A) A more stable carbonium ion
B) A more stable carbanion
C) A more stable free radical
D) None of the above being a concerted reaction
Correct Answer: A
Solution :
\[HI\to {{H}^{+}}+{{I}^{-}}\]\[C{{H}_{3}}-CH=C{{H}_{2}}+{{H}^{+}}\to \]\[\underset{\begin{smallmatrix} (\text{Minor)} \\ {{\text{1}}^{\text{o}}}\,\text{Carbonium ion} \\ \text{(Less stable)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-\overset{+}{\mathop{C}}\,{{H}_{2}}}}\,\,\,+\,\,\underset{\begin{smallmatrix} (\text{Major)} \\ {{\text{2}}^{\text{o}}}\,\text{Carbonium ion} \\ \text{(More stable)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{+\,\,\,}{\mathop{CH}}\,-C{{H}_{3}}}}\,\] \[\underset{{{2}^{o}}\,\text{Carbonium}\,\text{ion}}{\mathop{C{{H}_{3}}-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}+{{I}^{-}}}}\,\to \underset{\begin{smallmatrix} \text{Isopropyl iodide} \\ \text{(Major product)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\underset{I}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}}}\,\]
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