Question

The action of $AlC{l}_3$ in Friedel crafts reaction is

• A. $\mathrm{to}\mathrm{absorb}\mathrm{HCl}$
• B. $\mathrm{to}\mathrm{release}\mathrm{HCl}$
• C. $\mathrm{to}\mathrm{produce}\mathrm{electrophile}$
• D. $\mathrm{to}\mathrm{produce}\mathrm{nucleophile}$

Answer 1

The correct option is C $\mathrm{to}\mathrm{produce}\mathrm{electrophile}$

$(R{)}_{3}C-Cl+AlC{l}_3\to (R{)}_3{C}^{+}+[AlC{l}_4{]}^{-}$