Question

13. The generic resonance structures A and B can be drawn for all carboxylic acid derivatives, yet their contribution to the overall resonance hybrid differs for the derivatives. Explain why the contribution of resonance structure to the hybrid is: a. Poor for an acid chloride or bromide. b. Greater for an amide than an ester. 14. Listed are frequencies for the carbonyl stretching modes in each molecule's IR spectrum. Explain why one carbonyl absorbs IR at a higher frequency than the other. a. Acetyl chloride v= 1799 cm Ethyl acetate v= 1743 cm b. Butanoic acid v= 1717 cm Butanamide v= 1660 cm V = 1760 cm v = 1730 cm

Answer 1

DO RT Poot for acid chloride or bromide - Rh. f R -C R BY shows I effect because halogens are more elecbonegative. Moce elecbonegative atom does not gives lone pair & not form. double bond easily with carbonyl carbon. Amide 7 ester Rid_r slow, rto-R O-R Hore in amide nitrogen oxygon so it gives lone in ester is higher than form ion B. is less elecbonegative than pair easily - I effect amide so amide easily In case of carbonyl compounds various effects are present like inductive, Elsonance hyperconjugation & steric. Any effect which can cause conversion of double bond of carbonyl slightly to single bond can reduce / decrease carbonyl stretching frequonly while any effect which cause increase in double bond character can increases carbonyl stretching frequency

CLASSMATE Date Page (a Acetyl chloride: ar ligg cmt I effect group increases double bond character =O = 1743cmy I effect 17 17 cm) @ Butanoic acid on OLA + I as well as I effect Here HR effect of -on group is also present 6 Butanamide: cu 1660 cmt +R effect (Decreases double bond character) ③ u 1760 cmt R as well as - effect is present 173oomt o an che 1710 cent tr effed