resonance signals is displayed along the vertical axis of a spectrum, and is proportional to the molar concentration of the sample. Thus, a small or dilute sample will give a weak signal, and doubling or tripling the sample concentration increases the signal strength proportionally. If we take the nmr spectrum of equal molar amounts of benzene and cyclohexane in carbon tetrachloride solution, the resonance signal from cyclohexane will be twice as intense as that from benzene because cyclohexane has twice as many hydrogens per molecule. This is an important relationship when samples incorporating two or more different sets of hydrogen atoms are examined, since it allows the ratio of hydrogen atoms in each distinct set to be determined. To this end it is necessary to measure the relative strength as well as the chemical shift of the resonance signals that comprise an nmr spectrum. Two common methods of displaying the integrated intensities associated with a spectrum are illustrated by the following examples. In the three spectra in the top row, a horizontal integrator trace (light green) rises as it crosses each signal by a distance proportional to the signal strength. Alternatively, an arbitrary number, selected by the instrument&#8217;s computer to reflect the signal strength, is printed below each resonance peak, as shown in the three spectra in the lower row. From the relative intensities shown here, together with the previously noted chemical shift correlations, the reader should be able to assign the signals in these spectra to the set of hydrogens that generates each.&nbsp;If you click on one of the spectrum signals (colored red) or on hydrogen atom(s) in the structural formulas the spectrum will be enlarged and the relationship will be colored blue.<br><strong>Hint:</strong>&nbsp;When evaluating relative signal strengths, it is useful to set the smallest integration to unity and convert the other values proportionally.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nmrex11.gif" alt=""/></figure>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nmrex21.gif" alt=""/></figure>



<h4 class="wp-block-heading" id="hydroxyl-proton-exchange-and-the-influence-of-hydrogen-bonding">Hydroxyl Proton Exchange and the Influence of Hydrogen Bonding</h4>



<p>The last two compounds in the lower row are alcohols. The OH proton signal is seen at 2.37 δ in 2-methyl-3-butyne-2-ol, and at 3.87 δ in 4-hydroxy-4-methyl-2-pentanone, illustrating the wide range over which this chemical shift may be found. A six-membered ring intramolecular hydrogen bond in the latter compound is in part responsible for its low field shift, and will be shown by clicking on the hydroxyl proton. We can take advantage of&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alcohol1.htm#alcrx1" target="_blank" rel="noopener">rapid OH exchange</a>&nbsp;with the deuterium of heavy water to assign hydroxyl proton resonance signals . As shown in the following equation, this removes the hydroxyl proton from the sample and its resonance signal in the nmr spectrum disappears. Experimentally, one simply adds a drop of heavy water to a chloroform-d solution of the compound and runs the spectrum again. The result of this exchange is displayed below.</p>



<figure class="wp-block-table"><table><tbody><tr><td>R-O-H&nbsp;&nbsp; + &nbsp;&nbsp;D<sub>2</sub>O &nbsp;&nbsp;<img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/Images/arroweq3.gif">&nbsp;&nbsp; R-O-D&nbsp;&nbsp; + &nbsp;&nbsp;D-O-H</td></tr><tr><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/deutalc1.gif"></td></tr></tbody></table></figure>



<p><strong>Hydrogen bonding shifts the resonance signal of a proton to lower field ( higher frequency ).</strong>&nbsp;Numerous experimental observations support this statement, and a few of these will be described here.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>i) &nbsp;&nbsp;</strong>The chemical shift of the hydroxyl hydrogen of an alcohol varies with concentration. Very dilute solutions of 2-methyl-2-propanol, (CH<sub>3</sub>)<sub>3</sub>COH, in carbon tetrachloride solution display a hydroxyl resonance signal having a relatively high-field chemical shift (&lt; 1.0 δ ). In concentrated solution this signal shifts to a lower field, usually near 2.5 δ.</td></tr><tr><td><strong>ii) &nbsp;&nbsp;</strong>The more acidic hydroxyl group of phenol generates a lower-field resonance signal, which shows a similar concentration dependence to that of alcohols. OH resonance signals for different percent concentrations of phenol in chloroform-d are shown in the following diagram (C-H signals are not shown).</td></tr><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/phenol.gif"></th></tr><tr><td><strong>iii) &nbsp;&nbsp;</strong>Because of their favored&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm#crbacd4a" target="_blank" rel="noopener">hydrogen-bonded dimeric association</a>, the hydroxyl proton of carboxylic acids displays a resonance signal significantly down-field of other functions. For a typical acid it appears from 10.0 to 13.0 δ and is often broader than other signals. The spectra shown below for chloroacetic acid (left) and 3,5-dimethylbenzoic acid (right) are examples.</td></tr><tr><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/clacetac.gif"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/2mebzacd.gif"></td></tr><tr><td><strong>iv) &nbsp;&nbsp;</strong>Intramolecular hydrogen bonds, especially those defining a six-membered ring, generally display a very low-field proton resonance. The case of 4-hydroxypent-3-ene-2-one (the enol tautomer of 2,4-pentanedione) not only illustrates this characteristic, but also provides an instructive example of the sensitivity of the nmr experiment to dynamic change. In the nmr spectrum of the pure liquid, sharp signals from both the keto and enol tautomers are seen, their mole ratio being 4&nbsp;<strong>:</strong>&nbsp;21 (keto tautomer signals are colored purple). Chemical shift assignments for these signals are shown in the shaded box above the spectrum. The chemical shift of the hydrogen-bonded hydroxyl proton is δ 14.5, exceptionally downfield. We conclude, therefore, that the rate at which these tautomers interconvert is slow compared with the inherent time scale of nmr spectroscopy.</td></tr><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/acac2.gif"></th></tr><tr><td>Two structurally equivalent structures may be drawn for the enol tautomer (in magenta brackets). If these enols were slow to interconvert, we would expect to see two methyl resonance signals associated with each, one from the allylic methyl and one from the methyl ketone. Since only one strong methyl signal is observed, we must conclude that the interconversion of the enols is very fast-so fast that the nmr experiment detects only a single time-averaged methyl group (50% α-keto and 50% allyl).</td></tr></tbody></table></figure>



<p>Although hydroxyl protons have been the focus of this discussion, it should be noted that corresponding N-H groups in amines and amides also exhibit hydrogen bonding nmr shifts, although to a lesser degree. Furthermore, OH and NH groups can undergo rapid proton exchange with each other; so if two or more such groups are present in a molecule, the nmr spectrum will show a single signal at an average chemical shift. For example, 2-hydroxy-2-methylpropanoic acid, (CH<sub>3</sub>)<sub>2</sub>C(OH)CO<sub>2</sub>H, displays a strong methyl signal at δ 1.5 and a 1/3 weaker and broader OH signal at δ 7.3 ppm. Note that the average of the expected carboxylic acid signal (ca. 12 ) and the alcohol signal (ca. 2 ) is 7. Rapid exchange of these hydrogens with heavy water, as noted above, would cause the low field signal to disappear.</p>



<figure class="wp-block-table"><table><tbody><tr><th>For additional information about the influence of hydrogen bonding&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr15" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<h4 class="wp-block-heading" id="π-electron-functions">π-Electron Functions</h4>



<p>An examination of the proton chemical shift chart (<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr1.htm#nmr3bb" target="_blank" rel="noopener">above</a>) makes it clear that the inductive effect of substituents cannot account for all the differences in proton signals. In particular the low field resonance of hydrogens bonded to double bond or aromatic ring carbons is puzzling, as is the very low field signal from aldehyde hydrogens. The hydrogen atom of a terminal alkyne, in contrast, appears at a relatively higher field. All these anomalous cases seem to involve hydrogens bonded to pi-electron systems, and an explanation may be found in the way these pi-electrons interact with the applied magnetic field.<br>Pi-electrons are more polarizable than are sigma-bond electrons, as addition reactions of electrophilic reagents to alkenes testify. Therefore, we should not be surprised to find that field induced pi-electron movement produces strong secondary fields that perturb nearby nuclei. The pi-electrons associated with a benzene ring provide a striking example of this phenomenon, as shown below. The electron cloud above and below the plane of the ring circulates in reaction to the external field so as to generate an opposing field at the center of the ring and a supporting field at the edge of the ring. This kind of spatial variation is called&nbsp;<strong>anisotropy</strong>, and it is common to nonspherical distributions of electrons, as are found in all the functions mentioned above. Regions in which the induced field supports or adds to the external field are said to be&nbsp;<strong>deshielded</strong>, because a slightly weaker external field will bring about resonance for nuclei in such areas. However, regions in which the induced field opposes the external field are termed&nbsp;<strong>shielded</strong>&nbsp;because an increase in the applied field is needed for resonance. Shielded regions are designated by a&nbsp;<strong>plus sign</strong>, and deshielded regions by a&nbsp;<strong>negative sign</strong>.&nbsp;<br>The anisotropy of some important unsaturated functions will be displayed by clicking on the benzene diagram below. Note that the anisotropy about the triple bond nicely accounts for the relatively high field chemical shift of ethynyl hydrogens. The shielding &amp; deshielding regions about the carbonyl group have been described in two ways, which alternate in the display.</p>



<figure class="wp-block-image"><a href="javascript:chg5();"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/benzene.gif" alt=""/></a></figure>



<figure class="wp-block-table"><table><tbody><tr><th>For additional examples of chemical shift variation near strongly anisotropic groups&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr13" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<p>Sigma bonding electrons also have a less pronounced, but observable, anisotropic influence on nearby nuclei. This is seen in the small deshielding shift that occurs in the series CH<sub>3</sub>–R, R–CH<sub>2</sub>–R, R<sub>3</sub>CH; as well as the deshielding of equatorial versus axial protons on a fixed cyclohexane ring.</p>



<h4 class="wp-block-heading" id="solvent-effects">Solvent Effects</h4>



<p>Chloroform-d (CDCl<sub>3</sub>) is the most common solvent for nmr measurements, thanks to its good solubilizing character and relative unreactive nature ( except for 1º and 2º-amines). As noted earlier, other deuterium labeled compounds, such as deuterium oxide (D<sub>2</sub>O), benzene-d6 (C<sub>6</sub>D<sub>6</sub>), acetone-d6 (CD<sub>3</sub>COCD<sub>3</sub>) and DMSO-d6 (CD<sub>3</sub>SOCD<sub>3</sub>) are also available for use as nmr solvents. Because some of these solvents have π-electron functions and/or may serve as hydrogen bonding partners, the chemical shifts of different groups of protons may change depending on the solvent being used. The following table gives a few examples, obtained with dilute solutions at 300 MHz.</p>



<figure class="wp-block-table"><table><tbody><tr><td>SolventCompound</td><th>CDCl<sub>3</sub></th><th>C<sub>6</sub>D<sub>6</sub></th><th>CD<sub>3</sub>COCD<sub>3</sub></th><th>CD<sub>3</sub>SOCD<sub>3</sub></th><th>CD<sub>3</sub>C≡N</th><th>D<sub>2</sub>O</th></tr><tr><th>(CH<sub>3</sub>)<sub>3</sub>C–O–CH<sub>3</sub><br>C–CH<sub>3</sub><br>O–CH<sub>3</sub></th><td>1.19<br>3.22</td><td>1.07<br>3.04</td><td>1.13<br>3.13</td><td>1.11<br>3.03</td><td>1.14<br>3.13</td><td>1.21<br>3.22</td></tr><tr><th>(CH<sub>3</sub>)<sub>3</sub>C–O–H<br>C–CH<sub>3</sub><br>O–H</th><td>1.26<br>1.65</td><td>1.05<br>1.55</td><td>1.18<br>3.10</td><td>1.11<br>4.19</td><td>1.16<br>2.18</td><td>&#8212;<br>&#8212;</td></tr><tr><th>C<sub>6</sub>H<sub>5</sub>CH<sub>3</sub><br>CH<sub>3</sub><br>C<sub>6</sub>H<sub>5</sub></th><td>2.36<br>7.15-7.20</td><td>2.11<br>7.00-7.10</td><td>2.32<br>7.10-7.20</td><td>2.30<br>7.10-7.15</td><td>2.33<br>7.15-7.30</td><td>&#8212;<br>&#8212;</td></tr><tr><th>(CH<sub>3</sub>)<sub>2</sub>C=O</th><td>2.17</td><td>1.55</td><td>2.09</td><td>2.09</td><td>2.08</td><td>2.22</td></tr></tbody></table></figure>



<p>For most of the above resonance signals and solvents the changes are minor, being on the order of ±0.1 ppm. However, two cases result in more extreme changes and these have provided useful applications in structure determination. First, spectra taken in benzene-d<sub>6</sub>&nbsp;generally show small upfield shifts of most C–H signals, but in the case of acetone this shift is about five times larger than normal. Further study has shown that carbonyl groups form weak π–π collision complexes with benzene rings, that persist long enough to exert a significant shielding influence on nearby groups. In the case of substituted cyclohexanones, axial α-methyl groups are shifted upfield by 0.2 to 0.3 ppm; whereas equatorial methyls are slightly deshielded (shift downfield by about 0.05 ppm). These changes are all relative to the corresponding chloroform spectra.<br>The second noteworthy change is seen in the spectrum of tert-butanol in DMSO, where the hydroxyl proton is shifted 2.5 ppm down-field from where it is found in dilute chloroform solution. This is due to strong hydrogen bonding of the alcohol O–H to the sulfoxide oxygen, which not only de-shields the hydroxyl proton, but secures it from very rapid exchange reactions that prevent the display of spin-spin splitting. Similar but weaker hydrogen bonds are formed to the carbonyl oxygen of acetone and the nitrogen of acetonitrile. A useful application of this phenomenon is described&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr15" target="_blank" rel="noopener">elsewhere in this text</a>.</p>



<h4 class="wp-block-heading" id="spin-spin-interactions">Spin-Spin Interactions</h4>



<p>The nmr spectrum of 1,1-dichloroethane (below right) is more complicated than we might have expected from the previous examples. Unlike its 1,2-dichloro-isomer (below left), which displays a single resonance signal from the four structurally equivalent hydrogens, the two signals from the different hydrogens are split into close groupings of two or more resonances. This is a common feature in the spectra of compounds having different sets of hydrogen atoms bonded to adjacent carbon atoms. The signal splitting in proton spectra is usually small, ranging from fractions of a Hz to as much as 18 Hz, and is designated as&nbsp;<strong>J</strong>&nbsp;(referred to as the coupling constant). In the 1,1-dichloroethane example all the coupling constants are 6.0 Hz,&nbsp;as illustrated by clicking on the spectrum.</p>



<figure class="wp-block-table"><table><tbody><tr><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/12cleth.gif"></td><td></td><td><a href="javascript:chg6();"></a></td></tr><tr><th>1,2-dichloroethane</th><th></th><th>1,1-dichloroethane</th></tr></tbody></table></figure>



<p>The splitting patterns found in various spectra are easily recognized, provided the chemical shifts of the different sets of hydrogen that generate the signals differ by two or more ppm. The patterns are symmetrically distributed on both sides of the proton chemical shift, and the central lines are always stronger than the outer lines. The most commonly observed patterns have been given descriptive names, such as&nbsp;<strong>doublet</strong>&nbsp;(two equal intensity signals),&nbsp;<strong>triplet</strong>&nbsp;(three signals with an intensity ratio of 1:2:1) and&nbsp;<strong>quartet</strong>&nbsp;(a set of four signals with intensities of 1:3:3:1). Four such patterns are displayed in the following illustration. The line separation is always constant within a given multiplet, and is called the&nbsp;<strong>coupling constant (J)</strong>. The magnitude of J, usually given in units of Hz, is magnetic field independent.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/patterns.gif" alt=""/></figure>



<p>The splitting patterns shown above display the ideal or &#8220;<strong>First-Order</strong>&#8221; arrangement of lines. This is usually observed if the spin-coupled nuclei have very different chemical shifts (i.e. Δν is large compared to J). If the coupled nuclei have similar chemical shifts, the splitting patterns are distorted (second order behavior). In fact, signal splitting disappears if the chemical shifts are the same. Two examples that exhibit minor 2nd order distortion are shown below (both are taken at a frequency of 90 MHz). The ethyl acetate spectrum on the left displays the typical quartet and triplet of a substituted ethyl group. The spectrum of 1,3-dichloropropane on the right demonstrates that equivalent sets of hydrogens may combine their influence on a second, symmetrically located set.&nbsp;<br>Even though the chemical shift difference between the A and B protons in the 1,3-dichloroethane spectrum is fairly large (140 Hz) compared with the coupling constant (6.2 Hz), some distortion of the splitting patterns is evident. The line intensities closest to the chemical shift of the coupled partner are enhanced. Thus the B set triplet lines closest to A are increased, and the A quintet lines nearest B are likewise stronger. A smaller distortion of this kind is visible for the A and C couplings in the ethyl acetate spectrum.</p>



<figure class="wp-block-table"><table><tbody><tr><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/etoac1.gif"></td><td></td><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/13clprop.gif"></td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><th>For additional examples of&nbsp;<strong>Second Order</strong>&nbsp;splitting patterns&nbsp;<a href="javascript:chngtxt(xx)">Click Here</a>.</th></tr></tbody></table></figure>



<p><strong>What causes this signal splitting, and what useful information can be obtained from it ?</strong>&nbsp;<br>If an atom under examination is perturbed or influenced by a nearby nuclear spin (or set of spins), the observed nucleus responds to such influences, and its response is manifested in its resonance signal. This spin-coupling is transmitted through the connecting bonds, and it functions in both directions. Thus, when the perturbing nucleus becomes the observed nucleus, it also exhibits signal splitting with the same J. For spin-coupling to be observed, the sets of interacting nuclei must be bonded in relatively close proximity (e.g. vicinal and geminal locations), or be oriented in certain optimal and rigid configurations. Some spectroscopists place a number before the symbol J to designate the number of bonds linking the coupled nuclei (colored orange below). Using this terminology, a vicinal coupling constant is&nbsp;<sup>3</sup>J and a geminal constant is&nbsp;<sup>2</sup>J.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/gemvic.gif" alt=""/></figure>



<p><strong>The following general rules summarize important requirements and characteristics for spin 1/2 nuclei :</strong></p>



<p><strong>1)</strong>&nbsp;&nbsp; Nuclei having the same chemical shift (called&nbsp;<strong>isochronous</strong>) do not exhibit spin-splitting. They may actually be spin-coupled, but the splitting cannot be observed directly.<br><strong>2)</strong>&nbsp;&nbsp; Nuclei separated by three or fewer bonds (e.g. vicinal and geminal nuclei ) will usually be spin-coupled and will show mutual spin-splitting of the resonance signals (same J&#8217;s), provided they have different chemical shifts. Longer-range coupling may be observed in molecules having rigid configurations of atoms.<br><strong>3)</strong>&nbsp;&nbsp; The magnitude of the observed spin-splitting depends on many factors and is given by the coupling constant&nbsp;<strong>J</strong>&nbsp;(units of Hz). J is the same for both partners in a spin-splitting interaction and is independent of the external magnetic field strength.<br><strong>4)</strong>&nbsp;&nbsp; The splitting pattern of a given nucleus (or set of equivalent nuclei) can be predicted by the&nbsp;<strong>n+1 rule</strong>, where n is the number of neighboring spin-coupled nuclei with the same (or very similar) Js. If there are 2 neighboring, spin-coupled, nuclei the observed signal is a triplet ( 2+1=3 ); if there are three spin-coupled neighbors the signal is a quartet ( 3+1=4 ). In all cases the central line(s) of the splitting pattern are stronger than those on the periphery. The intensity ratio of these lines is given by the numbers in Pascal&#8217;s triangle. Thus a doublet has 1:1 or equal intensities, a triplet has an intensity ratio of 1:2:1, a quartet 1:3:3:1 etc. To see how the numbers in Pascal&#8217;s triangle are related to the Fibonacci series&nbsp;click on the diagram.</p>



<figure class="wp-block-table"><table><tbody><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/splitting.gif"></th><th><a href="javascript:chg8();"></a></th></tr><tr><td>If a given nucleus is spin-coupled to two or more sets of neighboring nuclei by different J values, the n+1 rule does not predict the entire splitting pattern. Instead, the splitting due to one J set is added to that expected from the other J sets. Bear in mind that there may be fortuitous coincidence of some lines if a smaller J is a factor of a larger J.</td></tr><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/splitting2.gif"></th></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/jconstnt.gif">&lt;</th></tr></tbody></table></figure>



<p>Spin 1/2 nuclei include&nbsp;<sup>1</sup>H,&nbsp;<sup>13</sup>C,&nbsp;<sup>19</sup>F &amp;&nbsp;<sup>31</sup>P. The spin-coupling interactions described above may occur between similar or dissimilar nuclei. If, for example, a&nbsp;<sup>19</sup>F is spin-coupled to a&nbsp;<sup>1</sup>H, both nuclei will appear as doublets having the same J constant.&nbsp;&nbsp;Spin coupling with nuclei having spin other than 1/2 is more complex and will not be discussed here.</p>



<p>To make use of a calculator that predicts first order splitting patterns&nbsp;<a href="http://www.colby.edu/chemistry/NMR/jmmset.html" target="_blank" rel="noopener">Click Here</a>. This application was developed at Colby College.</p>



<figure class="wp-block-table"><table><tbody><tr><th>For additional information about spin-spin coupling&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr16" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<h3 class="wp-block-heading" id="some-examples">Some Examples</h3>



<p>Test your ability to interpret&nbsp;<sup>1</sup>H nmr spectra by analyzing the seven examples presented below. The seven spectra may be examined in turn by clicking the &#8220;Toggle Spectra&#8221; button. Try to associate each spectrum with a plausible structural formula.&nbsp;<br>Although the first four cases are relatively simple, keep in mind that the integration values provide ratios, not absolute numbers. In two cases additional information from infrared spectroscopy is provided. When you have made an assignment you may check your answer by clicking on the spectrum itself. In the sixth example, a similar constitutional isomer cannot be ruled out by the data given.</p>



<figure class="wp-block-image"><a href="javascript:chg7();"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/nmrspc11.gif" alt=""/></a></figure>



<figure class="wp-block-table"><table><tbody><tr><th>For a challenging problem having many spin couplings&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#nmr18" target="_blank" rel="noopener">Click Here</a>.</th></tr></tbody></table></figure>



<p><strong>&nbsp; &nbsp; &nbsp; 3. Carbon NMR Spectroscopy</strong><br>The power and usefulness of&nbsp;<sup>1</sup>H nmr spectroscopy as a tool for structural analysis should be evident from the past discussion. Unfortunately, when significant portions of a molecule lack C-H bonds, no information is forthcoming. Examples include polychlorinated compounds such as chlordane, polycarbonyl compounds such as croconic acid, and compounds incorporating triple bonds (structures below, orange colored carbons).</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/c-struc1.gif" alt=""/></figure>



<p>Even when numerous C-H groups are present, an unambiguous interpretation of a proton nmr spectrum may not be possible. The following diagram depicts three pairs of isomers (A &amp; B) which display similar proton nmr spectra. Although a careful determination of chemical shifts should permit the first pair of compounds (blue box) to be distinguished, the second and third cases (red &amp; green boxes) might be difficult to identify by proton nmr alone.</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/c-struc2.gif" alt=""/></figure>



<p>These difficulties would be largely resolved if the carbon atoms of a molecule could be probed by nmr in the same fashion as the hydrogen atoms. Since the major isotope of carbon (<sup>12</sup>C) has no spin, this option seems unrealistic. Fortunately, 1.1% of elemental carbon is the&nbsp;<sup>13</sup>C isotope, which has a spin I = 1/2, so in principle it should be possible to conduct a carbon nmr experiment.&nbsp;It is worth noting here, that if much higher abundances of&nbsp;<sup>13</sup>C were naturally present in all carbon compounds, proton nmr would become much more complicated due to large one-bond coupling of&nbsp;<sup>13</sup>C and&nbsp;<sup>1</sup>H.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Many obstacles needed to be overcome before carbon nmr emerged as a routine tool :</strong><br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<strong>i)</strong>&nbsp;&nbsp; As noted, the abundance of&nbsp;<sup>13</sup>C in a sample is very low (1.1%), so higher sample concentrations are needed.<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<strong>ii)</strong>&nbsp;&nbsp; The&nbsp;<sup>13</sup>C nucleus is over fifty times less sensitive than a proton in the nmr experiment, adding to the previous difficulty.<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<strong>iii)</strong>&nbsp;&nbsp; Hydrogen atoms bonded to a&nbsp;<sup>13</sup>C atom split its nmr signal by 130 to 270 Hz, further complicating the nmr spectrum.</td></tr></tbody></table></figure>



<p>The most important operational technique that has led to successful and routine&nbsp;<sup>13</sup>C nmr spectroscopy is the use of high-field&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#pulse" target="_blank" rel="noopener">pulse technology</a>&nbsp;coupled with broad-band&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm#decoupl" target="_blank" rel="noopener">heteronuclear decoupling</a>&nbsp;of all protons. The results of repeated pulse sequences are accumulated to provide improved signal strength. Also, for reasons that go beyond the present treatment, the decoupling irradiation enhances the sensitivity of carbon nuclei bonded to hydrogen.&nbsp;<br>When acquired in this manner, the carbon nmr spectrum of a compound displays a single sharp signal for each structurally distinct carbon atom in a molecule (remember, the proton couplings have been removed). The spectrum of camphor, shown on the left below, is typical. Furthermore, a comparison with the&nbsp;<sup>1</sup>H nmr spectrum on the right illustrates some of the advantageous characteristics of carbon nmr. The dispersion of&nbsp;<sup>13</sup>C chemical shifts is nearly twenty times greater than that for protons, and this together with the lack of signal splitting makes it more likely that every structurally distinct carbon atom will produce a separate signal. The only clearly identifiable signals in the proton spectrum are those from the methyl groups. The remaining protons have resonance signals between 1.0 and 2.8 ppm from TMS, and they overlap badly thanks to spin-spin splitting.</p>



<figure class="wp-block-table"><table><tbody><tr><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/ccamphor.gif"></th><th><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/hcamphor.gif"></th></tr></tbody></table></figure>



<p>Unlike proton nmr spectroscopy,&nbsp;<strong>the relative strength of carbon nmr signals are not normally proportional to the number of atoms generating each one</strong>. Because of this, the number of discrete signals and their chemical shifts are the most important pieces of evidence delivered by a carbon spectrum. The general distribution of carbon chemical shifts associated with different functional groups is summarized in the following chart. Bear in mind that these ranges are approximate, and may not encompass all compounds of a given class. Note also that the over 200 ppm range of chemical shifts shown here is much greater than that observed for&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr1.htm#nmr3bb" target="_blank" rel="noopener">proton chemical shifts</a>.</p>



<figure class="wp-block-table"><table><tbody><tr><th><sup>13</sup>C Chemical Shift Ranges<sup>*</sup></th></tr><tr><th>Low Field<br>Region</th><td><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/cnmr1.gif"></td><th>High Field<br>Region</th></tr><tr><td></td><td><sup>*</sup>&nbsp;For samples in CDCl<sub>3</sub>&nbsp;solution. The δ scale is relative to TMS at δ=0.</td><td></td></tr></tbody></table></figure>



<p>The isomeric pairs previously cited as giving very similar proton nmr spectra are now seen to be distinguished by carbon nmr. In the example on the left below (blue box), cyclohexane and 2,3-dimethyl-2-butene both give a single sharp resonance signal in the proton nmr spectrum (the former at δ 1.43 ppm and the latter at 1.64 ppm). However, in its carbon nmr spectrum cyclohexane displays a single signal at δ 27.1 ppm, generated by the equivalent ring carbon atoms (colored blue); whereas the isomeric alkene shows two signals, one at δ 20.4 ppm from the methyl carbons (colored brown), and the other at 123.5 ppm (typical of the green colored sp<sup>2</sup>&nbsp;hybrid carbon atoms).</p>



<figure class="wp-block-image"><img decoding="async" src="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/Images/c-struc3.gif" alt=""/></figure>



<p>The C<sub>8</sub>H<sub>10</sub>&nbsp;isomers in the center (red) box have pairs of&nbsp;<a href="https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/suppmnt1.htm#nom1" target="_blank" rel="noopener">homotopic</a>&nbsp;carbons and hydrogens, so symmetry should simplify their nmr spectra. The fulvene (isomer A) has five structurally different groups of carbon atoms (colored brown, magenta, orange, blue and green respectively) and should display five&nbsp;<sup>13</sup>C nmr signals (one near 20 ppm and the other four greater than 100 ppm). Although ortho-xylene (isomer B) will have a proton nmr very similar to isomer A, it should only display four&nbsp;<sup>13</sup>C nmr signals, originating from the four different groups of carbon atoms (colored brown, blue, orange and green). The methyl carbon signal will appear at high field (near 20 ppm), and the aromatic ring carbons will all give signals having δ &gt; 100 ppm. Finally, the last isomeric pair, quinones A &amp; B in the green box, are easily distinguished by carbon nmr. Isomer A displays only four carbon nmr signals (δ 15.4, 133.4, 145.8 &amp; 187.9 ppm); whereas, isomer B displays five signals (δ 15.9, 133.3, 145.8, 187.5 &amp; 188.1 ppm), the additional signal coming from the non-identity of the two carbonyl carbon atoms (one colored orange and the other magenta).</p>
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<ol class="wp-block-list"><li>NMRshiftdb</li></ol>



<p>NMRshiftdb2 is a NMR database (web database) for organic structures and their nuclear magnetic resonance (nmr) spectra. It allows for spectrum prediction (<sup>13</sup>C,&nbsp;<sup>1</sup>H and other nuclei) as well as for searching spectra, structures and other properties. The nmrshiftdb2 software is open source, the data is published under an open content license. The core of nmrshitdb2 are fully assigned spectra with raw data and peak lists (we have pure peak lists as well). Those datasets are peer reviewed by a&nbsp;board of reviewers. The project is supported by a&nbsp;scientific advisory board.</p>



<span id="more-596"></span>



<p>nmrshiftdb2 is part of the&nbsp;NFDI4Chem initiative&nbsp;and will provide a component for a curated repository there. Please consult the&nbsp;documentation&nbsp;for more detailed information.</p>



<p class="has-text-align-center"><strong>See: https://nmrshiftdb.nmr.uni-koeln.de/portal</strong></p>



<p>2. ACD/NMR</p>



<p>ACD/NMR Workbook Suite is a comprehensive NMR software application with an intuitive interface. It features a full suite of advanced processing, analysis, and databasing functionalities for 1D and 2D NMR data from all major vendor formats. NMR Workbook Suite is built upon cutting-edge algorithms for the most reliable NMR data interpretation. It is designed to streamline routine NMR workflows, simplify structure characterization, and much more. </p>



<h5 class="wp-block-heading" id="powerful-nmr-interpretation-software-highlights">Powerful NMR Interpretation Software | Highlights</h5>



<ul class="wp-block-list"><li>Import and process 1D and 2D NMR data from all major instrument vendor formats in a single collaborative platform</li><li>Process NMR data manually or automate routine processing workflows—Fourier transformation, calibration, peak picking, integration, multiplet analysis, etc.</li><li>Synchronize peak picking and assignments across datasets within a project</li><li>Confidently verify structures with 3 different verification levels</li><li>Perform targeted analysis of known mixture components and optimize untargeted mixture analysis workflow</li><li>Perform Conformational Analysis using NOESY/ROESY spectra</li><li>Create comprehensive multiplet reports and publication-ready data</li><li>Store, manage, and share live NMR spectra</li></ul>



<p>Synchronize peak picking and assignments across NMR datasets using NMR<em>Sync</em>—our game-changing technology. Plus, the associated peaks from NMR<em>Sync</em>, NMR prediction, and connectivity-based algorithms are automatically used to only identify the assignments that match all data. This quick and accurate peak picking and assignment workflow helps you to maximize your productivity in the following ways:</p>



<ul class="wp-block-list"><li>Use any peak in any spectrum to initiate NMRSync</li><li>Integrate a peak in any spectrum and all related peaks in the 1D and 2D NMR spectra of that dataset will be identified and linked in real time</li><li>Automatically resolve overlapping&nbsp;<sup>1</sup>H and&nbsp;<sup>13</sup>C peaks from 2D NMR data</li><li>Receive immediate color-coded feedback on the best assignment for instant decision-making purposes</li></ul>



<p>NMR Workbook Suite includes three levels of structure verification that evaluate alternative structures to varying degrees for added flexibility in your NMR analysis. This ensures the best structure that matches the experimental NMR data is confirmed with much less time and effort than manual interpretation.</p>



<ul class="wp-block-list"><li>Determine how well your proposed structure matches the datasets in your NMR project with single structure verification</li><li>Generate a specified number of alternative structures, based on the user-defined proposed structure, and evaluate whether they are a better match to the NMR dataset using Combined and Concurrent Verification</li><li>Generate and view every alternative structural and&nbsp;<em>cis/trans</em>&nbsp;isomer that matches the experimental data in real-time using Unbiased Verification for an absolute level of confidence. This workflow eliminates the user bias and ensures the assigned structure is indeed the best structure that fits the experimental data.</li></ul>



<p class="has-text-align-center"><strong>See: https://www.acdlabs.com/products/spectrus/workbooks/nmr/</strong></p>



<p>3. <strong>See: http://www.cheminfo.org/Spectra/NMR/Predictions/1H_Prediction/index.html</strong></p>



<p>4. <strong>See: https://www.nmrprocflow.org/</strong></p>



<p>5. <strong>See: https://chem.washington.edu/facilities/data-processing</strong></p>



<p>6. <strong>See: https://www.cgl.ucsf.edu/home/sparky/</strong></p>



<p><strong>7. See: http://www.nmrdb.org/about/</strong></p>
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		<title>How to analyze Nuclear Magnetic Resonance (NMR) spectra</title>
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					<description><![CDATA[Click here to see more posts about NMR Only 15$ for interpretation of your NMR spectrum Payment Upon Completion Send your results... Nuclear Magnetic Resonance (NMR) interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors [&#8230;]]]></description>
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<pre class="wp-block-verse has-text-align-center"><span style="color:#ffffff" class="tadv-color">Only 15$ for interpretation of your NMR spectrum
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<p>Nuclear Magnetic Resonance (NMR) interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors such as chemical shift, spin multiplicity, coupling constants, and integration. This Module focuses on the most important&nbsp;<sup>1</sup>H and&nbsp;<sup>13</sup>C NMR spectra to find out structure even though there are various kinds of NMR spectra such as&nbsp;<sup>14</sup>N,&nbsp;<sup>19</sup>F, and&nbsp;<sup>31</sup>P. NMR spectrum shows that x- axis is chemical shift in ppm. It also contains integral areas, splitting pattern, and coupling constant.</p>



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<p>Strategy for Solving Structure</p>



<p>Here is the general strategy for solving structure with NMR:</p>



<ol class="wp-block-list"><li>Molecular formula is determined by chemical analysis such as elementary analysis</li><li><strong>Double-bond equivalent</strong>&nbsp;(also known as&nbsp;<a href="https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Degree_of_Unsaturation" target="_blank" rel="noopener">Degree of Unsaturation</a>) is calculated by a simple equation to estimate the number of the multiple bonds and rings. It assumes that oxygen (O) and sulfur (S) are ignored and halogen (Cl, Br) and nitrogen is replaced by CH. The resulting empirical formula is C<sub>a</sub>H<sub>b</sub></li></ol>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9380/Equation_1_(1).jpg?revision=1" alt="Equation 1 (1).jpg"/></figure>



<ol class="wp-block-list" start="3"><li>Structure fragmentation is determined by chemical shift, spin multiplicity, integral (peak area), and coupling constants (1J1J,&nbsp;2J2J)</li><li>Molecular skeleton is built up using 2-dimensional NMR spectroscopy.</li><li>Relative configuration is predicted by coupling constant (<sup>3</sup>J).</li></ol>



<h2 class="wp-block-heading" id="1h-nmr"><sup>1</sup>H NMR</h2>



<h3 class="wp-block-heading" id="chemical-shift">Chemical Shift</h3>



<p><a href="https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR_-_Theory" target="_blank" rel="noopener">Chemical shift</a>&nbsp;is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane (TMS,&nbsp;(CH3)4Si(CH3)4Si) is generally used as an internal standard to determine chemical shift of compounds: δ<sub>TMS</sub>=0 ppm. In other words, frequencies for chemicals are measured for a&nbsp;<sup>1</sup>H or&nbsp;<sup>13</sup>C nucleus of a sample from the&nbsp;<sup>1</sup>H or&nbsp;<sup>13</sup>C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane.&nbsp;<sup>1</sup>H chemical shift play a role in identifying many functional groups. Figure&nbsp;11. indicates important example to figure out the functional groups.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9384/Fig1.bmp?revision=1&amp;size=bestfit&amp;width=732&amp;height=323" alt="Fig1.bmp"/><figcaption>Figure&nbsp;11: 1H chemical shift ranges for organic compound</figcaption></figure>



<h3 class="wp-block-heading" id="chemical-equivalence">Chemical equivalence</h3>



<p>Protons with Chemical equivalence has the same chemical shift due to symmetry within molecule (CH3COCH3CH3COCH3) or fast rotation around single bond (-CH<sub>3</sub>; methyl groups).</p>



<h2 class="wp-block-heading" id="spin-spin-splitting">Spin-Spin Splitting</h2>



<p><a href="https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR_-_Theory" target="_blank" rel="noopener">Spin-Spin splitting</a>&nbsp;means that an absorbing peak is split by more than one “neighbor” proton. Splitting signals are separated to J Hz, where is called the coupling constant. The spitting is a very essential part to obtain exact information about the number of the neighboring protons. The maximum of distance for splitting is three bonds. Chemical equivalent protons do not result in spin-spin splitting. When a proton splits, the proton’s chemical shift is determined in the center of the splitting lines.</p>



<h4 class="wp-block-heading" id="spin-multiplicity-splitting-pattern">Spin Multiplicity (Splitting pattern)</h4>



<p>Spin Multiplicity plays a role in determining the number of neighboring protons. Here is a multiplicity rules: In case of&nbsp;AmBnAmBn&nbsp;system, the multiplicity rule is that Nuclei of&nbsp;BB&nbsp;element produce a splitting the&nbsp;AA&nbsp;signal into&nbsp;nB+1nB+1&nbsp;lines. The general formula which applies to all nuclei is&nbsp;2nI+12nI+1, where&nbsp;II&nbsp;is the spin quantum number of the coupled element. The relative intensities of the each lines are given by the coefficients of the Pascal’s triangle (Figure&nbsp;22).</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/24873/pascalstriangle8.png?revision=1&amp;size=bestfit&amp;width=550&amp;height=334" alt="Fig2.bmp"/><figcaption>Figure&nbsp;22: Pascal&#8217;s triangle</figcaption></figure>



<h4 class="wp-block-heading" id="first-order-splitting-pattern">First-order splitting pattern</h4>



<p>The chemical shift difference in Hertz between coupled protons in Hertz is much larger than the&nbsp;JJ&nbsp;coupling constant:ΔνJ≥8(1)(1)ΔνJ≥8</p>



<p>Where&nbsp;ΔνΔν&nbsp;is the difference of chemical shift. In other word, the proton is only coupled to other protons that are far away in chemical shift. The spectrum is called&nbsp;<em>first-order spectrum</em>. The splitting pattern depends on the magnetic field. The second-order splitting at the lower field can be resolved into first-order splitting pattern at the high field. The first-order splitting pattern is allowed to multiplicity rule (N+1) and Pascal’s triangle to determine splitting pattern and intensity distribution.</p>



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<p>Example&nbsp;11</p>



<p>The note is that structure system is A<sub>3</sub>M<sub>2</sub>X<sub>2</sub>. H<sub>a</sub>&nbsp;and H<sub>x</sub>&nbsp;has the triplet pattern by Hm because of N+1 rule. The signal of Hm is split into six peaks by H<sub>x</sub>&nbsp;and H<sub>a</sub>(Figure3) The First order pattern easily is predicted due to separation with equal splitting pattern.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9390/Fig3.bmp?revision=1&amp;size=bestfit&amp;width=561&amp;height=149" alt="Fig3.bmp"/><figcaption>Figure&nbsp;33: An example of splitting pattern</figcaption></figure>



<h4 class="wp-block-heading" id="high-order-splitting-pattern">High-order splitting pattern</h4>



<p>High-order splitting pattern takes place when chemical shift difference in Hertz is much less or the same that order of magnitude as the j coupling.ΔvJ≤10(2)(2)ΔvJ≤10</p>



<p>The second order pattern is observed as leaning of a classical pattern: the inner peaks are taller and the outer peaks are shorter in case of AB system (Figure&nbsp;44). This is called the&nbsp;<strong>roof effect</strong>.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9391/Fig4.bmp?revision=1&amp;size=bestfit&amp;width=646&amp;height=131" alt="Fig4.bmp"/><figcaption>Figure&nbsp;44: a) first-order pattern and b) second-order pattern of AB system</figcaption></figure>



<p>Here is other system as an example: A<sub>2</sub>B<sub>2</sub>&nbsp;(Figure&nbsp;55). The two triplet incline toward each other. Outer lines of the triplet are less than 1 in relative area and the inner lines are more than 1. The center lines have relative area 2.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9392/Fig5.bmp?revision=1&amp;size=bestfit&amp;width=720&amp;height=189" alt="Fig5.bmp"/><figcaption>Figure&nbsp;55: a) first-order pattern and b) second-order pattern of A<sub>2</sub>B<sub>2</sub>&nbsp;system</figcaption></figure>



<h4 class="wp-block-heading" id="coupling-constant-j-value">Coupling constant (J Value)</h4>



<p>Coupling constant is the strength of the spin-spin splitting interaction and the distance between the split lines. The value of distance is equal or different depending on the coupled nuclei. The coupling constants reflect the bonding environments of the coupled nuclei. Coupling constant is classified by the number of bonds:</p>



<h5 class="wp-block-heading" id="geminal-proton-proton-coupling-2jhh">Geminal proton-proton coupling (<sup>2</sup>J<sub>HH</sub>)</h5>



<p>Germinal coupling generates through two bonds (Figure&nbsp;66). Two proton having geminal coupling are not chemically equivalent. This coupling ranges from -20 to 40 Hz.&nbsp;<sup>2</sup>J<sub>HH</sub>depends on hybridization of carbon atom and the bond angle and the substituent such as electronegative atoms. When S-character is increased, Geminal coupling constant is increased:&nbsp;<sup>2</sup>J<sub>sp1</sub>&gt;<sup>2</sup>J<sub>sp2</sub>&gt;<sup>2</sup>J<sub>sp3</sub>&nbsp;The bond angle(HCH) gives rise to change&nbsp;<sup>2</sup>J<sub>HH</sub>&nbsp;value and depend on the strain of the ring in the cyclic systems. Geminal coupling constant determines ring size. When bond angle is decreased, ring size is decreased so that geminal coupling constant is more positive. If a atom is replace to an electronegative atom, Geminal coupling constant move to positive value.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9393/Fig6.bmp?revision=1&amp;size=bestfit&amp;width=408&amp;height=92" alt="Fig6.bmp"/><figcaption>Figure&nbsp;66: Geminal coupling</figcaption></figure>



<h5 class="wp-block-heading" id="vicinal-proton-proton-coupling-3jhh">Vicinal proton-proton coupling (<sup>3</sup>J<sub>HH</sub>)</h5>



<p>Vicinal coupling occurs though three bonds (Figure&nbsp;77.). The Vicinal coupling is the most useful information of dihedral angle, leading to stereochemistry and conformation of molecules. Vicinal coupling constant always has the positive value and is affected by the dihedral angle (?;HCCH), the valence angle (?; HCC), the bond length of carbon-carbon, and the effects of electronegative atoms. Vicinal coupling constant depending on the dihedral angle (Figure&nbsp;88) is given by the Karplus equation.3J=7.0−0.5cosϕ+4.5cos2ϕ(3)(3)3J=7.0−0.5cos⁡ϕ+4.5cos2⁡ϕ</p>



<p>When ? is the 90<sup>o</sup>, vicinal coupling constant is zero. In addition, vicinal coupling constant ranges from 8 to 10 Hz at the and ?=180<sup>o</sup>, where ?=0<sup>o</sup>&nbsp;and ?=180<sup>o</sup>&nbsp;means that the coupled protons have cis and trans configuration, respectively.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9394/Fig7.bmp?revision=1&amp;size=bestfit&amp;width=534&amp;height=111" alt="Fig7.bmp"/><figcaption>Figure&nbsp;77: Vicinal coupling</figcaption></figure>



<p>The valence angle(?;Figure&nbsp;88) also causes change of&nbsp;<sup>3</sup>J<sub>HH</sub>&nbsp;value. Valence angle is related with ring size. Typically, when the valence angle decreases, the coupling constant reduces. The distance between the carbons atoms gives influences to vicinal coupling constant</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9395/Fig8.bmp?revision=1&amp;size=bestfit&amp;width=299&amp;height=86" alt="Fig8.bmp"/><figcaption>Figure&nbsp;88: a) Dihedral angle and b) valence angle</figcaption></figure>



<p>The coupling constant increases with the decrease of bond length. Electronegative atoms affect vicinal coupling constants so that electronegative atoms decrease the vicinal coupling constants.</p>



<h4 class="wp-block-heading" id="integral">Integral</h4>



<p>Integral is referred to integrated peak area of 1H signals. The intensity is directly proportionally to the number of hydrogen.</p>



<h2 class="wp-block-heading" id="13c-nmr"><sup>13</sup>C NMR</h2>



<h4 class="wp-block-heading" id="chemical-shift">Chemical Shift</h4>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9378/1.jpg?revision=1" alt="1.jpg"/><figcaption>Figure&nbsp;99&nbsp;shows typical&nbsp;<sup>13</sup>C chemical shift regions of the major chemical class.</figcaption></figure>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9396/Fig9.bmp?revision=1&amp;size=bestfit&amp;width=719&amp;height=415" alt="Fig9.bmp"/><figcaption>Figure&nbsp;99:&nbsp;<sup>13</sup>C Chemical shift range for organic compound</figcaption></figure>



<h4 class="wp-block-heading" id="spin-spin-splitting">Spin-Spin splitting</h4>



<p>Comparing the&nbsp;<sup>1</sup>H NMR, there is a big difference thing in the&nbsp;<sup>13</sup>C NMR. The&nbsp;<sup>13</sup>C-&nbsp;<sup>13</sup>&nbsp;C spin-spin splitting rarely exit between adjacent carbons because&nbsp;<sup>13</sup>C is naturally lower abundant (1.1%)</p>



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<ul class="wp-block-list"><li><strong><sup>13</sup>C-<sup>1</sup>H Spin coupling</strong>:&nbsp;<sup>13</sup>C-<sup>1</sup>H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (<sup>1</sup>J<sub>CH</sub>), -CH, -CH<sub>2</sub>, and CH<sub>3</sub>&nbsp;have respectively doublet, triplet, quartets for the&nbsp;<sup>13</sup>C resonances in the spectrum. However,&nbsp;<sup>13</sup>C-<sup>1</sup>H Spin coupling has an disadvantage for&nbsp;<sup>13</sup>C spectrum interpretation.&nbsp;<sup>13</sup>C-<sup>1</sup>H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of&nbsp;<sup>1</sup>H.</li><li><strong>Decoupling</strong>: Decoupling is the process of removing&nbsp;<sup>13</sup>C-<sup>1</sup>H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling&nbsp;<sup>13</sup>C spectra shows only one peak(singlet) for each unique carbon in the molecule(Figure&nbsp;1010.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.</li></ul>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9385/Fig10.bmp?revision=1&amp;size=bestfit&amp;width=456&amp;height=143" alt="Fig10.bmp"/><figcaption>Figure&nbsp;1010. Decoupling in the&nbsp;<sup>13</sup>C&nbsp;NMR</figcaption></figure>



<ul class="wp-block-list"><li><strong>Distortionless enhancement by polarization transfer (DEPT):</strong>&nbsp;DEPT is used for distinguishing between a CH<sub>3</sub>&nbsp;group, a CH<sub>2</sub>&nbsp;group, and a CH group. The proton pulse is set at 45<sup>o</sup>, 90<sup>o</sup>, or 135<sup>o</sup>&nbsp;in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure&nbsp;1111. is an example about DEPT spectrum.</li></ul>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9386/Fig11.bmp?revision=1&amp;size=bestfit&amp;width=517&amp;height=430" alt="Fig11.bmp"/><figcaption>Figure&nbsp;1111. DEPT spectrum of n-isobutlybutrate</figcaption></figure>



<h3 class="wp-block-heading" id="2-dimensional-nmr-spectroscopy-cosy">2-dimensional NMR spectroscopy (COSY)</h3>



<p>COSY stands for COrrelation SpectroscopY. COSY spectrum is more useful information about what is being correlated.</p>



<h4 class="wp-block-heading" id="1h-1h-cosy-correlation-spectroscopy"><sup>1</sup>H-<sup>1</sup>H COSY (COrrelation SpectroscopY)</h4>



<p><sup>1</sup>H-<sup>1</sup>H COSY is used for clearly indicate correlation with coupled protons. A point of entry into a COSY spectrum is one of the keys to predict information from it successfully. Relation of Coupling protons is determined by cross peaks(correlation peaks) and in the COSY spectrum. In other words, Diagonal peaks by lines ar e coupled to each other. Figure&nbsp;1212&nbsp;indicates that there are correlation peaks between proton H<sub>1</sub>&nbsp;and H<sub>2</sub>&nbsp;as well as between H<sub>2</sub>&nbsp;and H<sub>4</sub>. This means the H<sub>2</sub>&nbsp;coupled to H<sub>1</sub>&nbsp;and H<sub>4</sub>.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9387/Fig12.bmp?revision=1&amp;size=bestfit&amp;width=455&amp;height=406" alt="Fig12.bmp"/><figcaption>Figure&nbsp;1212.&nbsp;<sup>1</sup>H-<sup>1</sup>H COSY spectrum</figcaption></figure>



<h4 class="wp-block-heading" id="1h-13c-cosy-hetcor"><sup>1</sup>H-<sup>13</sup>C COSY (HETCOR)</h4>



<p><sup>1</sup>H-<sup>13</sup>C COSY is the heteronuclear correlation spectroscopy. The HETCOR spectrum is correlated&nbsp;<sup>13</sup>C nuclei with directly attached protons.&nbsp;<sup>1</sup>H-<sup>13</sup>C coupling is one bond. The cross peaks mean correlation between a proton and a carbon (Figure&nbsp;1313). If a line does not have cross peak, this means that this carbon atoms has no attached proton (e.g. a quaternary carbon atom)</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9388/Fig13.bmp?revision=1&amp;size=bestfit&amp;width=512&amp;height=439" alt="Fig13.bmp"/><figcaption>Figure&nbsp;1313.&nbsp;<sup>1</sup>H-<sup>13</sup>C COSY spectrum</figcaption></figure>



<h2 class="wp-block-heading" id="references">References</h2>



<ol class="wp-block-list"><li>Balc*, M., Basic p1 sH- and p13 sC-NMR spectroscopy. 1st ed.; Elsevier: Amsterdam ; Boston, 2005; p xii, 427.</li><li>Breitmaier, E., Structure elucidation by NMR in organic chemistry : a practical guide. 3rd rev. ed.; Wiley: Chichester, West Sussex, England, 2002; p xii, 258.</li><li>Jacobsen, N. E., NMR spectroscopy explained : simplified theory, applications and examples for organic chemistry and structural biology. Wiley-Interscience: Hoboken, N.J., 2007; p xv, 668.</li><li>Silverstein, R. M.; Webster, F. X., Spectrometric identification of organic compounds. 6th ed.; Wiley: New York, 1998; p xiv, 482.</li></ol>



<h2 class="wp-block-heading" id="outside-links">Outside Links</h2>



<ul class="wp-block-list"><li>NMRShiftDB: a Free web database for NMR data : nmrshiftdb.chemie.uni-mainz.de/nmrshiftdb</li><li>NMR database from ACD/LAbs : www.acdlabs.com/products/spec_lab/exp_spectra/spec_libraries/aldrich.html</li><li>NMR database from John Crerar Library :&nbsp;<a href="http://crerar.typepad.com/crerar_lib...h_ir_nmr_.html" rel="noreferrer noopener" target="_blank">http://crerar.typepad.com/crerar_lib&#8230;h_ir_nmr_.html</a></li></ul>



<h2 class="wp-block-heading" id="problems">Problems</h2>



<p>Draw the 1H NMR spectrum for 2-Hydroxypropane in CDCl3. Assume sufficient resolution to provide a first-order spectrum and ignore vicinal proton-proton coupling(3JHH)</p>



<h2 class="wp-block-heading" id="solution">Solution</h2>



<p>1) the structure of 2-hydoroxyporpane is drawn</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9397/Pro1_1.bmp?revision=1" alt="Pro1_1.bmp"/></figure>



<p>Figure out which protons are chemically equivalent, i.e., two methyl (-CH<sub>3</sub>) groups are chemical equivalent.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9398/Pro1_2.bmp?revision=1" alt="Pro1_2.bmp"/><figcaption>Figure1): chemical shift of methyl groups (H<sub>a</sub>) : 1-2 ppm (?H<sub>a</sub>=1.1 ppm); chemical shift of -CH- groups (H<sub>b</sub>) moves to downfield due to effect on aldehyde groups:2-3ppm ( ?H<sub>b</sub>=2.4 ppm); chemical shift of aldehyde groups (H<sub>c</sub>):9-10 ppm (?H<sub>c</sub>=9.6 ppm)</figcaption></figure>



<p>4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by H<sub>b</sub>(#of proton=1). H<sub>b</sub>&nbsp;has the septet pattern by H<sub>a</sub>&nbsp;(#of proton=6). H<sub>c</sub>&nbsp;has one peak.(Note that H<sub>c</sub>&nbsp;has doublet pattern by H<sub>b</sub>&nbsp;due to vicinal proton-proton coupling.)</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/9379/Answer1.bmp?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/9379/Answer1.bmp?revision=1&amp;size=bestfit&amp;width=720&amp;height=308" alt="Answer1.bmp"/></a></figure>



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					<description><![CDATA[Click here to see more posts about NMR Only 15$ for interpretation of your NMR spectrum Payment Upon Completion Send your results... This handout relates the basic theory of NMR described on&#160;the theory web handout&#160;with spectra of real molecules and how to deduce structure from the spectra. Before reading this handout, you need to be [&#8230;]]]></description>
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<pre class="wp-block-verse has-text-align-center"><span style="color:#ffffff" class="tadv-color">Only 15$ for interpretation of your NMR spectrum
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<p>This handout relates the basic theory of NMR described on&nbsp;the theory web handout&nbsp;with spectra of real molecules and how to deduce structure from the spectra. Before reading this handout, you need to be thoroughly familiar with all of theory concepts that were described.</p>



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<p><strong>1.0&nbsp;</strong>The NMR spectrum.</p>



<p><strong>1.1</strong>&nbsp;Because different amounts of electron density are around different non-eqivalent nuclei, the different non-equivalent nuclei in a molecule are experiencing slightly different net magnetic fields in an NMR experiment<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">&nbsp;(Review Section&nbsp;<strong>5.2A</strong>&nbsp;of the theory handout)</a>. Recall also that the difference in energy between the two allowed spin states (+1/2 and -1/2 spin states) of a spin 1/2 nucleus (like in 1H and 13C nuclei) depends on the exact magnetic field felt by the nucleus&nbsp;<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">(Review Section&nbsp;<strong>2.3C</strong>in the theory handout)</a>. Recall further that in the NMR experiment, when and only when nuclei are irradiated with electromagnetic radiation of energy that<em>&nbsp;exactly</em>corresponds to the energy difference between the +1/2 and -1/2 spin states, the nuclei absorb the energy and the NMR spectrometer measures this absorbance&nbsp;<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">(Review section&nbsp;<strong>3.1</strong>&nbsp;of the theory handout)</a>. The absorbance of energy to convert a nucleus from a +1/2 to a -1/2 spin state is referred to as &#8220;resonance&#8221; of that nucleus.<strong>1.1A</strong>&nbsp;The key conclusion is that nuclei with different electron densities have +1/2 and -1/2 spin states that differ in energy by differing amounts, so these nuclei will absorb electromagnetic radiation of different frequencies in the NMR experiment.</p>



<p><strong>1.1B&nbsp;</strong>Nuclei surrounded by greater amounts of electron density will be more shielded from the external magnetic field, so they will absorb electromagnetic radiation of lower energy, that is, lower frequency&nbsp;(energy is proportional to frequency).&nbsp;You may want to review&nbsp;<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">Section&nbsp;<strong>5.2A</strong>&nbsp;of the theory handout again</a>.</p>



<p><strong>1.1C</strong>&nbsp;The converse is also true, namely that nuclei surrounded by lesser amounts of electron density will be less shielded (referred to as being &#8220;deshielded&#8221;) from the external magnetic field, so they will absorb electromagnetic radiation of higher energy, that is, higher frequency(energy is proportional to frequency).</p>



<p><strong>1.1D&nbsp;</strong>The three most important factors influencing the electron density around a hydrogen nucleus are: (i) adjacent electronegative atoms remove electron density; (ii) hybridization of the attached carbon atom, increasing shielding is observed in the order sp2, sp, sp3; (iii) adjacent pi bonds are deshielding, which relates to (ii).<strong>1.2</strong>&nbsp;An NMR spectrum is a plot of absorbance versus frequency.</p>



<p><strong>1.2A</strong>&nbsp;To make different spectra directly comparable, a standard is used for all NMR spectra. For 1H NMR spectra, the standard is called tetramethylsilane (TMS)&nbsp;and a small amount of TMS is usually added to any 1H NMR sample.</p>



<p><strong>1.2B</strong>&nbsp;Magnets of different strengths lead to absorbance of electromagnetic radiation at different frequencies for the same nucleus, meaning that if simple frequency were plotted in an NMR spectra, you could not compare spectra taken of the same sample on machines with different magnet strengths. To solve this problem,&nbsp;the frequency of absorption plotted on NMR spectra are corrected for the magnet strength. In addition, frequency is correlated to the reference compound TMS.&nbsp;The frequency at which TMS absorbs is defined as 0 frequency by convention. In the NMR spectrum, absorbance frequencies of electromagnetic radiation are plotted as chemical shift (d) listed in units called&nbsp;<strong>p</strong>arts&nbsp;<strong>p</strong>er&nbsp;<strong>m</strong>illion (ppm) that is defined by the following equation:</p>



<figure class="wp-block-image"><img decoding="async" src="http://iverson.cm.utexas.edu/courses/310N/GIFssp04/miscgifs/NMRhandout4.gif" alt=""/></figure>



<figure class="wp-block-image"><img decoding="async" src="http://iverson.cm.utexas.edu/courses/310N/GIFssp04/miscgifs/NMRhandout5.gif" alt=""/></figure>



<p><strong>1.3 The bottom line to this entire section is that the hydrogen atoms of different functional groups (methyl groups, -CH2- groups, aldehyde -C(O)H, alkene C-H, etc.) have characteristic chemical shifts, i.e. absorbance frequencies. These characteristic chemical shifts are collected in tables such as Fgure 13.8 and Appendix 4 of your book. From the chemical shift information, you thus know what functional groups are present in a molecule.</strong></p>



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<p><strong>1.4 Chemically equivalent hydrogen atoms will have the same chemical shift and therefore give rise to the same signal.</strong>&nbsp;This is why we defined equivalent atoms in<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">&nbsp;Section<strong>6.1</strong>&nbsp;of the theory handout.</a>&nbsp;Non-equivalent groups of hydrogens will have different chemical shifts.&nbsp;<strong>Thus, you will have as many different signals in an NMR spectrum as there are chemically non-equivalent groups of hydrogen atoms.</strong></p>



<p><strong>2.0</strong>&nbsp;The nuclear spin of hydrogen atoms creates a magnetic field that influences the chemical shift of nearby hydrogen atoms&nbsp;<a href="http://iverson.cm.utexas.edu/courses/310N/Handouts/NMRhandout.html" target="_blank" rel="noopener">(Review Sections 5.1 and 5.2)</a>.</p>



<blockquote class="wp-block-quote is-layout-flow wp-block-quote-is-layout-flow"><p><strong>2.1&nbsp;</strong>Nuclear spin magnetic fields will influence hydrogen atoms that are three or fewer bonds away from each other in the same molecule.&nbsp;Hydrogen atoms that are four or greater bonds away usually do not influence each other.</p><p><strong>2.2&nbsp;</strong>A hydrogen atom with a nuclecus in a spin state of +1/2 produces a slightly different magnetic field than a one in a –1/2 spin state.</p><p><strong>2.3&nbsp;</strong>Even in a strong magnetic field, across a population of molecules, there is only a very slight excess of nuclei in the +1/2 spin state.</p><p><strong>2.4&nbsp;</strong>Putting all of these ideas together means the following: Consider a hydrogen X adjacent (three bonds away) to another hydrogen Y in a molecule. In around half of the molecules in the NMR sample, hydrogen X feels the magnetic field from a Y with nuclear spin of +1/2. The other half feel from Y a nuclear spin of –1/2. Thus, when you look at the spectrum, there are actually two different, but closely spaced peaks as the signal for hydrogen X. This phenomenon is called “spin-spin” splitting, and the distance between the two signals for X is called the “coupling constant”, often denoted as “J”. Similarly, the signal for Y actually has two peaks because of spin-spin splitting by X.</p><p><strong>2.5&nbsp;</strong>Consider a –CH2- group adjacent to a hydrogen X.&nbsp;Both of the hydrogen atoms in the –CH2- are chemically equivalent and could be either in the +1/2 or –1/2 nuclear spin state. Thus, there are three situations possible:<strong>&nbsp;i)</strong>&nbsp;+1/2,+1/2;<strong>&nbsp;ii)</strong>&nbsp;+1/2,-1/2, which is the same as –1/2, +1/2 and<strong>&nbsp;iii)</strong>&nbsp;–1/2,-1/2. Thus, there are actually three different magnetic fields that are felt by X in molecules of the sample, in a 1:2:1 ratio.&nbsp;Thus, the signal for hydrogen X is split into three peaks in a 1:2:1 ratio.</p><p><strong>2.6</strong>&nbsp;The same holds for a –CH3&nbsp;group, that will split an adjacent hydrogen signal into four peaks, with a 1:3:3:1 ratio. You should verify this for yourself by making all the possible combinations of nuclear spins for the three equivalent hydrogen atoms of a methyl group.</p><p><strong>2.7&nbsp;</strong>In the general case, N equivalent hydrogen atoms will split an adjacent signal into (N+1) peaks, with relative ratios that are predicted by Pascal’s triangle (Figure 13.16 in the book).</p></blockquote>



<figure class="wp-block-image"><img decoding="async" src="http://iverson.cm.utexas.edu/courses/310N/GIFssp04/miscgifs/splitting.gif" alt=""/></figure>



<p><strong>3.0</strong>&nbsp;Following the same logic, the splitting should&nbsp;<strong>multiply</strong>&nbsp;if a single hydrogen atom is adjacent to hydrogen atoms on either side.&nbsp;Think about combining all the possible nuclear spin states for these nearby sets of hydrogen atoms. Thus, if you have a hydrogen atom X between one –CH2- and one –CH3&nbsp;group, it should be split into an amazing (2+1) x (3 + 1) = 12 signals because there are that many different combinations of +1/2 and -1/2 spins possible.</p>



<blockquote class="wp-block-quote is-layout-flow wp-block-quote-is-layout-flow"><p><strong>3.1&nbsp;</strong>Thus, if the coupling constants (J) from the –CH2- and –CH3&nbsp;groups are significantly different from each other, then 12 peaks will be observed as the signal for hydrogen X.</p><p><strong>3.2&nbsp;</strong>However, in practice, coupling constants (J) are pretty close to the same value for almost all sets of hydrogen atoms in organic molecules, simplifying the splitting pattern, since now many of the twelve peaks will overlap with each other. What this means is that for almost all the spectra you will see, if a hydrogen X is surrounded by N hydrogen atoms, the signal for X will be split into only (N+1) peaks, no matter how those N hydrogen atoms are grouped in terms of sets of equivalent atoms.&nbsp;Thus, what is actually seen for the example above is that the signal for X would&nbsp;<strong><em>appear</em></strong>&nbsp;in the spectrum to be split into 2 + 3 + 1 = 6 peaks, not 12, peaks.&nbsp;<strong>This is the so-called “N+1” rule.</strong></p><p><strong>3.3&nbsp;</strong>The diagram below shows these two different situations. When nuclei from hydrogen atoms Z and Y split the signal for hydrogen X with very different coupling constants (notice how the coupling constant J for the red Z hydrogen nuclei is larger than J for the blue Y hydrogen nuclei), all twelve peaks are spread out and identifiable. Below that is shown the situation in which the coupling constants are the same for nuclei of both Z and Y, so only 6 peaks are actually observed in the signal for hydrogen X due to extensive overlap. This latter case, with six peaks, is what you will almost always see in reality since coupling constants tend to be similar in organic molecules.</p></blockquote>



<figure class="wp-block-image"><img decoding="async" src="http://iverson.cm.utexas.edu/courses/310N/GIFssp04/miscgifs/splitting1.gif" alt=""/></figure>



<blockquote class="wp-block-quote is-layout-flow wp-block-quote-is-layout-flow"><p><strong>3.3&nbsp;</strong>The above explanation of splitting can confuse students for a while. The important point is that in the example given,&nbsp;you see 6 different peaks in the spectrum (N+1 rule) even though there are really 12 peaks produced, it is just that several of them are on top of each other because the coupling constants are the same. For alkyl groups in organic molecules, the coupling constants are generally the same so you will almost always see the fewer peaks, corresponding to the simple N+1 rule, rather than the greater number of peaks derived from the multiplication rule.</p><p><strong>3.4&nbsp;</strong>The bottom line here is that by seeing how a given signal is split, you can figure out how many hydrogen atoms are adjacent on the molecule, namely the number of peaks in the signal minus 1.&nbsp;<strong>From this information you can piece together what a molecule looks like</strong>&nbsp;if you know how many atoms of each type are present (i.e. the molecular formula such as C4H10N2O). You get the molecular formula information from something called a mass spectrum, described later in the text. Molecular formulas will be provided to you in homework or test questions.</p></blockquote>



<p><strong>4.0&nbsp;</strong>For a given signal, integrating the signal (include all splitting peaks for a given signal) gives you a relative value that is proportional to the number of equivalent hydrogen atoms that gave rise to the signal. Thus, by looking at the integration values, you can deduce how many of each type of equivalent hydrogen atoms are in the molecule. For example, a -CH3&nbsp;group would have a signal that integrates to a relative value of 3 (no matter how the signal is split), and a -CH2- group would have a relative integration of 2, etc. Note that sometimes integrations are simply given as absolute numbers, and you must find the common factor to deduce how many hydrogen atoms are represented by each integration value.</p>



<p><strong>5.0 Putting it all together: How to deduce a structure from an NMR spectrum. First, you must be given the molecular formula, so you know how many of each type of atom are present. Second, count the number of different signals and their relative integrations to see how many different sets of equivalent hydrogen atoms are in a molecule, and how many of each set are present. Compare the chemical shifts of each signal to tables to identify what functional groups are present. Finally, use the signal splittings to determine which hydrogen atoms must be no more than 3 bonds away from each other.</strong></p>



<p><strong>6.0&nbsp;</strong>For alkenes, the pi bond prevents bond rotation so the different hydrogen atoms on an&nbsp;<em>unsymmetrical</em>&nbsp;alkene are not equivalent, so they all have different signals, and splitting follows the multiplicative rule (the coupling constants are usually significantly different for geminal vs. cis. vs. trans relationships).</p>



<p><strong>7.0</strong>&nbsp;For hydrogens in a -CH2- group adjacent to a chiral center, the two different H atoms are no longer equivalent, because even with bond rotation, the two hydrogens are never in the same environment with respect to the groups on the adjacent chiral center. Thus, each H of -CH2- group adjacent to a chiral center usually has its own signal in the NMR spectrum.</p>



<p><strong>8.0</strong>&nbsp;There is a great deal more to NMR than this, I am only trying to give you the basics here.</p>



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					<description><![CDATA[Click here to see other posts about XPS Only 10 $ for interpretation of each element in your XPS spectrum Payment Upon Completion Send your spectra... XPST XPST is a program package for the analysis of X-ray Photoelectron Spectroscopy (XPS) data. It includes various graphical interfaces as well as commandline functions to facilitate the workup [&#8230;]]]></description>
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<pre class="wp-block-verse has-text-align-center"><span style="color:#ffffff" class="tadv-color">Only 10 $ for interpretation of each element in your XPS spectrum
</span><strong><mark>Payment Upon Completion
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<a href="http://www.analyzetest.com/index.php/contact-us/"><mark style="background-color:rgba(0, 0, 0, 0)" class="has-inline-color has-vivid-red-color">Send your spectra...</mark></a></pre>



<ol class="wp-block-list"><li>XPST</li></ol>



<p>XPST is a program package for the analysis of X-ray Photoelectron Spectroscopy (XPS) data. It includes various graphical interfaces as well as commandline functions to facilitate the workup of XPS data.</p>



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<p><br><br>When a XPST fit project is started, a corresponding subfolder with all required data is generated and saved within the Igor experiment. You can generate fit templates and you can export entire fit projects to share them with your co-workers. As a special feature, a flexible multiplet function was implemented to facilitate a convenient analysis of complex spectra. XPST can handle any number of peaks.<br><br>There is also a&nbsp;<a href="https://www.youtube.com/channel/UC5bNJCunFL4A08r6VktEBxQ" target="_blank" rel="noopener">youtube channel with tutorials.</a></p>



<p>Several changes in the newest version of XPST were made according to this&nbsp;<a href="https://www.amazon.com/Programming-Igor-Pro-Comprehensive-Introduction/dp/1985792613/ref=sr_1_1?ie=UTF8&amp;qid=1531396271&amp;sr=8-1&amp;keywords=Igor+Pro&amp;dpID=41Vt43IiGGL&amp;preST=_SY291_BO1,204,203,200_QL40_&amp;dpSrc=srch" target="_blank" rel="noopener">book about programming Igor</a>.</p>



<h4 class="wp-block-heading" id="installation">Installation</h4>



<p>XPST was initially developed with Igor 5, but a major revision was made with Igor 7. XPST works also nicely with Igor 8. The Igor 7/8 version is not compatible with Igor 6. If you still run Igor 6, you have to download a previous release.</p>



<ul class="wp-block-list"><li>Unpack the .zip file</li><li>Copy/Move the folder &#8216;XPST&#8217; to the folder &#8216;Igor Procedures&#8217; in Igor&#8217;s main folder</li><li>Copy/Move the folder &#8216;XPSTHelp&#8217; to the folder &#8216;Igor Help Files&#8217; in Igor&#8217;s main folder</li><li>Restart Igor</li></ul>



<p>Before you upgrade to a newer version of XPST, please remove all files associated with your old version from Igor&#8217;s folders.<br>&nbsp;</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<h4 class="wp-block-heading" id="stay-in-touch">Stay in touch &#8230;</h4>



<p>If you want to stay informed about updates or other issues, just send a blank mail to:&nbsp;<a href="mailto:xpst_update@freenet.de">xpst_update@freenet.de</a><br>This is not a support contact &#8211; it only serves to keep you informed about changes.<br>You can sign off anytime you want using the same address.<br>&nbsp;</p>



<h4 class="wp-block-heading" id="more-functions">More functions &#8230;</h4>



<p>Besides the graphical interfaces, XPST comes with several commandline functions. For example:</p>



<ul class="wp-block-list"><li>WhereIs() &#8230;. returns the absolute path of a selected wave (so it can be easily found)</li><li>WaveOverlap() &#8230;. computes the overlap of two selected waves and saves it to a new wave</li><li>CursorCut() &#8230;. cuts out regions from a selected wave</li></ul>



<h4 class="wp-block-heading" id="limitations">Limitations</h4>



<ul class="wp-block-list"><li>Data with &#8216;kinetic energy&#8217; as x-axis (in waveform format or not) can not be analyzed with the Fit Assistant. Only a positive &#8216;binding energy&#8217; scale works &#8211; however, this should be the most common case. If there is a strong demand for kinetic energies, it could be implemented in future versions.</li></ul>



<p class="has-text-align-center"><strong>See: https://www.wavemetrics.com/project/XPStools</strong></p>



<p>2. <strong>CasaXPS</strong></p>



<p>CasaXPS processing software offers powerful analysis techniques for both spectral and imaging data. The system originally designed for XPS and Auger data now offers features covering a wide range of analytical techniques including ToF SIMS, dynamic SIMS and many more.</p>



<p><strong>Features in CasaXPS include</strong>:</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Full quantification including transmission correction.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Configurable quantification reports.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Background type ranging from the standard Linear, Shirley and Tougaard to user-defined cubic-spline approximations.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Asymmetric and symmetric line-shapes:&nbsp;<a href="http://www.casaxps.com/help_manual/manual_updates/calib_asymmetric_peaks.pdf" target="_blank" rel="noopener">Doniach-Sunjic</a>, Voigt, Gaussian-Lorentzian and line-shapes defined from data.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Easy-to-use propagation of processing, annotation and peak models to other data.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Batch processing for repetitive tasks, including configurable processing, display layout with automatic printing and report generation.</p>



<p>·&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;State-of-the-art image processing for XPS spectromicroscopy offering quantified chemical-state XPS images.</p>



<p class="has-text-align-center"><strong>See: http://www.casaxps.com/</strong></p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<p>3. Originlab</p>



<p>Origin provides powerful and versatile tools such as&nbsp;<a rel="noreferrer noopener" href="https://www.originlab.com/doc/Origin-Help/PeakAnalyzer" target="_blank">Peak Analyzer</a>,&nbsp;<a rel="noreferrer noopener" href="https://www.originlab.com/doc/Origin-Help/Gadget-QuickPeaks" target="_blank">Quick Peaks Gadget</a>,&nbsp;<a rel="noreferrer noopener" href="https://www.originlab.com/doc/Origin-Help/Gadget-Integration" target="_blank">Integration Gadget</a>, etc. for baseline correction, peak detection, peak integration and peak fitting. </p>



<p>Origin provides baseline detection and subtraction. Key features include:</p>



<ul class="wp-block-list"><li>Create baseline with many baseline modes<ul><li>Commonly used baseline types such as Constant, Straight line, Use Existing Dataset, End Points Weighted, Min&amp;Max</li><li>Automatically find anchor points for baseline, modify the points, create baseline by interpolation or fitting</li><li>Create baseline for XPS using Shirley or Tougaard method and adjust corresponding parameters to optimize</li><li>Create baseline using Asymmetric Least Squares (ALS) Smoothing method and adjust corresponding parameters to optimize</li></ul></li><li>Subtract baseline</li></ul>



<p>Origin allows you to&nbsp;search for peaks including hidden (&#8220;convoluted&#8221;) peaks&nbsp;and&nbsp;filter out unwanted peaks&nbsp;or&nbsp;manually add or remove peaks.</p>



<ul class="wp-block-list"><li>Savitzky-Golay smoothing on the spectrum before peak finding</li></ul>



<p>In Origin, you can integrate data with multiple peaks, to obtain peak areas, FWHM and other peak characteristics. Baseline subtraction is supported before peak integration.&nbsp;<br><br>Available options include:&nbsp;</p>



<ul class="wp-block-list"><li>Directly select desired data range on graph</li><li>Instantly view results of peak area and FWHM</li><li>Subtract baseline from peak data</li><li>Auto detect peak positions</li><li>Auto determine peak widths for overlapped peaks</li><li>Individually set peak widths</li><li>Fit peaks and obtain fitted peak areas</li></ul>



<p>Origin provides many tools to perform peak fitting:&nbsp;</p>



<ul class="wp-block-list"><li><strong>Quick Peaks Gadget</strong>&nbsp;<a href="https://www.originlab.com/Index.aspx?go=Support/VideoTutorials&amp;pid=3164" target="_blank" rel="noopener"></a>&nbsp;: Visually correct baseline, find and fit peaks</li><li><strong>Multiple Peak Fit</strong>: Manually pick peak positions and fit peaks with same function. No baseline correction</li><li><strong>Peak Analyzer</strong>: Correct baseline, find peaks and fit by Peak Analyzer wizard</li><li><strong>Nonlinear Curve Fit Dialog</strong>: Fit multiple peaks with replicas in the nonlinear curve fit dialog</li></ul>



<p><br>Available options for peak fitting include:</p>



<ul class="wp-block-list"><li>Fit peaks with built-in or user-defined functions</li><li>Fit multiple peaks with different functions</li><li>Control fitting process using bounds and constraints</li><li>Fix or share peak parameters</li><li>Vary baseline parameters along with peak fitting</li></ul>



<p class="has-text-align-center"><strong>See: https://www.originlab.com/index.aspx?go=Products/Origin/DataAnalysis/PeakAnalysis#Peak_Fitting_PRO</strong></p>



<p>4. XPSPEAK</p>



<p>Free, fully featured, software for the analysis of XPS spectra written by Raymund Kwok.XPSPeak is a XPS Peak Fitting Program.The portable app creates a sandbox folder in its current location, where it stores all its settings and temporary files. Can be downloaded from the US, UK or Hong Kong.</p>



<p class="has-text-align-center"><strong>See: https://xpspeak.software.informer.com/4.1/</strong></p>



<p>5. Unifit</p>



<p class="has-text-align-center"><strong>See: https://home.uni-leipzig.de/unifit/downloads.htm</strong></p>



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		<title>Everythings about UV-Vis spectroscopy</title>
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					<description><![CDATA[Click here to see more posts about UV-Vis spectroscopy Only 10$ for interpretation of your UV-Vis spectrum Payment Upon Completion Send your UV-Vis spectra... In Figure 10.9 we examined Nessler’s original method for matching the color of a sample to the color of a standard. Matching the colors was a labor intensive process for the [&#8230;]]]></description>
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<p class="has-text-align-center"><strong><em><a href="http://www.analyzetest.com/index.php/category/analyzing/uv-vis/">Click here to see more posts about UV-Vis spectroscopy</a></em></strong></p>



<pre class="wp-block-verse has-text-align-center"><span style="color:#ffffff" class="tadv-color">Only 10$ for interpretation of your UV-Vis spectrum
</span><strong><mark>Payment Upon Completion
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<p>In Figure 10.9 we examined Nessler’s original method for matching the color of a sample to the color of a standard. Matching the colors was a labor intensive process for the analyst. Not surprisingly, spectroscopic methods of analysis were slow to develop. The 1930s and 1940s saw the introduction of photoelectric transducers for ultraviolet and visible radiation, and thermocouples for infrared radiation. As a result, modern instrumentation for absorption spectroscopy became routinely available in the 1940s—progress has been rapid ever since.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s7.picofile.com/file/8392387600/uv_vis_in.gif" alt=""/></a></figure>



<span id="more-508"></span>



<h2 class="wp-block-heading" id="10-3-1-instrumentation">10.3.1 Instrumentation</h2>



<p>Frequently an analyst must select—from among several instruments of different design—the one instrument best suited for a particular analysis. In this section we examine several different instruments for molecular absorption spectroscopy, emphasizing their advantages and limitations. Methods of sample introduction are also covered in this section.</p>



<h3 class="wp-block-heading" id="instrument-designs-for-molecular-uv-vis-absorption">Instrument Designs for Molecular UV/Vis Absorption</h3>



<p><em><strong>Filter Photometer.</strong></em>&nbsp;The simplest instrument for molecular UV/Vis absorption is a&nbsp;<strong>filter photometer</strong>&nbsp;(Figure 10.25), which uses an absorption or interference filter to isolate a band of radiation. The filter is placed between the source and the sample to prevent the sample from decomposing when exposed to higher energy radiation. A filter photometer has a single optical path between the source and detector, and is called a&nbsp;<strong>single-beam</strong>&nbsp;instrument. The instrument is calibrated to 0% T while using a shutter to block the source radiation from the detector. After opening the shutter, the instrument is calibrated to 100% T using an appropriate blank. The blank is then replaced with the sample and its transmittance measured. Because the source’s incident power and the sensitivity of the detector vary with wavelength, the photometer must be recalibrated whenever the filter is changed. Photometers have the advantage of being relatively inexpensive, rugged, and easy to maintain. Another advantage of a photometer is its portability, making it easy to take into the field. Disadvantages of a photometer include the inability to record an absorption spectrum and the source’s relatively large effective bandwidth, which limits the calibration curve’s linearity.</p>



<p>Note</p>



<p>The percent transmittance varies between 0% and 100%. As we learned in Figure 10.21, we use a blank to determine&nbsp;<em>P</em><sub>0</sub>, which corresponds to 100% T. Even in the absence of light the detector records a signal. Closing the shutter allows us to assign 0% T to this signal. Together, setting 0% T and 100% T calibrates the instrument. The amount of light passing through a sample produces a signal that is greater than or equal to that for 0% T and smaller than or equal to that for 100%T.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12706/Figure10.25.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12706/Figure10.25.jpg?revision=1&amp;size=bestfit&amp;width=474&amp;height=362" alt="Figure10.25.jpg"/></a></figure>



<p><strong>Figure 10.25</strong>&nbsp;Schematic diagram of a filter photometer. The analyst either inserts a removable filter or the filters are placed in a carousel, an example of which is shown in the photographic inset. The analyst selects a filter by rotating it into place.</p>



<p><em><strong>Single-Beam Spectrophotometer.&nbsp;</strong></em>An instrument that uses a monochromator for wavelength selection is called a&nbsp;<strong>spectrophotometer</strong>. The simplest spectrophotometer is a single-beam instrument equipped with a fixed-wavelength monochromator (Figure 10.26). Single-beam spectrophotometers are calibrated and used in the same manner as a photometer. One example of a single-beam spectrophotometer is Thermo Scientific’s Spectronic 20D+, which is shown in the photographic insert to Figure 10.26. The Spectronic 20D+ has a range of 340–625 nm (950 nm when using a red-sensitive detector), and a fixed effective bandwidth of 20 nm. Battery-operated, hand-held single-beam spectrophotometers are available, which are easy to transport into the field. Other single-beam spectrophotometers also are available with effective bandwidths of 2–8 nm. Fixed wavelength single-beam spectrophotometers are not practical for recording spectra because manually adjusting the wavelength and recalibrating the spectrophotometer is awkward and time-consuming. The accuracy of a single-beam spectrophotometer is limited by the stability of its source and detector over time.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12707/Figure10.26.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12707/Figure10.26.jpg?revision=1&amp;size=bestfit&amp;width=475&amp;height=349" alt="Figure10.26.jpg"/></a></figure>



<p><strong>Figure 10.26</strong>&nbsp;Schematic diagram of a fixed-wavelength single-beam spectrophotometer. The photographic inset shows a typical instrument. The shutter remains closed until the sample or blank is placed in the sample compartment. The analyst manually selects the wavelength by adjusting the wavelength dial. Inset photo modified from:&nbsp;<a href="http://commons.wikimedia.org/wiki/File:Spektrofotometr.JPG" rel="noreferrer noopener" target="_blank">Adi</a>&nbsp;(www.commons.wikipedia.org).</p>



<p><em><strong>Double-Beam Spectrophotometer.</strong></em>&nbsp;The limitations of fixed-wavelength, single-beam spectrophotometers are minimized by using a&nbsp;<strong>double-beam</strong>spectrophotometer (Figure 10.27). A chopper controls the radiation’s path, alternating it between the sample, the blank, and a shutter. The signal processor uses the chopper’s known speed of rotation to resolve the signal reaching the detector into the transmission of the blank,&nbsp;<em>P</em><sub>0</sub>, and the sample,&nbsp;<em>P</em><sub>T</sub>. By including an opaque surface as a shutter, it is possible to continuously adjust 0% T. The effective bandwidth of a double-beam spectrophotometer is controlled by adjusting the monochromator’s entrance and exit slits. Effective bandwidths of 0.2–3.0 nm are common. A scanning monochromator allows for the automated recording of spectra. Double-beam instruments are more versatile than single-beam instruments, being useful for both quantitative and qualitative analyses, but also are more expensive.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12708/Figure10.27.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12708/Figure10.27.jpg?revision=1&amp;size=bestfit&amp;width=700&amp;height=437" alt="Figure10.27.jpg"/></a></figure>



<p><strong>Figure 10.27</strong>&nbsp;Schematic diagram of a scanning, double-beam spectrophotometer. A chopper directs the source’s radiation, using a transparent window to pass radiation to the sample and a mirror to reflect radiation to the blank. The chopper’s opaque surface serves as a shutter, which allows for a constant adjustment of the spectrophotometer’s 0% T. The photographic insert shows a typical instrument. The unit in the middle of the photo is a temperature control unit that allows the sample to be heated or cooled.</p>



<p><em><strong>Diode Array Spectrometer.</strong></em>&nbsp;An instrument with a single detector can monitor only one wavelength at a time. If we replace a single photomultiplier with many photodiodes, we can use the resulting array of detectors to record an entire spectrum simultaneously in as little as 0.1 s. In a diode array spectrometer the source radiation passes through the sample and is dispersed by a grating (Figure 10.28). The photodiode array is situated at the grating’s focal plane, with each diode recording the radiant power over a narrow range of wavelengths. Because we replace a full monochromator with just a grating, a diode array spectrometer is small and compact.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12709/Figure10.28.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12709/Figure10.28.jpg?revision=1&amp;size=bestfit&amp;width=565&amp;height=431" alt="Figure10.28.jpg"/></a></figure>



<p><strong>Figure 10.28</strong>&nbsp;Schematic diagram of a diode array spectrophotometer. The photographic insert shows a typical instrument. Note that the 50-mL beaker provides a sense of scale.</p>



<p>One advantage of a diode array spectrometer is the speed of data acquisition, which allows to collect several spectra for a single sample. Individual spectra are added and averaged to obtain the final spectrum. This&nbsp;<strong>signal averaging</strong>&nbsp;improves a spectrum’s signal-to-noise ratio. If we add together&nbsp;<em>n</em>&nbsp;spectra, the sum of the signal at any point,&nbsp;<em>x</em>, increases as&nbsp;<em>nS<sub>x</sub></em>, where&nbsp;<em>S<sub>x</sub>&nbsp;</em>is the signal. The noise at any point,&nbsp;<em>N<sub>x</sub></em>, is a random event, which increases as&nbsp;√<em>nN<sub>x</sub></em>&nbsp;when we add together&nbsp;<em>n</em>spectra. The&nbsp;<strong>signal-to-noise ratio</strong>&nbsp;(S/N) after&nbsp;<em>n</em>&nbsp;scans isSN=nSxn−−√Nx=n−−√SxnNx(4.8.1)(4.8.1)SN=nSxnNx=nSxnNx</p>



<p>where&nbsp;<em>S<sub>x</sub></em>/<em>N<sub>x</sub></em>&nbsp;is the signal-to-noise ratio for a single scan. The impact of signal averaging is shown in Figure 10.29. The first spectrum shows the signal for a single scan, which consists of a single, noisy peak. Signal averaging using 4 scans and 16 scans decreases the noise and improves the signal-to-noise ratio. One disadvantage of a photodiode array is that the effective bandwidth per diode is roughly an order of magnitude larger than that for a high quality monochromator.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12710/Figure10.29.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12710/Figure10.29.jpg?revision=1&amp;size=bestfit&amp;width=275&amp;height=734" alt="Figure10.29.jpg"/></a></figure>



<p><strong>Figure 10.29</strong>&nbsp;Effect of signal averaging on a spectrum’s signal-to-noise ratio. From top to bottom: spectrum for a single scan; average spectrum after four scans; and average spectrum after adding 16 scans.</p>



<p><em><strong>Sample Cells.&nbsp;</strong></em>The sample compartment provides a light-tight environment that limits the addition of stray radiation. Samples are normally in the liquid or solution state, and are placed in cells constructed with UV/Vis transparent materials, such as quartz, glass, and plastic (Figure 10.30). A quartz or fused-silica cell is required when working at a wavelength &lt;300 nm where other materials show a significant absorption. The most common pathlength is 1 cm (10 mm), although cells with shorter (as little as 0.1 cm) and longer pathlengths (up to 10 cm) are available. Longer pathlength cells are useful when analyzing a very dilute solution, or for gas samples. The highest quality cells allow the radiation to strike a flat surface at a 90<sup>o</sup>&nbsp;angle, minimizing the loss of radiation to reflection. A test tube is often used as a sample cell with simple, single-beam instruments, although differences in the cell’s pathlength and optical properties add an additional source of error to the analysis.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12781/Figure_10.30.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12781/Figure_10.30.jpg?revision=1&amp;size=bestfit&amp;width=675&amp;height=262" alt="Figure 10.30.jpg"/></a></figure>



<p><strong>Figure 10.30</strong>&nbsp;Examples of sample cells for UV/Vis spectroscopy. From left to right (with path lengths in parentheses): rectangular plastic cuvette (10.0 mm), rectangular quartz cuvette (5.000 mm), rectangular quartz cuvette (1.000 mm), cylindrical quartz cuvette (10.00 mm), cylindrical quartz cuvette (100.0 mm). Cells often are available as a matched pair, which is important when using a double-beam instrument.</p>



<p>If we need to monitor an analyte’s concentration over time, it may not be possible to physically remove samples for analysis. This is often the case, for example, when monitoring industrial production lines or waste lines, when monitoring a patient’s blood, or when monitoring environmental systems. With a&nbsp;<strong>fiber-optic probe</strong>&nbsp;we can analyze samples in situ. An example of a remote sensing fiber-optic probe is shown in Figure 10.31. The probe consists of two bundles of fiber-optic cable. One bundle transmits radiation from the source to the probe’s tip, which is designed to allow the sample to flow through the sample cell. Radiation from the source passes through the solution and is reflected back by a mirror. The second bundle of fiber-optic cable transmits the nonabsorbed radiation to the wavelength selector. Another design replaces the flow cell shown in Figure 10.31 with a membrane containing a reagent that reacts with the analyte. When the analyte diffuses across the membrane it reacts with the reagent, producing a product that absorbs UV or visible radiation. The nonabsorbed radiation from the source is reflected or scattered back to the detector. Fiber optic probes that show chemical selectivity are called optrodes.<sup>6</sup></p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12711/Figure10.31.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12711/Figure10.31.jpg?revision=1&amp;size=bestfit&amp;width=500&amp;height=324" alt="Figure10.31.jpg"/></a></figure>



<p><strong>Figure 10.31</strong>&nbsp;Example of a fiber-optic probe. The inset photographs provide a close-up look at the probe’s flow cell and the reflecting mirror.</p>



<h3 class="wp-block-heading" id="instrument-designs-for-infrared-absorption">Instrument Designs for Infrared Absorption</h3>



<p><em><strong>Filter Photometer.</strong></em>&nbsp;The simplest instrument for IR absorption spectroscopy is a filter photometer similar to that shown in Figure 10.25 for UV/Vis absorption. These instruments have the advantage of portability, and typically are used as dedicated analyzers for gases such as HCN and CO.</p>



<p><em><strong>Double-beam spectrophotometer.</strong></em>&nbsp;Infrared instruments using a monochromator for wavelength selection use double-beam optics similar to that shown in Figure 10.27. Double-beam optics are preferred over single-beam optics because the sources and detectors for infrared radiation are less stable than those for UV/Vis radiation. In addition, it is easier to correct for the absorption of infrared radiation by atmospheric CO<sub>2</sub>&nbsp;and H<sub>2</sub>O vapor when using double-beam optics. Resolutions of 1–3 cm<sup>–1</sup>&nbsp;are typical for most instruments.</p>



<p><em><strong>Fourier transform spectrometer.</strong></em>&nbsp;In a Fourier transform infrared spectrometer, or FT–IR, the monochromator is replaced with an interferometer (Figure 10.13). Because an FT-IR includes only a single optical path, it is necessary to collect a separate spectrum to compensate for the absorbance of atmospheric CO<sub>2</sub>&nbsp;and H<sub>2</sub>O vapor. This is done by collecting a background spectrum without the sample and storing the result in the instrument’s computer memory. The background spectrum is removed from the sample’s spectrum by ratioing the two signals. In comparison to other instrument designs, an FT–IR provides for rapid data acquisition, allowing an enhancement in signal-to-noise ratio through signal-averaging.</p>



<p><em><strong>Sample Cells.</strong></em>&nbsp;Infrared spectroscopy is routinely used to analyze gas, liquid, and solid samples. Sample cells are made from materials, such as NaCl and KBr, that are transparent to infrared radiation. Gases are analyzed using a cell with a pathlength of approximately 10 cm. Longer pathlengths are obtained by using mirrors to pass the beam of radiation through the sample several times.</p>



<p>A liquid samples may be analyzed using a variety of different sample cells (Figure 10.32). For non-volatile liquids a suitable sample can be prepared by placing a drop of the liquid between two NaCl plates, forming a thin film that typically is less than 0.01 mm thick. Volatile liquids must be placed in a sealed cell to prevent their evaporation.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12712/Figure10.32.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12712/Figure10.32.jpg?revision=1&amp;size=bestfit&amp;width=700&amp;height=403" alt="Figure10.32.jpg"/></a></figure>



<p><strong>Figure 10.32</strong>&nbsp;Three examples of IR sample cells: (a) NaCl salts plates; (b) fixed pathlength (0.5 mm) sample cell with NaCl windows; (c) disposable card with a polyethylene window that is IR transparent with the exception of strong absorption bands at 2918 cm<sup>–1</sup>&nbsp;and 2849 cm<sup>–1</sup>.</p>



<p>The analysis of solution samples is limited by the solvent’s IR absorbing properties, with CCl<sub>4</sub>, CS<sub>2</sub>, and CHCl<sub>3</sub>&nbsp;being the most common solvents. Solutions are placed in cells containing two NaCl windows separated by a Teflon spacer. By changing the Teflon spacer, pathlengths from 0.015–1.0 mm can be obtained.</p>



<p>Transparent solid samples can be analyzed directly by placing them in the IR beam. Most solid samples, however, are opaque, and must be dispersed in a more transparent medium before recording the IR spectrum. If a suitable solvent is available, then the solid can be analyzed by preparing a solution and analyzing as described above. When a suitable solvent is not available, solid samples may be analyzed by preparing a mull of the finely powdered sample with a suitable oil. Alternatively, the powdered sample can be mixed with KBr and pressed into an optically transparent pellet.</p>



<p>The analysis of an aqueous sample is complicated by the solubility of the NaCl cell window in water. One approach to obtaining infrared spectra on aqueous solutions is to use&nbsp;<strong>attenuated total reflectance</strong>&nbsp;instead of transmission. Figure 10.33 shows a diagram of a typical attenuated total reflectance (ATR) FT–IR instrument. The ATR cell consists of a high refractive index material, such as ZnSe or diamond, sandwiched between a low refractive index substrate and a lower refractive index sample. Radiation from the source enters the ATR crystal where it undergoes a series of total internal reflections before exiting the crystal. During each reflection the radiation penetrates into the sample to a depth of a few microns. The result is a selective attenuation of the radiation at those wavelengths where the sample absorbs. ATR spectra are similar, but not identical, to those obtained by measuring the transmission of radiation.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12713/Figure10.33.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12713/Figure10.33.jpg?revision=1&amp;size=bestfit&amp;width=700&amp;height=327" alt="Figure10.33.jpg"/></a></figure>



<p><strong>Figure 10.33</strong>&nbsp;FT-IR spectrometer equipped with a diamond ATR sample cell. The inserts show a close-up photo of the sample platform, a sketch of the ATR’s sample slot, and a schematic showing how the source’s radiation interacts with the sample. The pressure tower is used to ensure the contact of solid samples with the ATR crystal.</p>



<p>Solid samples also can be analyzed using an ATR sample cell. After placing the solid in the sample slot, a compression tip ensures that it is in contact with the ATR crystal. Examples of solids that have been analyzed by ATR include polymers, fibers, fabrics, powders, and biological tissue samples. Another reflectance method is diffuse reflectance, in which radiation is reflected from a rough surface, such as a powder. Powdered samples are mixed with a non-absorbing material, such as powdered KBr, and the reflected light is collected and analyzed. As with ATR, the resulting spectrum is similar to that obtained by conventional transmission methods.</p>



<p>Note</p>



<p>Further details about these, and other methods for preparing solids for infrared analysis can be found in this chapter’s&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/Additional_Resources#Chapter_10" target="_blank" rel="noopener">additional resources</a>.</p>



<h2 class="wp-block-heading" id="10-3-2-quantitative-applications">10.3.2 Quantitative Applications</h2>



<p>The determination of an analyte’s concentration based on its absorption of ultraviolet or visible radiation is one of the most frequently encountered quantitative analytical methods. One reason for its popularity is that many organic and inorganic compounds have strong absorption bands in the UV/Vis region of the electromagnetic spectrum. In addition, if an analyte does not absorb UV/Vis radiation—or if its absorbance is too weak—we often can react it with another species that is strongly absorbing. For example, a dilute solution of Fe<sup>2</sup><sup>+</sup>&nbsp;does not absorb visible light. Reacting Fe<sup>2</sup><sup>+</sup>&nbsp;with&nbsp;<em>o</em>-phenanthroline, however, forms an orange–red complex of Fe(phen)<sub>3</sub><sup>2+</sup>&nbsp;that has a strong, broad absorbance band near 500 nm. An additional advantage to UV/Vis absorption is that in most cases it is relatively easy to adjust experimental and instrumental conditions so that Beer’s law is obeyed.</p>



<p>Note</p>



<p>Figure 10.18 shows the visible spectrum for Fe(phen)<sub>3</sub><sup>2+</sup>.</p>



<p>A quantitative analysis based on the absorption of infrared radiation, although important, is less frequently encountered than those for UV/Vis absorption. One reason is the greater tendency for instrumental deviations from Beer’s law when using infrared radiation. Because an infrared absorption band is relatively narrow, any deviation due to the lack of monochromatic radiation is more pronounced. In addition, infrared sources are less intense than UV/Vis sources, making stray radiation more of a problem. Differences in pathlength for samples and standards when using thin liquid films or KBr pellets are a problem, although an internal standard can be used to correct for any difference in pathlength. Finally, establishing a 100% T (<em>A</em>&nbsp;= 0) baseline is often difficult because the optical properties of NaCl sample cells may change significantly with wavelength due to contamination and degradation. We can minimize this problem by measuring absorbance relative to a baseline established for the absorption band. Figure 10.34 shows how this is accomplished.</p>



<p>Note</p>



<p>Another approach is to use a cell with a fixed pathlength, such as that shown in Figure 10.32b.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12714/Figure10.34.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12714/Figure10.34.jpg?revision=1&amp;size=bestfit&amp;width=450&amp;height=385" alt="Figure10.34.jpg"/></a></figure>



<p><strong>Figure 10.34</strong>&nbsp;Method for determining absorbance from an IR spectrum.</p>



<h3 class="wp-block-heading" id="environmental-applications">Environmental Applications</h3>



<p>The analysis of waters and wastewaters often relies on the absorption of ultraviolet and visible radiation. Many of these methods are outlined in Table 10.6. Several of these methods are described here in more detail.</p>



<figure class="wp-block-table"><table><thead><tr><th scope="col">Analyte</th><th scope="col">Method</th><th scope="col">λ&nbsp;(nm)</th></tr></thead><tbody><tr><td><em>T</em><em>r</em><em>ace Metals</em></td></tr><tr><td>aluminum</td><td>react with Eriochrome cyanide R dye at pH6; forms red to pink complex</td><td>535</td></tr><tr><td>arsenic</td><td>reduce to AsH<sub>3</sub>&nbsp;using Zn and react with silver diethyldithiocarbamate; forms red complex</td><td>535</td></tr><tr><td>cadmium</td><td>extract into CHCl<sub>3</sub>&nbsp;containing dithizone from a sample made basic with NaOH; forms pink to red complex</td><td>518</td></tr><tr><td>chromium</td><td>oxidize to Cr(VI) and react with diphenylcarbazide; forms red-violet product</td><td>540</td></tr><tr><td>copper</td><td>react with neocuprine in neutral to slightly acid solution and extract into CHCl<sub>3</sub>/CH<sub>3</sub>OH; forms yellow complex</td><td>457</td></tr><tr><td>iron</td><td>reduce to Fe<sup>2</sup><sup>+</sup>&nbsp;and react with&nbsp;<em>o</em>-phenanthroline; forms orange-red complex</td><td>510</td></tr><tr><td>lead</td><td>extract into CHCl<sub>3</sub>&nbsp;containing dithizone from sample made basic with NH<sub>3</sub>/NH<sub>4</sub><sup>+</sup>&nbsp;buffer; forms cherry red complex</td><td>510</td></tr><tr><td>manganese</td><td>oxidize to MnO<sub>4</sub><sup>–</sup>&nbsp;with persulfate; forms purple solution</td><td>525</td></tr><tr><td>mercury</td><td>extract into CHCl<sub>3</sub>&nbsp;containing dithizone from acidic sample; forms orange complex</td><td>492</td></tr><tr><td>zinc</td><td>react with zincon at pH 9; forms blue complex</td><td>620</td></tr><tr><td><em>I</em><em>norganic Nonmetals</em></td></tr><tr><td>ammonia</td><td>reaction with hypochlorite and phenol using a manganous salt catalyst; forms blue indophenol as product</td><td>630</td></tr><tr><td>cyanide</td><td>react with chloroamine-T to form CNCl and then with a pyridine-barbituric acid; forms a red-blue dye</td><td>578</td></tr><tr><td>fluoride</td><td>react with red Zr-SPADNS lake; formation of ZrF<sub>6</sub><sup>2–</sup>&nbsp;decreases color of the red lake</td><td>570</td></tr><tr><td>chlorine (residual)</td><td>react with leuco crystal violet; forms blue product</td><td>592</td></tr><tr><td>nitrate</td><td>react with Cd to form NO<sub>2</sub><sup>–</sup>&nbsp;and then react with sulfanilamide and&nbsp;<em>N</em>-(1-napthyl)-ethylenediamine; forms red azo dye</td><td>543</td></tr><tr><td>phosphate</td><td>react with ammonium molybdate and then reduce with SnCl<sub>2</sub>; forms molybdenum blue</td><td>690</td></tr><tr><td><em>O</em><em>rganics</em></td></tr><tr><td>phenol</td><td>react with 4-aminoantipyrine and K<sub>3</sub>Fe(CN)<sub>6</sub>; forms yellow antipyrine dye</td><td>460</td></tr><tr><td>anionic surfactant</td><td>react with cationic methylene blue dye and extract into CHCl<sub>3</sub>; forms blue ion pair</td><td>652</td></tr></tbody></table></figure>



<p><img decoding="async" alt="dithizone.png" width="68px" height="184px" src="https://chem.libretexts.org/@api/deki/files/12102/dithizone.png?revision=1&amp;size=bestfit&amp;width=68&amp;height=184">Although the quantitative analysis of metals in waters and wastewaters is accomplished primarily by atomic absorption or atomic emission spectroscopy, many metals also can be analyzed following the formation of a colored metal–ligand complex. One advantage to these spectroscopic methods is that they are easily adapted to the analysis of samples in the field using a filter photometer. One ligand that is used in the analysis of several metals is diphenylthiocarbazone, also known as dithizone. Dithizone is not soluble in water, but when a solution of dithizone in CHCl<sub>3</sub>&nbsp;is shaken with an aqueous solution containing an appropriate metal ion, a colored metal–dithizonate complex forms that is soluble in CHCl<sub>3</sub>. The selectivity of dithizone is controlled by adjusting the sample’s pH. For example, Cd<sup>2</sup><sup>+</sup>&nbsp;is extracted from solutions that are made strongly basic with NaOH, Pb<sup>2</sup><sup>+</sup>&nbsp;from solutions that are made basic with an NH<sub>3</sub>/NH<sub>4</sub><sup>+</sup>&nbsp;buffer, and Hg<sup>2</sup><sup>+</sup>&nbsp;from solutions that are slightly acidic.</p>



<p>Note</p>



<p>Atomic absorption is the subject of&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.4%3A_Atomic_Absorption_Spectroscopy" target="_blank" rel="noopener">Section 10.4</a>&nbsp;and atomic emission is the subject of&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.7%3A_Atomic_Emission_Spectroscopy" target="_blank" rel="noopener">Section 10.7</a>.</p>



<p>The structure of dithizone is shown to the right. See Chapter 7 for a discussion of extracting metal ions using dithizone.</p>



<p>When chlorine is added to water the portion available for disinfection is called the chlorine residual. There are two forms of chlorine residual. The free chlorine residual includes Cl<sub>2</sub>, HOCl, and OCl<sup>–</sup>. The combined chlorine residual, which forms from the reaction of NH<sub>3</sub>&nbsp;with HOCl, consists of monochloramine, NH<sub>2</sub>Cl, dichloramine, NHCl<sub>2</sub>, and trichloramine, NCl<sub>3</sub>. Because the free chlorine residual is more efficient at disinfection, there is an interest in methods that can distinguish between the different forms of the total chlorine residual. One such method is the leuco crystal violet method. The free residual chlorine is determined by adding leuco crystal violet to the sample, which instantaneously oxidizes to give a blue colored compound that is monitored at 592 nm. Completing the analysis in less than five minutes prevents a possible interference from the combined chlorine residual. The total chlorine residual (free + combined) is determined by reacting a separate sample with iodide, which reacts with both chlorine residuals to form HOI. When the reaction is complete, leuco crystal violet is added and oxidized by HOI, giving the same blue colored product. The combined chlorine residual is determined by difference.</p>



<p>Note</p>



<p>In Chapter 9 we explored how the total chlorine residual can be determined by a redox titration; see&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/09_Titrimetric_Methods/9.4%3A_Redox_Titrations#Representative_Method_9.3" target="_blank" rel="noopener">Representative Method 9.3</a>&nbsp;for further details. The method described here allows us to divide the total chlorine residual into its component parts.</p>



<p>The concentration of fluoride in drinking water may be determined indirectly by its ability to form a complex with zirconium. In the presence of the dye SPADNS, solutions of zirconium form a red colored compound, called a lake, that absorbs at 570 nm. When fluoride is added, the formation of the stable ZrF<sub>6</sub><sup>2–</sup>&nbsp;complex causes a portion of the lake to dissociate, decreasing the absorbance. A plot of absorbance versus the concentration of fluoride, therefore, has a negative slope.</p>



<p>Note</p>



<p>SPADNS, which is shown below, is an abbreviation for the sodium salt of 2-(4-sulfophenylazo)-1,8-dihydroxy-3,6-napthalenedisulfonic acid, which is a mouthful to say.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12103/SPADNS2.png?revision=1" alt="SPADNS2.png"/></figure>



<p><img decoding="async" alt="4-aminoantipyrene.png" src="https://chem.libretexts.org/@api/deki/files/12104/4-aminoantipyrene.png?revision=1">Spectroscopic methods also are used to determine organic constituents in water. For example, the combined concentrations of phenol, and ortho- and meta- substituted phenols are determined by using steam distillation to separate the phenols from nonvolatile impurities. The distillate reacts with 4-aminoantipyrine at pH 7.9 ± 0.1 in the presence of K<sub>3</sub>Fe(CN)<sub>6</sub>, forming a yellow colored antipyrine dye. After extracting the dye into CHCl<sub>3</sub>, its absorbance is monitored at 460 nm. A calibration curve is prepared using only the unsubstituted phenol, C<sub>6</sub>H<sub>5</sub>OH. Because the molar absorptivity of substituted phenols are generally less than that for phenol, the reported concentration represents the minimum concentration of phenolic compounds.</p>



<p>Molecular absorption also can be used for the analysis of environmentally significant airborne pollutants. In many cases the analysis is carried out by collecting the sample in water, converting the analyte to an aqueous form that can be analyzed by methods such as those described in Table 10.6. For example, the concentration of NO<sub>2</sub>&nbsp;can be determined by oxidizing NO<sub>2</sub>&nbsp;to NO<sub>3</sub><sup>–</sup>. The&nbsp;<img decoding="async" alt="red azo dye.png" src="https://chem.libretexts.org/@api/deki/files/12105/red_azo_dye.png?revision=1">concentration of NO<sub>3</sub><sup>–</sup>&nbsp;is then determined by first reducing it to NO<sub>2</sub><sup>–</sup>&nbsp;with Cd, and then reacting NO<sub>2</sub><sup>–</sup>&nbsp;with sulfanilamide and&nbsp;<em>N</em>-(1-naphthyl)-ethylenediamine to form a red azo dye. Another important application is the analysis for SO<sub>2</sub>, which is determined by collecting the sample in an aqueous solution of HgCl<sub>4</sub><sup>2–</sup>&nbsp;where it reacts to form Hg(SO<sub>3</sub>)<sub>2</sub><sup>2–</sup>. Addition of&nbsp;<em>p</em>-rosaniline and formaldehyde produces a purple complex that is monitored at 569 nm. Infrared absorption is useful for the analysis of organic vapors, including HCN, SO<sub>2</sub>, nitrobenzene, methyl mercaptan, and vinyl chloride. Frequently, these analyses are accomplished using portable, dedicated infrared photometers.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s7.picofile.com/file/8392387600/uv_vis_in.gif" alt=""/></a></figure>



<h3 class="wp-block-heading" id="clinical-applications">Clinical Applications</h3>



<p>The analysis of clinical samples is often complicated by the complexity of the sample matrix, which may contribute a significant background absorption at the desired wavelength. The determination of serum barbiturates provides one example of how this problem is overcome. The barbiturates are first extracted from a sample of serum with CHCl<sub>3</sub>&nbsp;and then extracted from the CHCl<sub>3</sub>&nbsp;into 0.45 M NaOH (pH ≈ 13). The absorbance of the aqueous extract is measured at 260 nm, and includes contributions from the barbiturates as well as other components extracted from the serum sample. The pH of the sample is then lowered to approximately 10 by adding NH<sub>4</sub>Cl and the absorbance remeasured. Because the barbiturates do not absorb at this pH, we can use the absorbance at pH 10,&nbsp;<em>A</em><sub>pH</sub><sub>&nbsp;10</sub>, to correct the absorbance at pH 13,&nbsp;<em>A</em><sub>pH</sub><sub>&nbsp;13</sub>Abarb=ApH 13−Vsamp+VNH4ClVsamp×ApH 10(4.8.2)(4.8.2)Abarb=ApH 13−Vsamp+VNH4ClVsamp×ApH 10</p>



<p>where&nbsp;<em>A</em><sub>barb</sub>&nbsp;is the absorbance due to the serum barbiturates, and&nbsp;<em>V</em><sub>samp</sub>&nbsp;and&nbsp;<em>V</em><sub>NH4Cl</sub>&nbsp;are the volumes of sample and NH<sub>4</sub>Cl, respectively. Table 10.7 provides a summary of several other methods for analyzing clinical samples.</p>



<figure class="wp-block-table"><table><thead><tr><th scope="col">Analyte</th><th scope="col">Method</th><th scope="col">λ&nbsp;(nm)</th></tr></thead><tbody><tr><td>total serum protein</td><td>react with NaOH and Cu<sup>2</sup><sup>+</sup>; forms blue-violet complex</td><td>540</td></tr><tr><td>serum cholesterol</td><td>react with Fe<sup>3</sup><sup>+</sup>&nbsp;in presence of isopropanol, acetic acid, and H<sub>2</sub>SO<sub>4</sub>; forms blue-violet complex</td><td>540</td></tr><tr><td>uric acid</td><td>react with phosphotungstic acid; forms tungsten blue</td><td>710</td></tr><tr><td>serum barbiturates</td><td>extract into CHCl<sub>3</sub>&nbsp;to isolate from interferents and then extract into 0.45 M NaOH</td><td>260</td></tr><tr><td>glucose</td><td>react with&nbsp;<em>o</em>-toludine at 100<sup>o</sup>C; forms blue-green complex</td><td>630</td></tr><tr><td>protein-bound iodine</td><td>decompose protein to release iodide, which catalyzes redox reaction between Ce<sup>3</sup><sup>+</sup>&nbsp;and As<sup>3</sup><sup>+</sup>; forms yellow colored Ce<sup>4</sup><sup>+</sup></td><td>420</td></tr></tbody></table></figure>



<h3 class="wp-block-heading" id="industrial-analysis">Industrial Analysis</h3>



<p>UV/Vis molecular absorption is used for the analysis of a diverse array of industrial samples including pharmaceuticals, food, paint, glass, and metals. In many cases the methods are similar to those described in Table 10.6 and Table 10.7. For example, the amount of iron in food can be determined by bringing the iron into solution and analyzing using the&nbsp;<em>o</em>-phenanthroline method listed in Table 10.6.</p>



<p>Many pharmaceutical compounds contain chromophores that make them suitable for analysis by UV/Vis absorption. Products that have been analyzed in this fashion include antibiotics, hormones, vitamins, and analgesics. One example of the use of UV absorption is in determining the purity of aspirin tablets, for which the active ingredient is acetylsalicylic acid. Salicylic acid, which is produced by the hydrolysis of acetylsalicylic acid, is an undesirable impurity in aspirin tablets, and should not be present at more than 0.01% w/w. Samples can be screened for unacceptable levels of salicylic acid by monitoring the absorbance at a wavelength of 312 nm. Acetylsalicylic acid absorbs at 280 nm, but absorbs poorly at 312 nm. Conditions for preparing the sample are chosen such that an absorbance of greater than 0.02 signifies an unacceptable level of salicylic acid.</p>



<h3 class="wp-block-heading" id="forensic-applications">Forensic Applications</h3>



<p>UV/Vis molecular absorption is routinely used for the analysis of narcotics and for drug testing. One interesting forensic application is the determination of blood alcohol using the Breathalyzer test. In this test a 52.5-mL breath sample is bubbled through an acidified solution of K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>, which oxidizes ethanol to acetic acid. The concentration of ethanol in the breath sample is determined by the decrease in absorbance at 440 nm where the dichromate ion absorbs. A blood alcohol content of 0.10%, which is above the legal limit, corresponds to 0.025 mg of ethanol in the breath sample.</p>



<h3 class="wp-block-heading" id="developing-a-quantitative-method-for-a-single-component">Developing a Quantitative Method for a Single Component</h3>



<p>In developing a quantitative analytical method, the conditions under which Beer’s law is obeyed must be established. First, the most appropriate wavelength for the analysis is determined from an absorption spectrum. In most cases the best wavelength corresponds to an absorption maximum because it provides greater sensitivity and is less susceptible to instrumental limitations. Second, if an instrument with adjustable slits is being used, then an appropriate slit width needs to be chosen. The absorption spectrum also aids in selecting a slit width. Usually we set the slits to be as wide as possible because this increases the throughput of source radiation, while also being narrow enough to avoid instrumental limitations to Beer’s law. Finally, a calibration curve is constructed to determine the range of concentrations for which Beer’s law is valid. Additional considerations that are important in any quantitative method are the effect of potential interferents and establishing an appropriate blank.</p>



<p>Note</p>



<p>The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of iron in water and wastewater provides an instructive example of a typical procedure. The description here is based on Method 3500- Fe B as published in&nbsp;<em>Standard Methods for the Examination of Water and Wastewater</em>, 20th Ed., American Public Health Association: Washington, D. C., 1998.</p>



<p>Representative Method 10.1</p>



<p><strong>Determination of Iron in Water and Wastewater</strong></p>



<p><em><strong>Description of Method</strong></em></p>



<p>Iron in the +2 oxidation state reacts with&nbsp;<em>o</em>-phenanthroline to form the orange-red Fe(phen)<sub>3</sub><sup>2+</sup>&nbsp;complex. The intensity of the complex’s color is independent of solution acidity between a pH of 3 and 9. Because the complex forms more rapidly at lower pH levels, the reaction is usually carried out within a pH range of 3.0–3.5. Any iron present in the +3 oxidation state is reduced with hydroxylamine before adding&nbsp;<em>o</em>-phenanthroline. The most important interferents are strong oxidizing agents, polyphosphates, and metal ions such as Cu<sup>2</sup><sup>+</sup>, Zn<sup>2</sup><sup>+</sup>, Ni<sup>2</sup><sup>+</sup>, and Cd<sup>2</sup><sup>+</sup>. An interference from oxidizing agents is minimized by adding an excess of hydroxylamine, and an interference from polyphosphate is minimized by boiling the sample in the presence of acid. The absorbance of samples and standards are measured at a wavelength of 510 nm using a 1-cm cell (longer pathlength cells also may be used). Beer’s law is obeyed for concentrations of within the range of 0.2–4.0 mg Fe/L.</p>



<p>(Figure 10.18 shows the visible spectrum for Fe(phen)<sub>3</sub><sup>2+</sup>.)</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12106/o-phenanthroline1.png?revision=1" alt="o-phenanthroline1.png"/></figure>



<p><em><strong>Procedure</strong></em></p>



<p>For samples containing less than 2 mg Fe/L, directly transfer a 50-mL portion to a 125-mL Erlenmeyer flask. Samples containing more than 2 mg Fe/L must be diluted before acquiring the 50-mL portion. Add 2 mL of concentrated HCl and 1 mL of hydroxylamine to the sample. Heat the solution to boiling and continue boiling until the solution’s volume is reduced to between 15 and 20 mL. After cooling to room temperature, transfer the solution to a 50-mL volumetric flask, add 10 mL of an ammonium acetate buffer, 2 mL of a 1000 ppm solution of&nbsp;<em>o</em>-phenanthroline, and dilute to volume. Allow 10–15 minutes for color development before measuring the absorbance, using distilled water to set 100% T. Calibration standards, including a blank, are prepared by the same procedure using a stock solution containing a known concentration of Fe<sup>2</sup><sup>+</sup>.</p>



<p><em><strong>Questions</strong></em></p>



<p>1. Explain why strong oxidizing agents are interferents, and why an excess of hydroxylamine prevents the interference.</p>



<p>A strong oxidizing agent oxidizes some Fe<sup>2</sup><sup>+</sup>&nbsp;to Fe<sup>3</sup><sup>+</sup>. Because Fe(phen)<sub>3</sub><sup>3+</sup>&nbsp;does not absorb as strongly as Fe(phen)<sub>3</sub><sup>2+</sup>, the absorbance decreases, producing a negative determinate error. The excess hydroxylamine reacts with the oxidizing agents, removing them from the solution.</p>



<p>2. The color of the complex is stable between pH levels of 3 and 9. What are some possible complications at more acidic or more basic pH’s?</p>



<p>Because&nbsp;<em>o</em>-phenanthroline is a weak base, its conditional formation constant for Fe(phen)<sub>3</sub><sup>2+</sup>&nbsp;is less favorable at more acidic pH levels, where&nbsp;<em>o</em>-phenanthroline is protonated. The result is a decrease in absorbance and a less sensitive analytical method.</p>



<p>(In&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/09_Titrimetric_Methods/9.3%3A_Complexation_Titrations" target="_blank" rel="noopener">Chapter 9</a>&nbsp;we saw the same effect of pH on the complexation reactions between EDTA and metal ions.)</p>



<p>When the pH is greater than 9, competition between OH<sup>–</sup>&nbsp;and&nbsp;<em>o</em>-phenanthroline for Fe<sup>2</sup><sup>+</sup>&nbsp;also decreased the absorbance. In addition, if the pH is sufficiently basic there is a risk that the iron will precipitate as Fe(OH)<sub>2</sub>.</p>



<p>3. Cadmium is an interferent because it forms a precipitate with&nbsp;<em>o</em>-phenanthroline. What effect would the formation of precipitate have on the determination of iron?</p>



<p>Because&nbsp;<em>o</em>-phenanthroline is present in large excess (2000 μg of&nbsp;<em>o</em>-phenanthroline for 100 μg of Fe<sup>2</sup><sup>+</sup>), it is not likely that the interference is due to an insufficient amount of&nbsp;<em>o</em>-phenanthroline being available to react with the Fe<sup>2</sup><sup>+</sup>. The presence of a precipitate in the sample cell results in the scattering of radiation, which causes an apparent increase in absorbance. Because the measured absorbance increases, the reported concentration is too high.</p>



<p>(Although scattering is a problem here, it can serve as the basis of a useful analytical method. See&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.8%3A_Spectroscopy_Based_on_Scattering" target="_blank" rel="noopener">Section 10.8</a>&nbsp;for further details.)</p>



<p>4. Even high quality ammonium acetate contains a significant amount of iron. Why is this source of iron not a problem?</p>



<p>Because all samples and standards are prepared using the same volume of ammonium acetate buffer, the contribution of this source of iron is accounted for by the calibration curve’s reagent blank.</p>



<h3 class="wp-block-heading" id="quantitative-analysis-for-a-single-analyte">Quantitative Analysis for a Single Analyte</h3>



<p>To determine the concentration of a an analyte we measure its absorbance and apply Beer’s law using any of the standardization methods described in Chapter 5. The most common methods are a normal calibration curve using external standards and the method of standard additions. A single point standardization is also possible, although we must first verify that Beer’s law holds for the concentration of analyte in the samples and the standard.</p>



<p>Example 10.5</p>



<p>The determination of Fe in an industrial waste stream was carried out by the&nbsp;<em>o</em>‑phenanthroline described in&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.3%3A_UV%2F%2FVis_and_IR_Spectroscopy#Representative_Method_10.1" target="_blank" rel="noopener">Representative Method 10.1</a>. Using the data in the following table, determine the mg Fe/L in the waste stream.</p>



<figure class="wp-block-table"><table><tbody><tr><td>mg Fe/L</td><td>absorbance</td></tr><tr><td>0.00</td><td>0.000</td></tr><tr><td>1.00</td><td>0.183</td></tr><tr><td>2.00</td><td>0.364</td></tr><tr><td>3.00</td><td>0.546</td></tr><tr><td>4.00</td><td>0.727</td></tr><tr><td>sample</td><td>0.269</td></tr></tbody></table></figure>



<p><strong><em>Solution</em></strong></p>



<p>Linear regression of absorbance versus the concentration of Fe in the standards gives a calibration curve with the following equation.A=0.0006+0.1817×(mgFe/L)(4.8.3)(4.8.3)A=0.0006+0.1817×(mgFe/L)</p>



<p>Substituting the sample’s absorbance into the calibration expression gives the concentration of Fe in the waste stream as 1.48 mg Fe/L.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12107/absorbance_vs_Fe.png?revision=1" alt="absorbance vs Fe.png"/></figure>



<p>Practice Exercise 10.5</p>



<p>The concentration of Cu<sup>2</sup><sup>+</sup>&nbsp;in a sample can be determined by reacting it with the ligand cuprizone and measuring its absorbance at 606 nm in a 1.00-cm cell. When a 5.00-mL sample is treated with cuprizone and diluted to 10.00 mL, the resulting solution has an absorbance of 0.118. A second 5.00-mL sample is mixed with 1.00 mL of a 20.00 mg/L standard of Cu<sup>2</sup><sup>+</sup>, treated with cuprizone and diluted to 10.00 mL, giving an absorbance of 0.162. Report the mg Cu<sup>2</sup><sup>+</sup>/L in the sample.</p>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.5" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<h3 class="wp-block-heading" id="quantitative-analysis-of-mixtures">Quantitative Analysis of Mixtures</h3>



<p>Suppose we need to determine the concentration of two analytes, X and Y, in a sample. If each analyte has a wavelength where the other analyte does not absorb, then we can proceed using the approach in Example 10.5. Unfortunately, UV/Vis absorption bands are so broad that it frequently is not possible to find suitable wavelengths. Because Beer’s law is additive the mixture’s absorbance,&nbsp;<em>A</em><sub>mix</sub>, is(Amix)λ1=(εX)λ1bCX+(εY)λ1bCY(10.11)(10.11)(Amix)λ1=(εX)λ1bCX+(εY)λ1bCY</p>



<p>where&nbsp;λ1 is the wavelength at which we measure the absorbance. Because equation 10.11 includes terms for the concentration of both X and Y, the absorbance at one wavelength does not provide enough information to determine either&nbsp;<em>C</em><sub>X</sub>&nbsp;or&nbsp;<em>C</em><sub>Y</sub>. If we measure the absorbance at a second wavelength(Amix)λ2=(εX)λ2bCX+(εY)λ2bCY(10.12)(10.12)(Amix)λ2=(εX)λ2bCX+(εY)λ2bCY</p>



<p>then&nbsp;<em>C</em><sub>X</sub>&nbsp;and&nbsp;<em>C</em><sub>Y</sub>&nbsp;can be determined by solving simultaneously equation10.11 and equation 10.12. Of course, we also must determine the value for ε<sub>X</sub>&nbsp;and ε<sub>Y</sub>&nbsp;at each wavelength. For a mixture of&nbsp;<em>n</em>&nbsp;components, we must measure the absorbance at&nbsp;<em>n</em>&nbsp;different wavelengths.</p>



<p>Example 10.6</p>



<p>The concentrations of Fe<sup>3</sup><sup>+</sup>&nbsp;and Cu<sup>2</sup><sup>+</sup>&nbsp;in a mixture can be determined following their reaction with hexacyanoruthenate (II), Ru(CN)<sub>6</sub><sup>4–</sup>, which forms a purple-blue complex with Fe<sup>3</sup><sup>+</sup>&nbsp;(λ<sub>max</sub>&nbsp;= 550 nm) and a pale-green complex with Cu<sup>2</sup><sup>+</sup>&nbsp;(λ<sub>max</sub>&nbsp;= 396 nm).<sup>7</sup>&nbsp;The molar absorptivities (M<sup>–1</sup>&nbsp;cm<sup>–1</sup>) for the metal complexes at the two wavelengths are summarized in the following table.</p>



<figure class="wp-block-table"><table><tbody><tr><td></td><td>ε<sub>550</sub></td><td>ε<sub>396</sub></td></tr><tr><td>Fe<sup>3</sup><sup>+</sup></td><td>9970</td><td>84</td></tr><tr><td>Cu<sup>2</sup><sup>+</sup></td><td>34</td><td>856</td></tr></tbody></table></figure>



<p>When a sample containing Fe<sup>3</sup><sup>+</sup>&nbsp;and Cu<sup>2</sup><sup>+</sup>&nbsp;is analyzed in a cell with a pathlength of 1.00 cm, the absorbance at 550 nm is 0.183 and the absorbance at 396 nm is 0.109. What are the molar concentrations of Fe<sup>3</sup><sup>+</sup>&nbsp;and Cu<sup>2</sup><sup>+</sup>&nbsp;in the sample?</p>



<p><strong><em>Solution</em></strong></p>



<p>Substituting known values into equations 10.11 and 10.12 givesA550=0.183=9970CFe+34CCu(4.8.4)(4.8.4)A550=0.183=9970CFe+34CCuA396=0.109=84CFe+856CCu(4.8.5)(4.8.5)A396=0.109=84CFe+856CCu</p>



<p>To determine&nbsp;<em>C</em><sub>Fe</sub>&nbsp;and&nbsp;<em>C</em><sub>Cu</sub>&nbsp;we solve the first equation for&nbsp;<em>C</em><sub>Cu</sub>CCu=0.183–9970CFe34(4.8.6)(4.8.6)CCu=0.183–9970CFe34</p>



<p>and substitute the result into the second equation.0.109=84CFe+856×0.183−9970CFe34=4.607–(2.51×105)CFe(4.8.7)(4.8.7)0.109=84CFe+856×0.183−9970CFe34=4.607–(2.51×105)CFe</p>



<p>Solving for&nbsp;<em>C</em><sub>Fe</sub>&nbsp;gives the concentration of Fe<sup>3+</sup>&nbsp;as 1.79 × 10<sup>–5</sup>&nbsp;M. Substituting this concentration back into the equation for the mixture’s absorbance at 396 nm gives the concentration of Cu<sup>2</sup><sup>+</sup>&nbsp;as 1.26 × 10<sup>–4</sup>&nbsp;M.</p>



<p>(Another approach is to multiply the first equation by 856/34 giving4.607=251009CFe+856CCu(4.8.8)(4.8.8)4.607=251009CFe+856CCu</p>



<p>Subtracting the second equation from this equation4.607=251009CFe+856CCu−0.109=84CFe+856CCu–––––––––––––––––––––––––––––––4.498=250925CFe(4.8.9)(4.8.10)(4.8.11)(4.8.9)−4.607=251009CFe+856CCu(4.8.10)−0.109=84CFe+856CCu6CCu_(4.8.11)−4.498=250925CFe</p>



<p>we find that&nbsp;<em>C</em><sub>Fe</sub>&nbsp;is 1.79×10<sup>–5</sup>. Having determined&nbsp;<em>C</em><sub>Fe</sub>&nbsp;we can substitute back into one of the other equations to solve for&nbsp;<em>C</em><sub>Cu</sub>, which is 1.26×10<sup>–5</sup>.)</p>



<p>Practice Exercise 10.6</p>



<p>The absorbance spectra for Cr<sup>3</sup><sup>+</sup>&nbsp;and Co<sup>2</sup><sup>+</sup>&nbsp;overlap significantly. To determine the concentration of these analytes in a mixture, its absorbance was measured at 400 nm and at 505 nm, yielding values of 0.336 and 0.187, respectively. The individual molar absorptivities (M<sup>–1</sup>&nbsp;cm<sup>–1</sup>) are</p>



<figure class="wp-block-table"><table><tbody><tr><td></td><td>ε<sub>400</sub></td><td>ε<sub>505</sub></td></tr><tr><td>Cr<sup>3</sup><sup>+</sup></td><td>15.2</td><td>0.533</td></tr><tr><td>Co<sup>2</sup><sup>+</sup></td><td>5.60</td><td>5.07</td></tr></tbody></table></figure>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.6" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<p>To obtain results with good accuracy and precision the two wavelengths should be selected so that ε<sub>X</sub>&nbsp;&gt; ε<sub>Y</sub>&nbsp;at one wavelength and ε<sub>X</sub>&nbsp;&lt; ε<sub>Y</sub>&nbsp;at the other wavelength. It is easy to appreciate why this is true. Because the absorbance at each wavelength is dominated by one analyte, any uncertainty in the concentration of the other analyte has less of an impact. Figure 10.35 shows that the choice of wavelengths for Practice Exercise 10.6 are reasonable. When the choice of wavelengths is not obvious, one method for locating the optimum wavelengths is to plot ε<sub>X</sub>/ε<sub>Y</sub>&nbsp;as function of wavelength, and determine the wavelengths where ε<sub>X</sub>/ε<sub>Y</sub>&nbsp;reaches maximum and minimum values.<sup>8</sup></p>



<p>Note</p>



<p>For example, in Example 10.6 the molar absorptivity for Fe<sup>3</sup><sup>+</sup>&nbsp;at 550 nm is 119× that for Cu<sup>2</sup><sup>+</sup>, and the molar absorptivity for Cu<sup>2+</sup>&nbsp;at 396 nm is 10.2× that for Fe<sup>3</sup><sup>+</sup>.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12715/Figure10.35.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12715/Figure10.35.jpg?revision=1&amp;size=bestfit&amp;width=450&amp;height=394" alt="Figure10.35.jpg"/></a></figure>



<p><strong>Figure 10.35</strong>&nbsp;Visible absorption spectra for 0.0250 M Cr<sup>3</sup><sup>+</sup>, 0.0750 M Co<sup>2</sup><sup>+</sup>, and for a mixture of Cr<sup>3</sup><sup>+</sup>&nbsp;and Co<sup>2</sup><sup>+</sup>. The two wavelengths used for analyzing the mixture of Cr<sup>3+</sup>&nbsp;and Co<sup>2</sup><sup>+</sup>&nbsp;are shown by the dashed lines. The data for the two standard solutions are from reference 7.</p>



<p>When the analyte’s spectra overlap severely, such that ε<sub>X</sub>&nbsp;≈ ε<sub>Y</sub>&nbsp;at all wavelength, other computational methods may provide better accuracy and precision. In a multiwavelength linear regression analysis, for example, a mixture’s absorbance is compared to that for a set of standard solutions at several wavelengths.<sup>9</sup>&nbsp;If&nbsp;<em>A</em><sub>SX</sub>and&nbsp;<em>A</em><sub>SY</sub>&nbsp;are the absorbance values for standard solutions of components X and Y at any wavelength, thenASX=εXbCSX(10.13)(10.13)ASX=εXbCSXASY=εYbCSY(10.14)(10.14)ASY=εYbCSY</p>



<p>where&nbsp;<em>C</em><sub>SX</sub>&nbsp;and&nbsp;<em>C</em><sub>SY</sub>&nbsp;are the known concentrations of X and Y in the standard solutions. Solving equation 10.13 and equation 10.14 for ε<sub>X</sub>&nbsp;and ε<sub>Y</sub>, substituting into equation 10.11, and rearranging, givesAmixASX=CXCSX+CYCSY×ASYASX(4.8.12)(4.8.12)AmixASX=CXCSX+CYCSY×ASYASX</p>



<p>To determine&nbsp;<em>C</em><sub>X</sub>&nbsp;and&nbsp;<em>C</em><sub>Y</sub>&nbsp;the mixture’s absorbance and the absorbances of the standard solutions are measured at several wavelengths. Graphing&nbsp;<em>A</em><sub>mix</sub>/<em>A</em><sub>SX</sub>&nbsp;versus&nbsp;<em>A</em><sub>SY</sub>/<em>A</em><sub>SX</sub>&nbsp;gives a straight line with a slope of&nbsp;<em>C</em><sub>Y</sub>/<em>C</em><sub>SY</sub>&nbsp;and a&nbsp;<em>y</em>-intercept of&nbsp;<em>C</em><sub>X</sub>/<em>C</em><sub>SX</sub>. This approach is particularly helpful when it is not possible to find wavelengths where ε<sub>X</sub>&nbsp;&gt; ε<sub>Y</sub>&nbsp;and ε<sub>X</sub>&nbsp;&lt; ε<sub>Y</sub>.</p>



<p>Note</p>



<p>The approach outlined here for a multiwavelength linear regression uses a single standard solution for each analyte. A more rigorous approach uses multiple standards for each analyte. The math behind the analysis of this data—what we call a multiple linear regression—is beyond the level of this text. For more details about multiple linear regression see Brereton, R. G.&nbsp;<em>Chemometrics: Data Analysis for the Laboratory and Chemical Plant</em>, Wiley: Chichester, England, 2003.</p>



<p>Example 10.7</p>



<p>Figure 10.35 shows visible absorbance spectra for a standard solution of 0.0250 M Cr<sup>3</sup><sup>+</sup>, a standard solution of 0.0750 M Co<sup>2+</sup>, and a mixture containing unknown concentrations of each ion. The data for these spectra are shown here.<sup>10</sup></p>



<figure class="wp-block-table"><table><tbody><tr><td>λ&nbsp;(nm)</td><td><em>A</em><sub>Cr</sub></td><td><em>A</em><sub>Co</sub></td><td><em>A</em><sub>mix</sub></td><td>λ&nbsp;(nm)</td><td><em>A</em><sub>Cr</sub></td><td><em>A</em><sub>Co</sub></td><td><em>A</em><sub>mix</sub></td></tr><tr><td>375</td><td>0.26</td><td>0.01</td><td>0.53</td><td>520</td><td>0.19</td><td>0.38</td><td>0.63</td></tr><tr><td>400</td><td>0.43</td><td>0.03</td><td>0.88</td><td>530</td><td>0.24</td><td>0.33</td><td>0.70</td></tr><tr><td>425</td><td>0.39</td><td>0.07</td><td>0.83</td><td>540</td><td>0.28</td><td>0.26</td><td>0.73</td></tr><tr><td>440</td><td>0.29</td><td>0.13</td><td>0.67</td><td>550</td><td>0.32</td><td>0.18</td><td>0.76</td></tr><tr><td>455</td><td>0.20</td><td>0.21</td><td>0.54</td><td>570</td><td>0.38</td><td>0.08</td><td>0.81</td></tr><tr><td>470</td><td>0.14</td><td>0.28</td><td>0.47</td><td>575</td><td>0.39</td><td>0.06</td><td>0.82</td></tr><tr><td>480</td><td>0.12</td><td>0.30</td><td>0.44</td><td>580</td><td>0.38</td><td>0.05</td><td>0.79</td></tr><tr><td>490</td><td>0.11</td><td>0.34</td><td>0.45</td><td>600</td><td>0.34</td><td>0.03</td><td>0.70</td></tr><tr><td>500</td><td>0.13</td><td>0.38</td><td>0.51</td><td>625</td><td>0.24</td><td>0.02</td><td>0.49</td></tr></tbody></table></figure>



<p>Use a multiwavelength regression analysis to determine the composition of the unknown.</p>



<p><em><strong>Solution</strong></em></p>



<p>First we need to calculate values for&nbsp;<em>A</em><sub>mix</sub>/<em>A</em><sub>SX</sub>&nbsp;and for&nbsp;<em>A</em><sub>SY</sub>/<em>A</em><sub>SX</sub>. Let’s define X as Co<sup>2</sup><sup>+</sup>&nbsp;and Y as Cr<sup>3</sup><sup>+</sup>. For example, at a wavelength of 375 nm&nbsp;<em>A</em><sub>mix</sub>/<em>A</em><sub>SX</sub>&nbsp;is 0.53/0.01, or 53 and&nbsp;<em>A</em><sub>SY</sub>/<em>A</em><sub>SX</sub>&nbsp;is 0.26/0.01, or 26. Completing the calculation for all wavelengths and graphing&nbsp;<em>A</em><sub>mix</sub>/<em>A</em><sub>SX</sub>&nbsp;versus&nbsp;<em>A</em><sub>SY</sub>/<em>A</em><sub>SX</sub>gives the result shown in Figure 10.36. Fitting a straight-line to the data gives a regression model ofAmixASX=0.636+2.01×ASYASX(4.8.13)(4.8.13)AmixASX=0.636+2.01×ASYASX</p>



<p>Using the&nbsp;<em>y</em>-intercept, the concentration of Co<sup>2</sup><sup>+</sup>&nbsp;isCXCSX=CCo0.0750M=0.636(4.8.14)(4.8.14)CXCSX=CCo0.0750M=0.636</p>



<p>or&nbsp;<em>C</em><sub>Co</sub>&nbsp;= 0.048 M, and using the slope the concentration of Cr<sup>3+</sup>&nbsp;isCYCSY=CCr0.0250M=2.01(4.8.15)(4.8.15)CYCSY=CCr0.0250M=2.01</p>



<p>or&nbsp;<em>C</em><sub>Cr</sub>&nbsp;= 0.050 M.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12716/Figure10.36.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12716/Figure10.36.jpg?revision=1&amp;size=bestfit&amp;width=330&amp;height=286" alt="Figure10.36.jpg"/></a></figure>



<p><strong>Figure 10.36</strong>&nbsp;Multiwavelength linear regression analysis for the data in Example 10.7.</p>



<p>Practice Exercise 10.7</p>



<p>A mixture of MnO<sub>4</sub><sup>–</sup>&nbsp;and Cr<sub>2</sub>O<sub>7</sub><sup>2–</sup>, and standards of 0.10 mM KMnO<sub>4</sub>&nbsp;and of 0.10 mM K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>&nbsp;gives the results shown in the following table. Determine the composition of the mixture. The data for this problem is from Blanco, M. C.; Iturriaga, H.; Maspoch, S.; Tarin, P.&nbsp;<em>J. Chem. Educ.&nbsp;</em><strong>1989</strong>,&nbsp;<em>66</em>, 178–180.</p>



<figure class="wp-block-table"><table><tbody><tr><td>λ&nbsp;(nm)</td><td><em>A</em><sub>Mn</sub></td><td><em>A</em><sub>Cr</sub></td><td><em>A</em><sub>mix</sub></td></tr><tr><td>266</td><td>0.042</td><td>0.410</td><td>0.766</td></tr><tr><td>288</td><td>0.082</td><td>0.283</td><td>0.571</td></tr><tr><td>320</td><td>0.168</td><td>0.158</td><td>0.422</td></tr><tr><td>350</td><td>0.125</td><td>0.318</td><td>0.672</td></tr><tr><td>360</td><td>0.056</td><td>0.181</td><td>0.366</td></tr></tbody></table></figure>



<p>(There are many additional ways to analyze mixtures spectrophotometrically, including generalized standard additions, H-point standard additions, principal component regression to name a few. Consult the chapter’s&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/Additional_Resources#Chapter_10" target="_blank" rel="noopener">additional resources</a>&nbsp;for further information.)</p>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.7" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<h2 class="wp-block-heading" id="10-3-3-qualitative-applications">10.3.3 Qualitative Applications</h2>



<p>As discussed earlier in&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.2%3A_Spectroscopy_Based_on_Absorption#10.2.1_Absorbance_Spectra" target="_blank" rel="noopener">Section 10.2.1</a>, ultraviolet, visible, and infrared absorption bands result from the absorption of electromagnetic radiation by specific valence electrons or bonds. The energy at which the absorption occurs, and its intensity is determined by the chemical environment of the absorbing moiety. For example, benzene has several ultraviolet absorption bands due to π → π* transitions. The position and intensity of two of these bands, 203.5 nm (ε = 7400 M<sup>–1</sup>&nbsp;cm<sup>–1</sup>) and 254 nm (ε = 204 M<sup>–1</sup>&nbsp;cm<sup>–1</sup>), are sensitive to substitution. For benzoic acid, in which a carboxylic acid group replaces one of the aromatic hydrogens, the two bands shift to 230 nm (ε = 11 600 M<sup>–1</sup>&nbsp;cm<sup>–1</sup>) and 273 nm (ε = 970 M<sup>–1</sup>&nbsp;cm<sup>–1</sup>). A variety of rules have been developed to aid in correlating UV/Vis absorption bands to chemical structure. Similar correlations have been developed for infrared absorption bands. For example a carbonyl’s C=O stretch is sensitive to adjacent functional groups, occurring at 1650 cm<sup>–1</sup>&nbsp;for acids, 1700 cm<sup>–1</sup>&nbsp;for ketones, and 1800 cm<sup>–1</sup>&nbsp;for acid chlorides. The interpretation of UV/Vis and IR spectra receives adequate coverage elsewhere in the chemistry curriculum, notably in organic chemistry, and is not considered further in this text.</p>



<p>With the availability of computerized data acquisition and storage it is possible to build digital libraries of standard reference spectra. The identity of an a unknown compound can often be determined by comparing its spectrum against a library of reference spectra, a process is known as&nbsp;<strong>spectral searching</strong>. Comparisons are made using an algorithm that calculates the cumulative difference between the sample’s spectrum and a reference spectrum. For example, one simple algorithm uses the following equationD=∑i=1n|(Asample)i−(Areference)i|(4.8.16)(4.8.16)D=∑i=1n|(Asample)i−(Areference)i|</p>



<p>where&nbsp;<em>D</em>&nbsp;is the cumulative difference,&nbsp;<em>A</em><sub>sample</sub>&nbsp;is the sample’s absorbance at wavelength or wavenumber&nbsp;<em>i</em>,&nbsp;<em>A</em><sub>reference</sub>&nbsp;is the absorbance of the reference compound at the same wavelength or wavenumber, and&nbsp;<em>n</em>&nbsp;is the number of digitized points in the spectra. The cumulative difference is calculated for each reference spectrum. The reference compound with the smallest value of&nbsp;<em>D</em>&nbsp;provides the closest match to the unknown compound. The accuracy of spectral searching is limited by the number and type of compounds included in the library, and by the effect of the sample’s matrix on the spectrum.</p>



<p>Another advantage of computerized data acquisition is the ability to subtract one spectrum from another. When coupled with spectral searching it may be possible, by repeatedly searching and subtracting reference spectra, to determine the identity of several components in a sample without the need of a prior separation step. An example is shown in Figure 10.37 in which the composition of a two-component mixture is determined by successive searching and subtraction. Figure 10.37a shows the spectrum of the mixture. A search of the spectral library selects cocaine&nbsp;<sup>.</sup>&nbsp;HCl (Figure 10.37b) as a likely component of the mixture. Subtracting the reference spectrum for cocaine&nbsp;<sup>.&nbsp;</sup>HCl from the mixture’s spectrum leaves a result (Figure 10.37c) that closely matches mannitol’s reference spectrum (Figure 10.37d). Subtracting the reference spectrum for leaves only a small residual signal (Figure 10.37e).</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12717/Figure10.37.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12717/Figure10.37.jpg?revision=1&amp;size=bestfit&amp;width=600&amp;height=838" alt="Figure10.37.jpg"/></a></figure>



<p><strong>Figure 10.37</strong>&nbsp;Identifying the components of a mixture by spectral searching and subtracting. (a) IR spectrum of the mixture; (b) Reference IR spectrum of cocaine<sup>.</sup>&nbsp;HCl; (c) Result of subtracting the spectrum of cocaine&nbsp;<sup>.</sup>&nbsp;HCl from the mixture’s spectrum; (d) Reference IR spectrum of mannitol; and (e) The residual spectrum after removing mannitol’s contribution to the mixture’s spectrum.</p>



<p>Note</p>



<p>IR spectra traditionally are displayed using percent transmittance, %T, along the<em>&nbsp;y</em>-axis (for example, see Figure 10.16). Because absorbance—not percent transmittance—is a linear function of concentration, spectral searching and spectral subtraction, is easier to do when displaying absorbance on the&nbsp;<em>y</em>-axis.</p>



<h2 class="wp-block-heading" id="10-3-4-characterization-applications">10.3.4 Characterization Applications</h2>



<p>Molecular absorption, particularly in the UV/Vis range, has been used for a variety of different characterization studies, including determining the stoichiometry of metal–ligand complexes and determining equilibrium constants. Both of these examples are examined in this section.</p>



<h3 class="wp-block-heading" id="stoichiometry-of-a-metal-ligand-complex">Stoichiometry of a Metal-Ligand Complex</h3>



<p>We can determine the stoichiometry of a metal–ligand complexation reactionM+yL⇋MLy(4.8.17)(4.8.17)M+yL⇋MLy</p>



<p>using one of three methods: the method of continuous variations, the mole-ratio method, and the slope-ratio method. Of these approaches, the&nbsp;<strong>method of continuous variations</strong>, also called Job’s method, is the most popular. In this method a series of solutions is prepared such that the total moles of metal and ligand,&nbsp;<em>n</em><sub>total</sub>, in each solution is the same. If (<em>n</em><sub>M</sub>)<sub><em>i</em></sub>&nbsp;and (<em>n</em><sub>L</sub>)<sub><em>i</em></sub>&nbsp;are, respectively, the moles of metal and ligand in solution&nbsp;<em>i</em>, thenntotal=(nM)i+(nL)i(4.8.18)(4.8.18)ntotal=(nM)i+(nL)i</p>



<p>The relative amount of ligand and metal in each solution is expressed as the mole fraction of ligand, (<em>X</em><sub>L</sub>)<sub><em>i</em></sub>, and the mole fraction of metal, (<em>X</em><sub>M</sub>)<sub><em>i</em></sub>,(XL)i=(nL)intotal(4.8.19)(4.8.19)(XL)i=(nL)intotal(XM)i=1−(nL)intotal=(nM)intotal(4.8.20)(4.8.20)(XM)i=1−(nL)intotal=(nM)intotal</p>



<p>The concentration of the metal–ligand complex in any solution is determined by the limiting reagent, with the greatest concentration occurring when the metal and the ligand are mixed stoichiometrically. If we monitor the complexation reaction at a wavelength where only the metal–ligand complex absorbs, a graph of absorbance versus the mole fraction of ligand will have two linear branches—one when the ligand is the limiting reagent and a second when the metal is the limiting reagent. The intersection of these two branches represents a stoichiometric mixing of the metal and the ligand. We can use the mole fraction of ligand at the intersection to determine the value of&nbsp;<em>y</em>&nbsp;for the metal–ligand complex ML<sub><em>y</em></sub>.y=nLnM=XLXM=XL1−XL(4.8.21)(4.8.21)y=nLnM=XLXM=XL1−XL</p>



<p>Note</p>



<p>You also can plot the data as absorbance versus the mole fraction of metal. In this case, y is equal to (1–<em>X</em><sub>M</sub>)/<em>X</em><sub>M</sub>.</p>



<figure class="wp-block-image size-large"><img decoding="async" src="http://s7.picofile.com/file/8392387600/uv_vis_in.gif" alt=""/></figure>



<p>Example 10.8</p>



<p>To determine the formula for the complex between Fe<sup>2+</sup>&nbsp;and&nbsp;<em>o</em>-phenanthroline, a series of solutions is prepared in which the total concentration of metal and ligand is held constant at 3.15 × 10<sup>–4</sup>&nbsp;M. The absorbance of each solution is measured at a wavelength of 510 nm. Using the following data, determine the formula for the complex.</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>X</em><sub>L</sub></td><td>absorbance</td><td><em>X</em><sub>L</sub></td><td>absorbance</td></tr><tr><td>0.000</td><td>0.000</td><td>0.600</td><td>0.693</td></tr><tr><td>0.100</td><td>0.116</td><td>0.700</td><td>0.809</td></tr><tr><td>0.200</td><td>0.231</td><td>0.800</td><td>0.693</td></tr><tr><td>0.300</td><td>0.347</td><td>0.900</td><td>0.347</td></tr><tr><td>0.400</td><td>0.462</td><td>1.000</td><td>0.000</td></tr><tr><td>0.500</td><td>0.578</td><td></td><td></td></tr></tbody></table></figure>



<p>(To prepare the solutions for this example I first prepared a solution of 3.15 × 10<sup>-4</sup>&nbsp;M Fe<sup>2</sup><sup>+</sup>&nbsp;and a solution of 3.15 × 10<sup>-4</sup>&nbsp;M&nbsp;<em>o</em>-phenanthroline. Because the two stock solutions are of equal concentration, diluting a portion of one solution with the other solution gives a mixture in which the combined concentration of&nbsp;<em>o</em>-phenanthroline and Fe<sup>2</sup><sup>+</sup>&nbsp;is 3.15 × 10<sup>-4</sup>&nbsp;M. If each solution has the same volume, then each solution contains the same total moles of metal and ligand.)</p>



<p><strong><em>Solution</em></strong></p>



<p>A plot of absorbance versus the mole fraction of ligand is shown in Figure 10.38. To find the maximum absorbance, we extrapolate the two linear portions of the plot. The two lines intersect at a mole fraction of ligand of 0.75. Solving for&nbsp;<em>y</em>&nbsp;givesy=XL1–XL=0.751–0.75=3(4.8.22)(4.8.22)y=XL1–XL=0.751–0.75=3</p>



<p>The formula for the metal–ligand complex is Fe(<em>o</em>-phenanthroline)<sub>3</sub><sup>2+</sup>.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12718/Figure10.38.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12718/Figure10.38.jpg?revision=1&amp;size=bestfit&amp;width=600&amp;height=565" alt="Figure10.38.jpg"/></a></figure>



<p><strong>Figure 10.38&nbsp;</strong>Continuous variations plot for Example 10.8. The photo shows the solutions used in gathering the data. Each solution is displayed directly below its corresponding point on the continuous variations plot.</p>



<p>Practice Exercise 10.8</p>



<p>Use the continuous variations data in the following table to determine the formula for the complex between Fe<sup>2</sup><sup>+</sup>&nbsp;and SCN<sup>–</sup>. The data for this problem is adapted from Meloun, M.; Havel, J.; Högfeldt, E.&nbsp;<em>Computation of Solution Equilibria</em>, Ellis Horwood: Chichester, England, 1988, p. 236.</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>X</em><sub>L</sub></td><td>absorbance</td><td><em>X</em><sub>L</sub></td><td>absorbance</td><td><em>X</em><sub>L</sub></td><td>absorbance</td><td><em>X</em><sub>L</sub></td><td>absorbance</td></tr><tr><td>0.0200</td><td>0.068</td><td>0.2951</td><td>0.670</td><td>0.5811</td><td>0.790</td><td>0.8923</td><td>0.325</td></tr><tr><td>0.0870</td><td>0.262</td><td>0.3887</td><td>0.767</td><td>0.6860</td><td>0.701</td><td>0.9787</td><td>0.071</td></tr><tr><td>0.1792</td><td>0.471</td><td>0.4964</td><td>0.807</td><td>0.7885</td><td>0.540</td><td></td><td></td></tr></tbody></table></figure>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.8" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<p>Several precautions are necessary when using the method of continuous variations. First, the metal and the ligand must form only one metal–ligand complex. To determine if this condition is true, plots of absorbance versus&nbsp;<em>X</em><sub>L</sub>&nbsp;are constructed at several different wavelengths and for several different values of&nbsp;<em>n</em><sub>total</sub>. If the maximum absorbance does not occur at the same value of&nbsp;<em>X</em><sub>L</sub>&nbsp;for each set of conditions, then more than one metal–ligand complex must be present. A second precaution is that the metal–ligand complex’s absorbance must obey Beer’s law. Third, if the metal–ligand complex’s formation constant is relatively small, a plot of absorbance versus&nbsp;<em>X</em><sub>L</sub>&nbsp;may show significant curvature. In this case it is often difficult to determine the stoichiometry by extrapolation. Finally, because the stability of a metal–ligand complex may be influenced by solution conditions, the composition of the solutions must be carefully controlled. When the ligand is a weak base, for example, the solutions must be buffered to the same pH.</p>



<p>In the&nbsp;<strong>mole-ratio method</strong>&nbsp;the amount of one reactant, usually the moles of metal, is held constant, while the amount of the other reactant is varied. The absorbance is monitored at a wavelength where the metal–ligand complex absorbs. A plot of absorbance as a function of the ligand-to-metal mole ratio,&nbsp;<em>n</em><sub>L</sub>/<em>n</em><sub>M</sub>, has two linear branches, which intersect at a mole–ratio corresponding to the complex’s formula. Figure 10.39a shows a mole-ratio plot for the formation of a 1:1 complex in which the absorbance is monitored at a wavelength where only the complex absorbs. Figure 10.39b shows a mole-ratio plot for a 1:2 complex in which all three species—the metal, the ligand, and the complex—absorb at the selected wavelength. Unlike the method of continuous variations, the mole-ratio method can be used for complexation reactions that occur in a stepwise fashion if there is a difference in the molar absorptivities of the metal–ligand complexes, and if the formation constants are sufficiently different. A typical mole-ratio plot for the step-wise formation of ML and ML<sub>2</sub>&nbsp;is shown in Figure 10.39c.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12719/Figure10.39.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12719/Figure10.39.jpg?revision=1&amp;size=bestfit&amp;width=700&amp;height=202" alt="Figure10.39.jpg"/></a></figure>



<p><strong>Figure 10.39</strong>&nbsp;Mole-ratio plots for: (a) a 1:1 metal–ligand complex in which only the complex absorbs; (b) a 1:2 metal–ligand complex in which the metal, the ligand, and the complex absorb; and (c) the stepwise formation of a 1:1 and a 1:2 metal–ligand complex.</p>



<p>In both the method of continuous variations and the mole-ratio method we determine the complex’s stoichiometry by extrapolating absorbance data from conditions in which there is a linear relationship between absorbance and the relative amounts of metal and ligand. If a metal–ligand complex is very weak, a plot of absorbance versus<em>&nbsp;X</em><sub>L</sub>&nbsp;or&nbsp;<em>n</em><sub>L</sub>/<em>n</em><sub>M</sub>&nbsp;may be so curved that it is impossible to determine the stoichiometry by extrapolation. In this case the slope-ratio may be used.</p>



<p>In the&nbsp;<strong>slope-ratio method</strong>&nbsp;two sets of solutions are prepared. The first set of solutions contains a constant amount of metal and a variable amount of ligand, chosen such that the total concentration of metal,&nbsp;<em>C</em><sub>M</sub>, is much larger than the total concentration of ligand,&nbsp;<em>C</em><sub>L</sub>. Under these conditions we may assume that essentially all the ligand reacts in forming the metal–ligand complex. The concentration of the complex, which has the general form M<sub><em>x</em></sub>L<sub><em>y</em></sub>, is[MxLy]=CLy(4.8.23)(4.8.23)[MxLy]=CLy</p>



<p>If we monitor the absorbance at a wavelength where only M<sub><em>x</em></sub>L<sub><em>y</em></sub>&nbsp;absorbs, thenA=εb[MxLy]=εbCLy(4.8.24)(4.8.24)A=εb[MxLy]=εbCLy</p>



<p>and a plot of absorbance versus&nbsp;<em>C</em><sub>L</sub>&nbsp;is linear with a slope,&nbsp;<em>s</em><sub>L</sub>, ofsL=εby(4.8.25)(4.8.25)sL=εby</p>



<p>A second set of solutions is prepared with a fixed concentration of ligand that is much greater than a variable concentration of metal; thus[MxLy]=CMx(4.8.26)(4.8.26)[MxLy]=CMxA=εb[MxLy]=εbCMx(4.8.27)(4.8.27)A=εb[MxLy]=εbCMxsM=εbx(4.8.28)(4.8.28)sM=εbx</p>



<p>A ratio of the slopes provides the relative values of&nbsp;<em>x</em>&nbsp;and&nbsp;<em>y</em>.sMsL=εb/xεb/y=yx(4.8.29)(4.8.29)sMsL=εb/xεb/y=yx</p>



<p>An important assumption in the slope-ratio method is that the complexation reaction continues to completion in the presence of a sufficiently large excess of metal or ligand. The slope-ratio method also is limited to systems in which only a single complex is formed and for which Beer’s law is obeyed.</p>



<h3 class="wp-block-heading" id="determination-of-equilibrium-constants">Determination of Equilibrium Constants</h3>



<p>Another important application of molecular absorption spectroscopy is the determination of equilibrium constants. Let’s consider, as a simple example, an acid–base reaction of the general formHIn(aq)+H2O(l)⇋H3O+(aq)+In−(aq)(4.8.30)(4.8.30)HIn(aq)+H2O(l)⇋H3O+(aq)+In−(aq)</p>



<p>where HIn and In<sup>–</sup>&nbsp;are the conjugate weak acid and weak base forms of an acid–base indicator. The equilibrium constant for this reaction isKa=[H3O+][In−][HIn](4.8.31)(4.8.31)Ka=[H3O+][In−][HIn]</p>



<p>To determine the equilibrium constant’s value, we prepare a solution in which the reaction is in a state of equilibrium and determine the equilibrium concentration of H<sub>3</sub>O<sup>+</sup>, HIn, and In<sup>–</sup>. The concentration of H<sub>3</sub>O<sup>+</sup>&nbsp;is easy to determine by simply measuring the solution’s pH. To determine the concentration of HIn and In<sup>–</sup>&nbsp;we can measure the solution’s absorbance.</p>



<p>If both HIn and In<sup>–</sup>&nbsp;absorb at the selected wavelength, then, from equation 10.6, we know thatA=εHInb[HIn]+εInb[In−](10.15)(10.15)A=εHInb[HIn]+εInb[In−]</p>



<p>where ε<sub>HIn</sub>&nbsp;and ε<sub>In</sub>&nbsp;are the molar absorptivities for HIn and In<sup>–</sup>. The total concentration of indicator,&nbsp;<em>C</em>, is given by a mass balance equationC=[HIn]+[In−](10.16)(10.16)C=[HIn]+[In−]</p>



<p>Solving equation 10.16 for [HIn] and substituting into equation 10.15 givesA=εHInb(C−[In−])+εInb[In−](4.8.32)(4.8.32)A=εHInb(C−[In−])+εInb[In−]</p>



<p>which we simplify toA=εHInbC−εHInb[In−]+εInb[In−](4.8.33)(4.8.33)A=εHInbC−εHInb[In−]+εInb[In−]A=AHIn+b[In−](εIn−εHIn)(10.17)(10.17)A=AHIn+b[In−](εIn−εHIn)</p>



<p>where&nbsp;<em>A</em><sub>HIn</sub>, which is equal to ε<sub>HIn</sub><em>bC</em>, is the absorbance when the pH is acidic enough that essentially all the indicator is present as HIn. Solving equation 10.17 for the concentration of In<sup>–</sup>&nbsp;gives[In−]=A−AHInb(εIn−εHIn)(10.18)(10.18)[In−]=A−AHInb(εIn−εHIn)</p>



<p>Proceeding in the same fashion, we can derive a similar equation for the concentration of HIn[HIn]=AIn−Ab(εIn−εHIn)(10.19)(10.19)[HIn]=AIn−Ab(εIn−εHIn)</p>



<p>where&nbsp;<em>A</em><sub>In</sub>, which is equal to ε<sub>In</sub><em>bC</em>, is the absorbance when the pH is basic enough that only In<sup>–</sup>&nbsp;contributes to the absorbance. Substituting equation 10.18 and equation 10.19 into the equilibrium constant expression for HIn givesKa=[H3O+]A−AHInAIn−A(10.20)(10.20)Ka=[H3O+]A−AHInAIn−A</p>



<p>We can use equation 10.20 to determine the value of&nbsp;<em>K</em><sub>a</sub>&nbsp;in one of two ways. The simplest approach is to prepare three solutions, each of which contains the same amount,&nbsp;<em>C</em>, of indicator. The pH of one solution is made sufficiently acidic such that [HIn] &gt;&gt; [In<sup>−</sup>]. The absorbance of this solution gives&nbsp;<em>A</em><sub>HIn</sub>. The value of&nbsp;<em>A</em><sub>In</sub>&nbsp;is determined by adjusting the pH of the second solution such that [In<sup>−</sup>] &gt;&gt; [HIn]. Finally, the pH of the third solution is adjusted to an intermediate value, and the pH and absorbance,&nbsp;<em>A</em>, recorded. The value of&nbsp;<em>K</em><sub>a</sub>&nbsp;is calculated using equation 10.20.</p>



<p>Example 10.9</p>



<p>The acidity constant for an acid–base indicator is determined by preparing three solutions, each of which has a total indicator concentration of 5.00 × 10<sup>–5</sup>&nbsp;M. The first solution is made strongly acidic with HCl and has an absorbance of 0.250. The second solution was made strongly basic and has an absorbance of 1.40. The pH of the third solution is 2.91 and has an absorbance of 0.662. What is the value of&nbsp;<em>K</em><sub>a</sub>&nbsp;for the indicator?</p>



<p><strong><em>Solution</em></strong></p>



<p>The value of&nbsp;<em>K</em><sub>a</sub>&nbsp;is determined by making appropriate substitutions into 10.20; thusKa=(1.23×10−3)×0.662−0.2501.40−0.662=6.87×10−4(4.8.34)(4.8.34)Ka=(1.23×10−3)×0.662−0.2501.40−0.662=6.87×10−4</p>



<p>Practice Exercise 10.9</p>



<p>To determine the&nbsp;<em>K</em><sub>a</sub>&nbsp;of a merocyanine dye, the absorbance of a solution of 3.5×10<sup>–4</sup>&nbsp;M dye was measured at a pH of 2.00, a pH of 6.00, and a pH of 12.00, yielding absorbances of 0.000, 0.225, and 0.680, respectively. What is the value of&nbsp;<em>K</em><sub>a</sub>&nbsp;for this dye? The data for this problem is adapted from Lu, H.; Rutan, S. C.&nbsp;<em>Anal. Chem.</em>,&nbsp;<strong>1996</strong>,&nbsp;<em>68</em>, 1381–1386.</p>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.9" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<p>A second approach for determining&nbsp;<em>K</em><sub>a</sub>&nbsp;is to prepare a series of solutions, each containing the same amount of indicator. Two solutions are used to determine values for&nbsp;<em>A</em><sub>HIn</sub>&nbsp;and&nbsp;<em>A</em><sub>In</sub>. Taking the log of both sides of equation 10.20 and rearranging leave us with the following equation.logA−AHInAIn−A=pH−pKa(10.21)(10.21)log⁡A−AHInAIn−A=pH−pKa</p>



<p>A plot of log[(<em>A</em>&nbsp;–&nbsp;<em>A</em><sub>HIn</sub>)/(<em>A</em><sub>In</sub>&nbsp;–&nbsp;<em>A</em>)] versus pH is a straight-line with a slope of +1 and a&nbsp;<em>y</em>-intercept of –p<em>K</em><sub>a</sub>.</p>



<p>Practice Exercise 10.10</p>



<p>To determine the&nbsp;<em>K</em><sub>a</sub>&nbsp;of the indicator bromothymol blue, the absorbance of a series of solutions containing the same concentration of the indicator was measured at pH levels of 3.35, 3.65, 3.94, 4.30, and 4.64, yielding absorbances of 0.170, 0.287, 0.411, 0.562, and 0.670, respectively. Acidifying the first solution to a pH of 2 changes its absorbance to 0.006, and adjusting the pH of the last solution to 12 changes its absorbance to 0.818. What is the value of&nbsp;<em>K</em><sub>a</sub>&nbsp;for this day? The data for this problem is from Patterson, G. S.<em>&nbsp;J. Chem. Educ.</em>,&nbsp;<strong>1999</strong>,&nbsp;<em>76</em>, 395–398.</p>



<p>Click&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/10_Spectroscopic_Methods/10.E%3A_Spectroscopic_Methods_(Exercises)#Practice_Exercise_10.10" target="_blank" rel="noopener">here</a>&nbsp;to review your answer to this exercise.</p>



<p>In developing these approaches for determining&nbsp;<em>K</em><sub>a</sub>&nbsp;we considered a relatively simple system in which the absorbance of HIn and In<sup>–</sup>&nbsp;are easy to measure and for which it is easy to determine the concentration of H<sub>3</sub>O<sup>+</sup>. In addition to acid–base reactions, we can adapt these approaches to any reaction of the general formX(aq)+Y(aq)⇋Z(aq)(4.8.35)(4.8.35)X(aq)+Y(aq)⇋Z(aq)</p>



<p>including metal–ligand complexation reactions and redox reactions, provided that we can determine spectrophotometrically the concentration of the product, Z, and one of the reactants, and that the concentration of the other reactant can be measured by another method. With appropriate modifications, more complicated systems, in which one or more of these parameters can not be measured, also can be treated.<sup>11</sup></p>



<h2 class="wp-block-heading" id="10-3-5-evaluation-of-uv-vis-and-ir-spectroscopy">10.3.5 Evaluation of UV/Vis and IR Spectroscopy</h2>



<h3 class="wp-block-heading" id="scale-of-operation">Scale of Operation</h3>



<p>Molecular UV/Vis absorption is routinely used for the analysis of trace analytes in macro and meso samples. Major and minor analytes can be determined by diluting the sample before analysis, while concentrating a sample may allow for the analysis of ultratrace analytes. The scale of operations for infrared absorption is generally poorer than that for UV/Vis absorption.</p>



<p>Note</p>



<p>See&nbsp;<a href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Analytical_Chemistry_2.0_(Harvey)/03_The_Vocabulary_of_Analytical_Chemistry/3.4%3A_Selecting_an_Analytical_Method#3D.6_Scale_of_Operation" target="_blank" rel="noopener">Figure 3.5</a>&nbsp;to review the meaning of macro and meso for describing samples, and the meaning of major, minor, and ultratrace for describing analytes.</p>



<h3 class="wp-block-heading" id="accuracy">Accuracy</h3>



<p>Under normal conditions a relative error of 1–5% is easy to obtained with UV/Vis absorption. Accuracy is usually limited by the quality of the blank. Examples of the type of problems that may be encountered include the presence of particulates in a sample that scatter radiation and interferents that react with analytical reagents. In the latter case the interferent may react to form an absorbing species, giving rise to a positive determinate error. Interferents also may prevent the analyte from reacting, leading to a negative determinate error. With care, it may be possible to improve the accuracy of an analysis by as much as an order of magnitude.</p>



<h3 class="wp-block-heading" id="precision">Precision</h3>



<p>In absorption spectroscopy, precision is limited by indeterminate errors—primarily instrumental noise—introduced when measuring absorbance. Precision is generally worse for low absorbances where&nbsp;<em>P</em><sub>0&nbsp;</sub>≈&nbsp;<em>P</em><sub>T</sub>, and for high absorbances when&nbsp;<em>P</em><sub>T</sub>&nbsp;approaches 0. We might expect, therefore, that precision will vary with transmittance.</p>



<p>We can derive an expression between precision and transmittance by applying the propagation of uncertainty as described in Chapter 4. To do so we rewrite Beer’s law asC=−1εblogT(10.22)(10.22)C=−1εblog⁡T</p>



<p>Table 4.10 in Chapter 4 helps us in completing the propagation of uncertainty for equation 10.22, giving the absolute uncertainty in the concentration,&nbsp;<em>s</em><sub>C</sub>, assC=−0.4343εb×sTT(10.23)(10.23)sC=−0.4343εb×sTT</p>



<p>where&nbsp;<em>s</em><sub>T</sub>&nbsp;is the absolute uncertainty in the transmittance. Dividing equation 10.23 by equation 10.22 gives the relative uncertainty in concentration,&nbsp;<em>s</em><sub>C</sub>/<em>C</em>, assCC=0.4343sTTlogT(4.8.36)(4.8.36)sCC=0.4343sTTlog⁡T</p>



<p>If we know the absolute uncertainty in transmittance, we can determine the relative uncertainty in concentration for any transmittance.</p>



<p>Determining the relative uncertainty in concentration is complicated because&nbsp;<em>s</em><sub>T</sub>&nbsp;may be a function of the transmittance. As shown in Table 10.8, three categories of indeterminate instrumental error have been observed.<sup>12</sup>&nbsp;A constant&nbsp;<em>s</em><sub>T</sub>&nbsp;is observed for the uncertainty associated with reading %T on a meter’s analog or digital scale. Typical values are ±0.2–0.3% (a&nbsp;<em>k</em><sub>1</sub>&nbsp;of ±0.002–0.003) for an analog scale, and ±0.001% a (<em>k</em><sub>1</sub>&nbsp;of ±0.000 01) for a digital scale. A constant&nbsp;<em>s</em><sub>T</sub>&nbsp;also is observed for the thermal transducers used in infrared spectrophotometers. The effect of a constant&nbsp;<em>s</em><sub>T</sub>&nbsp;on the relative uncertainty in concentration is shown by curve A in Figure 10.40. Note that the relative uncertainty is very large for both high and low absorbances, reaching a minimum when the absorbance is 0.4343. This source of indeterminate error is important for infrared spectrophotometers and for inexpensive UV/Vis spectrophotometers. To obtain a relative uncertainty in concentration of ±1–2%, the absorbance must be kept within the range 0.1–1.</p>



<figure class="wp-block-table"><table><thead><tr><th scope="col">Category</th><th scope="col">Sources of Indeterminate Error</th><th scope="col">Relative Uncertainty in Concentration</th></tr></thead><tbody><tr><td>sT=k1sT=k1</td><td>%T readout resolution noise in thermal detectors</td><td>sCC=0.4343k1TlogTsCC=0.4343k1Tlog⁡T</td></tr><tr><td>sT=k2T2+T−−−−−−√sT=k2T2+T</td><td>noise in photon detectors</td><td>sCC=0.4343k2logT1+1T−−−−−√sCC=0.4343k2log⁡T1+1T</td></tr><tr><td>sT=k3TsT=k3T</td><td>positioning of sample cell fluctuations in source intensity</td><td>sCC=0.4343k3logTsCC=0.4343k3log⁡T</td></tr></tbody></table></figure>



<p>Values of&nbsp;<em>s</em><sub>T</sub>&nbsp;are a complex function of transmittance when indeterminate errors are dominated by the noise associated with photon detectors. Curve B in Figure 10.40 shows that the relative uncertainty in concentration is very large for low absorbances, but is less at higher absorbances. Although the relative uncertainty reaches a minimum when the absorbance is 0.963, there is little change in the relative uncertainty for absorbances within the range 0.5–2. This source of indeterminate error generally limits the precision of high quality UV/Vis spectrophotometers for mid-to-high absorbances.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12720/Figure10.40.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12720/Figure10.40.jpg?revision=1&amp;size=bestfit&amp;width=450&amp;height=392" alt="Figure10.40.jpg"/></a></figure>



<p><strong>Figure 10.40</strong>&nbsp;Percent relative uncertainty in concentration as a function of absorbance for the categories of indeterminate errors in Table 10.8. A:&nbsp;<em>k</em><sub>1</sub>&nbsp;= ±0.0030; B:&nbsp;<em>k</em><sub>2</sub>&nbsp;= ±0.0030; C:<em>k</em><sub>3</sub>&nbsp;= ±0.0130. The dashed lines correspond to the minimum uncertainty for curve A (absorbance of 0.4343) and for curve B (absorbance of 0.963).</p>



<p>Finally, the value of&nbsp;<em>s</em><sub>T</sub>&nbsp;is directly proportional to transmittance for indeterminate errors resulting from fluctuations in the source’s intensity and from uncertainty in positioning the sample within the spectrometer. The latter is particularly important because the optical properties of any sample cell are not uniform. As a result, repositioning the sample cell may lead to a change in the intensity of transmitted radiation. As shown by curve C in Figure 10.40, the effect is only important at low absorbances. This source of indeterminate errors is usually the limiting factor for high quality UV/Vis spectrophotometers when the absorbance is relatively small.</p>



<p>When the relative uncertainty in concentration is limited by the %T readout resolution, the precision of the analysis can be improved by redefining 100% T and 0% T. Normally 100% T is established using a blank and 0% T is established while preventing the source’s radiation from reaching the detector. If the absorbance is too high, precision can be improved by resetting 100% T using a standard solution of the analyte whose concentration is less than that of the sample (Figure 10.41a). For a sample whose absorbance is too low, precision can be improved by redefining 0% T using a standard solution of the analyte whose concentration is greater than that of the analyte (Figure 10.41b). In this case a calibration curve is required because a linear relationship between absorbance and concentration no longer exists. Precision can be further increased by combining these two methods (Figure 10.41c). Again, a calibration curve is necessary since the relationship between absorbance and concentration is no longer linear.</p>



<figure class="wp-block-image"><a href="https://chem.libretexts.org/@api/deki/files/12721/Figure10.41.jpg?revision=1" target="_blank" rel="noopener"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/12721/Figure10.41.jpg?revision=1&amp;size=bestfit&amp;width=300&amp;height=330" alt="Figure10.41.jpg"/></a></figure>



<p><strong>Figure 10.41</strong>&nbsp;Methods for improving the precision of absorption methods: (a) high-absorbance method; (b) low-absorbance method; (c) maximum precision method.</p>



<h3 class="wp-block-heading" id="sensitivity">Sensitivity</h3>



<p>The sensitivity of a molecular absorption method, which is the slope of a Beer’s law calibration curve, is the product of the analyte’s absorptivity and the pathlength of the sample cell (ε<em>b</em>). You can improve a method’s sensitivity by selecting a wavelength where absorbance is at a maximum or by increasing the pathlength.</p>



<p>Note</p>



<p>See Figure 10.24 for an example of how the choice of wavelength affects a calibration curve’s sensitivity.</p>



<h3 class="wp-block-heading" id="selectivity">Selectivity</h3>



<p>Selectivity is rarely a problem in molecular absorption spectrophotometry. In many cases it is possible to find a wavelength where only the analyte absorbs. When two or more species do contribute to the measured absorbance, a multicomponent analysis is still possible, as shown in Example 10.6 and Example 10.7.</p>



<h3 class="wp-block-heading" id="time-cost-and-equipment">Time, Cost, and Equipment</h3>



<p>The analysis of a sample by molecular absorption spectroscopy is relatively rapid, although additional time may be required if we need to chemically convert a nonabsorbing analyte into an absorbing form. The cost of UV/Vis instrumentation ranges from several hundred dollars for a simple filter photometer, to more than $50,000 for a computer controlled high resolution, double-beam instrument equipped with variable slits, and operating over an extended range of wavelengths. Fourier transform infrared spectrometers can be obtained for as little as $15,000–$20,000, although more expensive models are available.</p>



<h2 class="wp-block-heading" id="contributors">Contributors</h2>



<p><a rel="noreferrer noopener" href="http://dpuadweb.depauw.edu/harvey_web/Index.html" target="_blank">David Harvey</a>&nbsp;<a rel="noreferrer noopener" href="http://www.depauw.edu/" target="_blank">(DePauw University)</a></p>



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		<title>A to Z of UV-Vis spectroscopy interpretation</title>
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<p>While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo&nbsp;<strong>electronic transitions</strong>. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s7.picofile.com/file/8392387600/uv_vis_in.gif" alt=""/></a></figure>



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<h2 class="wp-block-heading" id="electronic-transitions">Electronic transitions</h2>



<p>Let’s take as our first example the simple case of molecular hydrogen, H<sub>2</sub>. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding&nbsp;σ&nbsp;MO, and a higher energy antibonding&nbsp;σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding&nbsp;σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140660/image023.png?revision=1" alt="image024.png"/></figure>



<p>If the molecule is exposed to light of a wavelength with energy equal to&nbsp;<strong>Δ</strong>E, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the&nbsp;σ&nbsp;to the&nbsp;σ* orbital. This is referred to as a&nbsp;<strong>σ&nbsp;</strong><strong>&#8211;&nbsp;</strong><strong>σ</strong><strong>* transition</strong>.&nbsp;<strong>Δ</strong>E for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.</p>



<p>When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a&nbsp;<strong>π</strong><strong>&nbsp;&#8211;&nbsp;</strong><strong>π</strong><strong>* transition.&nbsp;</strong>Because&nbsp;π-&nbsp;π* energy gaps are narrower than&nbsp;σ&nbsp;<strong>&#8211;</strong>&nbsp;σ<strong>*&nbsp;</strong>gaps, ethene absorbs light at 165 nm &#8211; a longer wavelength than molecular hydrogen.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140652/image025.png?revision=1" alt="image026.png"/></figure>



<p>The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated&nbsp;ππsystems. In these groups, the energy gap for&nbsp;π&nbsp;-π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called&nbsp;<strong>chromophores</strong>.</p>



<p>Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2p<sub>z</sub>&nbsp;atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140644/image027.png?revision=1" alt="image028.png"/></figure>



<p>Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.</p>



<p>As conjugated pi systems become larger, the energy gap for a&nbsp;π&nbsp;&#8211;&nbsp;π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the&nbsp;π&nbsp;&#8211;&nbsp;π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a&nbsp;<strong>Δ</strong>E of 111 kcal/mol.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140637/image029.png?revision=1" alt="image030.png"/></figure>



<p>In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region &#8211; to be transmitted. This is why carrots are orange.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140631/image031.png?revision=1" alt="image032.png"/></figure>



<p>The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a&nbsp;π&nbsp;&#8211;&nbsp;π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a&nbsp;π* antibonding MO:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140625/image033.png?revision=1" alt="image034.png"/></figure>



<p>This is referred to as an&nbsp;<strong>n</strong><strong>&nbsp;&#8211;&nbsp;π</strong><strong>* transition</strong>. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an&nbsp;n→π∗n→π∗transition is smaller that that of a&nbsp;π&nbsp;&#8211;&nbsp;π* transition – and thus the n &#8211;&nbsp;π* peak is at a longer wavelength. In general, n &#8211;&nbsp;π* transitions are weaker (less light absorbed) than those due to&nbsp;π &#8211;&nbsp;π* transitions.</p>



<p><strong><u>Exercise 4.9</u></strong></p>



<p>What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the&nbsp;l<sub>max</sub>&nbsp;of&nbsp;b-carotene?</p>



<p><strong><u>Exercise 4.10</u></strong></p>



<p>Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140619/image035.png?revision=1&amp;size=bestfit&amp;width=306&amp;height=155" alt="image036.png"/></figure>



<p><a target="_blank" href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Solution_Manual/Chapter_04_Solutions" rel="noreferrer noopener">Solutions</a></p>



<p><strong>Protecting yourself from sunburn</strong></p>



<p>Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140772/sunscreen.png?revision=1&amp;size=bestfit&amp;width=461&amp;height=391" alt=""/></figure>



<p>Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (<em>para</em>-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s7.picofile.com/file/8392387600/uv_vis_in.gif" alt=""/></a></figure>



<h2 class="wp-block-heading" id="looking-at-uv-vis-spectra">Looking at UV-vis spectra</h2>



<p>We have been talking in general terms about how molecules absorb UV and visible light – now let&#8217;s look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.</p>



<p><img decoding="async" alt="" width="670px" height="376px" src="https://chem.libretexts.org/@api/deki/files/140776/Schematic_of_UV-_visible_spectrophotometer.png?revision=1&amp;size=bestfit&amp;width=670&amp;height=376"><strong>Schematic for a UV-Vis spectrophotometer</strong></p>



<p>(Image from&nbsp;<a target="_blank" href="https://upload.wikimedia.org/wikipedia/commons/9/95/Schematic_of_UV-_visible_spectrophotometer.png" rel="noreferrer noopener">Wikipedia Commons</a>)</p>



<p>Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD<sup>+</sup>. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140612/image037.png?revision=1" alt="image038.png"/></figure>



<p>You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm<sup>-1</sup>&nbsp;as is the convention in IR spectroscopy.</p>



<p>Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is&nbsp;λmaxλmax, which is the wavelength at maximal light absorbance. As you can see, NAD<sup>+</sup>&nbsp;has&nbsp;λmax=260nmλmax=260nm. We also want to record how much light is absorbed at&nbsp;λmaxλmax. Here we use a unitless number called&nbsp;<strong>absorbance</strong>, abbreviated &#8216;A&#8217;. This contains the same information as the &#8216;percent transmittance&#8217; number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength&nbsp;<em>before</em>&nbsp;it passes through the sample (I<sub>0</sub>), divides this value by the intensity of the same wavelength&nbsp;<em>after</em>&nbsp;it passes through the sample (I), then takes the log<sub>10</sub>&nbsp;of that number:A=logI0I(4.3.1)(4.3.1)A=log⁡I0I</p>



<p>You can see that the absorbance value at 260 nm (A<sub>260</sub>) is about 1.0 in this spectrum.</p>



<p><strong>Exercise 4.11</strong></p>



<p>Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).</p>



<p><a target="_blank" href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Solution_Manual/Chapter_04_Solutions" rel="noreferrer noopener">Solutions</a></p>



<p>Here is the absorbance spectrum of the common food coloring Red #3:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140602/image039.png?revision=1" alt="image040.png"/></figure>



<p>Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the&nbsp;<strong>λ</strong><sub>max&nbsp;</sub>of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140595/image041.png?revision=1" alt="image042.png"/></figure>



<p>Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.</p>



<h2 class="wp-block-heading" id="applications-of-uv-spectroscopy-in-organic-and-biological-chemistry">Applications of UV spectroscopy in organic and biological chemistry</h2>



<p>UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the&nbsp;<a href="https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_Basics/The_Beer-Lambert_Law" target="_blank" rel="noopener"><strong>Beer &#8211; Lambert Law</strong></a>&nbsp;to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.</p>



<p>If we divide the observed value of A at&nbsp;λ<sub>max</sub>&nbsp;by the concentration of the sample (<em>c</em>, in mol/L), we obtain the&nbsp;<strong>molar absorptivity</strong>, or&nbsp;<strong>extinction coefficient</strong>&nbsp;(<strong>ε</strong>), which is a characteristic value for a given compound.ϵ=Ac(4.3.2)(4.3.2)ϵ=Ac</p>



<p>The absorbance will also depend, of course, on the&nbsp;<strong>path length</strong>&nbsp;&#8211; in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L<sub>*</sub>&nbsp;mol<sup>-1</sup><sub>*</sub>cm<sup>-1</sup>. If we look up the value of e for our compound at&nbsp;λ<sub>max</sub>, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD<sup>+</sup>&nbsp;the literature value of&nbsp;ε&nbsp;at 260 nm is 18,000 L<sub>*</sub>&nbsp;mol<sup>-1</sup><sub>*</sub>cm<sup>-1</sup>. In our NAD<sup>+</sup>&nbsp;spectrum we observed A<sub>260</sub>&nbsp;= 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10<sup>-5</sup>&nbsp;M.</p>



<p>The bases of DNA and RNA are good chromophores:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140587/image043.png?revision=1" alt="image044.png"/></figure>



<p>Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of&nbsp;ε&nbsp;= 0.020 ng<sup>-1</sup>×mL for double-stranded DNA at its&nbsp;λ<sub>max</sub>&nbsp;of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).<strong><u>Exercise 4.12</u></strong>50&nbsp;microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in&nbsp;micrograms&nbsp;per microliter?<a target="_blank" href="https://chem.libretexts.org/Under_Construction/Purgatory/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Solution_Manual/Chapter_04_Solutions" rel="noreferrer noopener">Solutions</a></p>



<p>Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140579/image045.png?revision=1" alt="image046.png"/></figure>



<p>As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.</p>



<p>Later we will see how the Beer &#8211; Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD<sup>+</sup>, the compound whose spectrum we saw earlier in this section) to NADH:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140570/image047.png?revision=1" alt="image048.png"/></figure>



<p>Both NAD<sup>+</sup>&nbsp;and NADH absorb at 260 nm. However NADH, unlike NAD<sup>+</sup>, has a second absorbance band with&nbsp;λ<sub>max</sub>&nbsp;= 340 nm and&nbsp;ε&nbsp;= 6290 L<sub>*</sub>mol<sup>-1</sup><sub>*</sub>cm<sup>-1</sup>. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140562/image049.png?revision=1" alt="image050.png"/></figure>



<p>By monitoring the absorbance of a reaction mixture at 340 nm, we can &#8216;watch&#8217; NADH being formed as the reaction proceeds, and calculate the rate of the reaction.</p>



<p>UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.</p>



<figure class="wp-block-image"><img decoding="async" src="https://chem.libretexts.org/@api/deki/files/140551/image051.png?revision=1" alt="image052.png"/></figure>



<p>Biochemists frequently use UV spectroscopy to study conformational changes in proteins &#8211; how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.</p>



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		<title>Interpretation of XPS analysis</title>
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					<description><![CDATA[Click here to see other posts about XPS Only 10 $ for interpretation of each element in your XPS spectrum Payment Upon Completion Send your spectra... What is XPS? XPS is a surface-sensitive technique based on the photoelectric effect, which occurs when atoms or molecules are irradiated by photons of suitable energy, resulting in the [&#8230;]]]></description>
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<h2 class="wp-block-heading" id="what-is-xps">What is XPS?</h2>



<p>XPS is a surface-sensitive technique based on the photoelectric effect, which occurs when atoms or molecules are irradiated by photons of suitable energy, resulting in the ejection of electrons. The kinetic energy of the ejected electrons depends upon the elemental core level from which they originated.</p>



<p>Using this information, XPS data can be used to determine the elemental composition of the surface and, in most cases, the bonding environment of the elements.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<span id="more-487"></span>



<p>The schematic figure shown to the right, illustrates the XPS procedure, where x-rays are used to excite and eject photoelectrons from a sodium chloride molecule on a substrate.</p>



<p>XPS detects and quantifies the ejected photoelectrons, which are proportional to the amount present in the uppermost layers of the surface.</p>



<p>To understand more about these uppermost layers, further details on the XPS penetration depth and attenuation length of the X-ray photons are required.</p>



<p>Normal XPS can provide information from the top 10nm (approximately) layer of the surface.</p>



<p><img fetchpriority="high" decoding="async" alt="Schematic diagram" height="161" width="386" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/Schematic-diagram-Cropped-386x161.png">Schematic diagram of photoelectron emission from a sodium atom under x-ray exposure, and (right) example of XPS survey scan spectra of sodium chloride with the Na1s electron XPS peak highlighted in the spectra.</p>



<p><em>Image provided courtesy of Thermo Fisher Scientific.</em></p>



<p>Theoretically, XPS should be able to detect all elements. However, helium (<em>He</em>) does not readily form solid compounds and its 1s orbital has a tiny cross-section for photoemission.</p>



<p>Hydrogen (<em>H</em>) also has a tiny cross-section and suffers from having to share its only electron in forming compounds, which then resides in a valence-like orbital, the energy of which varies from compound to compound.</p>



<h2 class="wp-block-heading" id="features-of-the-thermo-fisher-escalab-250xi-xps-instrument">Features of the Thermo Fisher ESCALAB 250Xi XPS instrument</h2>



<p>The Thermo Fisher Scientific ESCALAB 250Xi is the most recent advancement in the ESCALAB series.</p>



<p>The instrument is an optimised multi-method platform that is expandable and comes with excellent flexibility and configurability. Its cutting edge technology is driven by smart software and hardware.</p>



<p>Equipped with a micro-focusing X-ray monochromator designed to deliver optimum XPS performance, the instrument ensures maximum sample throughput.</p>



<p>The multi-technique capability and availability of a range of preparation chambers and devices provides the solution to any surface analytical problem.</p>



<p><img decoding="async" alt="ESCALAB 250Xi XPS instrument" height="242" width="260" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/ESCALAB-250Xi-XPS-instrument-Cropped-260x242.png">ESCALAB 250Xi XPS instrument<img decoding="async" alt="ESCALAB standard sample loading chamber" height="272" width="398" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/ESCALAB-standard-sample-loading-chamber-Cropped-398x272.jpg">ESCALAB standard sample loading chamber</p>



<h3 class="wp-block-heading" id="some-notable-features-of-the-instrument-include">Some notable features of the instrument include:</h3>



<ul class="wp-block-list"><li>High sensitivity spectroscopy</li><li>Small area XPS</li><li>Depth profiling capability with MAGCIS (Monatomic &amp; Gas Cluster Ion Source)</li><li>Ion scattering spectroscopy (ISS)</li><li>Reflected electron energy loss spectroscopy (REELS)</li><li>Micro-focused X-ray spot</li><li>Efficient Charge Neutralisation</li><li>Angle Resolved Spectroscopy</li><li>Ultra-high sensitivity and energy resolution</li></ul>



<h2 class="wp-block-heading" id="xps-analysis-sample-requirement">XPS analysis sample requirement</h2>



<p>XPS is a surface analytical technique that can be used to study the surface properties of a range of sample types. This includes both inorganic and organic (ex-situ and samples suitable to be placed under ultra-high vacuum) materials, polymers, semiconductors, metals, composite materials, geological and archaeological samples, ceramics, and glasses amongst others.</p>



<ul class="wp-block-list"><li>The&nbsp;<strong>ideal dimension</strong>&nbsp;for an XPS analysis sample is 1cm X 1cm, with a maximum of 0.5cm thickness. Samples with lengths up to 5cm and widths up to 2.2cm can also be analysed. Samples below 0.5cm x 0.5cm dimension are difficult to mount on the sample stage. XPS analysis depends on the spot size. Ideally we can measure surface areas (spots) greater than 300µm X 300µm, as the smallest X-ray spot/exposure area is 200µm X 200µm for general analysis in our instrument. Please let us know your sample size and type in advance.</li><li><strong>Powder samples</strong>&nbsp;can also be measured using XPS. However, please discuss this with us in advance.</li><li><strong>Sample handling</strong>&nbsp;is crucial for XPS analysis. As XPS measures the top few atomic layers on the surface (different for depth profiling), it is very easy to contaminate samples with fingerprints. Ideally, XPS samples should be shipped and transferred inside a Wafer Carrier Box used in the semiconductor industry. However, any sealed container that doesn’t alter surface chemistry should be sufficient. Samples stored in zip-lock polymer bags are not ideal.</li></ul>



<figure class="wp-block-image"><img decoding="async" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/SampleNavigation-inside-XPS-Cropped-260x291.jpg" alt="Sample Navigation inside XPS"/></figure>



<p>XPS Instrument Analysis Chamber view.</p>



<p><em>Image provided courtesy of Thermo Fisher Scientific.</em></p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<h2 class="wp-block-heading" id="applications-of-xps">Applications of XPS</h2>



<p>XPS surface analysis can provide answers to a wide range of research problems. The following are examples of research questions addressed using XPS by researchers at the University of Brighton and elsewhere.</p>



<h2 class="wp-block-heading" id="what-material-did-we-make-purchase">What material did we make / purchase?</h2>



<p>XPS can provide data about the elemental distribution on the surface of a sample.</p>



<p>The limit of detection is 0.1 atomic per cent or better. In the figure (right), two XPS survey scans of gold coated QCM crystals are shown – the inset is a zoomed-in version of the Ti2p area of the spectra.</p>



<p>The two crystals are from two separate purchased batches. As can be seen, batch one (red) contains around 23 atomic per cent of titanium, making the crystal not fit for purpose, while batch two (green) is ‘pure’ as claimed by the manufacturer.</p>



<p><img loading="lazy" decoding="async" height="258" width="392" alt="Survey scan spectra of two different batch QCM crystals" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/Survey-scan-spectra-Cropped2-392x258.png"></p>



<p>Survey scan spectra of two different batch QCM crystals. (S. Ray, unpublished data)</p>



<h3 class="wp-block-heading" id="how-thick-is-the-coating-contamination-on-a-surface">How thick is the coating/contamination on a surface?</h3>



<p>XPS can measure precisely the thickness (below 10nm) of the surface adsorbed layer. This is useful in understanding the nature of surface contamination, and also in studying biomolecular adsorption on implants.</p>



<p>The spectra (right) is a comparison of XPS and Ellipsometric Thickness measurements of three different proteins adsorbed on hydrophobic surfaces, showing extremely close agreement.</p>



<p></p>



<p><img loading="lazy" decoding="async" alt="XPS and Ellipsometric thickness measurement comparison of three different proteins adsorbed on hydrophobic surfaces." height="272" width="271" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/XPS-and-Ellipsometric-thickness-measurement-Cropped-271x272.png">XPS and Ellipsometric Thickness measurement comparison of three different proteins adsorbed on hydrophobic surfaces.</p>



<h3 class="wp-block-heading" id="what-chemical-states-are-present-on-my-surface">What chemical states are present on my surface?</h3>



<p>XPS can identify and quantify the nature of the chemical states on a surface, and can help in visualising the surface functionalisation, essential for applications including bio-chemical sensors and chemical conjugations.</p>



<p>In the example (below left image), the experimental verification of graphene oxide reduction is shown, while the right image illustrates the success of a pluoronics (triblock copolymer) coating on reduced graphene oxide.</p>



<p><img loading="lazy" decoding="async" alt="C1s Narrow Scan XPS Data" height="258" width="797" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/Chemical-States3-Cropped-797x258.jpg">XPS C1s narrow scan spectra of (left) reduced graphene oxide and (right) pluronics coated reduced graphene oxide (S. Ray, unpublished data)</p>



<h3 class="wp-block-heading" id="how-can-we-analyse-and-quantify-contamination-or-doping-in-our-sample">How can we analyse and quantify contamination or doping in our sample?</h3>



<p>XPS can reveal surface or just below-surface contamination, and can successfully quantify any organic and inorganic contaminants and/or doping.</p>



<p>One example of doping quantification would be analysing the nitrogen doping effect on the properties of graphene. From the XPS spectra (right), the amount and bonding nature of nitrogen with graphene could be quantified.</p>



<p>Image on left: Nitrogen (N1s) narrow scan spectra and peak deconvolution to find the nature of nitrogen bonded to graphene (S. Ray, unpublished data).</p>



<figure class="wp-block-image"><img decoding="async" src="https://www.brighton.ac.uk/images/Business-and-community-images/Facilities/Capture2-Cropped-398x272.jpg" alt="Narrow Scan"/></figure>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<h3 class="wp-block-heading" id="can-xps-measure-the-thickness-of-coatings-on-nanoparticles">Can XPS measure the thickness of coatings on nanoparticles?</h3>



<p>In his 2017 publication, Professor David Castner (University of Washington) states,&nbsp;<em>“Single particle information from electron microscopy combined with XPS sensitivity in determining composition make a powerful combination for nanoparticle analysis” (&nbsp;<a href="http://apps.webofknowledge.com.ezproxy.brighton.ac.uk/full_record.do?product=WOS&amp;search_mode=GeneralSearch&amp;qid=1&amp;SID=D3ybybIWHY3opzCu5VN&amp;page=2&amp;doc=11" target="_blank" rel="noopener">Powell et al., 2017, J. Physical. Chemistry C</a>)</em>.</p>



<p>XPS can measure precisely the thickness of single layer or multiple layers of coatings on nano-micro particles. Currently there are numerous situations where nanoparticles are used (e.g. in targeted drug delivery, sunscreens, and antimicrobial socks).</p>



<p>To functionalise these particles according to their target use, a proper understanding of the coating on the nanoparticles is required. In the case of multifunction nanoparticle use for targeted drug delivery, quantification of the single, double or triple layers is necessary.</p>



<h3 class="wp-block-heading" id="xps-can-help-in-understanding-many-other-questions-including">XPS can help in understanding many other questions, including:</h3>



<ul class="wp-block-list"><li>What is the effect of heat, aging, chemical treatment, or real world use on my samples?</li><li>What coating is on the surface?</li><li>What is wrong inside my thick film?</li><li>Are the thicknesses of film layers correct?</li><li>Is the surface chemistry of a sample uniform?</li></ul>



<h2 class="wp-block-heading" id="accessing-the-surface-analysis-laboratory">Accessing the Surface Analysis Laboratory</h2>



<p>If you have queries about how XPS can help in your research and/or industrial project, please do not hesitate to get in touch by telephone or email.&nbsp;We are happy to discuss your research requirements and provide quotations as necessary.</p>



<p>We particularly welcome proposals for joint research funding applications.</p>



<p>We also offer an analysis-only service, but can also provide full interpretation of results on request.&nbsp;XPS analysis costs can be charged per sample, per day/half-day or according to your needs.</p>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>
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		<title>A to Z of X-ray photoelectron spectroscopy (XPS)</title>
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					<description><![CDATA[Click here to see other posts about XPS Only 10 $ for interpretation of each in element your XPS spectrum Payment Upon Completion Send your spectra... Introduction X-Ray photoelectron spectroscopy (XPS), also known as electron spectroscopy for chemical analysis (ESCA), is one of the most widely used surface techniques in materials science and chemistry. It [&#8230;]]]></description>
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<pre class="wp-block-verse has-text-align-center"><span style="color:#ffffff" class="tadv-color">Only 10 $ for interpretation of each in element your XPS spectrum
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<h4 class="wp-block-heading" id="introduction">Introduction</h4>



<p>X-Ray photoelectron spectroscopy (XPS), also known as electron spectroscopy for chemical analysis (ESCA), is one of the most widely used surface techniques in materials science and chemistry. It allows the determination of atomic composition of the sample in a non-destructive manner, as well as other chemical information, such as binding constants, oxidation states and speciation. The sample under study is subjected to irradiation by a high energy X-ray source. The X-rays penetrate only 5 – 20 Å into the sample, allowing for surface specific, rather than bulk chemical, analysis. As an atom absorbs the X-rays, the energy of the X-ray will cause a K-shell electron to be ejected, as illustrated by Figure&nbsp;1.13.11.13.1. The K-shell is the lowest energy shell of the atom. The ejected electron has a kinetic energy (KE) that is related to the energy of the incident beam (hν), the electron binding energy (BE), and the work function of the spectrometer (φ) (<a href="https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Book%3A_Physical_Methods_in_Chemistry_and_Nano_Science_(Barron)/01%3A_Elemental_Analysis/1.13%3A_X-ray_Photoelectron_Spectroscopy#mjx-eqn-1" target="_blank" rel="noopener">1.13.1</a>1.13.1). Thus, the binding energy of the electron can be calculated.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/c47411b4cd66e5a39e3039167c4d567619d16518/graphics1.jpg" alt=""/><figcaption>Figure&nbsp;1.13.11.13.1&nbsp;Excitation of an electron from an atom&#8217;s K-shell.</figcaption></figure>



<figure class="wp-block-image size-large"><a href="http://www.analyzetest.com/index.php/contact-us/"><img decoding="async" src="http://s6.picofile.com/file/8392388968/xps_in.gif" alt=""/></a></figure>



<span id="more-483"></span>



<p>BE&nbsp;=&nbsp;hν&nbsp;−&nbsp;KE&nbsp;−&nbsp;ψs(1.13.1)(1.13.1)BE&nbsp;=&nbsp;hν&nbsp;−&nbsp;KE&nbsp;−&nbsp;ψs</p>



<p>Table&nbsp;1.13.11.13.1&nbsp;shows the binding energy of the ejected electron, and the orbital from which the electron is ejected, which is characteristic of each element. The number of electrons detected with a specific binding energy is proportional to the number of corresponding atoms in the sample. This then provides the percent of each atom in the sample.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Element</strong></td><td><strong>Binding Energy (eV)</strong></td></tr><tr><td>Carbon (C) (1s)</td><td>284.5 &#8211; 285.1</td></tr><tr><td>Nitrogen (N) (1s)</td><td>396.1 &#8211; 400.5</td></tr><tr><td>Oxygen (O) (1s)</td><td>526.2 &#8211; 533.5</td></tr><tr><td>Silicon (Si) (2p)</td><td>98.8 &#8211; 99.5</td></tr><tr><td>Sulfur (S) (2p<sub>3/2</sub>)</td><td>164.0 &#8211; 164.3</td></tr><tr><td>Iron (Fe) (2p<sub>3/2</sub>)</td><td>706.8 &#8211; 707.2</td></tr><tr><td>Gold (Au) (4f<sub>7/2</sub>)</td><td>83.8 &#8211; 84.2</td></tr></tbody></table></figure>



<p>The chemical environment and oxidation state of the atom can be determined through the shifts of the peaks within the range expected (Table&nbsp;1.13.21.13.2). If the electrons are shielded then it is easier, or requires less energy, to remove them from the atom, i.e., the binding energy is low. The corresponding peaks will shift to a lower energy in the expected range. If the core electrons are not shielded as much, such as the atom being in a high oxidation state, then just the opposite occurs. Similar effects occur with electronegative or electropositive elements in the chemical environment of the atom in question. By synthesizing compounds with known structures, patterns can be formed by using XPS and structures of unknown compounds can be determined.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Compound</strong></td><td><strong>Binding Energy (eV)</strong></td></tr><tr><td>COH (C 1s)</td><td>286.01 &#8211; 286.8</td></tr><tr><td>CHF (C 1s)</td><td>287.5 &#8211; 290.2</td></tr><tr><td>Nitride (N 1s)</td><td>396.2 &#8211; 398.3</td></tr><tr><td>Fe<sub>2</sub>O<sub>3</sub>&nbsp;(from O, 1s)</td><td>529.5 &#8211; 530.2</td></tr><tr><td>Fe<sub>2</sub>O<sub>3</sub>&nbsp;(from Fe, 2p<sub>3/2</sub>)</td><td>710.7 &#8211; 710.9</td></tr><tr><td>FeO (from Fe 2p<sub>3/2</sub>)</td><td>709.1 &#8211; 709.5</td></tr><tr><td>SiO<sub>2</sub>&nbsp;(from O, 2s)</td><td>532.5 &#8211; 533.3</td></tr><tr><td>SiO<sub>2</sub>&nbsp;(from Si, 2p)</td><td>103.2 &#8211; 103.9</td></tr></tbody></table></figure>



<p>Sample preparation is important for XPS. Although the technique was originally developed for use with thin, flat films, XPS can be used with powders. In order to use XPS with powders, a different method of sample preparation is required. One of the more common methods is to press the powder into a high purity indium foil. A different approach is to dissolve the powder in a quickly evaporating solvent, if possible, which can then be drop-casted onto a substrate. Using sticky carbon tape to adhere the powder to a disc or pressing the sample into a tablet are an option as well. Each of these sample preparations are designed to make the powder compact, as powder not attached to the substrate will contaminate the vacuum chamber. The sample also needs to be completely dry. If it is not, solvent present in the sample can destroy the necessary high vacuum and contaminate the machine, affecting the data of the current and future samples.</p>



<h4 class="wp-block-heading" id="analyzing-functionalized-surfaces">Analyzing Functionalized Surfaces</h4>



<h6 class="wp-block-heading" id="depth-profiling">Depth Profiling</h6>



<p>When analyzing a sample (Figure&nbsp;1.13.21.13.2&nbsp;a) by XPS, questions often arise that deal with layers of the sample. For example, is the sample homogenous, with a consistent composition throughout, or layered, with certain elements or components residing in specific places in the sample? (Figure&nbsp;1.13.21.13.2&nbsp;b,c). A simple way to determine the answer to this question is to perform a depth analysis. By sputtering away the sample, data can be collected at different depths within the sample. It should be noted that sputtering is a destructive process. Within the XPS instrument, the sample is subjected to an Ar<sup>+</sup>&nbsp;ion beam that etches the surface. This creates a hole in the surface, allowing the X-rays to hit layers that would not have otherwise been analyzed. However, it should be realized that different surfaces and layers may be etched at different rates, meaning the same amount of etching does not occur during the same amount of time, depending on the element or compound currently being sputtered.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/f671713a23a91945f8b19797f7a1be7dbd70f314/graphics3.jpg" alt=""/><figcaption>Figure&nbsp;1.13.21.13.2&nbsp;Schematic representation of analysis of (a) an homogeneous sample, as compared to (b) an homogeneous layers in a sample, and (c) an inhomogeneous layers in a sample.</figcaption></figure>



<p>It is important to note that hydrocarbons sputter very easily and can contaminate the high vacuum of the XPS instrument and thus later samples. They can also migrate to a recently sputtered (and hence unfunctionalized) surface after a short amount of time, so it is imperative to sputter and take a measurement quickly, otherwise the sputtering may appear to have had no effect.</p>



<h6 class="wp-block-heading" id="functionalized-films">Functionalized Films</h6>



<p>When running XPS, it is important that the sample is prepared correctly. If it is not, there is a high chance of ruining not only data acquisition, but the instrument as well. With organic functionalization, it is very important to ensure the surface functional group (or as is the case with many functionalized nanoparticles, the surfactant) is immobile on the surface of the substrate. If it is removed easily in the vacuum chamber, it not only will give erroneous data, but it will contaminate the machine, which may then contaminate future samples. This is particularly important when studying thiol functionalization of gold samples, as thiol groups bond strongly with the gold. If there is any loose thiol group contaminating the machine, the thiol will attach itself to any gold sample subsequently placed in the instrument, providing erroneous data. Fortunately, with the above exception, preparing samples that have been functionalized is not much different than standard preparation procedures. However, methods for analysis may have to be modified in order to obtain good, consistent data.</p>



<p>A common method for the analysis of surface modified material is angle resolved X-ray photoelectron spectroscopy (ARXPS). ARXPS is a non-destructive alternative to sputtering, as it relies upon using a series of small angles to analyze the top layer of the sample, giving a better picture of the surface than standard XPS. ARXPS allows for the analysis of the topmost layer of atoms to be analyzed, as opposed to standard XPS, which will analyze a few layers of atoms into the sample, as illustrated in Figure&nbsp;1.13.31.13.3. ARXPS is often used to analyze surface contaminations, such as oxidation, and surface modification or passivation. Though the methodology and limitations are beyond the scope of this module, it is important to remember that, like normal XPS, ARXPS assumes homogeneous layers are present in samples, which can give erroneous data, should the layers be heterogeneous.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/1fd6c243cd6cccf395e2d3c6bd051a273e286f45/graphics4.jpg" alt=""/><figcaption>Figure&nbsp;1.13.31.13.3&nbsp;Schematic representation of (a) a standard XPS analysis and (b) ARXPS on a multilayer sample.</figcaption></figure>



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<h4 class="wp-block-heading" id="limitations-of-xps">Limitations of XPS</h4>



<p>There are many limitations to XPS that are not based on the samples or preparation, but on the machine itself. One such limitation is that XPS cannot detect hydrogen or helium. This, of course, leads to a ratio of elements in the sample that is not entirely accurate, as there is always some amount of hydrogen. It is a common fallacy to assume the percent of atoms obtained from XPS data are completely accurate due to this presence of undetected hydrogen (Table&nbsp;1.13.11.13.1).</p>



<p>It is possible to indirectly measure the amount of hydrogen in a sample using XPS, but it is not very accurate and has to be done in a roundabout, often time consuming manner. If the sample contains hydrogen with a partial positive charge (i.e. OH), the sample can be washed in sodium naphthalenide (C<sub>10</sub>H<sub>8</sub>Na). This replaces this hydrogen with sodium, which can then be measured. The sodium to oxygen ratio that is obtained infers the hydrogen to oxygen ratio, assuming that all the hydrogen atoms have reacted.</p>



<p>XPS can only give an average measurement, as the electrons lower down in the sample will lose more energy as they pass other atoms while the electrons on the surface retain their original kinetic energy. The electrons from lower layers can also undergo inelastic or elastic scattering, seen in Figure&nbsp;1.13.41.13.4. This scattering may have a significant impact on data at higher angles of emission. The beam itself is also relatively wide, with the smallest width ranging from 10 – 200 μm, lending to the observed average composition inside the beam area. Due to this, XPS cannot differentiate sections of elements if the sections are smaller than the size of the beam.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/8119cc0bb7a9efac71d35abea309c57b05059c79/graphics5.jpg" alt=""/><figcaption>Figure&nbsp;1.13.41.13.4&nbsp;Schematic representation of (a) no scattering, (b) inelastic scattering, and (c) elastic scattering.</figcaption></figure>



<p>Sample reaction or degredation are important considerations. Caution should be exercised when analyzing polymers, as they are often chemically active and X-rays will provide energy to start degrading the polymer, altering the properties of the sample. One method found to help overcome this particular limitation is to use angle-resolved X-ray photoelectron spectroscopy (ARXPS). XPS can often reduce certain metal salts, such as Cu<sup>2+</sup>. This reduction will give peaks that indicate a certain set of properties or chemical environments when it could be completely different. It needs to be understood that charges can build up on the surface of the sample due to a number of reasons, specifically due to the loss of electrons during the XPS experiment. The charge on the surface will interact with the electrons escaping from the sample, affecting the data obtained. If the charge collecting is positive, the electrons that have been knocked off will be attracted to the charge, slowing the electrons. The detector will pick up a lower kinetic energy of the electrons, and thus calculate a different binding energy than the one expected, giving peaks which could be labeled with an incorrect oxidation state or chemical environment. To overcome this, the spectra must be charge referenced by one of the following methods: using the naturally occurring graphite peak as a reference, sputtering with gold and using the gold peak as a reference or flooding the sample with the ion gun and waiting until the desired peak stops shifting.</p>



<h6 class="wp-block-heading" id="limitations-with-surfactants-and-sputtering">Limitations with Surfactants and Sputtering</h6>



<p>While it is known that sputtering is destructive, there are a few other limitations that are not often considered. As mentioned above, the beam of X-rays is relatively large, giving an average composition in the analysis. Sputtering has the same limitation. If the surfactant or layers are not homogeneous, then when the sputtering is finished and detection begins, the analysis will show a homogeneous section, due to the size of both the beam and sputtered area, while it is actually separate sections of elements.</p>



<p>The chemistry of the compounds can be changed with sputtering, as it removes atoms that were bonded, changing the oxidation state of a metal or the hybridization of a non-metal. It can also introduce charges if the sample is non-conducting or supported on a non-conducting surface.</p>



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<h2 class="wp-block-heading" id="using-xps-to-analyze-metal-nanoparticles">Using XPS to Analyze Metal Nanoparticles</h2>



<h5 class="wp-block-heading" id="introduction">Introduction</h5>



<p>X-ray photoelectron spectroscopy (XPS) is a surface technique developed for use with thin films. More recently, however, it has been used to analyze the chemical and elemental composition of nanoparticles. The complication of nanoparticles is that they are neither flat nor larger than the diameter of the beam, creating issues when using the data obtained at face value. Samples of nanoparticles will often be large aggregates of particles. This creates problems with the analysis acquisition, as there can be a variety of cross-sections, as seen in Figure&nbsp;1.13.51.13.5. This acquisition problem is also compounded by the fact that the surfactant may not be completely covering the particle, as the curvature of the particle creates defects and divots. Even if it is possible to create a monolayer of particles on a support, other issues are still present. The background support will be analyzed with the particle, due to their small size and the size of the beam and the depth at which it can penetrate.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/2ca29935d9f43c5b0f63611783361f16cf908567/graphics1.jpg" alt=""/><figcaption>Figure&nbsp;1.13.51.13.5&nbsp;Different cross-sections of analysis possible on a nanoparticle.</figcaption></figure>



<p>Many other factors can introduce changes in nanoparticles and their properties. There can be probe, environmental, proximity, and sample preparation effects. The dynamics of particles can wildly vary depending on the reactivity of the particle itself. Sputtering can also be a problem. The beam used to sputter will be roughly the same size or larger than the particles. This means that what appears in the data is not a section of particle, but an average composition of several particles.</p>



<p>Each of these issues needs to be taken into account and preventative measures need to be used so the data is the best representation possible.</p>



<h4 class="wp-block-heading" id="sample-preparation">Sample Preparation</h4>



<p>Sample preparation of nanoparticles is very important when using XPS. Certain particles, such as iron oxides without surfactants, will interact readily with oxygen in the air. This causes the particles to gain a layer of oxygen contamination. When the particles are then analyzed, oxygen appears where it should not and the oxidation state of the metal may be changed. As shown by these particles, which call for handling, mounting and analysis without exposure to air, knowing the reactivity of the nanoparticles in the sample is very important even before starting analysis. If the reactivity of the nanoparticle is known, such as the reactivity of oxygen and iron, then preventative steps can be taken in sample preparation in order to obtain the best analysis possible.</p>



<p>When preparing a sample for XPS, a powder form is often used. This preparation, however, will lead to aggregation of nanoparticles. If analysis is performed on such a sample, the data obtained will be an average of composition of each nanoparticle. If composition of a single particle is what is desired, then this average composition will not be sufficient. Fortunately, there are other methods of sample preparation. Samples can be supported on a substrate, which will allow for analysis of single particles. A pictorial representation in Figure&nbsp;1.13.61.13.6&nbsp;shows the different types of samples that can occur with nanoparticles.</p>



<figure class="wp-block-image"><img decoding="async" src="https://cnx.org/resources/881146419dd64772043e695259c1e8819886fb2f/graphics2.jpg" alt=""/><figcaption>Figure&nbsp;1.13.61.13.6&nbsp;Representation of (a) a theoretical isolated nanoparticles, (b) nanoparticles suspended on a substrate, (c) an aggregate of nanoparticles, and (d) a powdered form of nanoparticles.</figcaption></figure>



<h4 class="wp-block-heading" id="analysis-limitations">Analysis Limitations</h4>



<p>Nanoparticles are dynamic; their properties can change when exposed to new chemical environments, leading to a new set of applications. It is the dynamics of nanoparticles that makes them so useful and is one of the reasons why scientists strive to understand their properties. However, it is this dynamic ability that makes analysis difficult to do properly. Nanoparticles are easily damaged and can change properties over time or with exposure to air, light or any other environment, chemical or otherwise. Surface analysis is often difficult because of the high rate of contamination. Once the particles are inserted into XPS, even more limitations appear.</p>



<h6 class="wp-block-heading" id="probe-effects">Probe Effects</h6>



<p>There are often artifacts introduced from the simple mechanism of conducting the analysis. When XPS is used to analyze the relatively large surface of thin films, there is small change in temperature as energy is transferred. The thin films, however, are large enough that this small change in energy has to significant change to its properties. A nanoparticle is much smaller. Even a small amount of energy can drastically change the shape of particles, in turn changing the properties, giving a much different set of data than expected.</p>



<p>The electron beam itself can affect how the particles are supported on a substrate. Theoretically, nanoparticles would be considered separate from each other and any other chemical environments, such as solvents or substrates. This, however, is not possible, as the particles must be suspended in a solution or placed on a substrate when attempting analysis. The chemical environment around the particle will have some amount of interaction with the particle. This interaction will change characteristics of the nanoparticles, such as oxidation states or partial charges, which will then shift the peaks observed. If particles can be separated and suspended on a substrate, the supporting material will also be analyzed due to the fact that the X-ray beam is larger than the size of each individual particle. If the substrate is made of porous materials, it can adsorb gases and those will be detected along with the substrate and the particle, giving erroneous data.</p>



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<h6 class="wp-block-heading" id="environmental-effects">Environmental Effects</h6>



<p>Nanoparticles will often react, or at least interact, with their environments. If the particles are highly reactive, there will often be induced charges in the near environment of the particle. Gold nanoparticles have a well-documented ability to undergo plasmon interactions with each other. When XPS is performed on these particles, the charges will change the kinetic energy of the electrons, shifting the apparent binding energy. When working with nanoparticles that are well known for creating charges, it is often best to use an ion gun or a coating of gold. The purpose of the ion gun or gold coating is to try to move peaks back to their appropriate energies. If the peaks do not move, then the chance of there being no induced charge is high and thus the obtained data is fairly reliable.</p>



<h6 class="wp-block-heading" id="proximity-effects">Proximity Effects</h6>



<p>The proximity of the particles to each other will cause interactions between the particles. If there is a charge accumulation near one particle, and that particle is in close proximity with other particles, the charge will become enhanced as it spreads, affecting the signal strength and the binding energies of the electrons. While the knowledge of charge enhancement could be useful to potential applications, it is not beneficial if knowledge of the various properties of individual particles is sought.</p>



<p>Less isolated (i.e., less crowded) particles will have different properties as compared to more isolated particles. A good example of this is the plasmon effect in gold nanoparticles. The closer gold nanoparticles are to each other, the more likely they will induce the plasmon effect. This can change the properties of the particles, such as oxidation states and partial charges. These changes will then shift peaks seen in XPS spectra. These proximity effects are often introduced in the sample preparation. This, of course, shows why it is important to prepare samples correctly to get desired results.</p>



<h2 class="wp-block-heading" id="conclusions">Conclusions</h2>



<p>Unfortunately there is no good general procedure for all nanoparticles samples. There are too many variables within each sample to create a basic procedure. A scientist wanting to use XPS to analyze nanoparticles must first understand the drawbacks and limitations of using their sample as well as how to counteract the artifacts that will be introduced in order to properly use XPS.</p>



<p>One must never make the assumption that nanoparticles are flat. This assumption will only lead to a misrepresentation of the particles. Once the curvature and stacking of the particles, as well as their interactions with each other are taken into account, XPS can be run.</p>



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