Answers
Revision Questions #4

| Table of Contents | Return to Question Sheet |

1.

The signals in the spectrum can be readily assigned: d 9.2 is due to the proton of the aldehyde group, d 2.2 is due to the CH2 group adjacent to the aldehyde; d 1.6 is due to the CH of the isopropyl group and d 0.9 is due to the CH3 groups). The resonance at d 9.2 is a triplet because it is coupled to 2 protons on the adjacent carbon. The resonance at d 2.2 is a doublet of doublets because it is coupled to the aldehyde proton as well as the methine proton of the isopropyl group. The resonance at d 1.6 is a triplet of septets because it is coupled to the 2 protons at d 2.2 as well as the 6 equivalent protons in the isopropyl group. The resonance at d 0.9 is a doublet due to coupling with the methine proton at d 1.6. The effect of irradiation at each of these frequencies is to decouple the nuclei which have signals at these frequencies. Decoupling effectively removes the nucleus from the spin system.

i)	With irradiation at d 9.2, the multiplicity of the CH2 resonance at d 2.2 is simplified from doublet of doublets to a doublet. The remaining resonances are unchanged.
ii)	With irradiation at d 2.2, the multiplicity of the aldehyde resonance at d 9.2 is simplified to a singlet and the methine resonance at d 1.6 is simplified to a septet. The remaining resonance at d 0.9 is unchanged.
iii)	With irradiation at d 1.6, the multiplicity of the CH2 resonance at d 2.2 is simplified from a doublet of doublets to a doublet. The methyl resonances at d 0.9 collapse from a doublet to a singlet. The remaining resonances are unchanged.
iv)	With irradiation at d 0.9, the methine resonance at d 1.6 is simplified to a doublet. The remaining resonances are unchanged.

2.

The geometry of trimethyl phosphite is tetrahedral at phosphorus. All three of the CH3O groups are equivalent. There would be 1 resonance in the 13C NMR spectrum, 1 resonance in the 1H NMR spectrum and 1 resonance in the 31P NMR spectrum. Note that 31P is spin ½ and 100% abundant, 1H is spin ½ and effectively 100% abundant and 13C is spin ½ but only about 1% abundant.

i)	The 13C NMR spectrum is a doublet of quartets. The quartet splitting arises from the 3 protons directly attached to the carbon. The doublet splitting must arise from the 2-bond coupling between C and P. The 1H NMR spectrum is a doublet and the splitting must arise from the 3-bond coupling between the protons and phosphorus.
ii)	If the 1H NMR spectrum was irradiated with strong Rf radiation (ie decoupled), the quartet splitting would disappear in the 13C spectrum so the 13C spectrum would appear as a doublet.
iii)	If the 31P NMR spectrum was irradiated with strong Rf radiation (ie decoupled), the doublet splitting would disappear in the 1H spectrum so the 1H resonance would appear as a singlet.
iv)	If the 31P NMR spectrum was irradiated with strong Rf radiation (ie decoupled), the doublet splitting would disappear in the 13C spectrum so the 13C spectrum would appear as a quartet.
v)	If the 13C NMR spectrum was irradiated with strong Rf radiation (ie decoupled), there would be almost no effect on the 1H NMR spectrum. Almost all of the protons in the sample would be attached to the NMR-silent nucleus 12C (not 13C) because the natural abundance of 13C is only about 1%. If you could see the (small) 13C-satellites in the spectrum, the 13C splitting in these would be removed.

3.

Tetramethyllead [Me4Pb] is a tetrahedral molecule so all 4 of the methyl groups are equivalent. There will be only one resonance in the 1H NMR spectrum and this will be split by coupling to the Pb nucleus. Lead has 4 isotopes 204Pb (I=0, 1.5%), 206Pb (I=0, 24%), 207Pb (I=1/2, 22%) and 208Pb (I= 0, 52%). So for the 22% of molecules that contain 207Pb, the 1H resonance will be split into a doublet by coupling to 207Pb. For the remainder of the molecules that contain 204Pb, 206Pb and 208Pb the 1H resonance will not be split by coupling to the Pb nuclei and so will occur as a singlet. The observed spectrum will be the superposition of the contributing sub-spectra - one singlet (about 76%) and one doublet (splitting 60 Hz and about 24% of the signal intensity).

4.

This is a commonly encountered problem. You obtain a 1H and a 13C NMR spectrum and need to assign each of the resonances in both 1H and 13C spectra. It is generally straight-forward to assign the 1H spectrum using a combination of direct inspection (characteristic chemical shifts and multiplicities) in combination with homonuclear decoupling experiments. For 1H spectra, homonuclear decoupling gives you connectivity (which protons are on adjacent carbon atoms) because 1H is 100% abundant. Note that homonuclear decoupling cannot be used for 13C spectra because 13C is isotopically dilute there essentially no possibility that there will be 2 13C nuclei adjacent to each other in an organic molecule. For 2-hexanone, the 1H spectrum contains 5 resonances. You would expect to see one singlet (integral 3H) for the CH3 group attached to the ketone (ie at C1). The CH2 group adjacent to the ketone (ie at C3) would appear as a triplet (with coupling to the adjacent CH2. The CH2 at C4 would appear as a multiplet (a triplet of triplets with coupling to both the CH2 at C3 and the CH2 at C5). Likewise the CH2 at C5 would appear as a multiplet (a triplet of quartets with coupling to both the CH2 at C4 and the CH3 at C6). The resonance of the CH3 at C6 would be a triplet (integral 3H). Intuitively, you would also expect the CH3 and CH2 groups adjacent to the carbonyl to occur at low field (between 1.5 and 2.5 ppm) and the CH3 at C6 to be at high field (between 0.5 and 1 ppm). By inspection, you can assign the proton signals for C1 (d 1.7), C3 (d 1.9) and C6 (d 0.8). The multiplet signals at d 1.1 and d 1.3 must belong protons at C4 and C5 and to distinguish these you would use homonuclear decoupling experiments. If the protons at C3 (d 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 (d 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 1H spectrum, the 13C spectrum can be assigned using selective heteronuclear decoupling experiments. The 13C 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 CH3 and 3 x CH2 groups and in the absence of any 1H decoupling these would appear as 2 quartets and 3 triplets. In the heteronuclear decoupling experiment you would irradiate each of the resonances in the 1H spectrum and observe the 13C spectrum. As each of the 1H signals is irradiated, the resonance of the 13C coupled to it would collapse to a singlet - the multiplicity of the other signals would remain essentially unchanged. The correlation of 1H and 13C NMR spectra can also be achieved using two-dimensional NMR using a heteronuclear shift correlation (HSC) experiment.

| Table of Contents | Return to Question Sheet |

School of Chemistry
University of Sydney