A) \[{{C}_{6}}{{H}_{5}}-C{{H}_{2}}-OH\]
B) \[{{C}_{6}}{{H}_{5}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,{{H}_{2}}-{{C}_{2}}{{H}_{5}}\]
C)
D) \[{{C}_{6}}{{H}_{5}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-O{{C}_{2}}{{H}_{5}}\]
Correct Answer: C
Solution :
\[{{C}_{6}}{{H}_{5}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-OC{{H}_{3}}\xrightarrow[(ii)\,\,{{H}^{+}}/{{H}_{2}}O\,(hydrolysis)]{(i)\,\,{{C}_{2}}{{H}_{5}}MgBr\,\,(2\,\,mol)}\] \[{{C}_{6}}{{H}_{5}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ {{C}_{2}}{{H}_{5}} \\ \text{(}\,\,\,{{3}^{o}})\text{ }alcohol \end{smallmatrix}}{\mathop{C}}\,}}\,-{{C}_{2}}{{H}_{5}}\] Note Firstly ketone \[{{C}_{6}}{{H}_{5}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-{{C}_{2}}{{H}_{5}}\] is produced which on further reaction with \[{{C}_{2}}{{H}_{5}}MgBr\]gives tertiary alcohol.
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