Q. Identify X and Y in the following sequence $C_{2}H_{5}Br\ \overset{\ \ \ \ \ \ X\ \ \ \ \ }{\rightarrow}\ product\ \overset{\ \ \ \ \ \ Y\ \ \ \ \ \ \ }{\rightarrow}C_{3}H_{7}NH_{2}$
A) $X = KCN,\ \ \ \ \ \ \ \ \ Y = LiAlH_{4}$
B) $X = KCN,\ \ \ \ \ \ \ \ \ Y = H_{3}O^{+}$
C) $X = \ CH_{3}\ Cl,\ \ \ \ \ Y = AlCl_{3}/HCl$
D) $X = \ CH_{3}\ NH_{2},\ \ Y = HNO_{2}$
Explanation
(a)
Ethyl bromide on treating with KCN, gives ethyl cyanide, which on reduction gives propyl amine.
[image:10_image67.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.