3, 3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism.

[IIT, 1996]

Question

3, 3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism.

[IIT, 1996]

Accepted Answer

The mechanism is as follows:

(a) The protonation of hydroxyl group.

[latex] \begin{matrix} && CH_3 &&&&&&&& CH_3 \ && | &&&&&&&& |\ CH_3 & - & C & - & CH&-& CH_3 & \stackrel{ H^-}{\longrightarrow} & CH_3 &-&C& -&CH& - & CH_3 \ && | &&|&&&&&& |&& | \ && H_3C &&OH&&&&&& H_3C&& {}^{+}OH_2 \end{matrix} [/latex]

3,3-Dimethylbutan-2-ol

(b) The removal of [latex]H_2O [/latex] to form a secondary (2°) carbonium ion

[latex] \begin{matrix} && CH_3 &&&&&&&& CH_3 \ && | &&&&&&&& |\ CH_3 & - & C & - & CH&-& CH_3 & \stackrel{ -H_2O}{\longrightarrow} & CH_3 &-&C& -&{}^{+}CH& - & CH_3 \ && | &&|&&&&&& |&& \ && H_3C &&OH_2 &&&&&& CH_3&& \end{matrix} [/latex]

(c) The conversion of 2° carbonium to themore stable 3° carbonium ion by the shift of [latex] CH_3 [/latex] group

[latex] \begin{matrix} && CH_3 &&&&&&&& CH_3\ && | &&&&&&&&|\ CH _3 &-& C& -&{ }^{+} CH &-& CH _3 & \longrightarrow & CH _3&-&{ }^{+} C &-& CH &-& CH _3 \ && |&&&&&&&&&& | \ && CH_3&&&&&&&&&& CH_3 \end{matrix} [/latex]

(d) The removal of [latex] H^+ [/latex] to form a double bond

[latex] \begin{matrix} CH _3 & -& { }^{-} C &-& CH &-& CH _3& \overset{-H^-}{\xrightarrow {\hspace{2 cm}} } & CH _3 &-& C &=& C &-& CH _3 \ && | &&| &&&&&& |&&| \ && H_3C &&CH_3 &&&&&& H_3C&&CH_3\end{matrix} [/latex]

Tetramethyl ethylene

Question 5.SSE.11: 3, 3-Dimethylbutan-2-ol loses a molecule of water in the pre......

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