Question

An organic compound $A$ has the molecular formula ${C_7}{H_6}O$ . When $A$ is treated with $NaOH$ followed by acid hydrolysis, it gives two products $B$ and $C$. When $B$ is oxidized, it gives $A$ , when $A$ and $C$ are each treated separately with $PC{l_5}$ , they give two different products $D$ and $E$.
i.Identify $A$ , $B$ , $C$ , $D$ and $E$ .
ii.Give the chemical reaction when $A$ is treated with $NaOH$ and name the reaction.

Answer 1

Hint:This question gives the knowledge about cross cannizzaro reaction. Cannizarro reaction is very useful in increasing the yield of the chemical compound. The compounds which contain $\alpha $-hydrogen do not undergo a cannizzaro reaction.

Complete step-by-step answer: Cannizzaro reaction is the reaction between two different aldehydes treated with a strong base, results in the formation of alcohol and a carboxylic acid. The compounds which contain $\alpha $-hydrogen do not undergo a cannizzaro reaction.
i.Cannizzaro reaction is a redox process because in this reaction both the oxidation reaction and the reduction reactions are taking place simultaneously. One aldehyde reduces itself to produce an alcohol and the other aldehyde oxidizes itself to produce a carboxylic acid. The main advantage of Cannizzaro reaction is that it helps in increasing the yield of the desired chemical compound.
The mechanism of cannizzaro reaction is as follows:
Firstly, the hydroxide ion $H{O^ - }$of the strong base acts as a nucleophile and attacks the carbonyl group of the aldehyde which leads to accumulation of negative charge on the oxygen atom. In the next step, hydride shift takes place which then attacks on the carbonyl group of other aldehydes. In the next step one aldehyde is reduced to an alcohol and the other aldehyde is oxidized to the formic acid.
Compound $A$ is benzaldehyde.
When benzaldehyde is treated with sodium hydroxide, it leads to the formation of benzoic acid and benzyl alcohol.
$2{C_6}{H_5}CHO + NaOH \to {C_6}{H_5}C{H_2}OOH + {C_6}{H_5}COOH$
Therefore, compound $B$ is benzyl alcohol and compound $C$ is benzoic acid.
When compound $A$ is treated with $PC{l_5}$ , we have
${C_6}{H_5}CHO + PC{l_5} \to {C_6}{H_5}COCl$
Therefore, compound $D$ is benzoyl chloride.
When compound $C$ is treated with $PC{l_5}$ , we have
${C_6}{H_5}COOH + PC{l_5} \to POC{l_3}$
Therefore, compound $E$ is phosphoryl chloride.

ii.The chemical reaction is as follows:
$2{C_6}{H_5}CHO + NaOH \to {C_6}{H_5}C{H_2}OOH + {C_6}{H_5}COOH$
This reaction is a cannizzaro reaction.
Cannizarro reaction is a redox process because in this reaction both the oxidation reaction and the reduction reactions are taking place simultaneously. One aldehyde reduces itself to produce alcohol and the other aldehyde oxidizes itself to produce carboxylic acid. The main advantage of Cannizzaro reaction is that it helps in increasing the yield of the desired chemical compound.

Note: Cannizzaro reaction is a type of enolate reaction which involves the formation of enolate ion which further reacts with the reactant to produce the products. Cannizzaro reactions are very useful in increasing the yield of the desired chemical compound.\