Question

A compound A $\left(C_5{H}_{10}C{l}_2\right)$ on hydrolysis give $C_5{H}_{10}O$ which reacts with $N{H}_{2}OH$, forms iodoform but does not give Fehling test A is

• A. $C{H}_3-\underset{Cl}{\underset{|}{\stackrel{Cl}{\stackrel{|}{C}}}}-C{H}_{2}C{H}_{2}C{H}_3$
• B. $C{H}_{3}C{H}_2-\underset{Cl}{\underset{|}{\stackrel{Cl}{\stackrel{|}{C}}}}-C{H}_{2}C{H}_3$
• C. $C{H}_3-\underset{Cl}{\underset{|}{CH}}-\underset{Cl}{\underset{|}{CH}}-C{H}_{2}C{H}_3$
• D. $C{H}_{3}CH-2C{H}_{2}C{H}_2-\stackrel{Cl}{\stackrel{|}{CH}}-Cl$

Answer 1

The correct option is A $C{H}_3-\underset{Cl}{\underset{|}{\stackrel{Cl}{\stackrel{|}{C}}}}-C{H}_{2}C{H}_{2}C{H}_3$

$C{H}_3-\underset{\left(A\right)}{\underset{Cl}{\underset{|}{\stackrel{Cl}{\stackrel{|}{C}}}}}-C{H}_{2}C{H}_{2}C{H}_3\stackrel{Hydrolysis}{\to }$
$C{H}_3-\underset{methylketone}{\stackrel{O}{\stackrel{||}{C}}-C{H}_{2}C{H}_{2}C{H}_3}$
Since, compound A on hydrolysis gives $C_5{H}_{10}O$ which does rot reduce Fehling's solution (so not an aldehyde) but reacts with $N{H}_{4}OH$, forms iodofom, therefore, it must be methyl ketone.