| I. \[C{{H}_{2}}=CH-C{{H}_{3}}+{{H}_{2}}O\xrightarrow{H+}\] |
| II. \[C{{H}_{3}}-CHO\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,C{{H}_{3}}MgI}\] |
| III. \[C{{H}_{2}}O\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,{{C}_{2}}{{H}_{5}}MgI}\] |
| IV. \[C{{H}_{2}}=CH-C{{H}_{3}}\xrightarrow{Neutral\,KMn{{O}_{4}}}\] |
A) I and II
B) II and III
C) III and I
D) II and IV
Correct Answer: A
Solution :
[a] I. \[C{{H}_{3}}CH=C{{H}_{2}}+{{H}_{2}}O\xrightarrow{{{H}^{+}}}C{{H}_{3}}\overset{OH}{\mathop{\overset{|}{\mathop{C}}\,}}\,HC{{H}_{3}}\] II. \[C{{H}_{3}}CHO\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,C{{H}_{3}}MgI}C{{H}_{3}}+\overset{OH}{\mathop{\overset{|}{\mathop{C}}\,}}\,HC{{H}_{3}}\] III. \[C{{H}_{2}}O\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,{{C}_{2}}{{H}_{5}}MgI}{{C}_{2}}{{H}_{5}}C{{H}_{2}}OH\] IV.\[C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow[neutral]{KMn{{O}_{4}}}C{{H}_{3}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}\]
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