Exams › JEE Advanced › Chemistry
Correct answer: Only B gives a white crystalline precipitate with NaHSO3; A does not react
NaHSO3 forms a bisulfite addition product with carbonyl compounds only when steric hindrance is low: aldehydes and methyl ketones (RCOCH3) react readily. Compound B (pentan-2-one, CH3-CO-CH2CH2CH3) is a methyl ketone and reacts with NaHSO3 to give a white crystalline precipitate. Compound A (pentan-3-one, CH3CH2-CO-CH2CH3) is a diethyl ketone; the two ethyl groups create too much steric bulk around the carbonyl, preventing NaHSO3 addition. So B gives the test, A does not.