核磁共振部分习题及答案_2
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NMR problems 2nd
part
1.Below are the 1
H and 13
C NMR spectra of 2-hexanone (CH
3COCH
2CH
2CH
2CH
3).
Explain carefully how, using homonuclear and heteronuclear decoupling experiments,
you could assign the each of the resonances in the 1
H and 13
C spectra to which nuclei
give rise to them.
1
H NMR spectra of 2-hexanone
13
C NMR spectra of 2-hexanone
Answer This is a commonly encountered problem. You obtain a 1
H and a 13
C NMR
spectrum and need to assign each of the resonances in both 1
H and 13
C spectra.
It is generally straight-forward to assign the 1
H spectrum using a combination of
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direct inspection (characteristic chemical shifts and multiplicities) in combination
with homonuclear decoupling experiments. For 1
H spectra, homonuclear decoupling
gives you connectivity (which protons are on adjacent carbon atoms) because 1
H is
100% abundant. Note that homonuclear decoupling cannot be used for 13
C spectra
because 13
C is isotopically dilute there essentially no possibility that there will be 2
13
C nuclei adjacent to each other in an organic molecule.
For 2-hexanone, the 1
H spectrum contains 5 resonances. You would expect to see one
singlet (integral 3H) for the CH
3 group attached to the ketone (ie at C1). The CH
2
group adjacent to the ketone (ie at C3) would appear as a triplet (with coupling to the
adjacent CH
2. The CH
2 at C4 would appear as a multiplet (a triplet of triplets with
coupling to both the CH
2 at C3 and the CH
2 at C5). Likewise the CH
2 at C5 would
appear as a multiplet (a triplet of quartets with coupling to both the CH
2 at C4 and the
CH
3 at C6). The resonance of the CH
3 at C6 would be a triplet (integral 3H).
Intuitively, you would also expect the CH
3 and CH
2 groups adjacent to the carbonyl to
occur at low field (between 1.5 and 2.5 ppm) and the CH
3 at C6 to be at high field
(between 0.5 and 1 ppm). By inspection, you can assign the proton signals for C1
(δ 1.7), C3 (δ 1.9) and C6 (δ 0.8). The multiplet signals at δ1.1 and δ 1.3 must
belong protons at C4 and C5 and to distinguish these you would use homonuclear
decoupling experiments. If the protons at C3 (δ 1.9) are irradiated, the multiplet due
to protons at C4 would collapse from a triplet of triplets to a triplet and hence its shift
would be known. Similarly, if the protons at C6 (δ 0.8) were irradiated, the multiplet
due to protons at C5 would collapse from a triplet of quartets to a triplet and hence its
shift would be known.
Having assigned the 1
H spectrum, the 13
C spectrum can be assigned using selective
heteronuclear decoupling experiments.
The 13
C spectrum contains 6 resonances, the resonance due to the carbonyl carbon is
obvious from its shift. For the remaining 5 carbons you would expect to have signals
from 2 x CH
3 and 3 x CH
2 groups and in the absence of any 1
H decoupling these
would appear as 2 quartets and 3 triplets. In the heteronuclear decoupling experiment
you would irradiate each of the resonances in the 1
H spectrum and observe the 13
C
spectrum. As each of the 1
H signals is irradiated, the resonance of the 13
C coupled to it
would collapse to a singlet - the multiplicity of the other signals would remain
essentially unchanged.
The correlation of 1
H and 13
C NMR spectra can also be achieved using
two-dimensional NMR using a heteronuclear shift correlation (HSC) experiment.
2.The simple molecules below contain the NMR-active nuclei 14
N (I=1), 31
P (I=1/2),
1
H (I=1/2) and 13
C (I=1/2). Considering only the coupling constants through one bond
(1
J
ax) as being significant, construct diagrams which schematically represent the
splitting pattern you would see in :
1 The 1
H NMR spectrum of the ammonium ion [NH
4+
].
2 The 14
N NMR spectrum of the [NH
4+
].
3 The 13
C NMR spectrum of trimethylamine [(CH
3)
3N]
4 The 31
P NMR spectrum of phosphine [PH
3].
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5 The 1
H NMR spectrum of phosphine [PH
3].
6 The 13
C NMR spectrum of H
2P-CH
2-PH
2.
7 The 13
C NMR spectrum of [(PH
2)
3CH].
Answer This is simply an exercise in predicting the multiplicity observed using the
formula: multiplicity = 2nI + 1.
1 The 1
H NMR spectrum of CH
4 will be a singlet (no multiplicity) since the
molecule is tetrahedral in shape and all of the protons are equivalent. There is no
coupling from carbon since almost all C is 12
C and this is NMR silent.
2 The 13
C NMR spectrum of CH
4 is a quintet. The C is coupled to 4 equivalent
protons and the spin of 1
H is ½.
(2nI + 1) = (2 x 4 x ½) + 1 = 5
3 The 1
H NMR spectrum of NH
4+
will have 3 lines. The molecule is tetrahedral in
shape and all of the protons are equivalent and these will be coupled to 14
N (which
has a spin I = 1).
(2nI + 1) = (2 x 1 x 1) + 1 = 3
