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Alkene R−CH=CHX2 reacts with B2H6 in the presence of dilute H2O2, to give: -

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Question

Alkene \[\ce{R - CH = CH2}\] reacts with B2H6 in the presence of dilute H2O2, to give:

Options

  • \[\begin{array}{cc}
    \ce{O}\phantom{..}\\
    ||\phantom{..}\\
    \ce{R - C - CH3}
    \end{array}\]

  • \[\begin{array}{cc}
    \phantom{....}\ce{OH}\phantom{...}\ce{OH}\phantom{}\\
    \phantom{..}|\phantom{......}|\phantom{}\\
    \ce{R - CH - CH2}
    \end{array}\]

  • \[\ce{R - CH2 - CHO}\]

  • \[\ce{R - CH2 - CH2 - OH}\]

MCQ

Solution

\[\ce{R - CH2 - CH2 - OH}\]

Explanation:

\[\ce{\underset{Alkene}{6H - CH = CH2} + \underset{Diborane}{B2H6} -> \underset{Trialkylborane}{2(R - CH2 - CH2)3B}}\]

\[\ce{\underset{Trialkylborane}{(R - CH2 - CH2)3B} + 3H2O2 ->[OH^-] \underset{Alcohol}{3R - CH2 - CH2OH} + B(OH)3}\]

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