Chemistry - GOC Question with Solution | TestHub

The lone pair on N forms an $\mathrm{N}=\mathrm{C}$ pi bond, so N becomes ${\mathrm{N}}^{+}$. The $\mathrm{C}=\mathrm{C}$ pi pair moves onto the terminal carbon as a lone pair, giving a carbanion.

Product : $\mathrm{C}{\mathrm{H}}_2^{-}\mathrm{CH}={\mathrm{N}}^{+}(\mathrm{C}{\mathrm{H}}_3{)}_2$ (C and N atoms stay in place). Both charges must be opposite, so only (B) fits.