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(D) The reaction of benzonitrile $(C_6H_5CN)$ with phenyl magnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the ni

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Benzophenone

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(D) The reaction of benzonitrile $(C_6H_5CN)$ with phenyl magnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the nitrile group.$1$. The phenyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile group to form an imine magnesium bromide complex $(C_6H_5-C(C_6H_5)=NMgBr)$.$2$. Upon subsequent acid hydrolysis $(H_3O^+)$,the imine complex is converted into a ketone.$3$. The final product formed is benzophenone $(C_6H_5-CO-C_6H_5)$.

Item-$I$	Item-$II$
$I$. Benzene	$P$. $HCl$ and $SnCl_2, H_3O^{+}$
$II$. Benzonitrile	$Q$. $H_2, Pd-BaSO_4, S$ and quinoline
$III$. Benzoyl Chloride	$R$. $CO, HCl$ and $AlCl_3$

What is the final product obtained when benzonitrile reacts with phenyl magnesium bromide in equimolar proportion?

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In the following sequence of reactions,what is the end product $(D)$?
$C_2H_5Br$ $\xrightarrow{KCN} A$ $\xrightarrow{H_3O^{+}} B$ $\xrightarrow{LiAlH_4} C$ $\xrightarrow[573 \ K]{Cu} D$.

The correct match between Item-$I$ (starting material) and Item-$II$ (reagent) for the preparation of benzaldehyde is :
Item-$I$ Item-$II$
$I$. Benzene $P$. $HCl$ and $SnCl_2, H_3O^{+}$
$II$. Benzonitrile $Q$. $H_2, Pd-BaSO_4, S$ and quinoline
$III$. Benzoyl Chloride $R$. $CO, HCl$ and $AlCl_3$

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