Chapter 12, Problem 3PP

Interpretation Introduction

Interpretation:

For each transformation, the reducing agent used, out of ${\text{LiAlH}}_{\text{4}}$ and ${\text{NaBH}}_{\text{4}}$, is to be determined.

Concept introduction:

Lithium aluminium hydride and sodium borohydride are reducing agents.

${\text{LiAlH}}_{\text{4}}$ is a more powerful reducing agent than ${\text{NaBH}}_{\text{4}}$.

${\text{LiAlH}}_{\text{4}}$ reduces esters, acids, aldehydes and ketones to their corresponding alcohols.

${\text{NaBH}}_{\text{4}}$ reduces only aldehydes and ketones to their corresponding alcohols.

Answer to Problem 3PP

Solution:

(a) ${\text{LiAlH}}_{\text{4}}$

(b) ${\text{LiAlH}}_{\text{4}}$

(c) ${\text{NaBH}}_{\text{4}}$

(d) ${\text{LiAlH}}_{\text{4}}$

(e) ${\text{LiAlH}}_{\text{4}}$

(a)

Explanation of Solution

The given transformation is as follows:

In the above transformation, carboxylic acid is converted into alcohol. This transformation is done by using ${\text{LiAlH}}_{\text{4}}$ as lithium aluminium hydride is a strong reducing agent.

Hence, ${\text{LiAlH}}_{\text{4}}$ is used for the given transformation.

(b)

The given transformation is as follows;

In the above transformation, carboxylic acid and ketone are converted into their corresponding alcohols. This transformation is done by using ${\text{LiAlH}}_{\text{4}}$ as lithium aluminium hydride is a strong reducing agent and reduces both, carboxylic acid and ketone.

Hence, ${\text{LiAlH}}_{\text{4}}$ is used for the given transformation.

(c)

The given transformation is as follows:

In the above transformation, aldehyde is converted into its corresponding alcohol, but there is no effect on the ester group. This transformation is done by using ${\text{NaBH}}_{\text{4}}$ as sodiumborohydride reduces only aldehydes and ketones. Esters are not reduced using ${\text{NaBH}}_{\text{4}}$.

Hence, ${\text{NaBH}}_{\text{4}}$ is used for the given transformation

(d)

The given transformation is as follows:

In the above transformation, the bromine atom is replaced by the hydrogen atom. This transformation is done by using ${\text{LiAlH}}_{\text{4}}$ as lithium aluminium hydride is a strong reducing agent and reduces the halogen atom by the hydrogen atom.

Hence, ${\text{LiAlH}}_{\text{4}}$ is used for the given transformation.

(e)

The given transformation is as follows:

In the above transformation, the bromine atom is replaced by the hydrogen atom, and carboxylic acid is converted into its corresponding alcohol. This transformation is done by using ${\text{LiAlH}}_{\text{4}}$ as lithium aluminium hydride is a strong reducing agent and reduces the halogen atom by the hydrogen atom, and acids to alcohols.

Hence, ${\text{LiAlH}}_{\text{4}}$ is used for the given transformation.