A) Methyl amine
B) Acetamide
C) Ethyl cyanide
D) Ethyl amine
Correct Answer: D
Solution :
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.
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