Question

Hydrogenation of benzoyl chloride in the presence of $Pd$ and $BaS{O_4}$ followed by treatment with formaldehyde and $50\%$ $KOH$ gives
(A) Benzyl alcohol
(B) Benzaldehyde
(c) Benzoic acid
(D) Both benzyl alcohol and benzoic acid.

Answer 1

We know that Cannizzaro reaction is named after Stanislao Cannizzaro that includes the base-incited disproportionation of two particles of a non-enolizable aldehyde to yield a carboxylic acid and an alcohol.

Complete answer:
We must have to remember that the products that resulted in a Cannizzaro reaction incorporate essential alcohols and carboxylic acids. The carboxylate anion is protonated to manage the carboxylic corrosive while the alkoxide anion is protonated by water to yield alcohol.
$2{C_6}{H_5}CHO + KOH \to {C_6}{H_5}C{H_2}OH + {C_6}{H_5}COOK$
The resulting compound of hydrogenation of benzoyl chloride in the presence of $Pd$ and $BaS{O_4}$ followed by treatment with formaldehyde and $50\%$ $KOH$ is benzyl alcohol.

Hence option A is correct.

Note:
We have to remember that the Cannizzaro Reaction Mechanism subtleties the technique to get one atom of alcohol and one particle of carboxylic acid from two particles of a given aldehyde. Researcher Stanislao Cannizzaro, in $1853$ prevailing with regards to getting benzyl alcohol and potassium benzoate from Benzaldehyde. The reaction is executed by a nucleophilic acyl replacement on an aldehyde where the leaving bunch assaults another aldehyde. Tetrahedral moderate outcomes from the assault of hydroxide on a carbonyl. These tetrahedral moderate breakdowns, consequently transforming the carbonyl and moving a hydride which assaults another state.
Presently, a proton is traded by acid and alkoxide particles. At the point when a base of high focus is presented, the aldehyde frames an anion which has a charge of $2$. From this, a hydride particle is moved to a second atom of the aldehyde, framing carboxylate and alkoxide particles. The alkoxide particle likewise gets a proton from the dissolvable for the reaction.