Question

An optically active compound '$X$' has molecular formula $C_4{H}_8{O}_3$. It evolves $C{O}_2$ with aqueous $NaHC{O}_3$. '$X$' reacts with $LiAl{H}_4$ to give an achiral compound. '$X$' is :

• A. $C{H}_{3}C{H}_{2}CHCOOH$
  $|$ $OH$
• B. $C{H}_{3}CHCOOH$ $|$ $OMe$
• C. $C{H}_{3}CHCOOH$
  $|$ $C{H}_{2}OH$
• D. $C{H}_{3}CHC{H}_{2}COOH$
  $|$ $OH$

Answer 1

The correct option is C

$C{H}_{3}CHCOOH$

$|$

$C{H}_{2}OH$

Since, $X\stackrel{NaHC{O}_3}{\to }C{O}_2$

Hence, it must contain $-COOH$ group.