MCQ
$C{H_3}C{H_2}COOH\xrightarrow{{C{l_2}/Fe}}X\mathop {\xrightarrow{{Alcoholic}}}\limits_{K{O_4}} Y$ Compound Y is
- A$C{H_3}C{H_2}OH$
- B$C{H_3}C{H_2}CN$
- ✓$C{H_2} = CHCOOH$
- D$C{H_3}CHClCOOH$
✓
Answer
Correct option: C.
$C{H_2} = CHCOOH$
c
$C{H_3}C{H_2}COOH\xrightarrow{{C{l_2}/Fe}}$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$C{H_3}C{H_2}COOH\xrightarrow{{C{l_2}/Fe}}$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$\mathop {\xrightarrow{{Alcohol}}}\limits_{KOH} C{H_2} = CH - COOH$
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$\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
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\end{matrix}$
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||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
Ph-CH=CH-C-C{{H}_{3}}\to Ph-CH=CH-C{{O}_{2}}H \\
\end{matrix}$
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