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NEET Sample Paper NEET Sample Test Paper-6

  • question_answer
    An optically active compound 'X' has molecular formula \[{{C}_{4}}{{H}_{8}}{{O}_{3}}\]. It evolves \[C{{O}_{2}}\]with aq. \[NaHC{{O}_{3}}\]. 'X' reacts with \[LiAl{{H}_{4}}\] to give an achiral compound. The compound 'X' is

    A)  \[C{{H}_{3}}C{{H}_{2}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]

    B)  \[C{{H}_{3}}\underset{OMe}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]

    C)  \[C{{H}_{3}}\underset{C{{H}_{2}}OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]

    D)  \[C{{H}_{3}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HC{{H}_{2}}COOH\]

    Correct Answer: C

    Solution :

    \[X\xrightarrow{NaHC{{O}_{3}}}C{{O}_{2}}\] Hence, compound X has\[~-COOH\] group. \[\underset{\begin{smallmatrix}  \text{Chiral compound} \\  \text{optically active} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{2}}OH}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{|}{\mathop{C*}}\,}}\,}}\,}}\,-COOH}}\,\xrightarrow{LiAl{{H}_{4}}}\underset{\text{Achiral compound}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{2}}OH}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{2}}OH}}\,\]


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