In the hydroboration - oxidation reaction of propene with diborane, $\mathrm{H}_{2} \mathrm{O}_{2}$ and…

In the hydroboration - oxidation reaction of propene with diborane, $\mathrm{H}_{2} \mathrm{O}_{2}$ and $\mathrm{NaOH}$, the organic compound formed is:
  1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
  2. $\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}$
  3. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
  4. $\left(\mathrm{CH}_{3}ight)_{3} \mathrm{COH}$

Solution

In the hydroboration - oxidation reaction, an alkene reacts with $\mathrm{BH}_{3}$ to give alkyl boranes which on oxidation with alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$ gives alcohol. The product obtained in the accordance with the anti- Markovnikov's rule.\n$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{BH}_{3} ightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{BH}_{2} \xrightarrow[]{\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{OH}^{-}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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