${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CN}\left(\text{A}\right)\underset{ }{\overset{{\text{SnCl}}_{\text{2}}+\text{HCl}}{\to }}\text{B}\underset{ }{\overset{{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}}{\to }}\text{C}$.  Incorrect statement is

$\underset{\left(\text{A}\right)}{{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}-\text{C}\equiv \text{N}}\underset{ }{\overset{{\text{SnCl}}_{\text{2}}+\text{HCl}}{\to }}\underset{\left(\text{B}\right)}{{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}-\text{CH}=\text{NH}}\underset{ }{\overset{{\text{H}}_{\text{3}}{\text{O}}^{+}}{\to }}\underset{\left(\text{C}\right)}{{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CHO}+{\text{NH}}_3}$

‘B’ is Benzaldimine

– CHO  group is m – directing and deactivating towards ESR

${\mathrm{C}}_{\mathrm{6}}{\mathrm{H}}_{\mathrm{5}}-\mathrm{CH}=\mathrm{O}+{\mathrm{H}}_{\mathrm{2}}\mathrm{NH}\underset{ }{\overset{\mathrm{\Delta }}{\to }}\underset{\left(\mathrm{B}\right)}{{\mathrm{C}}_{\mathrm{6}}{\mathrm{H}}_{\mathrm{5}}-\mathrm{CH}=\mathrm{NH}}$