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

Answer 1

The Correct Option is C

To determine the products of compounds in a basic medium with NaOH and heat, we examine their reaction mechanisms:

• Ethanal: Undergoes aldol condensation, forming 3-Hydroxybutanal, which can dehydrate to But-2-enal. Matches with (II).
• Methanal: Oxidizes to formic acid and reduces to methanol in basic conditions. Corresponds to (IV).
• Benzenecarbaldehyde: Exhibits disproportionation (Cannizzaro reaction) in basic medium, yielding Benzoic acid and Phenyl methanol. Matches with (I).
• Acetone: Self-condenses in basic medium to form 4-Hydroxy-4-methylpentan-2-one and 4-Methylpent-3-en-2-one. Associates with (III).

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

The correct pairings are A-II, B-IV, C-I, D-III.