The correct order of decreasing reactivity towards nucleophilic addition of the following compounds are:
\(HCHO, C_6H_5CHO, CH_3COCH_3, C_2H_5 COC_2H_5\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(HCHO > C_6H_5CHO > CH_3COCH_3 > C_2H_5 COC_2H_5\)
In nucleophilic addition the nucleophile attacks the electrophilic carbonyl carbon. Aldehydes are more reactive than ketones for two reasons:
Steric: two relatively large groups in a ketone hinder the approach of the nucleophile more than the one group (and H) in an aldehyde.
Electronic: alkyl groups release electrons (+I) and reduce the positive charge on the carbonyl carbon; a ketone has two such groups.
Placing each compound, most reactive first:
HCHO — first. No alkyl group at all (two H atoms).
\( \ce{C6H5CHO} \) — second. An aldehyde, so more reactive than the ketones; the phenyl ring lowers its reactivity compared with HCHO by resonance, which makes the carbonyl carbon less electrophilic.
\( \ce{CH3COCH3} \) — third. A ketone with two small methyl groups.
\( \ce{C2H5COC2H5} \) — last. Two larger ethyl groups give more steric hindrance and a stronger +I effect.
NCERT (2026-27 edition):
Reactivity
