Question

In the reaction A and B respectively are:
\[C{{H}_{3}}COOH+\xrightarrow{LiAl{{H}_{4}}}A+C{{H}_{3}}COOH\xrightarrow{{{H}_{3}}{{O}^{+}}}B+{{H}_{2}}O\]

Answer 1

Hint : We know that $LiAl{{H}_{4}}$ is a reducing agent that transfers a hydride from aluminium. Lithium aluminium hydride is a strong donor reagent and rapidly reduces ester acids, nitriles, amides, as well as aldehydes and ketones. It does not reduce isolated carbon- carbon double bonds.

Complete Step By Step Answer:
Lithium aluminium hydride is nucleophilic reagent and they generally attack polarized multiple bonds such as $C=O,C\equiv N$ , by the transfer of hydride ions to the more positive atom. They do not reduce isolated carbon-carbon double bond $C=C$. Lithium aluminium hydride is a strong reducing agent and can reduce most of the commonly encountered organic functional groups. For example the aldehyde gets converted to primary alcohol, similarly Ketones on reaction with lithium aluminium hydride gets converted to secondary alcohol.
Ester gets converted to two different alcohols depending upon the alkyl group attached. Carboxylic acid gets converted to alcohol. Similarly amide converted to the amine. Thus we can say aldehydes , ketones , esters, carboxylic acids and lactones can all be reduced smoothly to corresponding alcohol under mild conditions. Carboxylic amides are converted to amines or aldehydes. Here in the given question, acetic acid is reacting with lithium aluminium hydride. Since carboxylic acids get converted to corresponding alcohol, then acetic acid will form ethyl alcohol or ethanol. The chemical reaction
\[C{{H}_{3}}COOH+4H\xrightarrow{LiAl{{H}_{4}}}\underset{\left( A \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,+{{H}_{2}}O\]
Thus, on further reaction with ${{H}_{3}}{{O}^{+}}.$
\[C{{H}_{3}}C{{H}_{2}}OH+C{{H}_{3}}COOH\xrightarrow{{{H}_{3}}{{O}^{+}}}\underset{\left( B \right)}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}\,+{{H}_{2}}O\]

Note :
Remember that Lithium aluminium hydride is a stronger reducing agent which directly converts carboxylic acid to alcohol. On the other hand sodium borohydride is also a hydride transfer reducing agent, but it is a very selective reducing agent. It does not reduce carboxylic acid. It does not react or slowly react with esters.