A) \[C{{H}_{3}}-\underset{OD}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\]
B) \[C{{H}_{3}}-\underset{O}{\mathop{\underset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}\]
C) \[C{{H}_{2}}=\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,-C{{H}_{2}}D\]
D) \[C{{D}_{2}}=\underset{OD}{\mathop{\underset{|}{\mathop{C}}\,}}\,-C{{D}_{3}}\]
Correct Answer: A
Solution :
After treatment with \[{{\text{D}}_{\text{2}}}\text{O,}\] the \[{{\text{H}}^{\text{+}}}\] ion of \[-\text{OH}\]group is replaced by \[{{\text{D}}^{\text{+}}}\] ion, because of being more reactive than deuterium. \[C{{H}_{3}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\xrightarrow{{{D}_{2}}O}C{{H}_{3}}-\underset{OD}{\mathop{\underset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\]
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