The above conversion can be done by:

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given reaction involves the reduction of both a ketone group $(-C=O)$ and a carboxylic acid group $(-COOH)$ to their corresponding alcohols. $

Vedclass · Accepted Answer

$LiAlH_4$

Vedclass · Answer

(B) The given reaction involves the reduction of both a ketone group $(-C=O)$ and a carboxylic acid group $(-COOH)$ to their corresponding alcohols. $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that can reduce both carboxylic acids to primary alcohols $(-CH_2OH)$ and ketones to secondary alcohols $(-CH(OH)-)$. $NaBH_4$ is a milder reducing agent that typically reduces ketones but not carboxylic acids. $PCC$ is an oxidizing agent,and $KMnO_4$ is a strong oxidizing agent. Therefore,$LiAlH_4$ is the correct reagent for this transformation.

Similar Questions

The increasing order of the acidity of the following carboxylic acids is:

The number of equivalents of Grignard reagent $(x)$ used in reaction $(1)$ is:

Which of the following tetracarboxylic acids forms a di-anhydride upon heating?

The product $D$ of the reaction is:
$CH_3Cl$ $\xrightarrow{KCN} (A)$ $\xrightarrow{H_2O} (B)$ $\xrightarrow{NH_3} (C)$ $\xrightarrow{\Delta} (D)$

Identify the product $A$ in the following reaction:
(Image of a tricyclic allylic alcohol reacting with $p$-benzoquinone in the presence of $[(CH_3)_3CO]_3Al$ to form $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module