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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

In the following reaction,$X$ and $Y$ are respectively
$CH_3 COOH + NH_3$ $\longrightarrow X \stackrel{\Delta}{}$ ${\longrightarrow} Y + H_2 O$

Write the chemical reactions for the following conversions:
$1.$ $1$-Bromopropane to Propan-$1$-ol and Propan-$2$-ol
$2.$ Ethanoic acid to Methanol

Compound $(B)$ is

In the given reaction sequence: $HOOC-CH_2-CH_2-COOH$ $\xrightarrow{\Delta} (A)$ $\xrightarrow{CH_3-CH_2-NH_2/\Delta} (B)$. $(B)$ will be:

$\text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{X} \xrightarrow{\text{PCl}_5} \text{Y} \xrightarrow{\text{Alcoholic KOH}} \text{Z}$. Find product '$Z$' in the above reaction.

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