| (a) | \[{{C}_{2}}{{H}_{5}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OC{{H}_{3}}\] |
| (b) | \[{{C}_{2}}{{H}_{5}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\] |
| (c) | \[{{C}_{2}}{{H}_{5}}C{{H}_{2}}=\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\] |
A) (a) and (b)
B) All of these
C) (a) and (c)
D) (c) only
Correct Answer: B
Solution :
\[{{C}_{2}}{{H}_{5}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\xrightarrow[C{{H}_{3}}OH]{NaOC{{H}_{3}}}\] possible mechanism which takes place is \[{{E}^{2}}\And S{{N}^{1}}\]mechanism. Hence possible products are. \[{{C}_{2}}{{H}_{5}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \\ (S{{N}^{1}}) \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OC{{H}_{3}}\]\[\underbrace{\begin{matrix} {{C}_{2}}{{H}_{5}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}} & {{C}_{2}}{{H}_{5}}CH=\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}} \\ \end{matrix}}_{({{E}^{2}})}\]
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