Major Chemistry B.Sc.III ( Semester V)
Name of paper- Spectral analysis
CHE/DSC/T/301
Proton magnetic resonance spectroscopy
Principle, chemical shift, shielding & deshielding effect. Measurement of chemical shift, TMS reference, peak area, integration, spin-spin coupling, coupling constants, J-value, Chemical shift values and correlation for protons bonded to carbon and other nuclei as in alcohols, phenols, carboxylic acids, amines, amides. Problems based on 1H NMR.
Principle
Proton magnetic resonance spectroscopy (PMR) is one of the most powerful analytical technique available for structure determination of compounds. PMR spectroscopy is used in conjunction with other spectroscopic techniques and chemical analysis to determine the structures of complicated organic molecules. It helps to identify the carbon–hydrogen framework of an organic compound. Proton magnetic resonance spectroscopy (PMR) has applications in agriculture, medicine, food technology. In medicine it is popularly known as MRI (magnetic resonance imaging) scan where it is used to identify tumor cells in the body. PMR spectroscopy is crucial technique for testing of food products especially, honey where the purity and adulterants in it could be easily identified using it. Historically, PMR was first used to study protons (the nuclei of hydrogen atoms), and proton magnetic resonance spectrometers have become most common. Proton magnetic resonance‖ is assumed to mean proton magnetic resonance‖ unless a different nucleus is specified. Spectrometers were later developed for 13C NMR, 15N NMR, 19F NMR, 31P NMR and other magnetic nuclei
Nuclear Spin States
Like an electron, the proton and neutron present in the nucleus of an atom also possess 1/2 spin states. Since, the proton and neutrons are constituents of the nucleus, the nuclear spin is the vector sum of the spin due to the protons and neutrons present in it.
Figure 1: Nuclear spin states
The spin of a nucleus is given by the symbol l and is called the spin quantum number. If the atomic nuclei has a non-zero or half integer value for spin quantum number (l≠0; l = 1/2, 3/2, 5/2, etc.) then it said to be NMR active. There is no direct method to arrive at the nuclear spin quantum numbers for the atomic nuclei but certain empirical rules are used to rationalize the spin values observed for the isotopes of various elements. When comparing isotopes of the same element, each isotope would have different nuclear spin quantum numbers since they are composed of different number of protons and neutrons. Eg.12C has 6 protons and 6 neutrons i.e., even number of protons and neutrons and so has l =0, whereas 13C has 6 protons and 7 neutrons i.e., odd mass number and so has l=1/2. Therefore, 13C is NMR active whereas 12C is not. The rules to find the nuclear spin quantum number of different nuclei are presented in the table below:
Rules to identify nuclear spin quantum number of atomic nuclei:
For every nucleus with spin, the number of allowed spin states it may adopt is quantized and is determined by its nuclear spin quantum number l. For each nucleus, the number l is a physical constant, and there are 2l + 1 allowed spin states with integral differences ranging from +l to -l. For instance, hydrogen nucleus (1H) has a proton and zero neutron and so has a spin quantum number l= ½. It has two allowed spin states ([2(1/2) +1] = 2 ) for its nucleus i.e., +1/2 and -1/2.
Nuclear Magnetic Moment:
A proton (1H) is the simplest nucleus that has both charge (since protons are positively charged species) and nuclear spin. A spinning charged nuclei (proton) can generate a magnetic field similar to that of a small bar magnet. Remember, a proton (hydrogen nucleus) has the spin quantum number l = 1/2 and has two allowed spin states ([2(1/2) +1] = 2) for its nucleus (+1/2 and -1/2). The nuclear spins are randomly oriented when there no external magnetic field is applied. However, when they are placed in an external magnetic field they either align themselves with or against the field of the external magnet (applied magnetic field). Protons that align with the field are said to be in the lower-energy α-spin state (+1/2) while protons that align against the field are said to be in the higher-energy β-spin state (-1/2).
Figure 2: α-spin state and β-spin state on application of external magnetic field
The energy difference (ΔE) between the α-spin and β-spin states depends on the strength of the applied magnetic field (B0). As the strength of the applied magnetic field to which the nucleus is exposed is increasedthe energy difference between the α-spin and β-spin states would also increase. It is noted that the major population of the nuclei are in the lower energy α-spin state than in the higher energy β-spin state.
Figure 3: Energy of α-spin state and β-spin state
The applied magnetic field could be produced by using superconducting magnets. Superconducting magnets can produce very strong magnetic field, on the order of 21 tesla (T). Lower field strengths can also be used, in the range of 4 - 7 T.
Nuclear Magnetic Resonance
Because the proton is behaving as a spinning magnet, not only can it align itself with or oppose an external magnetic field, but also it will move in a characteristic way under the influence of the external magnet. Consider the behavior of a spinning top: as well as describing its spinning motion, the top will (unless absolutely vertical) also perform a slower waltz-like motion, in which the spinning axis of the top moves slowly around the vertical. This is precessional motion, and the top is said to be precessing around the vertical axis of the earth's gravitational field. The precession arises from the interaction of spin-that is, gyroscopic motion-with the earth's gravity acting vertically downward. Only a spinning top will precess; a static top will merely fall over. As the proton is a spinning magnet, it will, like the top, precess around the axis of an applied external magnetic field, and can do so in two principal orientations, either aligned with the field (low energy) or opposed to the field (high energy).
We have seen that a proton, in an external magnetic field of 1.4T, will be precessing at a frequency of 60 MHz, and be capable of taking up one of two orientations with respect to the axis of the external field aligned or opposed, parallel or antiparallel. If a proton is precessing in the aligned orientation, it can absorb energy and pass into the opposed orientation; subsequently it can lose this extra energy and relax back into the aligned position. If we irradiate the precessing nuclei with a beam of radiofrequency energy of the correct frequency, the low-energy nuclei may absorb this energy and move to a higher energy state. The precessing proton will only absorb energy from the radiofrequency source if the precessing frequency is the same as the frequency of the radiofrequency beam; when this occurs, the nucleus and the radiofrequency beam are said to be in resonance; hence the term nuclear magnetic resonance.
Figure 4: Proton is behaving as a spinning magnet
When the magnetic field is applied, the nucleus begins to precess about its own axis of spin with angular frequency ω, which is called its Larmor frequency. The frequency at which a proton precesses is directly proportional to the strength of the applied magnetic field; the stronger the applied field, the higher the rate (angular frequency ω) of precession. For a proton, if the applied field is 1.41 Tesla, the frequency of precession is approximately 60 MHz. Since the nucleus has a charge, the precession generates an oscillating electric field of the same frequency. If radiofrequency waves of this frequency are supplied to the precessing proton, the energy can be absorbed. That is, when the frequency of the oscillating electric field component of the incoming radiation just matches the frequency of the electric field generated by the precessing nucleus, the two fields can couple, and energy can be transferred from the incoming radiation to the nucleus, thus causing a spin change. This condition is called resonance, and the nucleus is said to have resonance with the incoming electromagnetic wave.
Chemical shift
When a molecule is studied in the NMR spectrometer, different types of protons in a molecule absorb radiation at different frequencies. The differences in resonance frequency of these protons are very small. It is very difficult to measure the exact frequencies of the protons to that precision; hence, no attempt is made to measure the exact resonance frequency of any proton. A small amount of an inert reference compound is added to the substance whose NMR spectrum is to be measured, and the resonance frequency of each proton in the sample is measured relative to the resonance frequency of the protons of the reference substance. In other words, the frequency difference is measured directly and is presented as peaks in an NMR spectrum. The positions of the signals in an NMR spectrum are defined according to how far they are from the signal of the reference compound. The most commonly used reference compound is tetramethylsilane (TMS).
Silicon is electropositive and in TMS, silicon pushes electrons into the methyl groups of TMS by a +I inductive effect, and this powerful shielding effect means that the TMS protons come to resonance at low frequency (low δ value, defined as zero).
Advantages of TMS:
1. It is chemically inert and miscible with a large range of solvents.
2. It gives a sharp and intense single peak, since all 12 hydrogen in it are magnetically equivalent and hence absorb at exactly the same position.
3. Its resonance position is to high field of almost all other hydrogen resonances in organic molecules and hence can be easily recognised.
4. It is a low boiling point liquid and so can be readily removed from most samples after use.
The position at which a signal occurs in an NMR spectrum is called the chemical shift. The chemical shift is a measure of how far the signal is from the reference TMS signal. The most common scale for chemical shifts is the (delta) scale. The TMS signal is used to define the zero ppm on this scale. The chemical shift is determined by measuring the distance from the TMS peak (in hertz) and dividing by the operating frequency of the instrument (in megahertz). Because the units are Hz/MHz, a chemical shift has units of parts per million (ppm) of the operating frequency:
Figure 5: Chemical shift at 600MHz and 300MHz
For most proton chemical shifts fall in the range from 0 to 10 ppm. The advantage of the δ scale is that the chemical shift of a given nucleus is independent of the operating frequency of the NMR spectrometer. Thus, the chemical shift of the methyl protons of 1-bromo-2,2-dimethylpropane is at 1.05 ppm in both a 60-MHz and a 360-MHz instrument. In contrast, if the chemical shift were reported in hertz, it would be at 63 Hz in a 60 MHz instrument (63/60 = 1.05) and at 378 Hz in a 360-MHz instrument (378/360 = 1.05).
Shielding & de shielding effect
Nucleus is embedded in a cloud of electrons. These electrons are charged particles and circulate around the nucleus. They shield the nucleus from experiencing the applied magnetic field (Bexternal). Therefore, the effective magnetic field experienced by the nucleus (Beffective) will be decreased due to the shielding effect exerted by the electrons (Bshielding).
Beffective = Bexternal – Bshielding
When the electron density around the nucleus is high then shielding would also be high while a low electron density around the nucleus would exert poor shielding effect. Nuclei (proton) which is well shielded by electrons is called shielded nuclei (proton). A low frequency of Radiofrequency radiation is sufficient to flip the nuclei from the α-state to β-state since a shielded proton would experience only a small magnetic field resulting in a small ΔE between the two states. Whereas a deshielded proton would experience greater magnetic field leading to a larger ΔE between the α and β-states which in turn would require higher frequency of Radiofrequency radiation to cause resonance. In an NMR spectrum, the protons in electron-rich environments (more shielded) appear at lower frequencies-on the right-hand side of the spectrum. Protons in electron-poor environments (less shielded) appear at higher frequencies-on the left-hand side of the spectrum. The methyl protons of TMS are in a more electron-dense environment (shielded) than are most protons in organic molecules, because silicon is less electronegative than carbon (electronegativities of 1.8 and 2.5, respectively). Consequently, the signal for the methyl protons of TMS is at a lower frequency than most other signals (i.e., it appears to the right of the other signals). NMR signals occurring near the TMS resonance are said to be in an up field position while those shifted away by deshielding are said to be downfield.
Figure6: Shielding and deshielding of proton
Measurement of chemical shift
Chemical shift when measured in Hz is directly proportional to the strength of the applied magnetic field and therefore these values will vary with the strength of applied magnetic field. Hence if the chemical shift is given in hertz, the applied frequency should also be mentioned. The NMR instruments of different field strength are available hence It is advisable that the chemical shift should be expressed in the terms which are independent of field strength. One such unit is δ (delta). The chemical shifts are generally expressed in δ values whose units are ppm and it is independent of the field strength. The chemical shift of a particular proton is defined as the difference (in hertz) between the resonance frequency of the proton under observation and that of TMS, divided by the operating frequency of the spectrometer.
For example, when benzene is analyzed using an NMR spectrometer operating at 300 MHz, the protons of benzene absorb at a frequency that is 2181 Hz larger than the frequency of absorption of TMS. The chemical shift of these protons is then calculated in the following way:
If a 60 MHz spectrometer is used instead, the protons of benzene absorb a frequency of rf radiation that is 436 Hz larger than the frequency of absorption of TMS. The chemical shift of these protons is then calculated in the following way:
Thus the chemical shift value does not depend on the operating frequency and it is a dimension less. That is why chemical shifts have been defined in relative terms (δ-scale), rather than absolute terms (hertz). If signals were reported in hertz (the precise frequency of radiation absorbed), then the frequency of absorption would be dependent on the strength of the magnetic field and would not be a constant. For most organic compounds, the signals produced will fall in a range between 0 and 12 ppm. In rare cases, it is possible to observe a signal occurring at a chemical shift below 0 ppm, which results from a proton that absorbs a lower frequency than TMS. Most protons in organic compounds absorb a higher frequency than TMS, so most chemical shifts that we encounter will be positive numbers.
Chemical shift can also be measured on τ-scale in which TMS is given a value of 10 ppm. The values in δ can be converted into τ units as following
Ï„ = 10 ─ δ
Peak area integration
The area under a peak is proportional to the number of hydrogen’s contributing to that peak. We cannot simply compare peak heights, however; the area under the peak is proportional to the number of protons. An NMR spectrometer is equipped with a computer that calculates the integrals electronically. Modern spectrometers print out the integrals as numbers on the spectrum. The integrals can also be displayed by a line of integration superimposed on the original spectrum. The height of each integration step is proportional to the area under that signal, which, in turn, is proportional to the number of protons giving rise to the signal. By measuring the heights of the integration steps, you can determine number of protons by taking the ratio of the integrals. The integration tells us the relative number of protons that give rise to each signal, not the absolute number. For example, the NMR spectrum of 1-bromo-2,2-dimethylpropane shows two signals due to the presence of two types of chemically equivalent protons (three sets of methyl protons, one set of methylene protons) The signals of the compound (Figure below) are of different size and the area under each signal is proportional to the number of protons that gives rise to the signal. Between the two sets of chemically equivalent protons, the CH2 protons on C-1 are more deshielded, than the CH3 protons, because of the presence of electronegative Br atom is attached to it. Therefore, the signal near 3.5 ppm is due to the CH2 protons and the signal due to the CH3 protons is seen at 1.2 ppm. The evaluation of the area under each signal can be equated to the number of protons contributing to that signal.
Figure7: Peak area integration curve
By measuring the heights of the integration steps, one can determine that the ratio of the integrals is approximately 1.6: 7.0 = 1:4.4. The ratios are multiplied by a number that will cause all the numbers to be close to whole numbers—in this case, we multiply by 2—as there can be only whole numbers of protons. That means that the ratio of protons in the compound is 2:8.8 which is rounded to 2:9. In the present case it equals to the number of CH2 (2) and CH3 (3×3 =9) protons contributing to the signal.
Spin-spin coupling
To understand this concept, let’s consider the NMR spectrum of 1,1-dichloroethane and 1,2 dichloroethane. The NMR spectrum of 1,2-dichloroethane is more complicated than the 1,2 dichloro isomer which displays a single peak from the four equivalent hydrogens. The NMR spectrum of 1,1-dichloroethane shows two signals from the two different types of hydrogens which are further split into group of two or more peaks. 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 a small value to 18 Hz. It is designated as J and called as the coupling constant.
Figure8: 1H NMR spectrum of 1,2-dichloroethane and 1,1-dichloroethane
The separation between two consecutive lines is always constant within a given multiplet, and is called as coupling constant (J). The magnitude of J is usually expressed in units of Hz and it is independent of magnetic field.
If the chemical shifts of the two signals differ significantly (>2 ppm) then the splitting patterns found in spectrum is easily recognizable. 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) and so on. These patterns are shown in the picture given below.
Figure9: Signal pattern in NMR spectrum
Thus signal’s multiplicity is the result of the magnetic effects of neighboring protons and therefore indicates the number of neighboring protons. To illustrate this concept, consider the following example.
If Ha and Hb are not chemically equivalent, they will produce different signals. The chemical shift of Ha is affected by several electronic effects. All of these effects modify the magnetic field felt by Ha, thereby affecting the resonance frequency of Ha. The chemical shift of Ha is also affected by the presence of Hb, because Hb has a magnetic moment that can either be aligned with or against the external magnetic field. The chemical shift of Ha depends on the alignment of nuclear spin of Hb. In some molecules, Hb will be aligned with the field, while in other molecules Hb will be aligned against the field. As a result, the chemical shift of Ha in some molecules will be slightly different than the chemical shift of Ha in other molecules, resulting in the appearance of two peaks. In other words, the presence of Hb splits the signal for Ha into a doublet.
Figure10: Formation of doublet
Ha has the same effect on the signal of Hb, splitting the signal for Hb into a doublet. This phenomenon is called spin-spin splitting or coupling Now consider a scenario in which Ha has two neighboring protons.
The chemical shift of Ha is impacted by the presence of both Hb protons, each of which can be aligned either with or against the external field. Once again, each Hb is like a tiny magnet and has an impact on the chemical shift of Ha. In each molecule, Ha can find itself in one of three possible electronic environments, resulting in a triplet. If each peak of the triplet is separately integrated, a ratio of 1:2:1 is observed, consistent with statistical expectations.
Figure11: Formation of triplet
Now consider a scenario in which Ha has three neighbors.
The chemical shift of Ha is impacted by the presence of all three Hb protons, each of which can be aligned either with the field or against the field. Once again, each Hb is like a tiny magnet and has an impact on the chemical shift of Ha. In each molecule, Ha can find itself in one of four possible electronic environments, resulting in a quartet. If each peak of the quartet is integrated separately, a ratio of 1:3:3:1 is observed, consistent with statistical expectations.
Figure12: Formation of quartet
The table given below summarizes the splitting patterns and peak intensities for signals that result from coupling with neighbouring protons. When analyse this carefully, a uniform pattern emerges. If n is the number of neighbouring protons, then the multiplicity will be n+1. This observation is called the n+1 rule.
Figure13: Splitting pattern from coupling with neighbouring protons
This spin-coupling is transmitted through the connecting bonds. 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. There are two major factors that determine whether or not splitting occurs:
1. Equivalent protons do not split each other. Consider the two methylene groups in 1,2 dichloroethane. All four protons are chemically equivalent, and therefore, they do not split each other. In order for splitting to occur, the neighbouring protons must be different than the protons producing the signal.
2. Splitting is observed for non-equivalent protons on the same carbon or adjacent carbons or in other words we can say that splitting is observed when protons are separated by either two or three σ bonds; that is, when the protons are either diastereotopic protons on the same carbon atom (geminal) or when they are connected to adjacent carbon atoms (vicinal).
Figure14: Geminal and vicinal protons
When two protons are separated by more than three sigma bonds, splitting is generally not observed. Such long-range splitting is only observed in olefins, acetylenes, aromatics, hetero aromatics and in strained ring systems such as bicyclic compounds. This type of long range coupling will be discussed later in this article.
Coupling constants
The distance between the centers of two adjacent peaks in a multiplet is called coupling constant or spin-spin coupling constant (J). The magnitude of J is usually expressed in units of Hz and never in δ (ppm) values. The value of J remains constant and does not depend on magnetic field. Therefore the separation between two peaks in a multiplet remains always constant but if this separation changes that means that they represent different signals or does not belong to the same peak/multiplet. The separations of peaks (J value) in two coupled multiplets are exactly equal. In the 1H NMR spectrum of 1,1,2-trichloroethane, two multiplets, one doublet and one triplet are observed. The J value in the doublet is exactly the same as in the triplet (6.1 Hz).
Figure15: Coupling constant JAB
Since the coupling constant Jab quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Types of Coupling Constants
There are three types of coupling constants vicinal, geminal and long range based on the type of protons that are coupled in the compound.
Figure 16: Types of coupling constants
Vicinal Coupling
It is denoted as 3J. The coupling interaction is through three bonds as mentioned for 1,1,2 trichloroethane. The magnitude of the coupling constant depends on the dihedral angle of the coupled protons. The Newman projection could be used to identify the dihedral angles between the neighbouring protons. The coupling constant value is maximum (about 16 Hz) at dihedral angle 180° (staggered) and minimum (close to 0 Hz) at 90° while it is about 10 Hz at 0° (eclipsed).
Figure17: Vicinal coupling examples
Geminal Coupling
This kind of coupling is seen in terminal vinyl systems. It is denoted as 2J, which means the coupling is between non-equivalent protons on the same carbon atom. The J value depends on the angle of H-C-H bonds during coupling. Normal values are between 10-18 Hz. At angle of 125°, J=0 Hz, while at 100° the J value is as maximum as 35 Hz.
Figure18: Geminal coupling examples
The J value is usually very small and are unable to observe for nonequivalent hydrogens on the same sp2 carbon, whereas the J is usually large enough to be observed for nonequivalent hydrogens bonded to adjacent sp2 carbons.
Long Range Coupling
When the coupling between protons that are present beyond three bond distance (>3J) such a coupling is called long range coupling. The coupling is commonly observed up to 4 to 5 bond distances. In case of polyalkynes the coupling is also observed as far as 9 bonds too. The normal coupling constant values range from 0 to 4 Hz. Long-range couplings are common in allylic systems, aromatic rings, and rigid bicyclic systems. Long-range couplings are communicated through specific overlap of a series of orbitals and as a result have a stereochemical requirement. In alkenes, small couplings between the alkenyl hydrogens and protons on the carbon(s) α to the opposite end of the double bond are observed. This four-bond coupling (4J) is called allylic coupling.
Coupling Constant between Protons on sp2 Carbon
The coupling constants between proton sets on neighboring sp3 hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2 hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement. For vinylic hydrogens in a trans configuration, the coupling constants is in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine and is generally 5 Hz or lower.
Figure19: Coupling Constant between Protons on sp2 Carbon
Alcohols:
In alcohols, both the hydroxyl proton and the hydrogens (those on the same carbon as the hydroxyl group) have characteristic chemical shifts.
Spectral analysis box-Alcohols
The chemical shift of the -OH hydrogen is variable, its position depending on concentration, solvent, temperature, and presence of water or of acidic or basic impurities. This peak can be found anywhere in the range of 0.5–5.0 ppm. The variability of this absorption is dependent on the rates of –OH proton exchange and the amount of hydrogen bonding in the solution.
The -OH hydrogen is usually not split by hydrogens on the adjacent carbon (-CH-OH) because rapid exchange decouples this interaction
Exchange is promoted by increased temperature, small amounts of acid impurities, and the presence of water in the solution. In ultrapure alcohol samples, -CH-OH coupling is observed. A freshly purified and distilled sample, or a previously unopened commercial bottle, may show this coupling. On occasion, one may use the rapid exchange of an alcohol as a method for identifying the -OH absorption. In this method, a drop of D2O is placed in the NMR tube containing the alcohol solution. After shaking the sample and sitting for a few minutes, the -OH hydrogen is replaced by deuterium, causing it to disappear from the spectrum (or to have its intensity reduced).
The hydrogen on the adjacent carbon (-CH-OH) appears in the range 3.2–3.8 ppm, being deshielded by the attached oxygen. If exchange of the OH is taking place, this hydrogen will not show any coupling with the -OH hydrogen, but will show coupling to any hydrogens on the adjacent carbon located further along the carbon chain. If exchange is not occurring, the pattern of this hydrogen may be complicated by differently sized coupling constants for the –CH-OH and –CH-CH-O- couplings.
A spectrum of 2-methyl-1-propanol is shown in Figure. Note the large downfield shift (3.4 ppm) of the hydrogens attached to the same carbon as the oxygen of the hydroxyl group. The hydroxyl group appears at 2.4 ppm, and in this sample it shows some coupling to the hydrogens on the adjacent carbon. The methine proton at 1.75 ppm has been expanded and inset on the spectrum.
Figure20: 1H spectrum of 2-methyl-1-propanol (300 MHz).
There are nine peaks (nonet) in that pattern, suggesting coupling with the two methyl groups and one methylene group, n = (3 + 3 + 2) + 1 = 9.
Phenols
Protons attached to the aromatic ring in phenols show up near the aromatic region of an NMR spectrum (7-8 ppm). These peaks will have splitting typical for aromatic protons. The protons directly attached to the alcohol oxygen of phenols appear in the region of 3 to 8 ppm. These peaks tend to appear as short, broad singlets similarly to other alcohols.
Figure21: 1H spectrum of Phenol
Carboxylic acids:
Carboxylic acids have the acid proton (the one attached to the -COOH group) and the a hydrogens (those attached to the same carbon as the carboxyl group).
In carboxylic acids, the hydrogen of the carboxyl group (-COOH) has resonance in the range 11.0–12.0 ppm. With the exception of the special case of a hydrogen in an enolic OH group that has strong internal hydrogen bonding, no other common type of hydrogen appears in this region. A peak in this region is a strong indication of a carboxylic acid. Since the carboxyl hydrogen has no neighbors, it is usually unsplit; however, hydrogen bonding and exchange may cause the peak become broadened (become very wide at the base of the peak) and show very low intensity. Sometimes the acid peak is so broad that it disappears into the baseline. In that case, the acidic proton may not be observed. Infrared spectroscopy is very reliable for determining the presence of a carboxylic acid. As with alcohols, this hydrogen will exchange with water and D2O. In D2O, proton exchange will convert the group to -COOD, and the –COOH absorption near 12.0 ppm will disappear.
Carboxylic acids are often insoluble in CDCl3, and it is common practice to determine their spectra in D2O to which a small amount of sodium metal is added. This basic solution (NaOD, D2O) will remove the proton, making a soluble sodium salt of the acid. However, when this is done –COOH absorption will disappear from the spectrum.
A spectrum of ethylmalonic acid is shown in Figure. The -COOH absorption integrating for 2H is shown as an inset on the spectrum. Notice that this peak is very broad due to hydrogen bonding and exchange. Also notice that proton c is shifted downfield to 3.1 ppm, resulting from the effect of two neighboring carbonyl groups. The normal range for a proton next to just one carbonyl group would be expected to appear in the range 2.1 to 2.5 ppm.
Figure22: 1H spectrum of ethylmalonic acid (300 MHz)
Amines
Two characteristic types of hydrogens are found in amines: those attached to nitrogen (the hydrogens of the amino group) and those attached to the carbon (the same carbon to which the amino group is attached).
Spectral analysis box-Amines
Location of the –NH absorptions is not a reliable method for the identification of amines. These peaks are extremely variable, appearing over a wide range of 0.5–4.0 ppm, and the range is extended in aromatic amines. The position of the resonance is affected by temperature, acidity, amount of hydrogen bonding, and solvent. In addition to this variability in position, the –NH peaks are often very broad and weak without any distinct coupling to hydrogens on an adjacent carbon atom. This condition can be caused by chemical exchange of the –NH proton or by a property of nitrogen atoms called quadrupole broadening. The amino hydrogens will exchange with D2O, as already described for alcohols, causing the peak to disappear
The –NH peaks are strongest in aromatic amines (anilines), in which resonance appears to strengthen the NH bond by changing the hybridization. Although nitrogen is a spin-active element (l=1), coupling is usually not observed between either attached or adjacent hydrogen atoms,but it can appear in certain specific cases.
The hydrogens a to the amino group are slightly deshielded by the presence of the electronegative nitrogen atom, and they appear in the range 2.2–2.9 ppm. A spectrum of propylamine is shown in Figure. Notice the weak, broad NH absorptions at 1.8 ppm and that there appears to be a lack of coupling between the hydrogens on the nitrogen and those on the adjacent carbon atom.
Figure23: 1H spectrum of propylamine (300 MHz)
Amides
Amides have three distinct types of hydrogens: those attached to nitrogen, a hydrogens attached to the carbon atom on the carbonyl side of the amide group, and hydrogens attached to a carbon atom that is also attached to the nitrogen atom.
Spectral analysis box-Amides
The -NH absorptions of an amide group are highly variable, depending not only on their environment in the molecule, but also on temperature and the solvent used. Because of resonance between the unshared pairs on nitrogen and the carbonyl group, rotation is restricted in most amides. Without rotational freedom, the two hydrogens attached to the nitrogen in an unsubstituted amide are not equivalent, and two different absorption peaks will be observed, one for each hydrogen. Nitrogen atoms also have a quadrupole moment, its magni tude depending on the particular molecular environment. If the nitrogen atom has a large quadru pole moment, the attached hydrogens will show peak broadening (a widening of the peak at its base) and an overall reduction of its intensity. Hydrogens adjacent to a carbonyl group (regardless of type) all absorb in the same region of the NMR spectrum: 2.1–2.5 ppm.
The spectrum of butyramide is shown in Figure. Notice the separate absorptions for the two -NH hydrogens (6.6 and 7.2 ppm). This occurs due to restricted rotation in this compound. The hydrogens next to the C=O group appear characteristically at 2.1 ppm.
Figure24: 1H spectrum of butyramid
Problems based on 1H NMR:
1. Calculate the number of multiplets for each bond and their relative areas in ClCH2CH2CH2Cl.
Solution: The multiplicity of the band associated with the four equivalent protons on the two ends of the molecule would be determined by the number of protons on the central carbon. Thus multiplicity is (2 + 1) = 3 and areas would be 1 : 2 : 1. The multiplicity of two central methylene protons can be determined by the four equivalent protons at the ends and it is (4 + 1) = 5. Expansion of (x + 1)⁵ gives the coefficients in the ratio 1 : 4 : 6 : 4 : 1.
2. Calculate the number of multiplets in case of CH₃CH₂OCH₃.
Solution: The methyl protons on the extreme right are separated from the other protons by more than three bonds so that only a single peak will be observed for them. The protons on the central methylene group will have a multiplicity of (3 + 1) = 4 and a ratio of 1 : 3 : 3 : 1. The methyl protons at the extreme left will be split into three. The intensity ratio being 1 : 2 : 1.
3. Explain a complex splitting pattern when a set of protons is affected by two or more non-equivalent protons.
Solution. Non-equivalent protons have different chemical shifts. Spin-spin coupling takes place only between non-equivalent neighbouring protons. The complex splitting pattern would be illustrated by taking an example of 1-iodopropane, CH₃CH₂CH₂I.
If the three carbon atoms (on a, b, c) which resonate at 1.02, 1.86 and 3.17 respectively, the band at δ(a) = 1.02 will be split by the two methylene protons on (b) into (2 + 1) = 3. The areas being in the ratio 1 : 2 : 1.
Similarly δ(c) at 3.17 will also be split into three. The J for the two splittings being 7.3 (Jab) and 6.8 (Jbc) respectively. The band for methylene protons (b) is affected by two groups of protons which are magnetically non-equivalent (since Jab ≠ Jbc). The number of peaks will be (3 + 1)(2 + 1) = 12. The splitting pattern of methylene (b) proton is shown in Figure.
Fig. 15. Splitting of methylene (b) proton in 1-iodopropane.
In this Figure the effect of proton a is first shown. It leads to four peaks of relative areas
1 : 3 : 3 : 1 spaced at 7.3 Hz. Each of these further split into three new peaks spaced at 6.8 Hz, areas being 1 : 2 : 1. At very high resolution 1-iodopropane exhibits a series of peaks as shown at the top of Fig. 15. At lower resolution, only six peaks are observed with relative areas of 1 : 5 : 10 : 10 : 5 : 1.
4. Show that spin-spin coupling is a reciprocal phenomenon in 1,1,2-trichloroethane.
Solution. In 1,1,2-trichloroethane, the separation of peaks in a triplet (ratio 1 : 2 : 1) is exactly the same as that observed in a doublet (1 : 1). Since the effect of a kind of protons on b protons is exactly the same as the effect of b protons on a type of protons. So here spin-spin coupling is a reciprocal phenomenon.
5. Sometimes the peaks in a particular multiplet are not in the expected ratio. Why?
Solution. Sometimes the various multiplets do not show the symmetrical pattern due to the fact that the separation between the two signals is not very large relative to the separation of peaks within a particular multiplet. If the distance between the signals (multiplets) is very large i.e., when the chemical shift is much larger than the coupling constant we can expect perfectly symmetrical multiplets.
6. Explain splitting in 1,1-difluoro-1,2-dichloroethane.
Solution. The mass number as well as the atomic number of fluorine being odd, its nucleus also has magnetic properties of the same kind as those of the proton. Fluorine atoms can couple with each other as well as with proton and thus, the splitting of signal can be observed.
In this compound, the coupling of two equivalent protons with fluorine nuclei gives a triplet in the spectrum. It seems rather surprising, how a multiplet can appear in the spectrum since most commonly, it originates from a set of equivalent protons when these are under the influence of two equivalent nearby protons with different chemical shift.
7. Calculate signals for proton a in case of methyl cyclohexadiene.
Solution. In methyl cyclohexadiene,
Proton a is under the influence of three sets of protons (b, c, d) under different environments. Thus the signal for proton a is expected to be a complicated pattern consisting of
(n + 1)(n′ + 1)(n″ + 1) lines, i.e., (1 + 1)(1 + 1)(3 + 1) = 16 signals.
8. Find the number of signals for methylene protons in 1,1-dibromo-3,3-dichloropropane.
Solution. In 1,1-dibromo-3,3-dichloropropane,
Protons a and c are under different chemical shifts (non-equivalent). So these protons will influence the methylene (—CH2—, i.e., protons b) protons differently. The signal for —CH2— protons appears as a multiplet and consists of
(n + 1)(n′ + 1) = (1 + 1)(1 + 1) = 4 lines.
9. Calculate the number of peaks for methylene protons in case of 1-bromo-3-chloropropane.
Solution:
The multiplet for central (—CH2—) protons consists of
(n + 1)(n′ + 1) = (2 + 1) (2 + 1) = 9 peaks
where n and n′ are the number of protons a and c respectively. The —CH2— protons being under the influence of two kinds of protons with different chemical shifts. Hence asymmetrical pattern results for central methylene protons.
10. Why is splitting observed in 2-methyl propene (I) but not in neopentyl chloride (II)?
Solution. Although Hᵃ and Hᵇ in I are not on adjacent carbon atoms, they are close enough to couple because of the shorter C=C bond. In II, Hᵃ and Hᵇ on non-adjacent carbons are too far away to couple. The carbon atoms are joined by longer single bonds.
11. Distinguish between cis and trans-stilbenes on the basis of PMR spectroscopy.
Solution. In trans-stilbene (planar), each of the two olefinic protons is deshielded by both the aromatic rings, while in cis-stilbene, each H atom is deshielded by only one adjacent aromatic ring. Chemical shifts occur at 7.0 δ ppm in trans and 6.50 δ ppm in cis-stilbene.
MULTIPLE CHOICE QUESTIONS:
1. Only one signal is present in the PMR spectra of
(a) C₃H₄, C₃H₆ (b) C₄H₆, C₅H₁₂ (c) C₈H₁₈, C₂H₆O (d) All
2. A proton Hᵇ is coupled to four equivalent protons Hᵃ. The multiplicity and the relative intensity of lines in the signal Hᵇ is
(a) Doublet, 1:4 (b) Triplet, 1:4:6 (c) Quintet, 1:4:6:4:1 (d) Quartet, 1:4:6:4
3. The compounds showing only a single peak in its PMR spectrum are:
Acetone (I)
Ether (II)
Methyl acetate (III)
Dibromoethane (IV)
Chlorobromoethane (V)
(a) I, II, IV (b) II, III, V (c) I, III, V (d) III, IV, V
4. The actual value of nuclear spin depends on
(a) Mass number (b) Atomic number (c) Both (a) and (b) (d) Shielding effect
5. H¹, C¹³, F¹⁹, P³¹ have nuclear spin equal to
(a) 1/2 (b) 1 (c) 0 (d) 3/2
6. When the effective magnetic field experienced by the nucleus is less than that of the applied field (Hâ‚‘ff < H₀), the nucleus is said to be
(a) Deshielded (b) Shielded (c) Relaxed (d) None
7. In p-xylene, the ratio of methyl protons to ring protons is 6 : 4 while for mesitylene, it is
(a) 6 : 4 (b) 3 : 2 (c) 9 : 3 (d) 6 : 3
8. Compound C₄H₁₀O gave PMR spectrum consisting of two groups of lines (multiplets) with relative intensities in the ratio 3 : 2. Other compound of the same formula exhibited two lines with relative area of 9 : 1. Compounds are
(a) Diethyl ether (b) t-Butyl alcohol (c) Both (a) and (b) (d) None
9. The peaks expected in low-resolution NMR spectrum of vinyl chloride and ethyl cyclopropane are,
(a) 3, 5
(b) 5, 3
(c) 6, 3
(d) 3, 6
10. Four different types of protons are present in allyl bromide. The correct statement is/are
(a) Two enantiomeric identical allylic protons
(b) Two diastereotopic different protons
(c) Remaining protons
(d) All are correct
11. Most proton absorptions fall within a range of
(a) 600 Hz
(b) 200 Hz
(c) 100 Hz
(d) 60 Hz
12. Citric acid is a prochiral molecule. The molecule has
(a) A mirror plane of symmetry
(b) Two enantiotopic CH₂COOH chains
(c) Diastereotopic methylene protons
(d) All
13. Distance between the centres of the peaks of doublet is called
(a) Coupling constant
(b) Spin constant
(c) Spin-spin coupling
(d) None
14. Scalar coupling is also termed as
(a) Spin-spin coupling
(b) J-coupling
(c) Indirect coupling
(d) All
15. A compound show PMR peak at 240 Hz downfield from TMS peak and operating at 60 MHz. The value of Ï„ is
(a) 4 ppm
(b) 6 ppm
(c) 8 ppm
(d) 2 ppm
Mass Spectroscopy Part-I
Mass Spectroscopy Part-II
Mass Spectroscopy Problems
Mass Spectroscopy Multiple choice questions
Proton magnetic resonance spectroscopy
 Spectral analysis Proton magnetic resonance spectroscopy){kind=link}