Question

How is the following conversion carried out?

\[\ce{Propene -> Propan-2-ol}\]

Answer 1

If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

\[\begin{array}{cc}
\ce{\underset{Propene}{CH3 - CH = CH2}->[H2O/H+][\Delta]CH3 - CH - CH3}\\
\phantom{.........................}|\\
\phantom{............................}\ce{\underset{Propan-2-ol}{OH}}\
\end{array}\]

Answer 2

\[\begin{array}{cc}
\ce{\underset{Propene}{CH3 - CH = CH2} + Conc.H2SO4 -> CH3 - CH - CH3 ->[H2O, \Delta][-H2SO4] CH3 - CH - CH3}\\
\phantom{.....................................}|\phantom{..........................}|\\
\phantom{..........................................}\ce{OSO3H}\phantom{.................}\ce{\underset{Propan-2-ol}{OH}}\
\end{array}\]