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Old Tuesday, November 16, 2010
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Mass Spectrometry - Functional Groups

Alkanes: Simple alkanes tend to undergo fragmentation by the initial loss of a methyl group to form a (m-15) species. This carbocation can then undergo stepwise cleavage down the alkyl chain, expelling neutral two-carbon units (ethene). Branched hydrocarbons form more stable secondary and tertiary carbocations, and these peaks will tend to dominate the mass spectrum.





Aromatic Hydrocarbons: The fragmentation of the aromatic nucleus is somewhat complex, generating a series of peaks having m/e = 77, 65, 63, etc. While these peaks are difficult to describe in simple terms, they do form a pattern (the "aromatic cluster") that becomes recognizable with experience. If the molecule contains a benzyl unit, the major cleavage will be to generate the benzyl carbocation, which rearranges to form the tropylium ion. Expulsion of acetylene (ethyne) from this generates a characteristic m/e = 65 peak.


Aldehydes and Ketones: The predominate cleavage in aldehydes and ketones is loss of one of the side-chains to generate the substituted oxonium ion. This is an extremely favorable cleavage and this ion often represents the base peak in the spectrum. The methyl derivative (CH3CO+) is commonly referred to as the "acylium ion".


Another common fragmentation observed in carbonyl compounds (and in nitriles, etc.) involves the expulsion of neutral ethene via a process known as the McLafferty rearrangement, following the general mechanism shown below.



Esters, Acids and Amides: As with aldehydes and ketones, the major cleavage observed for these compounds involves expulsion of the "X" group, as shown below, to form the substituted oxonium ion. For carboxylic acids and unsubstituted amides, characteristic peaks at m/e = 45 and 44 are also often observed.



Alcohols: In addition to losing a proton and hydroxy radical, alcohols tend to lose one of the -alkyl groups (or hydrogens) to form the oxonium ions shown below. For primary alcohols, this generates a peak at m/e = 31; secondary alcohols generate peaks with m/e = 45, 59, 73, etc., according to substitution.



Ethers: Following the trend of alcohols, ethers will fragment, often by loss of an alkyl radical, to form a substituted oxonium ion, as shown below for diethyl ether.



Halides: Organic halides fragment with simple expulsion of the halogen, as shown below. The molecular ions of chlorine and bromine-containing compounds will show multiple peaks due to the fact that each of these exists as two isotopes in relatively high abundance. Thus for chlorine, the 35Cl/37Cl ratio is roughly 3.08:1 and for bromine, the 79Br/81Br ratio is 1.02:1. The molecular ion of a chlorine-containing compound will have two peaks, separated by two mass units, in the ratio 3:1, and a bromine-containing compound will have two peaks, again separated by two mass units, having approximately equal intensities.

[SIZE="7"]NMR[/SIZE]

Theoretical principles

Introduction
Nuclear Magnetic Resonance spectroscopy is a powerful and theoretically complex analytical tool. On this page, we will cover the basic theory behind the technique. It is important to remember that, with NMR, we are performing experiments on the nuclei of atoms, not the electrons. The chemical environment of specific nuclei is deduced from information obtained about the nuclei.


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Nuclear spin and the splitting of energy levels in a magnetic field
Subatomic particles (electrons, protons and neutrons) can be imagined as spinning on their axes. In many atoms (such as 12C) these spins are paired against each other, such that the nucleus of the atom has no overall spin. However, in some atoms (such as 1H and 13C) the nucleus does possess an overall spin. The rules for determining the net spin of a nucleus are as follows;

If the number of neutrons and the number of protons are both even, then the nucleus has NO spin.
If the number of neutrons plus the number of protons is odd, then the nucleus has a half-integer spin (i.e. 1/2, 3/2, 5/2)
If the number of neutrons and the number of protons are both odd, then the nucleus has an integer spin (i.e. 1, 2, 3)
The overall spin, I, is important. Quantum mechanics tells us that a nucleus of spin I will have 2I + 1 possible orientations. A nucleus with spin 1/2 will have 2 possible orientations. In the absence of an external magnetic field, these orientations are of equal energy. If a magnetic field is applied, then the energy levels split. Each level is given a magnetic quantum number, m.


When the nucleus is in a magnetic field, the initial populations of the energy levels are determined by thermodynamics, as described by the Boltzmann distribution. This is very important, and it means that the lower energy level will contain slightly more nuclei than the higher level. It is possible to excite these nuclei into the higher level with electromagnetic radiation. The frequency of radiation needed is determined by the difference in energy between the energy levels.


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Calculating transition energy
The nucleus has a positive charge and is spinning. This generates a small magnetic field. The nucleus therefore possesses a magnetic moment, m, which is proportional to its spin,I.


The constant, g, is called the magnetogyric ratioand is a fundamental nuclear constant which has a different value for every nucleus. h is Plancks constant.

The energy of a particular energy level is given by;

Where B is the strength of the magnetic field at the nucleus.

The difference in energy between levels (the transition energy) can be found from


This means that if the magnetic field, B, is increased, so is DE. It also means that if a nucleus has a relatively large magnetogyric ratio, then DE is correspondingly large.

If you had trouble understanding this section, try reading the next bit (The absorption of radiation by a nucleus in a magnetic field) and then come back.


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The absorption of radiation by a nucleus in a magnetic field
In this discussion, we will be taking a "classical" view of the behaviour of the nucleus - that is, the behaviour of a charged particle in a magnetic field.

Imagine a nucleus (of spin 1/2) in a magnetic field. This nucleus is in the lower energy level (i.e. its magnetic moment does not oppose the applied field). The nucleus is spinning on its axis. In the presence of a magnetic field, this axis of rotation will precess around the magnetic field;


The frequency of precession is termed the Larmor frequency, which is identical to the transition frequency.

The potential energy of the precessing nucleus is given by;

E = - m B cos q
where q is the angle between the direction of the applied field and the axis of nuclear rotation.

If energy is absorbed by the nucleus, then the angle of precession, q, will change. For a nucleus of spin 1/2, absorption of radiation "flips" the magnetic moment so that it opposes the applied field (the higher energy state).


It is important to realise that only a small proportion of "target" nuclei are in the lower energy state (and can absorb radiation). There is the possibility that by exciting these nuclei, the populations of the higher and lower energy levels will become equal. If this occurs, then there will be no further absorption of radiation. The spin system is saturated. The possibility of saturation means that we must be aware of the relaxation processes which return nuclei to the lower energy state.


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Relaxation processes
How do nuclei in the higher energy state return to the lower state? Emission of radiation is insignificant because the probability of re-emission of photons varies with the cube of the frequency. At radio frequencies, re-emission is negligible. We must focus on non-radiative relaxation processes (thermodynamics!).

Ideally, the NMR spectroscopist would like relaxation rates to be fast - but not too fast. If the relaxation rate is fast, then saturation is reduced. If the relaxation rate is too fast, line-broadening in the resultant NMR spectrum is observed.

There are two major relaxation processes;

Spin - lattice (longitudinal) relaxation
Spin - spin (transverse) relaxation

Spin - lattice relaxation
Nuclei in an NMR experiment are in a sample. The sample in which the nuclei are held is called the lattice. Nuclei in the lattice are in vibrational and rotational motion, which creates a complex magnetic field. The magnetic field caused by motion of nuclei within the lattice is called the lattice field. This lattice field has many components. Some of these components will be equal in frequency and phase to the Larmor frequency of the nuclei of interest. These components of the lattice field can interact with nuclei in the higher energy state, and cause them to lose energy (returning to the lower state). The energy that a nucleus loses increases the amount of vibration and rotation within the lattice (resulting in a tiny rise in the temperature of the sample).

The relaxation time, T1 (the average lifetime of nuclei in the higher energy state) is dependant on the magnetogyric ratio of the nucleus and the mobility of the lattice. As mobility increases, the vibrational and rotational frequencies increase, making it more likely for a component of the lattice field to be able to interact with excited nuclei. However, at extremely high mobilities, the probability of a component of the lattice field being able to interact with excited nuclei decreases.

Spin - spin relaxation
Spin - spin relaxation describes the interaction between neighbouring nuclei with identical precessional frequencies but differing magnetic quantum states. In this situation, the nuclei can exchange quantum states; a nucleus in the lower energy level will be excited, while the excited nucleus relaxes to the lower energy state. There is no net change in the populations of the energy states, but the average lifetime of a nucleus in the excited state will decrease. This can result in line-broadening.


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Chemical shift
The magnetic field at the nucleus is not equal to the applied magnetic field; electrons around the nucleus shield it from the applied field. The difference between the applied magnetic field and the field at the nucleus is termed the nuclear shielding.

Consider the s-electrons in a molecule. They have spherical symmetry and circulate in the applied field, producing a magnetic field which opposes the applied field. This means that the applied field strength must be increased for the nucleus to absorb at its transition frequency. This upfield shift is also termed diamagnetic shift.


Electrons in p-orbitals have no spherical symmetry. They produce comparatively large magnetic fields at the nucleus, which give a low field shift. This "deshielding" is termed paramagnetic shift.

In proton (1H) NMR, p-orbitals play no part (there aren't any!), which is why only a small range of chemical shift (10 ppm) is observed. We can easily see the effect of s-electrons on the chemical shift by looking at substituted methanes, CH3X. As X becomes increasingly electronegative, so the electron density around the protons decreases, and they resonate at lower field strengths (increasing dH values).

Chemical shift is defined as nuclear shielding / applied magnetic field. Chemical shift is a function of the nucleus and its environment. It is measured relative to a reference compound. For 1H NMR, the reference is usually tetramethylsilane, Si (CH3)4.


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Spin - spin coupling
Consider the structure of ethanol;


The 1H NMR spectrum of ethanol (below) shows the methyl peak has been split into three peaks (a triplet) and the methylene peak has been split into four peaks (a quartet). This occurs because there is a small interaction (coupling) between the two groups of protons. The spacings between the peaks of the methyl triplet are equal to the spacings between the peaks of the methylene quartet. This spacing is measured in Hertz and is called the coupling constant, J.


To see why the methyl peak is split into a triplet, let's look at the methylene protons. There are two of them, and each can have one of two possible orientations (aligned with or opposed against the applied field). This gives a total of four possible states;


In the first possible combination, spins are paired and opposed to the field. This has the effect of reducing the field experienced by the methyl protons; therefore a slightly higher field is needed to bring them to resonance, resulting in an upfield shift. Neither combination of spins opposed to each other has an effect on the methyl peak. The spins paired in the direction of the field produce a downfield shift. Hence, the methyl peak is split into three, with the ratio of areas 1:2:1.

Similarly, the effect of the methyl protons on the methylene protons is such that there are eight possible spin combinations for the three methyl protons;


Out of these eight groups, there are two groups of three magnetically equivalent combinations. The methylene peak is split into a quartet. The areas of the peaks in the quartet have the ration 1:3:3:1.

In a first-order spectrum (where the chemical shift between interacting groups is much larger than their coupling constant), interpretation of splitting patterns is quite straightforward;

The multiplicity of a multiplet is given by the number of equivalent protons in neighbouring atoms plus one, i.e. the n + 1 rule

Equivalent nuclei do not interact with each other. The three methyl protons in ethanol cause splitting of the neighbouring methylene protons; they do not cause splitting among themselves

The coupling constant is not dependant on the applied field. Multiplets can be easily distinguished from closely spaced chemical shift peaks.


Supplemental NMR Topics

Spin Properties of Nuclei
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner:
Odd mass nuclei (i.e. those having an odd number of nucleons) have fractional spins. Examples are I = 1/2 ( 1H, 13C, 19F ), I = 3/2 ( 11B ) & I = 5/2 ( 17O ).
Even mass nuclei composed of odd numbers of protons and neutrons have integral spins. Examples are I = 1 ( 2H, 14N ).
Even mass nuclei composed of even numbers of protons and neutrons have zero spin ( I = 0 ). Examples are 12C, and 16O.
Spin 1/2 nuclei have a spherical charge distribution, and their nmr behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments (μ), but the nonspherical nuclei also have an electric quadrupole moment (eQ). Some characteristic properties of selected nuclei are given in the following table.

Isotope
Natural %
Abundance
Spin (I)
Magnetic
Moment (μ)*
Magnetogyric
Ratio (γ)†

1H
99.9844 1/2 2.7927 26.753
2H
0.0156 1 0.8574 4,107
11B
81.17 3/2 2.6880 --
13C
1.108 1/2 0.7022 6,728
17O
0.037 5/2 -1.8930 -3,628
19F
100.0 1/2 2.6273 25,179
29Si
4.700 1/2 -0.5555 -5,319
31P
100.0 1/2 1.1305 10,840
* μ in units of nuclear magnetons = 5.05078•10-27 JT-1
† γ in units of 107rad T-1 sec-1





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A Model for NMR Spectroscopy
The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of nmr phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: ωo = γBo. The frequency ωo is called the Larmor frequency and has units of radians per second. The proportionality constant γ is known as the gyromagnetic ratio and is proportional to the magnetic moment (γ = 2pm/hI). Some characteristic γ's were listed in a preceding table of nuclear properties.

magnetic moment
μ
A Spinning Gyroscope
in a Gravity Field
A Spinning Charge
in a Magnetic Field



If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = _1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2π.



The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and _1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (Bo). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of _1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis.


Net Macroscopic Magnetization of a Sample in an External Magnetic Field Bo
Excitation by RF Energy and Subsequent Relaxation




The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the "Introduce RF Energy" button the animation will begin, and will repeat five times.

• First, the net magnetization shifts away from the z-axis and toward the y-axis. This occurs because some of the +1/2 nuclei are excited to the _1/2 state, and the precession about the z-axis becomes coherent (non-random), generating a significant y component to the net magnetization (M). The animation pauses at this stage.
• After irradiation the nuclear spins return to equilibrium in a process called relaxation. As the xy coherence disappears and the population of the +1/2 state increases, energy is released and detected by the receiver. The net magnetization spirals back, and eventually the equilibrium state is reestablished.

An inherent problem of the nmr experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where A is a proportionality constant. If the mole fractions of the spin states are equal (η+ = η- ) then the population difference is zero and no absorption will occur. If the rf energy used in an nmr experiment is too high this saturation of the higher spin state will result and useful signals will disappear.






Relaxation Mechanisms
For nmr spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy _1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H2) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). Nmr spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called Spin-Lattice Relaxation (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated T1, the spin-lattice relaxation time. In non-viscous liquids at room temperature T1 ranges from 0.1 to20 sec. A larger T1 indicates a slower or more inefficient spin relaxation.
Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T2. This process, which is actually a spin exchange, will not be discussed here.




Pulsed Fourier Transform Spectroscopy



In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. The drawing on the right depicts a set of four chimes, with the frequency of each designated by a colored sine wave. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified. The button beneath the drawing activates an animation showing this for one chime. Clicking on the illustration returns the original drawing.
An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized..



A CW nmr spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited.


The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds.
Because the relaxation mechanism is a first order process, the rf signal emitted by the sample decays exponentially. This is called a free induction decay signal, abbreviated FID. An example of a FID signal may be seen by .

Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the nmr signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as 13C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively.



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Examples of Anisotropy Influences on Chemical Shift
The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called paracyclophanes. The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a nmr shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region).




The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent (homotopic), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal.
A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the nmr spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an nmr signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts. By clicking on the above structure the dianion data will appear.

Compounds in which two or more benzene rings are fused together were described in an earlier section. Examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure.




Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the 1Hnmr spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned.

The much larger C48H24 fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges.



The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths.
If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds

The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution nmr spectra. These problems were eventually solved by using high boiling solvents, the 1Hnmr spectrum being taken at 150 to 200 C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence will be shown above by clicking on the diagram. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds.

It is important to understand that the shielding and deshielding terms used throught our discussion of relative chemical shifts are themselves relative. Indeed, compared to a hypothetical isolated proton, all the protons in a covalent compound are shielded by the electrons in nearby sigma and pi-bonds. Consequently, it would be more accurate to describe chemical shift differences in terms of the absolute shielding experienced by different groups of hydrogens. There is, in fact, good evidence that the anisotropy of neighboring C-H and C-C sigma bonds, together with that of the bond to the observed hydrogen, are the dominate shielding factors influencing chemical shifts. The anisotropy of pi-electron systems augments this sigma skeletal shielding.
Nevertheless, the deceptive focus on anisotropic pi-electron influences is so widely and commonly used that this view has been retained and employed in these pages.





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Hydrogen Bonding Influences
Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton nmr spectrum of pure (neat) methanol exhibits two signals, as expected. At 30 C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the nmr spectrum, which will be illustrated by clicking the "Cool the Sample" button, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here.







Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm.
The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.

1H Chemical Shifts of Methanol in Selected Solvents Solvent CDCl3 CD3COCD3 CD3SOCD3 CD3C≡N
CH3–O–H
CH3
O–H 3.40
1.10 3.31
3.12 3.16
4.01 3.28
2.16


The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH2 signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d6 solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the solvent give a small broad signal near δ 2.5 ppm.



For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d6 as a solvent, not only shifts this signal to a lower field, but permits 1-, 2 - & 3 -alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d6.


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Spin-Spin Coupling



The spin-spin interaction of neighboring hydrogens takes place through the covalent bonds that join them. The most common bonding relationship is vicinal (joined by three sigma bonds). In this case a neighboring proton having a +1/2 spin shifts the resonance frequency of the proton being observed to a slightly higher value (up to 7 Hz), and a _1/2 neighboring spin shifts it to a lower frequency. Remember that the total population of these two spin states is roughly equal, differing by only a few parts per million in a strong magnetic field. If several neighboring spins are present, their effect is additive.
In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. By clicking on one of these signals, the spin relationship that leads to the coupling pattern will be displayed. Clicking elsewhere in the picture will return the original spectrum.
The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all _1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz.

Multiplicity Relative Line Intensity
singlet
doublet
triplet
quartet
quintet

A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities.

Coupling constants are independent of the external magnetic field, and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH3CH2X or YCH2CH2X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied.



The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the Karplus equation shown below.





Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens. An example illustrating geminal couplings is shown elsewhere.




Spin Decoupling
We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a _1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such nmr experiments are possible, and are called spin decoupling.
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. By clicking on one of the three signals in the spectrum, the results of decoupling at that frequency will be displayed.




In this example, the nuclei being decoupled and the nuclei being observed by the spectrometer are of the same kind (both protons). We call this homonuclear decoupling. It is also possible to decouple different kinds of nuclei. For example, a compound having both hydrogen and fluorine as part of its molecular composition may exhibit spin-coupling between their nuclei, and one may be decoupled while the other is observed. This is termed heteronuclear decoupling. Heteronuclear decoupling is very important in 13C nmr spectroscopy.



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The Influence of Magnetic Field Strength
The presence of symmetrical, easily recognized first-order splitting patterns in a nmr spectrum depends on the relative chemical shifts of the spin-coupled nuclei and the magnitude of the coupling constant. If the chemical shift difference (i.e. Δδ in Hz) is large compared to J the splitting patterns will be nearly first order. If, on the other hand, the difference is relatively small (less than 10 J) second order distortion of the signal splitting will be observed. One important advantage in using very high field magnets for nmr is that the separation (or dipersion) of different sets of protons is proportional to field strength, whereas coupling constants do not change.
It is important to remember that structurally different sets of nuclei do not always produce distinctly different signals in an nmr spectrum. For example, the hydrocarbon octane has four different sets of protons, as shown in the following formula:


CH3CH2CH2CH2CH2CH2CH2CH3

Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many nmr spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an nmr spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement).



Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. Spectra of anisole will be displayed by clicking on the benzyl alcohol diagram. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored.


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An Example of Extensive Spin-Splitting
Not all simple compounds have simple proton nmr spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum.
The 100 MHz 1H nmr spectrum of a C3H5ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present. By clicking the "500 MHz Spectrum" button beneath the spectrum, a higher field spectrum will appear. Here it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. Clicking in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred. The structure and signal assignments for this compound will be disclosed by clicking the " Show Solution " button.










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Additional Information from 13C NMR Spectroscopy
Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon nmr spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum.
Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The 13C nmr spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented by clicking the "High Field Expansion" button. The two lowest field signals are missing in the expanded display.





Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by off-resonance decoupling. In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. By clicking the "Off-Resonance Decouple" button the results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors.
A better way for classifying the carbon signals is by a technique called INEPT (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on 13C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal. An INEPT spectrum of camphor will be displayed by clicking the "INEPT Spectrum" button.



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Properties of Some Deuterated NMR Solvents
Solvent B.P. C Residual
1H signal (δ) Residual
13C signal (δ)
acetone-d6 55.5 2.05 ppm 206 & 29.8 ppm
acetonitrile-d3 80.7 1.95 ppm 118 & 1.3 ppm
benzene-d6 79.1 7.16 ppm 128 ppm
chloroform-d 60.9 7.27 ppm 26.4 ppm
cyclohexane-d12 78.0 1.38 ppm 26.4 ppm
dichloromethane-d2 40.0 5.32 ppm 53.8 ppm
dimethylsulfoxide-d6 190 2.50 ppm 39.5 ppm
nitromethane-d3 100 4.33 ppm 62.8 ppm
pyridine-d5 114 7.19, 7.55 & 8.71 ppm 150, 135.5 & 123.5 ppm
tetrahydrofuran-d8 65.0 1.73 & 3.58 ppm 67.4 & 25.2 ppm
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