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An Introduction to NMR Spectroscopy - DOC - DOC

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					                         An Introduction to NMR Spectroscopy

                            Shelby Feinberg and Steven Zumdahl

       Nuclear magnetic resonance spectroscopy (NMR) has risen to the same level of

importance as electronic and vibrational spectroscopy as a tool for studying molecular properties,

particularly structural properties. Although the following discussion of NMR specifically deals

with hydrogen atom nuclei in organic molecules, the principles described here apply to other

types of molecules as well. Many other types of nuclei ( 13C,      19
                                                                        F,   31
                                                                               P, etc.) have nuclear spins

and thus can be studied using NMR techniques.

       Certain nuclei, such as the hydrogen nucleus (but not carbon-12 or oxygen-16), have a

nuclear spin. The spinning nucleus generates a small magnetic field termed . When placed in a

strong external magnetic field, called H o, the nucleus can exist in two distinct spin states: a low

energy state A, in which  is aligned with the external magnetic field, H o , and a high energy state

B, in which  is opposed to the external magnetic field, Ho (figure 1). Alignment of  with Ho is

the more stable and lower energy state.

                       Ho                                                         Ho


Figure 1                         




                                                            

                 Spin + 1/2                                          Spin  1/2
                  Parallel                                          Anti-parallel
                     A                                                   B
                Low Energy                                          High Energy



       In NMR, transitions from the more stable alignment, A, (with the field) to the less stable

alignment, B, (against the field) occur when the nucleus absorbs electromagnetic energy that is
exactly equal to the energy separation between the states (E). This amount of energy is usually

found in the radiofrequency range. The condition for absorption of energy is called the condition

of resonance. It can be calculated as the following:

                                                       γh
                                              ΔE         H  hυ
                                                       2π

       h = Planck’s constant

       H = the strength of the applied magnetic field, H o , at the nucleus

        = the gyromagnetic ratio (a constant that is characteristic of a particular nucleus)

        = the frequency of the electromagnetic energy absorbed that causes the change in
               spin states

       There are three features of NMR spectra that we will focus on: the number and size of

signals, the chemical shift, and spin-spin coupling.



Number and Size of Signals

       Let’s consider how the NMR spectrometer can distinguish between hydrogen nuclei and

produce multiple signals. Magnetically equivalent hydrogen nuclei produce one signal. These

hydrogen nuclei experience the same local environment. For example, in a molecule su ch as

diethyl ether (Figure 2), there are two sets of magnetically equivalent hydrogens. The hydrogens

labeled a are six magnetically equivalent methyl hydrogens, while the hydrogens labeled b are

four magnetically equivalent methylene hydrogens. Notice that the methyl (a) hydrogens are all

located adjacent to a carbon containing two hydrogen atoms. Additionally, the methylene (b)

hydrogens are all located adjacent to an oxygen atom and a carbon atom containing three

hydrogen atoms.

Figure 2
                          CH3     CH2     O      CH2       CH3
                            a      b              b        a
Because of rapid rotations about sigma bonds and molecular symmetry, the six methyl hydrogens

(a) and the four methylene hydrogens (b) comprise two individual magnetically equivalent

groups of hydrogens. The methyl hydrogens (a) experience a different total magnetic field than

the methylene hydrogens (b), because of different local magnetic fields.           As a result, the

resonance energy, E, corresponding to the frequency of absorption, , will be different for these

two groups of hydrogen nuclei. Thus, the NMR spectrometer can distinguish between groups of

hydrogen nuclei which experience different local magnetic fields. Different frequencies are

required for the two different groups of hydrogens, thereby producing two distinct signals.

         The areas of the signals are directly proportional to the number of hydrogens. So, in the

case of diethyl ether, the two peaks have an area ratio of 3:2 (or 6:4), because there are six

methyl hydrogens (a) and four methylene hydrogens (b). We will look at an actual spectrum

presently.

The Chemical Shift

         When an organic molecule is placed in an external magnetic field, H o , each hydrogen

nucleus experiences a total field that is the sum of H o and two other local magnetic fields: one

produced by bonding and non-bonding electrons, He, and the other produced by neighboring

protons which possess a nuclear spin, H h. Thus, the applied magnetic field (H o ) is constant,

while the magnetic fields produced as H e and H h are not constant. The magnetic field produced

by the neighboring electrons (He) determines the position (relative frequency) of an NMR peak

(called the chemical shift). The magnetic fields produced by neighboring nuclei (H h) are smaller

than He and cause splitting of an NMR peak (called spin-spin coupling). This will be discussed

later.

         For the moment, let’s assume that H h is zero. There is no spin-spin coupling. Under

these conditions, the total magnetic field, H, experienced by a particular nucleus is given by Ho +

He. That is, we are assuming that the resonance energy is only dependent on the sum of the
external magnetic field and the magnetic field produced by neighboring electrons. A mathe-

matical representation of the value of this resonance energy is the following:

                                                     γh
                                              ΔE       (Ho  He)  hυ
                                                     2π

If we vary the frequency () of the electromagnetic energy until h = E, absorption will cause a

transition in spin states, and a signal will be recorded by the NMR spectrometer.

       The position of an NMR signal is recorded relative to the position of the signal of an

internal standard. This standard is commonly tetramethyl silane (TMS), (CH 3)4Si. If TMS

absorbs at frequency s, and the hydrogen nucleus of interest absorbs at , the spectrometer will

                           ( υ  υs )
record a signal at                     10 6  δ , where  o is the spectrometer frequency (commonly 60
                               υo

megacycles per second).  is called the position of absorption, or the chemical shift, expressed

as parts per million (ppm) on a scale of 0-10 ppm. TMS absorbs at 0.0 ppm, while most nuclei

absorb downfield towards 10 ppm. The value of  is independent of Ho , but it DOES depend on

He. Figure 3 gives a few examples of how neighboring atoms affect the chemical shifts of

hydrogen atoms.

Figure 3

                                     H        H                H    H          H   H       H
  O               O
                                              C                  
                                                               C   C  C   C   C
RCOH            RCH
                                              C                O     X         N   C           (satd.)

                                                                                   C

                                                                                   C (or O)              TMS


10         9           8         7        6           5         4          3           2            1          0 (ppm)

           Downfield                       Ho                            Upfield

               Deshielding                                      Shielding
       Hydrogen nuclei which absorb at large  are said to be deshielded from the external

magnetic field. These signals appear downfield (towards 10 ppm) from TMS because the

frequency at which they absorb differs greatly from the frequency of the TMS hydrogens. As a

result, the value of   s is large. Deshielding is caused by adjacent atoms which are strongly

electronegative (e.g. oxygen, nitrogen, halogen) or groups of atoms which possess -electron

clouds (e.g. C=O, C=C, aromatics).

       Let’s now look at the NMR spectrum of benzyl acetate (Figure 4).

Figure 4


                                O

                    CH2 O       C    CH3
                     b                a
                                                             c
           c

                                                                                   a
                                                                       b

                                                                                            TM S

                                                   600               300                0




Note that there are three absorptions corresponding to the three sets of magnetically equivalent

hydrogens. Compare the chemical shifts with those given in Figure 3. The relative area of the

peaks corresponds to the number of hydrogens in each set (3:2:5).



Spin-Spin Coupling

       We have discussed how the position of the NMR absorption is affected by the magnetic

field produced by neighboring electrons (H e). Now, let’s examine how the magnetic field

produced by neighboring protons (H h) affects the splitting of the NMR absorption. The peaks
can appear as singlets, doublets, triplets, etc… Let’s first consider the absorption of a hydrogen

atom (H x) with only one neighboring hydrogen atom (H y) (Figure 5).

Figure 5


       *Hx =

                *Hx             Hy              Hy =    = A = Ho + He + Hh (low energy)

                 C              C
                                                Hy =    = B = Ho + He + Hh (high energy)
    Ho




nmr spectrum for *Hx =                       intensity ratio = 1:1

If we assume that H x is aligned with Ho , then the neighboring hydrogen nucleus will have

approximately equal probability of existing in either the low energy state, A, or the high energy

state, B. Refer to Figure 1 for a clear picture of these energy states. For those molecules in

which the neighboring hydrogen nucleus exists in the low energy state, its magnetic field (H h)

will add to the magnetic field (Ho + He) and for those molecules in which the neighboring

hydrogen nucleus exists in the high energy state, its magnetic field (H h) will subtract from the

magnetic field (Ho + He). As a result, two different peaks are observed. A higher frequency is

required for the low energy state, while a lower frequency is required for the high energy state.

This two line pattern is called a doublet, and we say that the neighboring hydrogen nucleus (H y)

splits the absorption of H x into a doublet. The intensity of the two lines will be equal, because

the probability of the neighboring hydrogen nucleus (H y) existing in either spin state A or B is

approximately equal.

         When there are two magnetically equivalent adjacent hydrogen nuclei, three possibilities

exist for their combined magnetic fields (Figure 6).
Figure 6


         *Hx =
                                                      Hy =           (low energy)
                 *Hx               Hy
                                                      Hy =           (middle energy)
                  C                C      Hy
                                                      Hy =           (middle energy)
    Ho
                                                      Hy =           (high energy)




nmr spectrum for *Hx =                            intensity ratio = 1:2:1

Note that both can have spin state A, one can have spin state A and the other have spin state B,

or both can have spin state B. These three possibilities have a probability ratio of 1:2:1, which

correspond to three different absorption frequencies. The appearance of the resulting NMR

signal is a three line pattern, called a triplet, with intensities 1:2:1.

         When there are three magnetically equivalent neighboring hydrogen nuclei (Figure 7), the

absorption splits into a quartet with intensity pattern 1:3:3:1 for similar reasons to those

aforementioned.

Figure 7


       *Hx =
                                                    Hy =              (very low energy)
                 *Hx              Hy
                                                    Hy =                            (med. low energy)
                 C                C      Hy
                                                    Hy =                            (med. high energy)
    Ho                            Hy
                                                    Hy =              (very high energy)




nmr spectrum for *Hx =                          intensity ratio = 1:3:3:1
       In general, N magnetically equivalent neighboring hydrogen nuclei will split the

absorption into N + 1 lines. The intensity ratios are equal to the coefficients of the expansion (a

+ b)N, or pascal’s triangle.

       The spacing between the lines of a doublet, triplet, or quartet is dependent on the

coupling constant. It is given the symbol J, and it is measured in units of cycles per second.

Coupling constants range in magnitude from 0 to 20 cps. Observable coupling will generally

occur between hydrogen nuclei which are separated by no more than three sigma bonds. For

example, in the molecule 2-butanone (Figure 8), hydrogens from groups a and b split one

another because they are only separated by three bonds. However, hydrogens from group c are

not split by any other hydrogens because they are separated from other hydrogens by more than

three sigma bonds.



Figure 8


                O

      CH3CH2CCH3
       a    b       c




                                                                                    TM S

                                         600                300                 0




       Examples of typical coupling constants are given in Figure 9.
Figure 9

     H                             H
            C       12-15 cps          C     C           13-18 cps   H   C    O     H      0 cps
     H                                               H

      H         H
                                   H                                               H

      C         C        6-8 cps       C     C           0-3 cps     H   C    C          1-3 cps
                                   H                                               O
                                        H
 H                  H                            H

      C         C       7-12 cps                         6-9 cps




          Note that coupling is never observed between magnetically equivalent hydrogen atoms.

          Let’s now consider the coupling of hydroxylic protons (OH). Hydroxylic protons do

not split other protons, nor are they split by other protons. These protons undergo exchange

between two molecules of alcohol. This exchange is so fast that these protons sample molecules

with all possible spin states, giving rise to a singlet in the NMR spectrum at a position which is

the average for all the spin states of the other hydrogens.

          For example, ethanol gives an NMR spectrum consisting of a triplet, quartet, and a

singlet, as seen in the idealized spectrum in Figure 10. Even though the hydroxylic proton is

separated from a methylene hydrogen by three bonds, the signal given by the hydroxylic proton

is not split by the methylene protons, nor are the methylene hydrogens split by the hydroxylic

proton because of proton exchange.

Figure 10




     CH3CH2OH
        Now that you’ve learned some of the main principles necessary to examine NMR spectra,

try to draw an idealized spectra for each of the structures given below.



1.            O    CH3
                                                      5        O


  H3C         C    C       CH3                         H3C     C     O        CH2       CH2       CH3


                   CH3
                                                                   n-propyl acetate
        tert-butyl-acetate
                                                                                              O
                                                                         O          CH3
2.        O                         O                 6.                                      COCH2CH3
                                                             CH3CH2OC

 H3C       C      CH2       CH2     C      CH3

                                                                    CH3             N         CH3
              acetonylacetone

                                                      diethyl 2,4,6-trimethyl-3,5-pyridinedicarboxylate
3.
         O                 CH3             O

 HO      C        CH2      C      CH2      C     OH
                                                       7.             H3C        CH2      NH2
                           CH3

         3,3-dimethylglutaric acid                                           ethylamine


                        CH2CH3



4.


     CH3CH2                       CH2CH3


           1,3,5-triethylbenzene

				
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