Correct Answer - A
The primary amine undergoes exhaustive methylation forming quaternary ammonium iodie `CH_(3)CH_(2)CH_(2)overset(+)(Nme_(3)I`
(A) which is converted to the quaternary ammonium hydroxide `CH_(3)CH_(2)CH_(2)overset(+)Nme_(3)OH bar` (B), on treatment with silver hydroxide When heated a Hofmann elimination (also called Hofmann degradation) occurs to form an alkene
` CH_(3) CH_(2)underset((B))(CH_(2)overset(+)NMe_(3))barOH overset(heat)rarrCH_(3)CH=underset(C)(CH_(2))+H_(2)O+NMe_(3)`
The complete sequence is
`CH_(3)CH_(2)CH_(2)underset(excess)overset(MeI)rarrCH_(3)CH_(2)overset(+)NMe_(3)Ibar`
`overset(AgOH)rarrCH_(3)underset((B))(CH_(2))CH_(2)overset(+)NM_(3)barOH overset(heat)rarrunderset((C))(CH_(3)CH=CH_(2))`
The lest step in an `E2` elimination reaction the leaving group is `NMe_(3)` Note that `CH_(3)CH_(2)CH_(2)NH_(2)` cannot undergo an `E2` elimination because `NHbar_(2)` is a very poor leaving group .