Question

What is the action of the following reagent on ethanoic acid?
$LiAl{H_4}/{H_3}{O^ + }$.

Answer 1

Hint: Alcohols when oxidized convert to aldehydes or ketones. Aldehydes when oxidized convert to carboxylic acids. The reducing agents such as Lithium Aluminum hydride reduce carboxylic acids to alcohols straight away because they have strong reducing properties. Addition of hydrogen is referred to as reduction.

Complete answer:
Lithium aluminium hydride in the presence of an acid is a strong reducing agent that will reduce a carboxylic acid to corresponding alcohol with the same carbon number. No aldehyde will be formed because of the strong reducing action of this particular reducing agent. In this case the hydrogens will be added to the carbon and so the basic definition of reduction, which is related to the direct addition of hydrogens is sufficient.
Since ethanoic acid is a two carbon carboxylic acid, a two carbon alcohol will be formed with - OH attached to the same carbon atom to which carboxylic acid group - COOH was attached.
The reaction is shown below:
$C{H_3} - COOH\mathop {\mathop \to \limits_{{H_3}{O^ + }} }\limits^{LiAl{H_4}} C{H_3} - C{H_2} - OH$
By successive replacement of oxygen atoms attached to the carbon by hydrogen atoms, the $ - COOH$ group has been reduced to $ - C{H_2} - OH$ group by $LiAl{H_4}$.

Therefore, ethanol is formed as a result of the reduction of ethanoic acid by Lithium aluminium hydride in the presence of an acid.

Note: Lithium aluminium hydride, commonly abbreviated to LAH, is an inorganic compound with the chemical formula $LiAl{H_4}$. Acetaldehyde is not formed because Lithium aluminium hydride is a strong reducing agent.