MCQ
In the given reaction

${C_6}{H_5}CHO+X\mathop {\xrightarrow{{(i)\,Zn\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}\limits_{(ii)\,HOH/N{H_4}Cl} \begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {{C_6}{H_5} - CH - C{H_2} - COO{C_2}{H_5}} 
\end{array}$

$[X]$ will be:

  • A
    $CH_3-COOC_2H_5$
  • B
    $CH_3-CH_2-COOC_2H_5$
  • $Br-CH_2-COOC_2H_5$
  • D

Answer

Correct option: C.
$Br-CH_2-COOC_2H_5$
c

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Similar questions

Arrange acidity of given four compounds in decreasing order :

$(I)$  $\begin{array}{*{20}{c}}
  {O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_3} - C - C{H_2} - C - C{H_3}} 
\end{array}$

$(II)$ $\begin{array}{*{20}{c}}
  O \\ 
  {||} \\ 
  {C{H_3} - C - C{H_3}} 
\end{array}$

$(III)$ $CH \equiv CH$

$(IV)$ $CH_3-CHO$

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