question_answer

Benzene reacts with \[{{H}_{2}}S{{O}_{4}},\] in the presence of anhydrous \[BrO_{3}^{-}(aq)+5B{{r}^{-}}(aq)+6{{H}^{+}}\xrightarrow[{}]{{}}\] to form

A) toluene               

B)        chlorobenzene                

C) benzylchloride 

D) xylene

Correct Answer: A

Solution :

Key Idea \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{a}^{2}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{b}\]in the presence of anhydrous \[+\frac{1}{4\pi {{\varepsilon }_{0}}},\frac{\sigma 4\pi {{c}^{2}}}{c}\]acts as alkylation agent and introduces an alkyl group This reaction is known as Friedel-Craft's alkylation of benzene.