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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.

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.

Step 1: ¹H-NMR

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.

To begin, let’s start with proton 3. 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.

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 proton 4! The integration and chemical shift support the assignment, as proton 4 is the only aromatic proton in the structure.

There is only one singlet in the ¹H-NMR spectrum. The only proton that should show up as a singlet is proton 6, 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.

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

Step 2: ¹H-¹H COSY

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.

Two types of COSY coupling: 3-bond short range coupling between protons 7 and 8 (red) and 4-bond long range coupling between protons 3 and 4 (blue).

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 threeother protons by COSY. The only proton expected to correlate with three nonequivalent protons is proton 9!

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 proton 8 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 proton 10 as 5.209 ppm and proton 11 as 3.754 ppm.

Once proton 8 has been assigned, we can easily assign proton 7 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.

All that remains are protons 12 and 13. We can assign proton 12 (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 proton 13; this is confirmed by COSY correlation with proton 12, triplet multiplicity based on splitting by proton 12, and integration of one proton.

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

Step 3: ¹³C-NMR

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.

Step 4: DEPT-45, 90, and 135

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 carbon 3-it is the only CH₃ peak. I always go back and use DEPT to confirm the carbons I assigned by HSQC.

• DEPT-45 shows CH, CH₂, and CH₃ carbons as positive peaks. Carbons with no protons are not visible.
• DEPT-90 shows only CH peaks as positive peaks. Carbons with no protons, CH₂, and CH₃ carbons are not visible.
• DEPT-135 shows CH and CH₃ carbons as positive peaks and CH₂ carbons as negative peaks. Carbons with no protons are not visible.

Step 5: ¹H-¹³C HSQC

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.

For example, HSQC shows a correlation between proton 4 and the carbon at 136.113 ppm; this carbon is now assigned as carbon 4. Carbons 3, 4, 7, 8, 9, 11, and 12 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.

An example correlation between proton and carbon 4 is observed by HSQC.

Step 6: ¹H-¹³C HMBC

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.

HMBC interactions between proton 4 and carbons 1, 2, 3, 5, and 7.

HMBC is incredibly useful for assigning carbons that have no protons attached. In this example, carbons 1, 2, and 5 have no protons attached. Carbon 1 is assigned by HMBC interactions with protons 3, 4, and 6; carbon 2 by interaction with protons 3, 4, 6, and 7; and carbon 5 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.

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 carbons 8, 9, and 11, while proton 13 is confirmed by interactions with 11 and 12. HMBC supports all proton and all carbon assignments, unambiguously confirming both the structure and analysis of thymidine.

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. 

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].

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].

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.

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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. To be successful in using nmr as an analytical tool, it is necessary to understand the physical principles on which the methods are based.

The nuclei of many elemental isotopes have a characteristic spin (I). Some nuclei have integral spins (e.g. I = 1, 2, 3 ….), some have fractional spins (e.g. I = 1/2, 3/2, 5/2 ….), and a few have no spin, I = 0 (e.g. ¹²C, ¹⁶O, ³²S, ….). Isotopes of particular interest and use to organic chemists are ¹H, ¹³C, ¹⁹F and ³¹P, 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.

The following features lead to the nmr phenomenon:

1. A spinning charge generates a magnetic field, as shown by the animation on the right. The resulting spin-magnet has a magnetic moment (μ) proportional to the spin.
2. In the presence of an external magnetic field (B0), two spin states exist, +1/2 and -1/2. 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.
3. 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 Bx a formula for the energy difference is given (remember I = 1/2 and μ is the magnetic moment of the nucleus in the field).
Strong magnetic fields are necessary for nmr spectroscopy. The international unit for magnetic flux is the tesla (T). The earth’s magnetic field is not constant, but is approximately 10-4 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. For nmr purposes, this small energy difference (ΔE) is usually given as a frequency in units of MHz (106 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 radio and television broadcast spectrum. Nmr spectroscopy is therefore the energetically mildest probe used to examine the structure of molecules.  The nucleus of a hydrogen atom (the proton) has a magnetic moment μ = 2.7927, and has been studied more than any other nucleus. The previous diagram may be changed to display energy differences for the proton spin states (as frequencies) by mouse clicking anywhere within it.
4. 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: 1H μ = 2.7927, 19F μ = 2.6273, 31P μ = 1.1305 & 13C μ = 0.7022. These moments are in nuclear magnetons, which are 5.05078•10-27 JT-1. 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’s constant = 6.626069•10-34 Js).

      2. Proton NMR Spectroscopy
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 continuous wave (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.

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 Show Field Sweep 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.

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 Show Different Protons 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.

Why should the proton nuclei in different compounds behave differently in the nmr experiment ? 
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_o) so as to generate a secondary field that opposes the much stronger applied field. This secondary field shields the nucleus from the applied field, so B_o must be increased in order to achieve resonance (absorption of rf energy). As illustrated in the drawing on the right, B_o 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 (CH₄, HCl, HBr and HI) have proton nuclei that are more shielded than those on the lower field (left) side of the diagram. 
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 parts per million (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. 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.

Chemical Shift

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.

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. Tetramethylsilane, (CH₃)₄Si, usually referred to as TMS, meets all these characteristics, and has become the reference compound of choice for proton and carbon nmr.
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. 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), 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_ref is the resonant frequency of the reference signal and ν_samp is the frequency of the sample signal. This operation gives a locator number called the Chemical Shift, having units of parts-per-million (ppm), and designated by the symbol δ   Chemical shifts for all the compounds in the original display will be presented by a third click on the diagram.

The compounds referred to above share two common characteristics:

• The hydrogen atoms in a given molecule are all structurally equivalent, averaged for fast conformational equilibria. 
• The compounds are all liquids, save for neopentane which boils at 9 °C and is a liquid in an ice bath.

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, CCl₄ is a poor solvent for many polar compounds and is also toxic. Deuterium labeled compounds, such as deuterium oxide (D₂O), chloroform-d (DCCl₃), benzene-d₆(C₆D₆), acetone-d₆ (CD₃COCD₃) and DMSO-d₆ (CD₃SOCD₃) 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.

From the previous discussion and examples we may deduce that one factor contributing to chemical shift differences in proton resonance is the inductive effect. 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_o) 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 shielded. 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 deshielded). 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.

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.

Proton Chemical Shift Ranges*
Low Field Region		High Field Region
	* For samples in CDCl3 solution. The δ scale is relative to TMS at δ = 0.

To make use of a calculator that predicts aliphatic proton chemical shifts Click Here. This application was developed at Colby College.

Signal Strength

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’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. 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.
Hint: When evaluating relative signal strengths, it is useful to set the smallest integration to unity and convert the other values proportionally.

Hydroxyl Proton Exchange and the Influence of Hydrogen Bonding

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 rapid OH exchange 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.

R-O-H   +   D2O      R-O-D   +   D-O-H

Hydrogen bonding shifts the resonance signal of a proton to lower field ( higher frequency ). Numerous experimental observations support this statement, and a few of these will be described here.

i)   The chemical shift of the hydroxyl hydrogen of an alcohol varies with concentration. Very dilute solutions of 2-methyl-2-propanol, (CH3)3COH, in carbon tetrachloride solution display a hydroxyl resonance signal having a relatively high-field chemical shift (< 1.0 δ ). In concentrated solution this signal shifts to a lower field, usually near 2.5 δ.
ii)   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).
iii)   Because of their favored hydrogen-bonded dimeric association, 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.
iv)   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 : 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.
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).

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, (CH₃)₂C(OH)CO₂H, 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.

π-Electron Functions

An examination of the proton chemical shift chart (above) 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.
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 anisotropy, 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 deshielded, 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 shielded because an increase in the applied field is needed for resonance. Shielded regions are designated by a plus sign, and deshielded regions by a negative sign. 
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 & deshielding regions about the carbonyl group have been described in two ways, which alternate in the display.

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 CH₃–R, R–CH₂–R, R₃CH; as well as the deshielding of equatorial versus axial protons on a fixed cyclohexane ring.

Solvent Effects

Chloroform-d (CDCl₃) 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 (D₂O), benzene-d6 (C₆D₆), acetone-d6 (CD₃COCD₃) and DMSO-d6 (CD₃SOCD₃) 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.

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-d₆ 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.
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 elsewhere in this text.

Spin-Spin Interactions

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 J (referred to as the coupling constant). In the 1,1-dichloroethane example all the coupling constants are 6.0 Hz, as illustrated by clicking on the spectrum.

1,2-dichloroethane		1,1-dichloroethane

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 doublet (two equal intensity signals), triplet (three signals with an intensity ratio of 1:2:1) and quartet (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 coupling constant (J). The magnitude of J, usually given in units of Hz, is magnetic field independent.

The splitting patterns shown above display the ideal or “First-Order” 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. 
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.

What causes this signal splitting, and what useful information can be obtained from it ? 
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 ³J and a geminal constant is ²J.

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

1)   Nuclei having the same chemical shift (called isochronous) do not exhibit spin-splitting. They may actually be spin-coupled, but the splitting cannot be observed directly.
2)   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’s), provided they have different chemical shifts. Longer-range coupling may be observed in molecules having rigid configurations of atoms.
3)   The magnitude of the observed spin-splitting depends on many factors and is given by the coupling constant J (units of Hz). J is the same for both partners in a spin-splitting interaction and is independent of the external magnetic field strength.
4)   The splitting pattern of a given nucleus (or set of equivalent nuclei) can be predicted by the n+1 rule, 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’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’s triangle are related to the Fibonacci series click on the diagram.

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.

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

To make use of a calculator that predicts first order splitting patterns Click Here. This application was developed at Colby College.

Some Examples

Test your ability to interpret ¹H nmr spectra by analyzing the seven examples presented below. The seven spectra may be examined in turn by clicking the “Toggle Spectra” button. Try to associate each spectrum with a plausible structural formula. 
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.

      3. Carbon NMR Spectroscopy
The power and usefulness of ¹H 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).

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 & 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 & green boxes) might be difficult to identify by proton nmr alone.

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 (¹²C) has no spin, this option seems unrealistic. Fortunately, 1.1% of elemental carbon is the ¹³C isotope, which has a spin I = 1/2, so in principle it should be possible to conduct a carbon nmr experiment. It is worth noting here, that if much higher abundances of ¹³C were naturally present in all carbon compounds, proton nmr would become much more complicated due to large one-bond coupling of ¹³C and ¹H.

The most important operational technique that has led to successful and routine ¹³C nmr spectroscopy is the use of high-field pulse technology coupled with broad-band heteronuclear decoupling 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. 
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 ¹H nmr spectrum on the right illustrates some of the advantageous characteristics of carbon nmr. The dispersion of ¹³C 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.

Unlike proton nmr spectroscopy, the relative strength of carbon nmr signals are not normally proportional to the number of atoms generating each one. 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 proton chemical shifts.

13C Chemical Shift Ranges*
Low Field Region		High Field Region
	* For samples in CDCl3 solution. The δ scale is relative to TMS at δ=0.

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 sp² hybrid carbon atoms).

The C₈H₁₀ isomers in the center (red) box have pairs of homotopic 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 ¹³C 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 ¹³C 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 > 100 ppm. Finally, the last isomeric pair, quinones A & 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 & 187.9 ppm); whereas, isomer B displays five signals (δ 15.9, 133.3, 145.8, 187.5 & 188.1 ppm), the additional signal coming from the non-identity of the two carbonyl carbon atoms (one colored orange and the other magenta).

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1. NMRshiftdb

NMRshiftdb2 is a NMR database (web database) for organic structures and their nuclear magnetic resonance (nmr) spectra. It allows for spectrum prediction (¹³C, ¹H 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 board of reviewers. The project is supported by a scientific advisory board.

nmrshiftdb2 is part of the NFDI4Chem initiative and will provide a component for a curated repository there. Please consult the documentation for more detailed information.

See: https://nmrshiftdb.nmr.uni-koeln.de/portal

2. ACD/NMR

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.

Powerful NMR Interpretation Software | Highlights

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

Synchronize peak picking and assignments across NMR datasets using NMRSync—our game-changing technology. Plus, the associated peaks from NMRSync, 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:

• Use any peak in any spectrum to initiate NMRSync
• 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
• Automatically resolve overlapping ¹H and ¹³C peaks from 2D NMR data
• Receive immediate color-coded feedback on the best assignment for instant decision-making purposes

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.

• Determine how well your proposed structure matches the datasets in your NMR project with single structure verification
• 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
• Generate and view every alternative structural and cis/trans 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.

See: https://www.acdlabs.com/products/spectrus/workbooks/nmr/

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

4. See: https://www.nmrprocflow.org/

5. See: https://chem.washington.edu/facilities/data-processing

6. See: https://www.cgl.ucsf.edu/home/sparky/

7. See: http://www.nmrdb.org/about/

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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 ¹H and ¹³C NMR spectra to find out structure even though there are various kinds of NMR spectra such as ¹⁴N, ¹⁹F, and ³¹P. NMR spectrum shows that x- axis is chemical shift in ppm. It also contains integral areas, splitting pattern, and coupling constant.

Strategy for Solving Structure

Here is the general strategy for solving structure with NMR:

1. Molecular formula is determined by chemical analysis such as elementary analysis
2. Double-bond equivalent (also known as Degree of Unsaturation) 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_aH_b

3. Structure fragmentation is determined by chemical shift, spin multiplicity, integral (peak area), and coupling constants (1J1J, 2J2J)
4. Molecular skeleton is built up using 2-dimensional NMR spectroscopy.
5. Relative configuration is predicted by coupling constant (³J).

¹H NMR

Chemical Shift

Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane (TMS, (CH3)4Si(CH3)4Si) is generally used as an internal standard to determine chemical shift of compounds: δ_TMS=0 ppm. In other words, frequencies for chemicals are measured for a ¹H or ¹³C nucleus of a sample from the ¹H or ¹³C 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. ¹H chemical shift play a role in identifying many functional groups. Figure 11. indicates important example to figure out the functional groups.

Chemical equivalence

Protons with Chemical equivalence has the same chemical shift due to symmetry within molecule (CH3COCH3CH3COCH3) or fast rotation around single bond (-CH₃; methyl groups).

Spin-Spin Splitting

Spin-Spin splitting 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.

Spin Multiplicity (Splitting pattern)

Spin Multiplicity plays a role in determining the number of neighboring protons. Here is a multiplicity rules: In case of AmBnAmBn system, the multiplicity rule is that Nuclei of BB element produce a splitting the AA signal into nB+1nB+1 lines. The general formula which applies to all nuclei is 2nI+12nI+1, where II 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 22).

First-order splitting pattern

The chemical shift difference in Hertz between coupled protons in Hertz is much larger than the JJ coupling constant:ΔνJ≥8(1)(1)ΔνJ≥8

Where ΔνΔν 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 first-order spectrum. 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.

Example 11

The note is that structure system is A₃M₂X₂. H_a and H_x has the triplet pattern by Hm because of N+1 rule. The signal of Hm is split into six peaks by H_x and H_a(Figure3) The First order pattern easily is predicted due to separation with equal splitting pattern.

High-order splitting pattern

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

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 44). This is called the roof effect.

Here is other system as an example: A₂B₂ (Figure 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.

Coupling constant (J Value)

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:

Geminal proton-proton coupling (²J_HH)

Germinal coupling generates through two bonds (Figure 66). Two proton having geminal coupling are not chemically equivalent. This coupling ranges from -20 to 40 Hz. ²J_HHdepends 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: ²J_sp1>²J_sp2>²J_sp3 The bond angle(HCH) gives rise to change ²J_HH 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.

Vicinal proton-proton coupling (³J_HH)

Vicinal coupling occurs though three bonds (Figure 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 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

When ? is the 90^o, vicinal coupling constant is zero. In addition, vicinal coupling constant ranges from 8 to 10 Hz at the and ?=180^o, where ?=0^o and ?=180^o means that the coupled protons have cis and trans configuration, respectively.

The valence angle(?;Figure 88) also causes change of ³J_HH 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

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.

Integral

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

¹³C NMR

Chemical Shift

Spin-Spin splitting

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

• ¹³C-¹H Spin coupling: ¹³C-¹H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (¹J_CH), -CH, -CH₂, and CH₃ have respectively doublet, triplet, quartets for the ¹³C resonances in the spectrum. However, ¹³C-¹H Spin coupling has an disadvantage for ¹³C spectrum interpretation. ¹³C-¹H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of ¹H.
• Decoupling: Decoupling is the process of removing ¹³C-¹H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling ¹³C spectra shows only one peak(singlet) for each unique carbon in the molecule(Figure 1010.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.

• Distortionless enhancement by polarization transfer (DEPT): DEPT is used for distinguishing between a CH₃ group, a CH₂ group, and a CH group. The proton pulse is set at 45^o, 90^o, or 135^o in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure 1111. is an example about DEPT spectrum.

2-dimensional NMR spectroscopy (COSY)

COSY stands for COrrelation SpectroscopY. COSY spectrum is more useful information about what is being correlated.

¹H-¹H COSY (COrrelation SpectroscopY)

¹H-¹H 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 1212 indicates that there are correlation peaks between proton H₁ and H₂ as well as between H₂ and H₄. This means the H₂ coupled to H₁ and H₄.

¹H-¹³C COSY (HETCOR)

¹H-¹³C COSY is the heteronuclear correlation spectroscopy. The HETCOR spectrum is correlated ¹³C nuclei with directly attached protons. ¹H-¹³C coupling is one bond. The cross peaks mean correlation between a proton and a carbon (Figure 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)

References

1. Balc*, M., Basic p1 sH- and p13 sC-NMR spectroscopy. 1st ed.; Elsevier: Amsterdam ; Boston, 2005; p xii, 427.
2. Breitmaier, E., Structure elucidation by NMR in organic chemistry : a practical guide. 3rd rev. ed.; Wiley: Chichester, West Sussex, England, 2002; p xii, 258.
3. 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.
4. Silverstein, R. M.; Webster, F. X., Spectrometric identification of organic compounds. 6th ed.; Wiley: New York, 1998; p xiv, 482.

Outside Links

• NMRShiftDB: a Free web database for NMR data : nmrshiftdb.chemie.uni-mainz.de/nmrshiftdb
• NMR database from ACD/LAbs : www.acdlabs.com/products/spec_lab/exp_spectra/spec_libraries/aldrich.html
• NMR database from John Crerar Library : http://crerar.typepad.com/crerar_lib…h_ir_nmr_.html

Problems

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)

Solution

1) the structure of 2-hydoroxyporpane is drawn

Figure out which protons are chemically equivalent, i.e., two methyl (-CH₃) groups are chemical equivalent.

4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by H_b(#of proton=1). H_b has the septet pattern by H_a (#of proton=6). H_c has one peak.(Note that H_c has doublet pattern by H_b due to vicinal proton-proton coupling.)

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This handout relates the basic theory of NMR described on the theory web handout 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.

1.0 The NMR spectrum.

1.1 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 (Review Section 5.2A of the theory handout). 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 (Review Section 2.3Cin the theory handout). Recall further that in the NMR experiment, when and only when nuclei are irradiated with electromagnetic radiation of energy that exactlycorresponds 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 (Review section 3.1 of the theory handout). The absorbance of energy to convert a nucleus from a +1/2 to a -1/2 spin state is referred to as “resonance” of that nucleus.1.1A 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.

1.1B 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 (energy is proportional to frequency). You may want to review Section 5.2A of the theory handout again.

1.1C The converse is also true, namely that nuclei surrounded by lesser amounts of electron density will be less shielded (referred to as being “deshielded”) from the external magnetic field, so they will absorb electromagnetic radiation of higher energy, that is, higher frequency(energy is proportional to frequency).

1.1D 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).1.2 An NMR spectrum is a plot of absorbance versus frequency.

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

1.2B 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, the frequency of absorption plotted on NMR spectra are corrected for the magnet strength. In addition, frequency is correlated to the reference compound TMS. 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 parts per million (ppm) that is defined by the following equation:

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.

1.4 Chemically equivalent hydrogen atoms will have the same chemical shift and therefore give rise to the same signal. This is why we defined equivalent atoms in Section6.1 of the theory handout. Non-equivalent groups of hydrogens will have different chemical shifts. Thus, you will have as many different signals in an NMR spectrum as there are chemically non-equivalent groups of hydrogen atoms.

2.0 The nuclear spin of hydrogen atoms creates a magnetic field that influences the chemical shift of nearby hydrogen atoms (Review Sections 5.1 and 5.2).

2.1 Nuclear spin magnetic fields will influence hydrogen atoms that are three or fewer bonds away from each other in the same molecule. Hydrogen atoms that are four or greater bonds away usually do not influence each other.

2.2 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.

2.3 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.

2.4 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.

2.5 Consider a –CH2- group adjacent to a hydrogen X. 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: i) +1/2,+1/2; ii) +1/2,-1/2, which is the same as –1/2, +1/2 and iii) –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. Thus, the signal for hydrogen X is split into three peaks in a 1:2:1 ratio.

2.6 The same holds for a –CH3 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.

2.7 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).

3.0 Following the same logic, the splitting should multiply if a single hydrogen atom is adjacent to hydrogen atoms on either side. 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 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.

3.1 Thus, if the coupling constants (J) from the –CH2- and –CH3 groups are significantly different from each other, then 12 peaks will be observed as the signal for hydrogen X.

3.2 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. Thus, what is actually seen for the example above is that the signal for X would appear in the spectrum to be split into 2 + 3 + 1 = 6 peaks, not 12, peaks. This is the so-called “N+1” rule.

3.3 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.

3.3 The above explanation of splitting can confuse students for a while. The important point is that in the example given, 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.

3.4 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. From this information you can piece together what a molecule looks like 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.

4.0 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 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.

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.

6.0 For alkenes, the pi bond prevents bond rotation so the different hydrogen atoms on an unsymmetrical 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).

7.0 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.

8.0 There is a great deal more to NMR than this, I am only trying to give you the basics here.

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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.

10.3.1 Instrumentation

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.

Instrument Designs for Molecular UV/Vis Absorption

Filter Photometer. The simplest instrument for molecular UV/Vis absorption is a filter photometer (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 single-beam 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.

Note

The percent transmittance varies between 0% and 100%. As we learned in Figure 10.21, we use a blank to determine P₀, 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.

Figure 10.25 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.

Single-Beam Spectrophotometer. An instrument that uses a monochromator for wavelength selection is called a spectrophotometer. 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.

Figure 10.26 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: Adi (www.commons.wikipedia.org).

Double-Beam Spectrophotometer. The limitations of fixed-wavelength, single-beam spectrophotometers are minimized by using a double-beamspectrophotometer (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, P₀, and the sample, P_T. 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.

Figure 10.27 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.

Diode Array Spectrometer. 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.

Figure 10.28 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.

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 signal averaging improves a spectrum’s signal-to-noise ratio. If we add together n spectra, the sum of the signal at any point, x, increases as nS_x, where S_x is the signal. The noise at any point, N_x, is a random event, which increases as √nN_x when we add together nspectra. The signal-to-noise ratio (S/N) after n scans isSN=nSxn−−√Nx=n−−√SxnNx(4.8.1)(4.8.1)SN=nSxnNx=nSxnNx

where S_x/N_x 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.

Figure 10.29 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.

Sample Cells. 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 <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^o 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.

Figure 10.30 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.

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 fiber-optic probe 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.⁶

Figure 10.31 Example of a fiber-optic probe. The inset photographs provide a close-up look at the probe’s flow cell and the reflecting mirror.

Instrument Designs for Infrared Absorption

Filter Photometer. 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.

Double-beam spectrophotometer. 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 CO₂ and H₂O vapor when using double-beam optics. Resolutions of 1–3 cm^–1 are typical for most instruments.

Fourier transform spectrometer. 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 CO₂ and H₂O 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.

Sample Cells. 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.

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.

Figure 10.32 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^–1 and 2849 cm^–1.

The analysis of solution samples is limited by the solvent’s IR absorbing properties, with CCl₄, CS₂, and CHCl₃ 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.

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.

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 attenuated total reflectance 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.

Figure 10.33 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.

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.

Note

Further details about these, and other methods for preparing solids for infrared analysis can be found in this chapter’s additional resources.

10.3.2 Quantitative Applications

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 Fe²⁺ does not absorb visible light. Reacting Fe²⁺ with o-phenanthroline, however, forms an orange–red complex of Fe(phen)₃²⁺ 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.

Note

Figure 10.18 shows the visible spectrum for Fe(phen)₃²⁺.

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 (A = 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.

Note

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

Figure 10.34 Method for determining absorbance from an IR spectrum.

Environmental Applications

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.

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 CHCl₃ is shaken with an aqueous solution containing an appropriate metal ion, a colored metal–dithizonate complex forms that is soluble in CHCl₃. The selectivity of dithizone is controlled by adjusting the sample’s pH. For example, Cd²⁺ is extracted from solutions that are made strongly basic with NaOH, Pb²⁺ from solutions that are made basic with an NH₃/NH₄⁺ buffer, and Hg²⁺ from solutions that are slightly acidic.

Note

Atomic absorption is the subject of Section 10.4 and atomic emission is the subject of Section 10.7.

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

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 Cl₂, HOCl, and OCl^–. The combined chlorine residual, which forms from the reaction of NH₃ with HOCl, consists of monochloramine, NH₂Cl, dichloramine, NHCl₂, and trichloramine, NCl₃. 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.

Note

In Chapter 9 we explored how the total chlorine residual can be determined by a redox titration; see Representative Method 9.3 for further details. The method described here allows us to divide the total chlorine residual into its component parts.

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 ZrF₆^2– 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.

Note

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.

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 K₃Fe(CN)₆, forming a yellow colored antipyrine dye. After extracting the dye into CHCl₃, its absorbance is monitored at 460 nm. A calibration curve is prepared using only the unsubstituted phenol, C₆H₅OH. Because the molar absorptivity of substituted phenols are generally less than that for phenol, the reported concentration represents the minimum concentration of phenolic compounds.

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 NO₂ can be determined by oxidizing NO₂ to NO₃^–. The concentration of NO₃^– is then determined by first reducing it to NO₂^– with Cd, and then reacting NO₂^– with sulfanilamide and N-(1-naphthyl)-ethylenediamine to form a red azo dye. Another important application is the analysis for SO₂, which is determined by collecting the sample in an aqueous solution of HgCl₄^2– where it reacts to form Hg(SO₃)₂^2–. Addition of p-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, SO₂, nitrobenzene, methyl mercaptan, and vinyl chloride. Frequently, these analyses are accomplished using portable, dedicated infrared photometers.

Clinical Applications

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 CHCl₃ and then extracted from the CHCl₃ 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 NH₄Cl and the absorbance remeasured. Because the barbiturates do not absorb at this pH, we can use the absorbance at pH 10, A_pH 10, to correct the absorbance at pH 13, A_pH 13Abarb=ApH 13−Vsamp+VNH4ClVsamp×ApH 10(4.8.2)(4.8.2)Abarb=ApH 13−Vsamp+VNH4ClVsamp×ApH 10

where A_barb is the absorbance due to the serum barbiturates, and V_samp and V_NH4Cl are the volumes of sample and NH₄Cl, respectively. Table 10.7 provides a summary of several other methods for analyzing clinical samples.

Industrial Analysis

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 o-phenanthroline method listed in Table 10.6.

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.

Forensic Applications

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 K₂Cr₂O₇, 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.

Developing a Quantitative Method for a Single Component

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.

Note

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 Standard Methods for the Examination of Water and Wastewater, 20th Ed., American Public Health Association: Washington, D. C., 1998.

Representative Method 10.1

Determination of Iron in Water and Wastewater

Description of Method

Iron in the +2 oxidation state reacts with o-phenanthroline to form the orange-red Fe(phen)₃²⁺ 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 o-phenanthroline. The most important interferents are strong oxidizing agents, polyphosphates, and metal ions such as Cu²⁺, Zn²⁺, Ni²⁺, and Cd²⁺. 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.

(Figure 10.18 shows the visible spectrum for Fe(phen)₃²⁺.)

Procedure

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 o-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 Fe²⁺.

Questions

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

A strong oxidizing agent oxidizes some Fe²⁺ to Fe³⁺. Because Fe(phen)₃³⁺ does not absorb as strongly as Fe(phen)₃²⁺, the absorbance decreases, producing a negative determinate error. The excess hydroxylamine reacts with the oxidizing agents, removing them from the solution.

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?

Because o-phenanthroline is a weak base, its conditional formation constant for Fe(phen)₃²⁺ is less favorable at more acidic pH levels, where o-phenanthroline is protonated. The result is a decrease in absorbance and a less sensitive analytical method.

(In Chapter 9 we saw the same effect of pH on the complexation reactions between EDTA and metal ions.)

When the pH is greater than 9, competition between OH^– and o-phenanthroline for Fe²⁺ also decreased the absorbance. In addition, if the pH is sufficiently basic there is a risk that the iron will precipitate as Fe(OH)₂.

3. Cadmium is an interferent because it forms a precipitate with o-phenanthroline. What effect would the formation of precipitate have on the determination of iron?

Because o-phenanthroline is present in large excess (2000 μg of o-phenanthroline for 100 μg of Fe²⁺), it is not likely that the interference is due to an insufficient amount of o-phenanthroline being available to react with the Fe²⁺. 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.

(Although scattering is a problem here, it can serve as the basis of a useful analytical method. See Section 10.8 for further details.)

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

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.

Quantitative Analysis for a Single Analyte

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.

Example 10.5

The determination of Fe in an industrial waste stream was carried out by the o‑phenanthroline described in Representative Method 10.1. Using the data in the following table, determine the mg Fe/L in the waste stream.

Solution

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)

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

Practice Exercise 10.5

The concentration of Cu²⁺ 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 Cu²⁺, treated with cuprizone and diluted to 10.00 mL, giving an absorbance of 0.162. Report the mg Cu²⁺/L in the sample.

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Quantitative Analysis of Mixtures

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, A_mix, is(Amix)λ1=(εX)λ1bCX+(εY)λ1bCY(10.11)(10.11)(Amix)λ1=(εX)λ1bCX+(εY)λ1bCY

where 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 C_X or C_Y. 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

then C_X and C_Y can be determined by solving simultaneously equation10.11 and equation 10.12. Of course, we also must determine the value for ε_X and ε_Y at each wavelength. For a mixture of n components, we must measure the absorbance at n different wavelengths.

Example 10.6

The concentrations of Fe³⁺ and Cu²⁺ in a mixture can be determined following their reaction with hexacyanoruthenate (II), Ru(CN)₆^4–, which forms a purple-blue complex with Fe³⁺ (λ_max = 550 nm) and a pale-green complex with Cu²⁺ (λ_max = 396 nm).⁷ The molar absorptivities (M^–1 cm^–1) for the metal complexes at the two wavelengths are summarized in the following table.

When a sample containing Fe³⁺ and Cu²⁺ 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 Fe³⁺ and Cu²⁺ in the sample?

Solution

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

To determine C_Fe and C_Cu we solve the first equation for C_CuCCu=0.183–9970CFe34(4.8.6)(4.8.6)CCu=0.183–9970CFe34

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

Solving for C_Fe gives the concentration of Fe³⁺ as 1.79 × 10^–5 M. Substituting this concentration back into the equation for the mixture’s absorbance at 396 nm gives the concentration of Cu²⁺ as 1.26 × 10^–4 M.

(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

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

we find that C_Fe is 1.79×10^–5. Having determined C_Fe we can substitute back into one of the other equations to solve for C_Cu, which is 1.26×10^–5.)

Practice Exercise 10.6

The absorbance spectra for Cr³⁺ and Co²⁺ 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^–1 cm^–1) are

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To obtain results with good accuracy and precision the two wavelengths should be selected so that ε_X > ε_Y at one wavelength and ε_X < ε_Y 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_X/ε_Y as function of wavelength, and determine the wavelengths where ε_X/ε_Y reaches maximum and minimum values.⁸

Note

For example, in Example 10.6 the molar absorptivity for Fe³⁺ at 550 nm is 119× that for Cu²⁺, and the molar absorptivity for Cu²⁺ at 396 nm is 10.2× that for Fe³⁺.

Figure 10.35 Visible absorption spectra for 0.0250 M Cr³⁺, 0.0750 M Co²⁺, and for a mixture of Cr³⁺ and Co²⁺. The two wavelengths used for analyzing the mixture of Cr³⁺ and Co²⁺ are shown by the dashed lines. The data for the two standard solutions are from reference 7.

When the analyte’s spectra overlap severely, such that ε_X ≈ ε_Y 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.⁹ If A_SXand A_SY 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

where C_SX and C_SY are the known concentrations of X and Y in the standard solutions. Solving equation 10.13 and equation 10.14 for ε_X and ε_Y, substituting into equation 10.11, and rearranging, givesAmixASX=CXCSX+CYCSY×ASYASX(4.8.12)(4.8.12)AmixASX=CXCSX+CYCSY×ASYASX

To determine C_X and C_Y the mixture’s absorbance and the absorbances of the standard solutions are measured at several wavelengths. Graphing A_mix/A_SX versus A_SY/A_SX gives a straight line with a slope of C_Y/C_SY and a y-intercept of C_X/C_SX. This approach is particularly helpful when it is not possible to find wavelengths where ε_X > ε_Y and ε_X < ε_Y.

Note

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. Chemometrics: Data Analysis for the Laboratory and Chemical Plant, Wiley: Chichester, England, 2003.

Example 10.7

Figure 10.35 shows visible absorbance spectra for a standard solution of 0.0250 M Cr³⁺, a standard solution of 0.0750 M Co²⁺, and a mixture containing unknown concentrations of each ion. The data for these spectra are shown here.¹⁰

Use a multiwavelength regression analysis to determine the composition of the unknown.

Solution

First we need to calculate values for A_mix/A_SX and for A_SY/A_SX. Let’s define X as Co²⁺ and Y as Cr³⁺. For example, at a wavelength of 375 nm A_mix/A_SX is 0.53/0.01, or 53 and A_SY/A_SX is 0.26/0.01, or 26. Completing the calculation for all wavelengths and graphing A_mix/A_SX versus A_SY/A_SXgives 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

Using the y-intercept, the concentration of Co²⁺ isCXCSX=CCo0.0750M=0.636(4.8.14)(4.8.14)CXCSX=CCo0.0750M=0.636

or C_Co = 0.048 M, and using the slope the concentration of Cr³⁺ isCYCSY=CCr0.0250M=2.01(4.8.15)(4.8.15)CYCSY=CCr0.0250M=2.01

or C_Cr = 0.050 M.

Figure 10.36 Multiwavelength linear regression analysis for the data in Example 10.7.

Practice Exercise 10.7

A mixture of MnO₄^– and Cr₂O₇^2–, and standards of 0.10 mM KMnO₄ and of 0.10 mM K₂Cr₂O₇ 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. J. Chem. Educ. 1989, 66, 178–180.

(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 additional resources for further information.)

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10.3.3 Qualitative Applications

As discussed earlier in Section 10.2.1, 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^–1 cm^–1) and 254 nm (ε = 204 M^–1 cm^–1), 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^–1 cm^–1) and 273 nm (ε = 970 M^–1 cm^–1). 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^–1 for acids, 1700 cm^–1 for ketones, and 1800 cm^–1 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.

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 spectral searching. 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|

where D is the cumulative difference, A_sample is the sample’s absorbance at wavelength or wavenumber i, A_reference is the absorbance of the reference compound at the same wavelength or wavenumber, and n 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 D 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.

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 ^. HCl (Figure 10.37b) as a likely component of the mixture. Subtracting the reference spectrum for cocaine ^. 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).

Figure 10.37 Identifying the components of a mixture by spectral searching and subtracting. (a) IR spectrum of the mixture; (b) Reference IR spectrum of cocaine^. HCl; (c) Result of subtracting the spectrum of cocaine ^. 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.

Note

IR spectra traditionally are displayed using percent transmittance, %T, along the y-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 y-axis.

10.3.4 Characterization Applications

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.

Stoichiometry of a Metal-Ligand Complex

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

using one of three methods: the method of continuous variations, the mole-ratio method, and the slope-ratio method. Of these approaches, the method of continuous variations, 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, n_total, in each solution is the same. If (n_M)_i and (n_L)_i are, respectively, the moles of metal and ligand in solution i, thenntotal=(nM)i+(nL)i(4.8.18)(4.8.18)ntotal=(nM)i+(nL)i

The relative amount of ligand and metal in each solution is expressed as the mole fraction of ligand, (X_L)_i, and the mole fraction of metal, (X_M)_i,(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

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 y for the metal–ligand complex ML_y.y=nLnM=XLXM=XL1−XL(4.8.21)(4.8.21)y=nLnM=XLXM=XL1−XL

Note

You also can plot the data as absorbance versus the mole fraction of metal. In this case, y is equal to (1–X_M)/X_M.

Example 10.8

To determine the formula for the complex between Fe²⁺ and o-phenanthroline, a series of solutions is prepared in which the total concentration of metal and ligand is held constant at 3.15 × 10^–4 M. The absorbance of each solution is measured at a wavelength of 510 nm. Using the following data, determine the formula for the complex.

(To prepare the solutions for this example I first prepared a solution of 3.15 × 10^-4 M Fe²⁺ and a solution of 3.15 × 10^-4 M o-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 o-phenanthroline and Fe²⁺ is 3.15 × 10^-4 M. If each solution has the same volume, then each solution contains the same total moles of metal and ligand.)

Solution

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 y givesy=XL1–XL=0.751–0.75=3(4.8.22)(4.8.22)y=XL1–XL=0.751–0.75=3

The formula for the metal–ligand complex is Fe(o-phenanthroline)₃²⁺.

Figure 10.38 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.

Practice Exercise 10.8

Use the continuous variations data in the following table to determine the formula for the complex between Fe²⁺ and SCN^–. The data for this problem is adapted from Meloun, M.; Havel, J.; Högfeldt, E. Computation of Solution Equilibria, Ellis Horwood: Chichester, England, 1988, p. 236.

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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 X_L are constructed at several different wavelengths and for several different values of n_total. If the maximum absorbance does not occur at the same value of X_L 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 X_L 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.

In the mole-ratio method 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, n_L/n_M, 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 ML₂ is shown in Figure 10.39c.

Figure 10.39 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.

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 X_L or n_L/n_M may be so curved that it is impossible to determine the stoichiometry by extrapolation. In this case the slope-ratio may be used.

In the slope-ratio method 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, C_M, is much larger than the total concentration of ligand, C_L. 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_xL_y, is[MxLy]=CLy(4.8.23)(4.8.23)[MxLy]=CLy

If we monitor the absorbance at a wavelength where only M_xL_y absorbs, thenA=εb[MxLy]=εbCLy(4.8.24)(4.8.24)A=εb[MxLy]=εbCLy

and a plot of absorbance versus C_L is linear with a slope, s_L, ofsL=εby(4.8.25)(4.8.25)sL=εby

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

A ratio of the slopes provides the relative values of x and y.sMsL=εb/xεb/y=yx(4.8.29)(4.8.29)sMsL=εb/xεb/y=yx

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.

Determination of Equilibrium Constants

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)

where HIn and In^– 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]

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 H₃O⁺, HIn, and In^–. The concentration of H₃O⁺ is easy to determine by simply measuring the solution’s pH. To determine the concentration of HIn and In^– we can measure the solution’s absorbance.

If both HIn and In^– 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−]

where ε_HIn and ε_In are the molar absorptivities for HIn and In^–. The total concentration of indicator, C, is given by a mass balance equationC=[HIn]+[In−](10.16)(10.16)C=[HIn]+[In−]

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−]

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)

where A_HIn, which is equal to ε_HInbC, 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^– gives[In−]=A−AHInb(εIn−εHIn)(10.18)(10.18)[In−]=A−AHInb(εIn−εHIn)

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)

where A_In, which is equal to ε_InbC, is the absorbance when the pH is basic enough that only In^– 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

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

Example 10.9

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^–5 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 K_a for the indicator?

Solution

The value of K_a 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

Practice Exercise 10.9

To determine the K_a of a merocyanine dye, the absorbance of a solution of 3.5×10^–4 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 K_a for this dye? The data for this problem is adapted from Lu, H.; Rutan, S. C. Anal. Chem., 1996, 68, 1381–1386.

Click here to review your answer to this exercise.

A second approach for determining K_a is to prepare a series of solutions, each containing the same amount of indicator. Two solutions are used to determine values for A_HIn and A_In. 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

A plot of log[(A – A_HIn)/(A_In – A)] versus pH is a straight-line with a slope of +1 and a y-intercept of –pK_a.

Practice Exercise 10.10

To determine the K_a 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 K_a for this day? The data for this problem is from Patterson, G. S. J. Chem. Educ., 1999, 76, 395–398.

Click here to review your answer to this exercise.

In developing these approaches for determining K_a we considered a relatively simple system in which the absorbance of HIn and In^– are easy to measure and for which it is easy to determine the concentration of H₃O⁺. 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)

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.¹¹

10.3.5 Evaluation of UV/Vis and IR Spectroscopy

Scale of Operation

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.

Note

See Figure 3.5 to review the meaning of macro and meso for describing samples, and the meaning of major, minor, and ultratrace for describing analytes.

Accuracy

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.

Precision

In absorption spectroscopy, precision is limited by indeterminate errors—primarily instrumental noise—introduced when measuring absorbance. Precision is generally worse for low absorbances where P₀ ≈ P_T, and for high absorbances when P_T approaches 0. We might expect, therefore, that precision will vary with transmittance.

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

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, s_C, assC=−0.4343εb×sTT(10.23)(10.23)sC=−0.4343εb×sTT

where s_T is the absolute uncertainty in the transmittance. Dividing equation 10.23 by equation 10.22 gives the relative uncertainty in concentration, s_C/C, assCC=0.4343sTTlogT(4.8.36)(4.8.36)sCC=0.4343sTTlog⁡T

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

Determining the relative uncertainty in concentration is complicated because s_T may be a function of the transmittance. As shown in Table 10.8, three categories of indeterminate instrumental error have been observed.¹² A constant s_T 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 k₁ of ±0.002–0.003) for an analog scale, and ±0.001% a (k₁ of ±0.000 01) for a digital scale. A constant s_T also is observed for the thermal transducers used in infrared spectrophotometers. The effect of a constant s_T 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.

Values of s_T 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.

Figure 10.40 Percent relative uncertainty in concentration as a function of absorbance for the categories of indeterminate errors in Table 10.8. A: k₁ = ±0.0030; B: k₂ = ±0.0030; C:k₃ = ±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).

Finally, the value of s_T 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.

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.

Figure 10.41 Methods for improving the precision of absorption methods: (a) high-absorbance method; (b) low-absorbance method; (c) maximum precision method.

Sensitivity

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 (εb). You can improve a method’s sensitivity by selecting a wavelength where absorbance is at a maximum or by increasing the pathlength.

Note

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

Selectivity

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.

Time, Cost, and Equipment

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.

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Raman Spectroscopy is a non-destructive chemical analysis technique which provides detailed information about chemical structure, phase and polymorphy, crystallinity and molecular interactions. It is based upon the interaction of light with the chemical bonds within a material.

Raman is a light scattering technique, whereby a molecule scatters incident light from a high intensity laser light source. Most of the scattered light is at the same wavelength (or color) as the laser source and does not provide useful information – this is called Rayleigh Scatter. However a small amount of light (typically 0.0000001%) is scattered at different wavelengths (or colors), which depend on the chemical structure of the analyte – this is called Raman Scatter.

A Raman spectrum features a number of peaks, showing the intensity and wavelength position of the Raman scattered light. Each peak corresponds to a specific molecular bond vibration, including individual bonds such as C-C, C=C, N-O, C-H etc., and groups of bonds such as benzene ring breathing mode, polymer chain vibrations, lattice modes, etc.

Fig. 2: A typical Raman spectrum, in this case, of aspirin (4-acetylsalicylic acid). The inset image shows the detail which is present in the spectrum

Information provided by Raman spectroscopy

Fig. 3: Raman spectra of ethanol and methanol, showing the significant spectral differences which allow the two liquids to be distinguished.

Raman spectroscopy probes the chemical structure of a material and provides information about:

• Chemical structure and identity
• Phase and polymorphism
• Intrinsic stress/strain
• Contamination and impurity

Typically a Raman spectrum is a distinct chemical fingerprint for a particular molecule or material, and can be used to very quickly identify the material, or distinguish it from others. Raman spectral libraries are often used for identification of a material based on its Raman spectrum – libraries containing thousands of spectra are rapidly searched to find a match with the spectrum of the analyte.

Fig. 4: Mineral distribution

In combination with mapping (or imaging) Raman systems, it is possible to generate images based on the sample’s Raman spectrum. These images show distribution of individual chemical components, polymorphs and phases, and variation in crystallinity.

Raman spectroscopy is both qualitative and quantitative.

The general spectrum profile (peak position and relative peak intensity) provides a unique chemical fingerprint which can be used to identify a material, and distinguish it from others. Often the actual spectrum is quite complex, so comprehensive Raman spectral libraries can be searched to find a match, and thus provide a chemical identification.

The intensity of a spectrum is directly proportional to concentration. Typically, a calibration procedure will be used to determine the relationship between peak intensity and concentration, and then routine measurements can be made to analyze for concentration. With mixtures, relative peak intensities provide information about the relative concentration of the components, while absolute peak intensities can be used for absolute concentration information.

Raman is used for microscopic analysis

Fig. 5: A modern Raman microscope system

Raman spectroscopy can be used for microscopic analysis, with a spatial resolution in the order of 0.5-1 µm. Such analysis is possible using a Raman microscope.

A Raman microscope couples a Raman spectrometer to a standard optical microscope, allowing high magnification visualization of a sample and Raman analysis with a microscopic laser spot. Raman micro-analysis is easy: simply place the sample under the microscope, focus, and make a measurement.

A true confocal Raman microscope can be used for the analysis of micron size particles or volumes. It can even be used for the analysis of different layers in a multilayered sample (e.g., polymer coatings), and of contaminants and features beneath the surface of a transparent sample (e.g., impurities within glass, and fluid/gas inclusions in minerals).

Motorized mapping stages allow Raman spectral images to be generated, which contain many thousands of Raman spectra acquired from different positions on the sample. False color images can be created based on the Raman spectrum – these show the distribution of individual chemical components, and variation in other effects such as phase, polymorphism, stress/strain, and crystallinity.

Type of samples analyzed with Raman

Raman can be used to analyze many different samples. In general it is suitable for analysis of:

• Solids, powders, liquids, gels, slurries and gases
• Inorganic, organic and biological materials
• Pure chemicals, mixtures and solutions
• Metallic oxides and corrosion

In general it is not suitable for analysis of:

• Metals and their alloys

Typical examples of where Raman is used today include:

• Art and archaeology – characterization of pigments, ceramics and gemstones
• Carbon materials – structure and purity of nano-tubes, defect/disorder characterization
• Chemistry – structure, purity, and reaction monitoring
• Geology – mineral identification and distribution, fluid inclusions and phase transitions
• Life sciences – single cells and tissue, drug interactions, disease diagnosis
• Pharmaceutics – content uniformity and component distribution
• Semiconductors – purity, alloy composition, intrinsic stress/strain microscope.

Analysis of solids, liquids and gases

Raman spectra can be acquired from nearly all samples which contain true molecular bonding. This means that solids, powders, slurries, liquids, gels and gases can be analyzed using Raman spectroscopy.

Although gases can be analyzed using Raman spectroscopy, the concentration of molecules in a gas is typically very low, so the measurement is often more challenging. Usually specialized equipment such as higher powered lasers and long path length sample cells are necessary. In some cases where gas pressures are high (such as gas inclusions in minerals) standard Raman instrumentation can easily be used.

Analysis from a mixture of materials

The Raman spectrum from a material will contain Raman information about all of the molecules which are within the analysis volume of the system. Thus, if there is a mixture of molecules, the Raman spectrum will contain peaks representing all of the different molecules. If the components are known, the relative peak intensities can be used to generate quantitative information about the mixture’s composition. In case of complex matrixes, chemometrics methods might also be employed to build quantitative methods.