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
$C{H_3}C{H_2}COOH\xrightarrow{{C{l_2}/Fe}}$ $\begin{array}{*{20}{c}}
  {C{H_3} - CH - COOH} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {Cl\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

$\mathop {\xrightarrow{{Alcohol}}}\limits_{KOH} C{H_2} = CH - COOH$

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