Question

The enol form of acetone, after treatment with $D_2,O,$ gives

• A. $C{H}_3-\stackrel{\stackrel{OD}{|}}{C}=C{H}_2$
• B. $C{D}_3-\stackrel{\stackrel{O}{||}}{C}=C{D}_3$
• C. $C{H}_2=\stackrel{\stackrel{OH}{|}}{C}=C{H}_{2}D$
• D. $C{D}_2=\stackrel{\stackrel{OD}{|}}{C}=C{D}_2$

Answer 1

The correct option is B $C{D}_3-\stackrel{\stackrel{O}{||}}{C}=C{D}_3$

$\underset{Acetone\left(ketoform\right)}{C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_3}\left[\underset{enolform}{C{H}_3-\stackrel{\stackrel{OH}{|}}{C}=C{H}_2}\right]\rightleftharpoons C{H}_2-\stackrel{\stackrel{OD}{|}}{C}=C{H}_2\rightleftharpoons C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_{2}D\rightleftharpoons C{H}_2=\stackrel{\stackrel{OH}{|}}{C}-C{H}_{2}D\stackrel{D_{2}O}{\rightleftharpoons }C{H}_2=\stackrel{\stackrel{OD}{|}}{C}=C{H}_{2}D\rightleftharpoons DC{H}_2-\stackrel{\stackrel{O}{||}}{C}=C{H}_{2}D\rightleftharpoons D_{3}C-\stackrel{\stackrel{O}{||}}{C}=C{D}_3$
The enol form of acetone on treatment with $D_{2}O$ undergoes enolisation, deutration (addition of ( $D_{2}O$) and dehydration (removal of ( $H_{2}O$). The repeated enolisation, deutration and dehydration ultimately gives ( $C{D}_3.COC{D}_3\left(B\right)$.