A) \[{{\operatorname{CH}}_{2}} = CHCl\]
B) \[\operatorname{ClCH} = CHCl\]
C) \[{{\operatorname{CH}}_{3}}- CH = C{{H}_{2}}\]
D) \[{{(C{{H}_{3}})}_{2}} C = CH,\]
Correct Answer: D
Solution :
\[{{\left( C{{H}_{3}} \right)}_{2}}C=C{{H}_{2}}\] Greater the number of electron - donating group attached to the double bond, more is the k- electron density and hence easiest is the addition. \[\begin{align} & {{\left( C{{H}_{3}} \right)}_{2}}C=C{{H}^{2}}~\xrightarrow[\left( Mark\,addn. \right)]{HBr} C{{H}_{3}}\overset{\overset{C{{H}_{3}}}{\mathop{|}}\,}{\mathop{\underset{\underset{Br}{\mathop{|}}\,}{\mathop{C}}\,}}\,C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,Tert-butyl bromide \\ \end{align}\]
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