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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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