In the reaction,C_6H_5CN xrightarrow[(ii) H_3 | vedclass

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(D) The given reaction is: $C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2 + HCl} C_6H_5CHO (X)$ $\xrightarrow{conc. KOH} C_6H_5COOK (Y) + C_6H_5CH_2

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Stephen reaction,Cannizzaro reaction

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(D) The given reaction is: $C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2 + HCl} C_6H_5CHO (X)$ $\xrightarrow{conc. KOH} C_6H_5COOK (Y) + C_6H_5CH_2OH (Z)$.$1$. The conversion of nitrile $(C_6H_5CN)$ to aldehyde $(C_6H_5CHO)$ using $SnCl_2 + HCl$ followed by hydrolysis is known as the Stephen reaction.$2$. Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms,so it undergoes the Cannizzaro reaction in the presence of concentrated alkali $(KOH)$ to form a mixture of alcohol $(C_6H_5CH_2OH)$ and salt of carboxylic acid $(C_6H_5COOK)$.

In the reaction,$C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2 + HCl} X$ $\xrightarrow{conc. KOH} Y + Z$,the formation of $X, Y$ and $Z$ are known by:

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