Question

In the following sequence of reactions
C₂H₅Br □(→┴AgCN) X □(→┴Reduction )Y ; Y is
(a) n-propyl amine (b) isopropylamine
(c) ethylamine (d) ethylmethyl amine

Answer 1

Answer:

In the following sequence of reactions

C₂H₅Br □(→┴AgCN) X □(→┴Reduction )Y ; Y is

(a) n-propyl amine (b) isopropylamine

(c) ethylamine

(d) ethylmethyl amine

Answer 2

Y is

(d) ethylmethyl amine

When silver cyanide (AgCN) reacts with alkyl bromide, alkyl isocyanide is formed.

C₂H₅Br + AgCN → C₂H₅NC + AgBr

When C₂H₅NC is reduced (Hydrogenation), N atom attaches itself to one 1 H atom and the C atom to 3 H atoms to complete their octet.

So, the product formed is → C₂H₅NHCH₃, which is ethylmethyl amine

C₂H₅NC + Reduction → C₂H₅NHCH₃

Option (d) is correct.