Question
An ester $$(A)$$ with molecular formula $${C_9}{H_{10}}{O_2}$$ was treated with excess of $$C{H_3}MgBr$$ and the complex so formed was treated with $${H_2}S{O_4}$$ to give an olefin $$(B).$$ Ozonolysis of $$(B)$$ gave a ketone with molecular formula $${C_8}{H_8}O$$ which shows $$+ve$$ iodoform test. The structure of $$(A)$$ is
A.
$${C_6}{H_5}COO{C_2}{H_5}$$
B.
$$C{H_3}COC{H_2}CO{C_6}{H_5}$$
C.
$$p{\text{ - }}C{H_3}O{\text{ - }}{C_6}{H_4}{\text{ - }}COC{H_3}$$
D.
$${C_6}{H_5}COO{C_6}{H_5}$$
Answer :
$${C_6}{H_5}COO{C_2}{H_5}$$
Solution :
Since ketone with molecular formula $${C_8}{H_8}O$$ shows $$+ve$$ iodoform test, therefore, it must be a methyl ketone, i.e., $${C_6}{H_5}COC{H_3}.$$ Since this ketone is obtained by the ozonolysis of an olefin $$(B)$$ which is obtained by the addition of excess of $$C{H_3}MgBr$$ to an ester $$(A)$$ with molecular formula $${C_9}{H_{10}}{O_2},$$ therefore, ester $$(A)$$ is $${C_6}{H_5}COO{C_2}{H_5}$$ and the olefin $$(B)$$ is
\[\overset{\begin{smallmatrix}
C{{H}_{3}} \\
|\,\,\,\,\,
\end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}C=C{{H}_{2}}}}\,\] as explained below :
\[\underset{A\left( M.F.\,{{C}_{9}}{{H}_{10}}{{O}_{2}} \right)}{\mathop{{{C}_{6}}{{H}_{5}}COO{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[\left( \text{ii} \right)\,\frac{{{H}^{+}}}{{{H}_{2}}O}]{\left( \text{i} \right)\,2C{{H}_{3}}MgBr}\] \[{{C}_{6}}{{H}_{5}}\underset{\begin{smallmatrix}
| \\
\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,\,\,\,\,C{{H}_{3}} \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,OH\xrightarrow[-{{H}_{2}}O]{{{H}_{2}}S{{O}_{4}}}\]
\[{{C}_{6}}{{H}_{5}}\underset{\left( B \right)}{\overset{\begin{smallmatrix}
\,\,\,\,\,C{{H}_{3}} \\
|
\end{smallmatrix}}{\mathop{-C=}}}\,C{{H}_{2}}\xrightarrow[Zn]{{{O}_{3}}}\underset{\text{M}\text{.F}\text{.}\,\,{{C}_{8}}{{H}_{8}}O}{\mathop{{{C}_{6}}{{H}_{5}}COC{{H}_{3}}}}\,+HCHO\]