Question

An optically active compound 'X'has molecule formula C4H8O3. It invloes CO2 with NaHCO3. 'X' react with LiAIH 4 to gives an achiral compounds

• A. $C{H}_{3}C{H}_2\underset{OH}{\underset{|}{CH}}COOH$
• B. $C{H}_{3}C{H}_2\underset{Me}{\underset{|}{CH}}COOH$
• C. $C{H}_{3}C{H}_2\underset{C{H}_{2}OH}{\underset{|}{CH}}COOH$
• D. $C{H}_3\underset{OH}{\underset{|}{C}}HC{H}_{2}COOH$

Answer 1

Answer: (C)

$C{H}_{3}C{H}_2\underset{C{H}_{2}OH}{\underset{|}{CH}}COOH$

$X\to$ optically active compound

$x\stackrel{NaHC{O}_3}{\to }\underset{\left(3\right)}{C{O}_2}\uparrow$

Hence, it must contain $-COOH$ group

$\underset{\left(x\right)}{C{H}_3-C{H}_2-\underset{C{H}_{2}OH}{\underset{|}{\stackrel{H}{\stackrel{|}{C}}}}-COOH}\underset{hydrogenation}{\overset{LiAI{H}_4}{\to }}\underset{aciralcompound}{C{H}_3-C{H}_2-\stackrel{H}{\stackrel{|}{\underset{C{H}_{2}OH}{\underset{|}{C}}}}-C{H}_{2}OH}$