Question

What is the end product in the following reaction? ${{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow{HN{{O}_{2}}}A\xrightarrow{PC{{l}_{5}}}B\xrightarrow{N{{H}_{3}}}C$

• A. Methyl amine
• B. Acetamide
• C. Ethyl cyanide
• D. Ethyl amine

Answer 1

Answer: (D)

Ethyl amine

Answer 2

The reactions occur in the following way ${{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow[-{{N}_{2}}]{HN{{O}_{2}}}\underset{\text{(A)}}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{PC{{l}_{5}}}$ $\underset{(B)}{\mathop{{{C}_{2}}{{H}_{5}}Cl}}\,\xrightarrow{N{{H}_{3}}}\,\underset{(C)\,\,Ethyl\,\,a\min e}{\mathop{{{C}_{2}}{{H}_{5}}N{{H}_{2}}}}\,$ Thus, ethyl amine is obtained back.