question_answer

\[C{{H}_{3}}-CH=C{{H}_{2}}\xrightarrow[(ii)\,NaB{{H}_{4}}]{(i)\,Hg{{(OAc)}_{2}}/{{H}_{2}}O}X+Na\] \[\xrightarrow{{}}Y+C{{H}_{3}}Cl\xrightarrow{{}}Z+HI\xrightarrow{0{}^\circ C}A+B\] What are A and B?

A) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-OH\And C{{H}_{3}}I\]

B) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-OH\And C{{H}_{3}}I\]

C) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-I\And C{{H}_{3}}OH\]

D) \[C{{H}_{3}}-C{{H}_{2}}C{{H}_{2}}I\And C{{H}_{3}}OH\]

Correct Answer: D

Solution :

[d]

Going backward from A and B
	\[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow{{}}C{{H}_{3}}C{{H}_{2}}\underset{(X)}{\mathop{C{{H}_{2}}}}\,OH\]