MCQ
An optically active compound $X$ has molecular formula $C_4H_8O_3.$ It evolves $CO_2$ with $NaHCO_3.$  $'X' $ Reacts with $LiAlH_4$ to give an achiral compound. $'X'$ is
  • A
    $\begin{array}{*{20}{c}}
    {C{H_3} - C{H_2} - CH - COOH}\\
    {\,\,\,\,|}\\
    {\,\,\,\,\,\,\,OH}
    \end{array}$
  • B
    $\begin{array}{*{20}{c}}
    {C{H_3} - CH - COOH}\\
    {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
    {\,\,\,\,\,\,\,\,\,C{H_2} - OH}
    \end{array}$
  • $\begin{array}{*{20}{c}}
    {C{H_3} - CH - COOH}\\
    \,\,\,\,{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
    \,\,\,\,{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
    {C{H_3} - CH - C{H_2} - COOH}\\
    {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
    {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
    \end{array}$

Answer

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

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