MHT CET · Chemistry · Hydrocarbons
What is the product formed when \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\) is treated with \(\mathrm{B}_2 \mathrm{H}_6\) followed by the action of \(\mathrm{H}_2 \mathrm{O}_2\)
- A \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3\)
- B \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)
- C \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)
- D \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3\)
Answer & Solution
Correct Answer
(C) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \frac{{ }^{(\mathrm{i})} \mathrm{B}_2 \mathrm{H}_6}{\underset{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}}{\longrightarrow}} \mathrm{CH}_3-\mathrm{CH}_2\) \(-~\mathrm{CH}_2-\mathrm{OH}\)
It is H.B.O reaction
Anti-markowinikoff addition of \(\mathrm{H}^{+} / \mathrm{OH}^{-}\)takes place
It is H.B.O reaction
Anti-markowinikoff addition of \(\mathrm{H}^{+} / \mathrm{OH}^{-}\)takes place
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