Minggu, 11 Oktober 2009

1H-NMR Spectroscopy

Introduction

The language of organic chemistry is highly symbolic. Organic chemists use pictures such as Lewis structures to describe molecules. All organic chemists have picture of molecules in their heads. A picture of acetone might look like this:

But where did that picture come from? How do chemists know what a molecule looks like? There is no single answer to these questions. Chemists have developed their insights into molecular structure from many sources. However, no source has proven more insightful than spectroscopy, especially nuclear magnetic resonance (NMR) spectroscopy. In this topic we will briefly examine the theory of NMR. At this level that theory comprises three fundamental components, the chemical shift, integration, and spin-spin coupling. The information that each of these components provides is like a piece to puzzle. It is the job of the spectroscopist to put those pieces together. The discussion that follows focuses on proton NMR, abbreviated 1H-NMR. The principles presented apply equally well to other magnetic nuclei.

1H-NMR Spectroscopy

Like all spectroscopic methods, NMR spectroscopy involves the interaction of electromagnetic radiation with matter. Any atom whose nucleus contains an odd number of protons and/or neutrons behaves like a tiny bar magnet. Common nuclei that display this behavior include 1H, 2H, 13C, 15N, and 19F. Nuclei which contain an even number of protons and neutrons are non-magnetic and are not NMR active.


Exercise 1 The table below provides information about several different nuclei. Complete the table.
Nucleus
Isotope
% Natural Abundance
Protons
Neutrons
NMR Active
Hydrogen
1H
99.985
1
0
Yes
Hydrogen
2H
0.015
1
1
Yes
Boron
11B
80
Carbon
12C
99
6
6
No
Carbon
13C
1
Nitrogen
14N
99.63
Nitrogen
15N
0.37
Oxygen
16O
99.759
Oxygen
17O
0.037
Fluorine
19F
100
Silicon
28Si
92.2
Phosphorous
31P
100
Sulfur
32S
95
Chlorine
35Cl
75.5


Bar magnets have magnetic moments, which are analogous to dipole moments in chemical bonds. Figure 1 compares these two phenomena.

Figure 1

Let's Take a Moment

The magnetic moment associated with a single nucleus is extremely small. In the absence of an external magnetic field, the magnetic moments of a collection of nuclei are randomly oriented and all the nuclei have the same energy. However, when a sample is placed in an external magnetic field, Bo, the magnetic moments of those nuclei adopt specific orientations with respect to the applied field. In the case of 1H nuclei only two orientations are allowed; the nuclear magnetic moments may be aligned with or aligned against the direction of the applied magnetic field. Furthermore, the energy of those nuclei whose magnetic moments are aligned with the applied field is less than that of those whose nuclei are aligned against the field. The lower energy state is labeled a while the higher is designated b. The difference in energy between the two states increases as the strength of the applied magnetic field increases as shown in Figure 2.

Figure 2

Spread 'em

A fundamental equation of spectroscopy is DE = hn, where DE represents the difference in energy between two states of a system, n symbolizes frequency of electromagnetic radiation, and h is a proportionality constant. This equation says that a sample will absorb electromagnetic radiation when the frequency of that radiation matches the difference in energy between two energy states of the system. For NMR spectroscopy the frequencies of interest are in the radio frequency (RF) range, typically 60-500 MHz depending upon the strength of Bo. This is in the radio frequency range and an RF transmitter is the source of the electromagnetic radiation. The condition where DE = hn is referred to as resonance. The absorption of energy creates an excited state of the system. The process whereby the system returns to its lowest energy state, i.e. its ground state, is called relaxation. One way for the system to relax to the ground state is for it to emit radiation. If a suitable detector is available, e.g. an RF receiver, the emitted radiation may be recorded as a peak on a graph. Figure 3 provides a schematic diagram of the apparatus that is required for NMR spectroscopy.

Figure 4 animates the processes that occur during an NMR experiment.

Figure 4

A Simple NMR Experiment

Chemical Shifts

During our discussion of polarity, we considered the 1H-NMR spectra of several compounds with the general formula CH3X. Figure 5 reiterates some of that data for CH3F, CH3Cl, CH3Br, and CH3I. You should recall that an NMR spectum is a plot of signal intensity (Y-axis) as a function of the frequency of emitted radiation (X-axis).

Figure 5

Simple 1H-NMR Spectra

The spectrum of each methyl halide contains a single peak since the three hydrogen atoms of a methyl group are identical. However the frequency at which each peak occurs depends upon the halogen atom that is attached to the carbon. The change in frequency is called the chemical shift. There is an inverse correlation between chemical shift and the electron density around the hydrogen atoms absorbing (and emitting) the electromagnetic radiation; the higher the electron density, the lower its chemical shift value.

To appreciate the origins of chemical shifts you must understand that the resonance frequency for a given hydrogen depends upon the effective magnetic field strength, Beff, experienced by that hydrogen. There is a magnetic field associated with an electron just as there is with a proton, except that the direction of the magnetic moment is opposite to that of the proton. Recall that magnetic moments are vector quantities. Hence the interaction between the magnetic field associated with an electron, Be, and the applied magnetic field reduces the magnitude of Bo as shown in Figure 6. The magnitude of Beff, therefore, depends upon the electron density around the hydrogen. Since Be reduces the magnitude of Bo, electrons are said to shield protons from the applied magnetic field. The higher the electron density around a hydrogen atom, the greater the shielding, and the smaller the chemical shift.

Figure 6

Keep Up Your Guard

Figure 6 grossly exaggerates the magnitude of Be. For a magnetic field strength, Bo, of 1.90 Tesla, DE equals 100 MHz (100,000,000 Hz or 100, 000,000 cycles/second). The range of proton chemical shifts caused by electronic shielding is approximately 2,000 Hz. In other words, Beff = 100,000,000 ± 1000 Hz. In order to avoid dealing with large numbers such as 100,000,500, chemists developed a chemical shift scale in which the RF frequency is expressed as a fraction of the absolute frequency. For example, 500/100,000,000 = 5/1,000,000. The fraction 5/1,000,000 is read as 5 parts per million or 5 ppm. When expressed this way the chemical shift axis is labeled d, ppm.


Exercise 2 Express the following chemical shifts in ppm.

a. 850/100,000,000 = ppm b. 300/60,000,000 = c. 600/300,000,000 =


Figure 7 indicates the different ways in which chemists view the d scale of an NMR spectrum.

Figure 7

A Vocabulary Lesson

Bottom line: The chemical shift of a peak in an NMR spectrum tells you something about the electronic environment in the vicinity of the atom(s) that give rise to that peak. Tables correlating chemical shifts with structure are readily available.

Integration

The 1H-NMR spectrum of chloromethyl methyl ether, ClCH2OCH3, contains two peaks as shown in Figure 8. Note that the peaks are not the same size. The ratio of the area of the peak at 3.5 ppm to the peak at 5.5 ppm is 1.5/1 or 3/2. In other words, the area of each peak is proportional to the number of hydogens absorbing the electromagnetic radiation of a particular frequency. NMR spectrometers are equipped with automatic integrators to measure peak areas.

Figure 8

Everything is Relative

Bottom Line: The integration of an NMR spectrum tells you the relative numbers of hydrogen atoms that give rise to each peak.

Spin-Spin Coupling

Consider the molecular fragment HA-C-C-HX, where the subscripts A and X indicate that the electronic environment around HA is very different than that around HX, i.e. the chemical shifts of these two hydrogens are very different. We have seen that the application of an external magnetic field causes the magnetic moments of a collection of hydrogen nuclei to adopt one of two orientations with respect to the magnetic moment of the applied field. In the case of a molecule containing an HA-C-C-HX fragment such a magnetic field generates four spin states. In the lowest energy state some of the HA and some of the HX nuclei have their spins aligned with the applied field. In the highest energy state some of the HA and some of the HX nuclei have their spins aligned against the applied field. There are two states of intermediate energy. In one, some of the HA nuclei have their spins aligned with the applied field while some of the HX nuclei have their spins aligned against the applied field. In the other some of the HA nuclei have their spins aligned against the applied field while some of the HX nuclei have their spins aligned with the applied field. Figure 9 diagrams this situation.

Figure 9

This Will Make Your Head Spin

As you can see from the figure, there are two transitions from an a spin state to a b spin state involving HA nuclei and two transitions from a to b involving HX nuclei. Consequently the signals arising from these two types of hydrogens appear as two lines. The two lines comprise a single signal that is called a doublet. The spectrum shown in Figure 9 contains two signals, both doublets. The difference in the frequency of the two HA transitions is the same as that for the two HX transitions. This difference is called the coupling constant, J. The magnitude of J typically ranges from 0 to approximately 15 Hz. In the vernacular of the NMR spectroscopist HA is coupled to HX with a coupling constant of J Hz. The coupling is called spin-spin coupling. The convention for describing spin-spin coupling in a fragment like HA-C-C-HX is 3JHH, where the superscript 3 indicates that the coupling occurs through 3 bonds and the subscript HH says that it is between two hydrogen nuclei.


Exercise 3 Draw the structures of three molecules that contain the molecular fragment HA-C-C-HX.

Exercise 4 Draw the molecular fragments implied by the following coupling constants: a. 2JHH b. 4JHH c. 1JCH d. 3JHF


Before proceeding we need to emphasize two points. The first is that spin-spin coupling arises because a hydrogen atom attached to a carbon can "sense" the magnetic state of hydrogen atoms attached adjacent carbons. This information is transmitted through sigma bonds. It is a short range effect. The magnitude of JAX generally drops to zero when there are more than 3 sigma bonds separating A and X. In other words, spin-spin coupling between two nuclei requires that those nuclei be attached to adjacent atoms. Thus HA and HX are not coupled in the molecular fragment HA-C-C-C-HX. The molecule chloromethyl methyl ether contains the molecular fragment HA-C-O-C-HX, and, as Figure 8 demonstrates, there is no spin-spin coupling between HAand HX.

The n+1 Rule

Spin-spin coupling arises from the interactions of nuclear spin states. Various combinations of spin states are possible depending upon the number of interacting nuclei. Figure 10 presents some common molecular fragments and their associated spin-spin coupling patterns.

Figure 10

More Pattern Recognition

Note that the number of lines in each blue signal is one more than the number of red hydrogens. Similarly, the number of lines in each red signal is one more than the number of blue hydrogens. The number of lines in a signal is called the multiplicity of the signal. In the ideal case, the multiplicity of a signal arising from a set of hydrogen atoms is one more than the number of hydrogen atoms 3 bonds away.

Bottom line: When there are n hydrogen atoms separated by 3 bonds from a set of hydrogen atoms that gives a signal, the multiplicity of the signal will equal n+1.


Exercise 5 Draw structures of two molecules that fit each pattern A-D in Figure 10.

Exercise 6 Which of the following compounds would produce an NMR spectrum that includes spin-spin splitting pattern B?

A
B
C
D
E

Exercise 7 Select the compound that is most consistent with the following data from the alternative structures shown below.

a. b. c.

The structure most consistent with the data in problem a is number .

The structure most consistent with the data in problem b is number .

The structure most consistent with the data in problem c is number .

If you'd like to learn more about spin-spin coupling and the interpretation of NMR spectra, take a look at the HNMR Tutorial. After you've looked at the tutorial, you might want to try some Practice Problems.

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