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Identify 'X' in the following reaction. $Br-C_6H_4-Cl\ [1.0\ mol]\xrightarrow[dry\ ether]{Mg\ [1.0\ mol]}$Intermediate$\xrightarrow{D_2O}X$
A
$Cl-C_6H_4-D$
B
$DO-C_6H_4-OD$
C
$D-C_6H_4-D$
D
$D-C_6H_4-Br$
Detailed Solution
With only 1.0 mol of Mg for 1.0 mol of the dihalide (Br and Cl on the ring), Mg preferentially inserts into the more reactive C–Br bond (aryl bromides are generally more reactive than aryl chlorides in this context) to form the Grignard reagent $BrMg-C_6H_4-Cl$. Reaction with $D_2O$ replaces the $-MgBr$ group with $-D$, giving $D-C_6H_4-Cl$ (chlorine remaining unreacted).