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Question
How is the following conversion carried out?
\[\ce{Propene -> Propan-2-ol}\]
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Solution 1
If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.
\[\begin{array}{cc}
\ce{\underset{Propene}{CH3 - CH = CH2}->[H2O/H+][\Delta]CH3 - CH - CH3}\\
\phantom{.........................}|\\
\phantom{............................}\ce{\underset{Propan-2-ol}{OH}}\
\end{array}\]
Solution 2
\[\begin{array}{cc}
\ce{\underset{Propene}{CH3 - CH = CH2} + Conc.H2SO4 -> CH3 - CH - CH3 ->[H2O, \Delta][-H2SO4] CH3 - CH - CH3}\\
\phantom{.....................................}|\phantom{..........................}|\\
\phantom{..........................................}\ce{OSO3H}\phantom{.................}\ce{\underset{Propan-2-ol}{OH}}\
\end{array}\]
Notes
Students can refer to the provided solutions based on their preferred marks.
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