A) \[C{{H}_{3}}-\overset{OD\,\,\,\,\,\,\,\,\,\,}{\mathop{\overset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C=C{{H}_{2}}}}\,}}\,\]
B) \[C{{D}_{3}}-\overset{O\,\,}{\mathop{\overset{||\,\,\,\,}{\mathop{C-}}\,}}\,C{{D}_{3}}\]
C) \[C{{H}_{2}}=\overset{\,OH}{\mathop{\overset{|\,\,\,\,\,}{\mathop{C-}}\,}}\,C{{H}_{2}}D\]
D) \[C{{D}_{2}}=\overset{OD}{\mathop{\overset{|\,\,\,\,\,}{\mathop{C-}}\,}}\,CD\]
Correct Answer: B
Solution :
\[\underset{\text{Acetone}\,\text{(ketoform)}}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}}}\,\]\[\left[ \underset{\text{enol form}}{\mathop{C{{H}_{3}}-\overset{OH}{\mathop{\overset{|\,\,\,\,\,}{\mathop{C\,\,}}\,}}\,=C{{H}_{2}}}}\, \right]\] ⇌ \[C{{H}_{3}}-\overset{OD}{\mathop{\overset{|}{\mathop{C}}\,=}}\,C{{H}_{2}}\] ⇌ \[C{{H}_{3}}-\overset{O\ \ \ \ \ \ \ \ \ \ }{\mathop{\overset{||}{\mathop{C}}\,-C{{H}_{2}}D}}\,\]
\[C{{H}_{2}}=\overset{OH\ \ \ \ \ \ \ \,}{\mathop{\overset{|}{\mathop{C}}\,-C{{H}_{2}}D}}\,\]
\[C{{H}_{2}}=\overset{OD\ \ \ \ \ \ \ \,}{\mathop{\overset{|}{\mathop{C}}\,-C{{H}_{2}}D}}\,\] ⇌ \[DC{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{2}}D\]⇌\[{{D}_{3}}C-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{D}_{3}}\] 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 \[C{{D}_{3}}\,.\,COC{{D}_{3}}\,(\text{B})\].
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