In the reaction:C_6H_5-H xrightarrow{[X] / Al | vedclass

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(C) The given reaction is a Friedel-Crafts acylation reaction.In Friedel-Crafts acylation,benzene reacts with an acylating agent (such as an acid chlo

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

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(C) The given reaction is a Friedel-Crafts acylation reaction.In Friedel-Crafts acylation,benzene reacts with an acylating agent (such as an acid chloride or an acid anhydride) in the presence of a Lewis acid catalyst like $AlCl_3$ to form an aromatic ketone.For the product $C_6H_5-C(=O)-CH_2-C_6H_5$ (phenyl benzyl ketone),the acylating group is $C_6H_5-CH_2-CO-$.Option $(A)$ is $C_6H_5CH_2COCl$ (phenylacetyl chloride),which is an acid chloride.Option $(B)$ is $(C_6H_5-CH_2CO)_2O$ (phenylacetic anhydride),which is an acid anhydride.Both acid chlorides and acid anhydrides can act as acylating agents in the presence of $AlCl_3$ to produce the same ketone.Therefore,both $(A)$ and $(B)$ are correct.

In the reaction:
$C_6H_5-H \xrightarrow{[X] / AlCl_3} C_6H_5-C(=O)-CH_2-C_6H_5$
$[X]$ will be:

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In the reaction:$C_6H_5-H \xrightarrow{[X] / AlCl_3} C_6H_5-C(=O)-CH_2-C_6H_5$$[X]$ will be: