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 solve the problem of matching compounds with their products in a basic medium, we need to understand the reaction mechanisms involved. The compounds listed in List-I undergo specific reactions when treated with NaOH and heat:

• Ethanal: This compound undergoes an aldol condensation reaction in a basic medium, leading to the formation of 3-Hydroxybutanal, which can dehydrate to form But-2-enal. Therefore, Ethanal matches with (II).
• Methanal: Under basic conditions, methanal oxidizes to formic acid and reduces to form methanol. Hence, it corresponds to (IV).
• Benzenecarbaldehyde: Given its structure, it undergoes disproportionation (Cannizzaro reaction) in the basic medium to yield Benzoic acid and Phenyl methanol. This matches with (I).
• Acetone: In basic medium, acetone undergoes self-condensation to form products like 4-Hydroxy-4-methylpentan-2-one and 4-Methylpent-3-en-2-one. Hence, it 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

Therefore, the correct matches are A-II, B-IV, C-I, D-III.

Answer 2

Aldehydes and ketones undergo different reactions in basic medium, often leading to the formation of various products. These reactions are typically nucleophilic, and the nature of the product depends on the structure of the starting compound and the reaction conditions.

Ethanal (acetaldehyde) undergoes aldol condensation in basic medium. This reaction involves the enolate ion formed from ethanal, which attacks another molecule of ethanal. The result is the formation of 3-Hydroxybutanal (an aldol product) and But-2-enal (the α,β-unsaturated carbonyl compound), depending on the reaction conditions.

Methanal (formaldehyde) reacts in basic medium to form formic acid and methanol. The reaction proceeds through the oxidation of methanal, where it undergoes nucleophilic attack by hydroxide ions, eventually forming the carboxylate ion of formic acid.

Benzenecarbaldehyde (benzaldehyde) undergoes oxidation in a basic medium to give benzoic acid and phenylmethanol. The aldehyde group of benzenecarbaldehyde is oxidized, leading to the formation of benzoic acid, while the remaining carbonyl group can lead to the production of phenylmethanol through reduction.

Acetone undergoes a reaction in basic medium to yield two products: 4-Hydroxy-4-methylpentan-2-one and 4-Methylpent-3-en-2-one. The reaction involves the enolate ion formed from acetone, which reacts with another acetone molecule, leading to the formation of the aldol product, followed by dehydration to form the α,β-unsaturated carbonyl compound.

In conclusion, aldehydes and ketones undergo various types of reactions in basic medium, such as aldol condensation, oxidation, and reduction, with different products being formed depending on the compound and the specific conditions.