Identify the reagents used for the following conversion: — Solution | JEE Chemistry

MCQEasyJEE 2024Aldehydes & Ketones

Identify the reagents used for the following conversion:

A
$A = \text{LiAlH}_4$ , $B = \text{NaOH(aq)}$ , $C = \text{NH}_2\text{NH}_2/\text{KOH}$
B
$A = \text{LiAlH}_4$ , $B = \text{NaOH(alc)}$ , $C = \text{Zn/HCl}$
C
$A = \text{DIBAL-H}$ , $B = \text{NaOH(aq)}$ , $C = \text{NH}_2\text{NH}_2/\text{KOH}$
D
$A = \text{DIBAL-H}$ , $B = \text{NaOH(alc)}$ , $C = \text{Zn/HCl}$

Answer

Correct answer:D

Standard Method

Given: A sequence of reagents $A$ , $B$ , and $C$ is to be identified for the conversion.

Find: The correct combination of reagents.

From the solution:

Therefore, the required set of reagents is:

A = \text{DIBAL-H}, \qquad B = \text{NaOH(alc)}, \qquad C = \text{Zn/HCl}

Hence, the correct option is D.

Confusing $\text{LiAlH}_4$ with $\text{DIBAL-H}$ is a common mistake. $\text{LiAlH}_4$ generally reduces esters further to alcohols, whereas $\text{DIBAL-H}$ can stop at the aldehyde stage. Use the selective reducing agent required by the conversion.
Using aqueous $\text{NaOH}$ instead of alcoholic $\text{NaOH}$ is incorrect here. Aldol condensation is typically facilitated in alcoholic medium. Check the reaction condition, not only the base.
Mixing up Clemmensen reduction and Wolff-Kishner reduction leads to the wrong reagent choice. $\text{Zn/HCl}$ is Clemmensen reduction, while $\text{NH}_2\text{NH}_2/\text{KOH}$ is Wolff-Kishner reduction. Match the named reduction with the required conditions.

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