A) \[C{{H}_{3}}\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}\,HC{{H}_{2}}C{{H}_{2}}NHCHO\]
B) \[C{{H}_{3}}CH=CH-C{{H}_{2}}N{{H}_{2}}\]
C)
D) \[C{{H}_{3}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}\,H-CH=C{{H}_{2}}\]
Correct Answer: A
Solution :
\[C{{H}_{3}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}-C{{H}_{2}}-N{{H}_{2}}\xrightarrow[triethyla\min e]{ethyl\,formate\,(lequiv.)}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{2}}-NH-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H\] as \[N{{H}_{2}}\] is a better nucleophile than OH.
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