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CHX3CHX2CH=CHX2→HX2O,HX2OX2,OHX−BX2HX6X What is X? -

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प्रश्न

\[\ce{CH3CH2CH = CH2 ->[B2H6][H2O, H2O2, OH^-] X}\]

What is X?

विकल्प

  • \[\ce{CH3CH2CH2CH3}\]

  • \[\ce{CH3CH2CH2CH2OH}\]

  • \[\begin{array}{cc}
    \ce{CH3CH2CHCH3}\\
    \phantom{...}|\\
    \phantom{.....}\ce{OH}
    \end{array}\]

  • \[\ce{CH3CH2CH2CHO}\]

MCQ

उत्तर

\[\ce{CH3CH2CH2CH2OH}\]

Explanation:

According to Anti-rule, Markovnikov's hydroboration-oxidation process involves the addition of −H and −OH to unsymmetrical alkenes. As a result, butan-1-ol is the end product.

\[\ce{\underset{But-1-ene}{6CH3CH2CH = CH2} + \underset{Diborane}{B2H6} -> 2(CH3CH2CH2CH2)3B (CH3CH2CH2CH2)3B + 3H2O2 ->[OH^-][-B(OH)3] \underset{Butan-1-ol}{3CH3CH2CH2CH2OH}}\]

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Alkenes
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