Which of the following catalyst can carry out  the conversion of carboxylic acid to alcohol?

LiAlH₄/ H₂O

The correct answer is option 3. \(LiAlH_4 / H_2O\).

Let us go through the different options to understand why lithium aluminum hydride \((LiAlH_4)\) is the correct choice for reducing a carboxylic acid to an alcohol.

Carboxylic Acid Reduction

Reducing a carboxylic acid to a primary alcohol involves breaking the strong \(C=O\) bond of the carboxylic acid group \((–COOH)\) and adding hydrogen atoms to form a \(–CH_2OH\) group. This process requires a powerful reducing agent.

Analysis of the Given Options

1. \(NaBH_4/H_2O\) (Sodium borohydride in water):

Sodium borohydride \((NaBH_4)\) is a mild reducing agent. It is effective for reducing aldehydes and ketones to alcohols. However, \(NaBH_4\) is not strong enough to reduce carboxylic acids. The reduction of a carboxylic acid requires a much stronger reducing agent.

2. \(Pd-C/H_2\) (Palladium on carbon with hydrogen):

Palladium on carbon (Pd-C) with hydrogen is used in catalytic hydrogenation. It is effective for reducing alkenes, alkynes, and nitro groups to corresponding alkanes and amines. This catalyst does not typically reduce carboxylic acids to alcohols under standard conditions.

3. \(LiAlH_4/H_2O\) (Lithium aluminum hydride with water):

Lithium aluminum hydride \((LiAlH_4)\) is a very strong reducing agent. It is capable of reducing carboxylic acids, esters, aldehydes, ketones, and other carbonyl-containing compounds to alcohols. The reduction of a carboxylic acid to a primary alcohol using \(LiAlH_4\) proceeds as follows:

Reaction: The carboxylic acid reacts with \(LiAlH_4\) in an inert solvent like ether.

Hydrolysis: The reaction mixture is then hydrolyzed with water to yield the primary alcohol.

The overall reaction is:

\(LiAlH_4\) is effective because it can deliver multiple hydride ions \((H^−)\) necessary for the reduction of the carboxyl group.

4. \(Zn/CH_3OH\) (Zinc in methanol):

Zinc in methanol is used in various reduction reactions, such as the Clemmensen reduction, which converts ketones and aldehydes to alkanes. It is not typically used for reducing carboxylic acids to alcohols.

Conclusion: \(LiAlH_4\) is the correct choice for reducing carboxylic acids to primary alcohols due to its strong reducing power and ability to provide the necessary hydride ions. The process involves two main steps: reduction in an inert solvent and subsequent hydrolysis.

Therefore, the correct answer is: \(LiAlH_4/H_2O\).