Question

Consider the following set of reactions
$C{H_3}COOH\xrightarrow{{SOC{l_2}}}A\xrightarrow[{Benzene}]{{Anhyd.AlC{l_3}}}B\xrightarrow{{HCN}}C\xrightarrow{{HOH}}D$
The structure $D$ will be
A.)

B.)

C.)

D.)

Answer 1

Hint: When thionyl chloride that is $SOC{l_2}$ reacts with a carboxylic acid then the $ - OH$ of the carboxylic acid replaces with the chlorine to form an acid chloride or acyl chloride in the product. The benzene with $AlC{l_3}$ reacts with this acyl chloride then chlorine is replaced by the benzene group.

Complete answer:
In the given set of reactions, the first reactant is a carboxylic acid. The $SOC{l_2}$ has the chemical name as Thionyl Chloride. When thionyl chloride ($SOC{l_2}$) reacts with carboxylic acid then the product formed is acid chloride or we can say it is acyl chloride with general chemical formulas as $R - OCl$. In this reaction, $ - OH$ of the carboxylic is replaced by the $ - Cl$ . Hence, this reaction can be written as:
$C{H_3}COOH\xrightarrow{{SOC{l_2}}}C{H_3}COCl$
When the acyl chloride that is formed in the above reaction reacts with the benzene in the presence of aluminium trichloride ($AlC{l_3}$) then the benzene ring will attach in place of chlorine. This reaction is known as Friedel craft acylation. In Friedel crafts acylation, when acyl group that is $C{H_3}COCl$ reacts with an aromatic compound then the chlorine of the acyl chloride gets replaced with the aromatic compound that is benzene here. So, the reaction can be shown as:
$C{H_3}COCl\xrightarrow[{Anhyd.AlC{l_3}}]{{Benzene}}{C_6}{H_5}COC{H_3}$
When ${C_6}{H_5}COC{H_3}$ reacts with $HCN$ then the $CN$ combines with the carbon of carboxylic group and $H$ combines with oxygen of the carboxylic group to form ${C_6}{H_5}C(OH)(C{H_3})CN$. This reaction can be represented as:
${C_6}{H_5}COC{H_3}\xrightarrow{{HCN}}{C_6}{H_5}C(OH)(C{H_3})CN$

When the formed product that is ${C_6}{H_5}C(OH)(C{H_3})CN$ reacts with water then $-CN$ will be replaced by the $-COOH$ group. This reaction can be represented as:
${C_6}{H_5}C(OH)(C{H_3})CN\xrightarrow{{HOH}}{C_6}{H_5}C(OH)(C{H_3})COOH$
Therefore, the set of reactions can be written as:

$ C{H_3}COOH\xrightarrow{{SOC{l_2}}}C{H_3}COCl\xrightarrow[{Benzene}]{{Anhyd.AlC{l_3}}}{C_6}{H_5}COC{H_3} \\
  \xrightarrow{{HCN}}{C_6}{H_5}C(OH)(C{H_3})CN\xrightarrow{{HOH}}{C_6}{H_5}C(OH)(C{H_3})COOH \\
 $
Therefore, the product $D$ is ${C_6}{H_5}C(OH)(C{H_3})COOH$

Hence, the option A.) is the correct answer.

Note: Always remember that when any acyl chloride that is of general formula as $R - OCl$ reacts in the presence of $AlC{l_3}$ with an aromatic compound then it is Friedel craft acylation and when an alkyl halide reacts with these reagents the it is called as Friedel craft alkylation.