Question

Consider the following reaction \[\ce{CH3 - CH = CH2 ->[1. HBr][2. aq. KOH]}\] The major end product is

पर्याय

• \[\begin{array}{cc}
  \ce{CH3 - CH - CH3}\\
  |\phantom{..}\\
  \ce{OH}\\
  \end{array}\]

• \[\begin{array}{cc}
  \ce{CH3 - CH - CH3}\\
  |\phantom{..}\\
  \ce{Br}\\
  \end{array}\]

• \[\ce{CH3 - CH2 - CH2 - OH}\]

• \[\ce{CH3 - CH2 - CH2 - Br}\]

Answer 1

\[\begin{array}{cc}
\ce{CH3 - CH - CH3}\\
|\phantom{..}\\
\ce{OH}\\
\end{array}\]

Explanation:

Propene produces two products, although only one prevails according to Markovnikov's rule: 2-Bromopropane. When heated with aqueous \[\ce{KOH}\], it produces secondary alcohol. Because aqueous \[\ce{KOH}\] is alkaline, it produces hydroxide ions, which are nucleophilic. It substitutes halide (bromide) ions to produce alcohols.