JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Identify the reagents used for the following conversion

(A) $A = LiAlH_{4}, B = NaOH_{(aq)}, C = NH_{2} - NH_{2} / KOH$ , ethylene glycol
(B) $A = LiAlH_{4}, B = NaOH_{(alc)}, C = Zn / HCl$
(C) $A =$ DIBAL-H, $B = NaOH_{(aq)}$ , $C = NH_{2} - NH_{2} / KOH$ , ethylene glycol
(D) $A =$ DIBAL-H, $B = NaOH_{(alc)}, C = Zn / HCl$ ✓ Correct

Answer: (D)

Solution

The sequence converts an ester group to an aldehyde, carries out an intramolecular aldol condensation, and then removes the newly formed carbonyl completely.

Step A: DIBAL-H at low temperature is a selective reducing agent that reduces an ester only as far as the aldehyde, stopping there rather than over-reducing to the alcohol the way

LiAl H_{4}

would.

Step B: alcoholic NaOH promotes an intramolecular aldol condensation between the aldehyde groups now present on the molecule, forming a new C=C bond with loss of water; alcoholic base favours this condensation (elimination) product over the simple aldol addition product that aqueous base would give.

Step C:

Zn/HCl

(Clemmensen reduction) reduces the remaining carbonyl completely to a

C H_{2}

group under acidic conditions, which is required here since the molecule carries functionality that basic Wolff–Kishner conditions would not leave intact.

The reagent set is A = DIBAL-H, B = alcoholic NaOH, C = Zn/HCl.

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JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

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