Question

In the following reactions:
1. $ PC{{l}_{5}}+S{{O}_{2}}\to A+B. $
2. $ A+MeCOOH\to C+S{{O}_{2}}+HCl. $
3. $ 2C+M{{e}_{2}}Cd\to 2D+CDC{{l}_{2}}. $
The compound C is:
(A) $ Me-CH-C{{l}_{2}} $
(B) $ Me-C{{H}_{2}}-Cl $
(C) $ Me-C(=O)-Cl $
(D) $ Me-C-C{{l}_{3}} $

Answer 1

Hint: We know that Friedel-Crafts reaction is an organic coupling reaction involving an electrophilic aromatic substitution that is used for the attachment of substituents to aromatic rings. There are two primary types of Friedel-Crafts reactions- the alkylation and acylation reactions.

Complete step by step solution:
The reaction that is provided in the question is Friedel-Crafts acylation. Friedel-Crafts acylation reaction involves the addition of an acyl group to an aromatic ring. We will do it by using an acid chloride $ \left( R-COCl \right) $ and a Lewis acid catalyst Such as $ AlC{{l}_{3}}. $ In a Friedel-Crafts acylation reaction, the aromatic ring is transformed into an aryl ketone. It is known that the nucleophilic reaction of the nitrogen of aromatic amines which bears a lone pair of electrons with haloalkanes and with the acid chlorides is regarded as the nucleophilic substitution reaction.
Here we have, Phosphorus pentachloride reacts with sulphur dioxide to form thionyl chloride and phosphorus oxychloride.
 $ PC{{l}_{5}}+S{{O}_{2}}\to \underset{A}{\mathop{SOC{{l}_{2}}}}\,+\underset{B}{\mathop{POC{{l}_{3}}}}\, $
Thionyl chloride reacts with acetic acid to form acetyl chloride, sulphur dioxide and hydrogen chloride gas. $ \underset{A}{\mathop{SOC{{l}_{2}}}}\,+C{{H}_{3}}COOH\to \underset{C}{\mathop{C{{H}_{3}}COCl}}\,+S{{O}_{2}}+HCl $
Acetyl chloride reacts with dimethyl cadmium to form acetone and cadmium (II) chloride. $ \underset{C}{\mathop{C{{H}_{3}}COCl}}\,+{{\left( C{{H}_{3}} \right)}_{2}}Cd\to 2\underset{D}{\mathop{C{{H}_{3}}COC{{H}_{3}}}}\,+CdC{{l}_{2}} $
It is known that the reaction of aromatic amine with aliphatic acid chloride or the aliphatic acid anhydride results in the formation of amides. The acetylation reaction is best carried out with acetic anhydride rather than acetyl chloride.
Therefore the correct answer is option C $ \underset{{}}{\mathop{C{{H}_{3}}COCl}}\, $ i.e. $ Me-C(=O)-Cl $ .

Note:
Remember that we cannot use any inorganic acid in place of Lewis acid here in Friedel crafts’ reaction. Note that the alkoxy group is always an ortho-para director in electrophilic aromatic substitution reactions.