A) sec-butyl
B) ter-butyl
C) n-butyl
D) None of these
Correct Answer: B
Solution :
\[\underset{\,\,\,\,\,\,\,\,\,{{3}^{o}}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{\,\,C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,}{\overset{|\,\,\,\,}{\mathop{{{C}^{+}}\,}}}\,}}}\,>}}\,\,\underset{\,\,\,\,\,\,\,\,\,\,\,2{}^\circ }{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,}{\mathop{CH{{_{{}}^{{}}}^{+}}}}\,}}\,>}}\,\,\underset{\begin{matrix} \begin{matrix} {} \\ \begin{matrix} {} \\ {{1}^{o}} \\ \end{matrix} \\ \end{matrix} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{matrix}}{\mathop{C{{H}_{3}}-CH{{_{2}^{{}}}^{+}}}}\,>\underset{\begin{matrix} {} \\ \begin{matrix} {} \\ \begin{smallmatrix} \text{Methyl } \\ \text{carbocation} \end{smallmatrix} \\ \end{matrix} \\ \end{matrix}}{\mathop{CH{{_{3}^{{}}}^{+}}\,\,\,\,\,\,\,}}\,\] Greater the no. of alkyl groups, greater would be the dispersal of the charge and hence more stable will be the carbonium ion.
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