A) I and III
B) II and III
C) III and IV
D) II and IV
Correct Answer: A
Solution :
\[I.\,C{{H}_{3}}CHO\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,C{{H}_{3}}Mgl}\]\[\begin{align} & OH \\ & | \\ & C{{H}_{3}}-CH-C{{H}_{3}} \\ \end{align}\] \[II.\,C{{H}_{2}}O\xrightarrow[(ii)\,{{h}_{2}}O]{(i){{C}_{2}}{{H}_{5}}Mgl}{{C}_{2}}{{H}_{5}}C{{H}_{2}}OH\] \[III.\,\,C{{H}_{3}}CH=C{{H}_{2}}+{{H}_{2}}O\xrightarrow{{{H}^{+}}}\] \[\begin{align} & H \\ & | \\ & C{{H}_{3}}-CH-C{{H}_{3}} \\ \end{align}\] \[\begin{align} & \\ & IV.\,C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[Neutral]{KMn{{O}_{4}}}C{{H}_{3}}CH-C{{H}_{2}} \\ & |\,\,| \\ & OH\,\,OH \\ \end{align}\]You need to login to perform this action.
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