What is the correct order of reactivity of (I) HCHO, (II) CH3CHO and (III) C6H5CHO towards nucleophilic addition reactions?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → I > II > III
In a nucleophilic addition the nucleophile attacks the carbonyl carbon. The reaction is faster when the carbonyl carbon is more electrophilic and less hindered; alkyl groups and resonance with a benzene ring lower its reactivity.
(I) \( \ce{HCHO} \) — first. Two H atoms on the carbonyl carbon: no alkyl group to reduce its electrophilicity and the least crowding, so it is the most reactive.
(II) \( \ce{CH3CHO} \) — second. One \( \ce{CH3} \) group reduces the electrophilicity of the carbonyl carbon and adds some bulk.
(III) \( \ce{C6H5CHO} \) — third. The C=O group is in resonance with the benzene ring, which reduces its polarity; the carbonyl carbon is the least electrophilic, so benzaldehyde is the least reactive.
NCERT (2026-27 edition):
8.4 Chemical Reactions: (ii) Reactivity – “Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former.”
“… The polarity of the carbonyl group is reduced in benzaldehyde due to resonance … and hence it is less reactive than propanal.”