An optically active compound $'X'$ having molecular formula $C_4H_8O_3$ evolves $CO_2$ with $NaHCO_3$. $'X'$ on treatment with $LiAlH_4$ gives an achiral compound. Then $'X'$ is

  • A
    $CH_3-CH(OH)-CH_2-COOH$
  • B
    $CH_3-CH_2-CH(OH)-COOH$
  • C
    $HO-CH_2-CH(CH_3)-COOH$
  • D
    $CH_3-CH(OCH_3)-COOH$

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Complete the following reaction:
$[C]$ is $........$

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