An amine on treatment with `HNO_2` evolved `N_2` . The amine on exhaustive methylation with `CH_3 I` formed a quaternary salt containing 59.07% iodine. The amine is likely to be:

Options

A. `CH_3 NH_2`
B. `(CH_3)_2 NH`
C. `C_2 H_5 NH_2`
D. `(CH_3)_2 N`

Answer

C

Explanation

Evolution of `N_2` with `HNO_2` suggests that `1^@` amine [either (A) or (C)].
59.07 g of (I) is present in 100 g of compound
127 g of I (1 mol) is present in `(100x127)/(59.07) =215,` Molecular mass = 215
A) `CH_3 NH_2 overset(3CH_3 I)to CH_3 - overset(o+)N (CH_3)_3]I^- , ` (Molecular mass = 201)
C) `C_2 H_5 NH_2 overset(3CH_3 I)to C_2 H_5 - overset(o+)(CH_3)_3]I^- ,` (Molecular mass = 215)
Hence , the amine is `C_2 H_5 NH_2 ` (C).