A) butene-1
B) butene-2
C) ethylene
D) propylene
Correct Answer: B
Solution :
A carbonium ion is stabilised by the presence of electron repelling groups on it. The \[+ve\] \[IE\] of methyl group transmits \[-ve\] charge on \[{{C}^{+}}\] to disperse its \[+ve\] charge and stabilise it. \[\underset{methyl\,{{1}^{o}}}{\mathop{H-{{\underset{H}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,}^{\oplus }}}}\,\] \[\underset{carbonium\,ion\,{{2}^{o}}}{\mathop{H-{{\underset{H}{\mathop{\underset{\begin{smallmatrix} | \\ \uparrow \end{smallmatrix}}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,}^{\oplus }}}}\,\] \[\underset{carbonium\,\,ion\,{{3}^{o}}}{\mathop{H-\underset{C{{H}_{3}}}{\mathop{\underset{\begin{smallmatrix} | \\ \uparrow \end{smallmatrix}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{\begin{smallmatrix} \downarrow \\ | \end{smallmatrix}}{\mathop{{{C}^{\oplus }}}}\,}}\,}}\,}}\,}}\,\] \[\underset{carbonium\,\,ion\,{{4}^{o}}}{\mathop{C{{H}_{3}}\to -\underset{C{{H}_{3}}}{\mathop{\underset{\begin{smallmatrix} | \\ \uparrow \end{smallmatrix}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{\begin{smallmatrix} \downarrow \\ | \end{smallmatrix}}{\mathop{{{C}^{\oplus }}}}\,}}\,}}\,}}\,}}\,\] \[\because \]Stability order: \[{{3}^{o}}>{{2}^{o}}>1%>\]methyl carbonium or\[tert-butyl>iso-propyl>ethyl>methyl\]and reactivity order is\[\,{{3}^{o}}<{{2}^{o}}<{{1}^{o}}<methylcarbonium\text{ }ion\]
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