• # question_answer The major product of the following reaction is: $C{{H}_{3}}\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}\,HC{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\xrightarrow[triethyla\min e]{ehtyl\,formate\,(lequiv.)}$ [JEE Main 10-4-2019 Morning] 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) $C{{H}_{3}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}\,H-CH=C{{H}_{2}}$

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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