question_answer

In the reaction below, X is Neo-pentyl alcohol \[\xrightarrow{{{H}_{2}}S{{O}_{4}}}\,\,X\]

A)  2-methylpentane

B)  2-methylpent-2-ene

C)  2-methylbut-2-ene

D)  neo-pentane

Correct Answer: C

Solution :

\[\xrightarrow[-{{H}_{2}}O]{}C{{H}_{3}}-\overset{\begin{smallmatrix}  C{{H}_{3}} \\  | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix}  | \\  C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-\overset{+}{\mathop{C}}\,{{H}_{3}}\xrightarrow{Methyl\,\,shift}\]\[C{{H}_{3}}-\overset{\oplus }{\mathop{\underset{\begin{smallmatrix}  | \\  C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{2}}C{{H}_{3}}\xrightarrow[-{{H}^{+}}]{}\]\[C{{H}_{3}}-\underset{\begin{smallmatrix}  C{{H}_{3}} \\  2-methylbut-2-ene \end{smallmatrix}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=CHC{{H}_{3}}\]