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In the following reaction
$C_6H_5CH_2Br \xrightarrow[2.\ H_3O^+]{1.\ Mg,\ Ether}$ X,
the product 'X' is
$C_6H_5CH_2Br \xrightarrow[2.\ H_3O^+]{1.\ Mg,\ Ether}$ X,
the product 'X' is
A
$C_6H_5CH_2OCH_2C_6H_5$
B
$C_6H_5CH_2OH$
C
$C_6H_5CH_3$
D
$C_6H_5CH_2CH_2C_6H_5$
Detailed Solution
Step 1: benzyl bromide reacts with magnesium in dry ether to form the Grignard reagent: $C_6H_5CH_2Br + Mg \xrightarrow{ether} C_6H_5CH_2MgBr$
In a Grignard reagent the carbon bonded to magnesium is strongly nucleophilic and basic (carbanion-like).
Step 2: with $H_3O^+$ it takes up a proton to form the hydrocarbon: $C_6H_5CH_2MgBr + H_2O \rightarrow C_6H_5CH_3 + Mg(OH)Br$
So the net change is replacement of Br by H.
Hence X is toluene, $C_6H_5CH_3$.
In a Grignard reagent the carbon bonded to magnesium is strongly nucleophilic and basic (carbanion-like).
Step 2: with $H_3O^+$ it takes up a proton to form the hydrocarbon: $C_6H_5CH_2MgBr + H_2O \rightarrow C_6H_5CH_3 + Mg(OH)Br$
So the net change is replacement of Br by H.
Hence X is toluene, $C_6H_5CH_3$.