Question

Identify the reagents A and B used in the following reactions:
${C_6}{H_5}OH\xrightarrow{A}{C_6}{H_6}\xrightarrow{B}{C_6}{H_5}Cl$
A. $Sn/HCl,C{l_2}/hv$
B. $C{l_2}/hv,Zn$
C. $C{l_2}/Fe,FeC{l_3}$
D. $Zn,C{l_2}/FeC{l_3}$

Answer 1

${C_6}{H_5}OH$ is the formula of phenol. In this reaction, the reduction of phenol occurs and it gets converted into benzene. The formula of benzene is ${C_6}{H_6}$ . Benzene further reacts and one hydrogen gets replaced by chlorine, a halogen to form chlorobenzene.

Complete step by step answer:
The reaction is completed in 2 steps: first is reduction reaction (removal of oxygen or addition of hydrogen to a carbon atom) and the second is electrophilic substitution reaction (electrophile displaces the functional group in a compound, mainly hydrogen atom). Phenol reacts with $Zn$ dust, gets reduced, and converted into benzene because $Zn$ dust acts as a reducing agent. Then benzene further reacts with $C{l_2}$ (Chlorine) and $FeC{l_3}$(Ferric chloride) to form (Chlorobenzene ) ${C_6}{H_5}Cl$ . It is an electrophilic substitution reaction.
${C_6}{H_5}OH\xrightarrow{{Zn}}{C_6}{H_6}\xrightarrow[{FeC{l_3}}]{{C{l_2}}}{C_6}{H_5}Cl$
A = $Zn$ dust
B = $C{l_2}/FeC{l_3}$

So, the correct answer is Option D .

Additional Information:
Mechanism of the reaction is $Zn$ shows an oxidation state of $+ 2$. It itself oxidized into $ZnO$ and phenol converted into phenoxide ion due to removal of hydrogen ion and proton thus releasing an electron from zinc to form $H$ radical. After that carbon and oxygen bond breaks to form phenyl radical which combines with hydrogen radical to benzene. Thereafter, benzene reacts with chlorine. The $C{l_2}$ breaks into $C{l^ - }$ and $C{l^ + }$ where $C{l^ - }$ attacks ferric chloride to form $FeCl_4^ -$ and $C{l^ + }$ attacks benzene. The addition of chlorine to the ring of the benzene results in the formation of chlorobenzene and released hydrogen from ring attacks $FeCl_4^ -$ to form $FeC{l_3}$ and $HCl$ .

Note:
We use zinc dust instead of zinc granules because it can produce hydrogen gas at a faster rate. It happens due to an increase in surface areas results in increases in the rate of reaction. It is a bluish-gray powder with odorless and insoluble water.