Question

Reaction of ${C_6}{H_5}C{H_2}Br$ with aqueous sodium hydroxide follows:
A.${S_N}1$ mechanism
B.${S_N}2$ mechanism
C.Any of the above two depending upon the temperature of the reaction
D.Saytzeff rule

Answer 1

When the reactant benzyl bromide ${C_6}{H_5}C{H_2}Br$ is reacted with an aqueous sodium hydroxide solution the carbocation is formed as a reaction intermediate. So the reaction will follow the mechanism which leads to the formation of the most stable carbocation.

Complete step by step answer:
First, we will understand the concept of reaction mechanism in organic chemistry. The reaction mechanism is the sequential representation of all the elementary reactions that occurs during the complete reaction. All the basic steps and properties of the reaction are explained step by step to go from the reactants to products. During the reaction, an intermediate is formed in one step which is also called a reaction intermediate. The mechanism followed by reaction depends upon the stability of the intermediate formed during the reaction.
Now we are aware of the basic concepts of the reaction mechanism. Now we can easily solve our question. The question says that benzyl bromide ${C_6}{H_5}C{H_2}Br$ is reacted with the aqueous sodium hydroxide solution. The reaction between the reactants is given below.
${C_6}{H_5}C{H_2}Br\xrightarrow{{aq.NaOH}}{C_6}{H_5}C{H_2}^ + \xrightarrow{{O{H^ - }}}{C_6}{H_5}C{H_2}OH$
From the above reaction, we can observe that the intermediate formed ${C_6}{H_5}C{H_2}^ +$ is carbocation. Now we will check the stability of carbocation. So, the carbocation ${C_6}{H_5}C{H_2}^ +$ is taking part in resonance. So due to the more number of resonance structures the carbocation is stable. We know that the stability of carbocation is directly proportional to the ${S_N}1$ mechanism. Hence, the reaction will follow ${S_N}1$ the mechanism.
Therefore, the correct option is (A).

Note:
The ${S_N}1$ reaction involves two steps. The first step involves the separation of halogen to form a stable carbocation and the second step involves an attack on the stable carbocation ion from both sides to form the final product. Here, the final product was benzyl alcohol.