Benzonitrile on reduction with stannous chlor | vedclass

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Question

(D) The reaction described is the $Stephen$ reduction. Benzonitrile $(C_6H_5CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine hydrochloride interme

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Benzaldehyde

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(D) The reaction described is the $Stephen$ reduction. Benzonitrile $(C_6H_5CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine hydrochloride intermediate $(C_6H_5CH=NH \cdot HCl)$. This intermediate,upon subsequent acid hydrolysis $(H_3O^+)$,yields benzaldehyde $(C_6H_5CHO)$ and ammonium chloride $(NH_4Cl)$. The overall reaction is: $C_6H_5CN + 2[H] \xrightarrow{SnCl_2, HCl} C_6H_5CH=NH \cdot HCl$ $C_6H_5CH=NH \cdot HCl + H_2O \xrightarrow{H_3O^+} C_6H_5CHO + NH_4Cl$

Benzonitrile on reduction with stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$ followed by acid hydrolysis forms:

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