Hydrocarbons Ques 85

85. When $CH_3 Cl$ and $AlCl_3$ are used in Friedel-Crafts reaction, the electrophile is

[1994]

(a) $Cl^{+}$

(b) $AlCl_4^{-}$

(c) $CH_3^{+}$

(d) $AlCl_2^{+}$

Show Answer

Solution:

$ \text{ (c) } CH_3 Cl+AlCl_3 \longrightarrow \underset{\text{ Electrophile }}{CH_3^{\oplus}}+AlCl_4^{-} $



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