Question

The carbonyl compound producing an optically active product by reaction with $LiAl{H}_4$ is:

• A. propanone
• B. butanone
• C. 3-pentanone
• D. benzophenone.

Answer 1

The correct option is B butanone

$C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3\stackrel{LiAI{H}_4}{\to }C{H}_3\stackrel{OH}{\stackrel{|}{C}H}C{H}_3$

$C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_{2}C{H}_3\stackrel{LiAl{H}_4}{\to }C{H}_3-\underset{\underset{Opticallyactive}{H}}{\underset{|}{\stackrel{OH}{\stackrel{\star |}{C}}}}-C{H}_{2}C{H}_3$

$H_{3}CC{H}_2\stackrel{O}{\stackrel{||}{C}}-C{H}_{2}C{H}_3\stackrel{LiAl{H}_4}{\to }{H}_{3}CC{H}_2\underset{H}{\underset{|}{\stackrel{OH}{\stackrel{|}{C}}}}-C{H}_{2}C{H}_3$

Option B is correct.