JEE Main · 2021 · 22 Jul 2021 · Shift 2 · General Organic Chemistry

Resonance needs a lone pair or a $\pi$ bond next to another $\pi$ bond or an empty orbital, so that the electrons can be delocalised. In $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CON}{\mathrm{H}}_2$ the nitrogen lone pair is conjugated with the carbonyl $\pi$ bond. In $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OCH}=\mathrm{C}{\mathrm{H}}_2$ the oxygen lone pair is conjugated with the vinyl double bond. Benzyl alcohol carries a benzene ring, which is itself a resonance hybrid. In $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CHC}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$ an $\mathrm{s}{\mathrm{p}}^3$ methylene group stands between the nitrogen lone pair and the double bond and blocks overlap, while the isolated double bond has nothing to delocalise into. The compound that shows no resonance is $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CHC}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$.