question_answer

\[{{N}_{2}}\]gas is liberated when\[[HCl+NaN{{O}_{2}}]\]reacts with the following compound \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]           Urea \[C{{H}_{3}}CON{{H}_{2}}\]         \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\] The answer is

A)  a, b, c                                  

B)  a, b, d

C)  a, c, d                                  

D)  b, c, d

Correct Answer: A

Solution :

                 \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}C{{H}_{3}}c{{H}_{2}}OH\] \[+{{N}_{2}}+NaCl+{{H}_{2}}O\]  \[N{{H}_{2}}CON{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}\] \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}C{{H}_{3}}COOH+{{N}_{2}}\]          \[+{{H}_{2}}O+NaCl\]