Q. $N_{2}$ gas is liberated when $\left\lbrack HCl + NaNO_{2} \right\rbrack\$reacts with the following compounds (A)$CH_{3}CH_{2}NH_{2}$ (B) Urea (C)$CH_{3}CONH_{2}$ (D)$C_{6}H_{5}NH_{2}$ The answer is
A) A, B, C
B) B, C, D
C) A, C, D
D) A, B, D
Explanation
(a)
$CH_{3}CH_{2}NH_{2}\overset{NaNO_{2} + HCl\ \ \ }{\rightarrow}CH_{3}\ CH_{2}\ OH + N_{2}\ + H_{2}O + NaCl$
Ethylamine
$NH_{2}CONH_{2}\overset{NaNO_{2} + HCl\ \ \ }{\rightarrow}2N_{2} + H_{2}O + CO_{2} + NaCl$
Urea
$CH_{3}CONH_{2}\overset{NaNO_{2} + HCl\ \ \ }{\rightarrow}CH_{3}\ COOH + N_{2}\ + H_{2}O + NaCl$
Acetamide
$C_{6}H_{5}NH_{2}\overset{NaNO_{2} + HCl\ \ \ }{\rightarrow}C_{6}H_{5}N_{2}^{+}Cl^{-}H_{2}O + NaCl$
Aniline
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.