Question

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

\[\begin{array}{cc}
\phantom{.......................}\ce{Br}\\
\phantom{......................}|\\
\ce{CH3 - CH - CH - CH3 ->[HBr] CH3 - C - CH2 - CH3}\\
\phantom{.}|\phantom{......}|\phantom{......................}|\phantom{........}\\
\phantom{}\ce{CH3}\phantom{...}\ce{OH}\phantom{...................}\ce{CH3}\phantom{.....}
\end{array}\]

Give a mechanism for this reaction.

(Hint: The secondary carbocation formed in step II rearranges to a more

stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Answer 1

The described reaction is an example of carbocation rearrangement that occurs via hydride shift. The mechanism for it is –

Step 1: Formation of carbocation: Protonation of alcohol.

\[\ce{HBr -> H+ + B\overset{—}{r}}\]

Step 2: 1, 2-hydride shift: Formation of a more stable, 3° carbocation.

Initially, a 2° carbocation (I) was produced. However, the more stable 3° counterpart causes a hydride shift, forming the more stable carbocation (II).

Step 3: Attack of nucleophile: Generation of product.