A) \[\underset{\underset{Cl}{\mathop{|}}\,}{\mathop{C{{H}_{2}}}}\,-\underset{\underset{Cl}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}\]
B) \[\underset{\underset{Cl}{\mathop{|}}\,}{\mathop{C{{H}_{2}}}}\,-\underset{\underset{Cl}{\mathop{|}}\,}{\mathop{C{{H}_{2}}}}\,-C{{H}_{3}}\]
C) \[CH-\underset{\underset{Cl}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}\]
D)
Correct Answer: D
Solution :
[d] \[\xrightarrow{KOH}C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{\underset{(II)}{\mathop{C}}\,}}\,}}\,-C{{H}_{3}}\xrightarrow[(II)\,\,{{H}_{2}}O/{{H}^{+}}]{(I)\,\,C{{H}_{3}}MgBr}\] \[C{{H}_{3}}-\overset{OH}{\mathop{\overset{|}{\mathop{\underset{\underset{3{}^\circ \,\,alcohol}{\mathop{C{{H}_{3}}}}\,}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{3}}\xrightarrow[Lucas\,\,Test]{Anhy.\,\,ZnC{{l}_{3}}+HCl}Gives\,\,turbidity\,\,immediately\]You need to login to perform this action.
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