Question

Match List-I with List-II:

List-I Compound		List-II Product in Basic Medium (in NaOH + Heat)
A	Ethanal	(I)	Benzoic acid + Phenyl methanol
B	Methanal	(II)	3-Hydroxybutanal + But-2-enal
C	Benzenecarbaldehyde	(III)	4-Hydroxy-4-methylpentan-2-one + 4-Methylpent-3-en-2-one
D	Acetone	(IV)	Formic acid + Methanol

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-II, B-IV, C-I, D-III
• C. A-I, B-II, C-IV, D-III
• D. A-III, B-IV, C-I, D-II

Answer 1

The Correct Option is C

Aldehydes and ketones undergo different reactions in basic medium: Ethanal undergoes aldol condensation to give 3-Hydroxybutanal and But-2-enal. Methanal reacts to form formic acid and methanol. Benzenecarbaldehyde forms benzoic acid and phenylmethanol. Acetone undergoes a reaction to give 4-Hydroxy-4-methylpentan-2-one and 4-Methylpent-3-en-2-one.