A) 2-methyl-1-butene
B) 2-methyl-2-butene
C) 2, 2-dimethyl-1-butene
D) 2-butene
Correct Answer: B
Solution :
\[\begin{array}{*{35}{l}} \,\,\,\,\,\,\,\,\,\,\ \ C{{H}_{3}} \\ C{{H}_{3}}-\underset{|}{\overset{|}{\mathop{C}}}\,\ -C{{H}_{2}}-Br+\underset{\text{(alc)}}{\mathop{KOH}}\,\xrightarrow{{}} \\ \ \ \ \ \ \ \ \ C{{H}_{3}} \\ \end{array}\]\[\begin{array}{*{35}{l}} \,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ C{{H}_{3}}-\overset{|}{\mathop{C}}\,=\ CH-C{{H}_{3}}+KBr+{{H}_{2}}O \\ \end{array}\] In this reaction \[{{1}^{o}}\] carbonium ion is formed which rearranges to form \[{{3}^{o}}\] carbonium ion from which base obstruct proton. Hence 2-methyl-2-butene is formed as a main product. \[\underset{{{\text{1}}^{\text{o }}}\text{carbonium less stable}}{\mathop{\begin{array}{*{35}{l}} \,\,\,\,\,\,\,\,\,\,\ \ C{{H}_{3}} \\ C{{H}_{3}}-\underset{|}{\overset{|}{\mathop{C}}}\,\ -\ \overset{+}{\mathop{C}}\,{{H}_{2}} \\ \ \ \ \ \ \ \ \ C{{H}_{3}} \\ \end{array}}}\,\xrightarrow{\text{Methyl shift}}C{{H}_{3}}-\underset{{}}{\overset{C{{H}_{3}}}{\mathop{\underset{+\,\,\,\,}{\overset{|\,\,\,\,}{\mathop{C\,-\,}}}\,}}}\,C{{H}_{2}}-C{{H}_{3}}\]
\[\underset{\text{2-Methyl-2-Butene}}{\mathop{C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|\,\,\,\,\,\,\,\,}{\mathop{C\,\,=\,\,}}\,}}\,CH-C{{H}_{3}}}}\,\]
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