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- (a) The NMR of I-bromopropane w ould have three sets of signals, whereas the NMR of 2-bromopropane would have only two sets (a septet and a doublet, the typical isopropyl pattern). (b) Each spectrum would have a methyl singlet at 8 2. The left structure would show an ethyl pattern (a 2H quartet and a 3H triplet) , whereas the ri ght structure would exhibit an isopropyl pattern (a IH septet and a 6H doublet) . (c ) The most obvious difference i s the c hemical shift of the CH3 singlet. In the compound on the left, the CH3 singlet would appear at 8 2. 1 , while the compound on the right would show the CH3 singlet at 8 3.8. Refer to the sol ution of 13-22 for the spectrum of the second compound. (d) The spl itting and integration for the peaks in these two compounds would be identical, so the chemical shift must make the difference. As described in text section 1 3-SB, the alkyne is not nearly as deshielding as a carbon y l , so the protons in pent-2-yne would be farther upfield than the protons in butan-2-one, by about O.S ppm. For example, the methyl on the carbonyl would appear near 8 2. 1 while the methyl on the alkyne would appear about 8 1 .7. 13-4 1 The multiplicity in the off-resonance decoupled spectrum is gi ven below each chemical shift: s = s i nglet; d = doublet; t = t ri plet; q = q uartet. It is often difficult to predict exact chemical shift values; your predictions should be in the ri ght vicinity. There should be no question about the multiplicity and DEPT spectra, however. (a) estimated q
.!! s t 8 1 70
c I 200 1 80 I I 1 60 I 1 40 I 80 1 00 1 20 Carbon - 1 3 8 ( ppm) I I I 60 I 40 I I I 20 o The DEPT-90 spectrum for ethyl acetate would have no peaks because there are no C H groups. DEPT-135 292 ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.
- Spring '10