Answer:
Dehydration of 2, 3-dimethylbutan-2-ol occurs according to Saytzeff rule as a result of which more highly substituted alkene, i.e., 2, 3-dimethylbut-2-ene is formed as the major product.
\[\underset{\begin{smallmatrix} \\ \text{2,}\,\text{3-Dimethylbutan-2-ol} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} {{H}_{3}}C \\ | \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\xrightarrow[-{{H}_{2}}O]{Conc.\,{{H}_{2}}S{{O}_{4}}}}}\,\]\[\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\text{2,}\,\text{3-Dimethylbut-2-ene}\, \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(major}\,\text{product)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{{}}{\overset{\begin{smallmatrix} {{H}_{3}}C \\ | \end{smallmatrix}}{\mathop{C}}}\,=\underset{{}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\]\[\,+\,\,\,\,\,\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\text{2,}\,\text{3-Dimethylbut-1-ene}\, \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(minor}\,\text{product)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{{}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\,\,\,CH}}}\,-\underset{{}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,=C{{H}_{2}}}}\,\]
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