MCQ
Which optically active compound on reduction with $LiAlH_4$ will give optically inactive compound ?
  • A
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - COOH} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      \,\,{OC{H_3}} 
    \end{array}$
  • B
    $\begin{array}{*{20}{c}}
      {C{H_3} - C{H_2} - CH - COOH} \\ 
      {\,\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} 
    \end{array}$
  • $\begin{array}{*{20}{c}}
      {C{H_3} - C{H_2} - CH - COOH} \\ 
      {\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}OH} 
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - C{H_2} - COOH} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$

Answer

Correct option: C.
$\begin{array}{*{20}{c}}
  {C{H_3} - C{H_2} - CH - COOH} \\ 
  {\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}OH} 
\end{array}$
c

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$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} \\ 
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\end{array}$

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$\begin{matrix}
   O\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   ||\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{3}}-C-Cl-N{{H}_{3}}  \\
\end{matrix}$ $\to $ Inter mediate $\to $ product