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