Simplest alkene (A) reacts with HCl to form compound (B).Conmpound (B) reacts with ammonia to form compound (C) of molecular formula C_2H_7N. Conpound (C) undergoes carbylamine test. Identify (A). (B), and (C).

Simplest alkene (A) reacts with HCl to form compound (B).Conmpound (B)

1.	Simplest alkene (A) reacts with HCl to form compound (B).Conmpound (B) reacts with ammonia to form compound (C) of molecular formula C_2H_7N. Conpound (C) undergoes carbylamine test. Identify (A). (B), and (C).
Answer» Solution : (i)The simplest alkene (A) is `CH_2 = CH_2`, ethene. (II) Ethene reacts with HCl to GIVE Chloroethane `CH_3 - CH_2Cl` as (B) by addition reaction. `CH_2 = CH_2 + HCl to underset("Chloroethane (B)")(CH_3 - CH_2Cl)` (iii) Chloroethane reacts with ammonia to give Ethylamine `CH_3 - CH_2" " NH_2` as (C ). It is a primary amine and Carbylamine test is the characteristic test for `1^@` amine. `CH_3-CH_2Cl + NH_3 to underset("ETHYL amine (or) Aminocthane")(CH_3 - CH_2NH_2) + HCl`