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The CORRECT order of carbonyl stretching frequencies for the following compounds is

(Structures: I is benzoyl chloride, C6H5COCl; II is phenyl acetate, CH3COOC6H5; III is N,N-dimethylbenzamide, C6H5CON(CH3)2; IV is benzaldehyde, C6H5CHO.)

  • A.II \(\lt\) I \(\lt\) III \(\lt\) IV
  • B.I \(\lt\) III \(\lt\) II \(\lt\) IV
  • C.IV \(\lt\) II \(\lt\) III \(\lt\) I
  • D.III \(\lt\) IV \(\lt\) II \(\lt\) I

Answer: D

The order of C=O stretching frequency is III < IV < II < I.

The frequency rises with the double-bond character of the carbonyl. Groups that donate electrons by resonance lower it; electronegative groups that withdraw inductively raise it.

  • III, N,N-dimethylbenzamide: the nitrogen lone pair donates strongly into C=O, giving it much single-bond character, so it absorbs lowest (about 1630-1650 cm−1).
  • IV, benzaldehyde: conjugation with the ring lowers it to about 1700 cm−1.
  • II, phenyl acetate: the ester oxygen's lone pair is partly drawn into the ring, so its inductive pull wins and the band rises to about 1765 cm−1.
  • I, benzoyl chloride: chlorine withdraws strongly and donates poorly, giving the highest frequency (about 1775 cm−1).

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