A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\]
B) \[C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\]
C) \[C{{H}_{3}}C{{H}_{2}}NHC{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]
Correct Answer: D
Solution :
For reaction, \[\underset{\begin{smallmatrix} \\ 'A' \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\underset{\begin{smallmatrix} -S{{O}_{2}} \\ -HCl \end{smallmatrix}}{\mathop{\xrightarrow{SOC{{l}_{2}}}}}\,\underset{\begin{smallmatrix} \\ 'B' \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}COCl}}\,\] \[\underset{-NHCl}{\mathop{\xrightarrow{N{{H}_{3}}}}}\,\underset{'C'}{\mathop{C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}}}\,\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+B{{r}_{2}} \\ (-2KBr-{{K}_{2}}C{{O}_{3}}, \\ +2{{H}_{2}}O) \end{smallmatrix}}{\mathop{\xrightarrow{4KOH}}}\,\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,'D' \\ (Ethyl\,a\min e) \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}}}\,\] (Hofmann-Bromamide reaction) Hence, compound ?D? is \[C{{H}_{3}}-C{{H}_{2}}-N{{H}_{2}}.\]
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