Question

CHOICES: Resonance Form: RF A, RF B, RF C, or, RF A and B

The H NMR spectra of amides (1) have some interesting quirks. One quirk is that the N-H protons of primary and secondary amides (e.g. 2) are quite broad and appear at relatively high but variable chemical shifts (5-9 ppm). Another "quirk" is that identical N-R groups (e.g R2 groups in 3) can appear at different chemical shifts even though the same groups in a ketone analog (e.g. R2 groups in 3') are equivalent and therefore appear at identical chemical shift NR2 R- R. 2 R. 3" 1 amide N-H broad, 8 5-9 ppm different 8 possible same 8 For instance, the two methyl groups of N,N-dimethylformamide (below) appear as two singlets in the H NMR spectrum. нЬ CНCЗ На Нс CHb3 N,N-dimethylformamide Ha 11 10 ppm HSP-04-529 Which resonance form and corresponding explanation best justify why the methyl groups of dimethylformamide are inequivalent? Resonance form: [Select Explanation: [Select] өо -CH CH3 CHз HO H1 A в CH3 CH3 Which resonance form and corresponding explanation best justify why the methyl groups of dimethylformamide are inequivalent? Resonance form: [Select] Explanati Select] Contribution of RF A and RF B facilitates rotation around the C=O pi bond Donation of the N lone pair into the C=O pi system facilitates rotation around the C-N bond of the O=C-N system Donation of the N lone pair into the C-O pi system restricts rotation around the C-N bond of the O=C-N system A

Answer 1

Resonance form : RF C.

CH ch 3 H CH

Explanation.

Donation of the lone pair into C=O pi system restricts rotation around the C-N bond of the O=C-N system.

This restriction in rotation of C=N bond makes difference in the environment of two methyl groups.