Question

An alkene $C{H_3}CH = C{H_2}$ is treated with ${B_2}{H_6}$ in the presence of ${H_2}{O_2}$. The final product formed is:
A.$C{H_3}C{H_2}CHO$
B.$C{H_3}CH(OH)C{H_3}$
C.$C{H_3}C{H_2}C{H_2}OH$
D.${\left( {C{H_3}C{H_2}C{H_2}} \right)_3}B$

Answer 1

Hint: The given reaction is known as hydroboration-oxidation reaction. This reaction follows Anti-Markovnikov’s rule in which addition of $H - OH$ takes place across the $C = C$. The reaction is completed in two steps. The chemical hydrogen peroxide is used as an oxidizing agent.

Complete step by step answer:
Hydroboration oxidation reaction is a chemical reaction used for the conversation of alkene into alcohols.
In this reaction ${B_2}{H_6}$ known as diborane is used as a boronating agent and ${H_2}{O_2}$ known as hydrogen peroxide is used as the oxidizing agent. Net addition of water across the double bond takes place

This reaction involves two steps.
STEP-1
Hydroboration step
In the hydroboration step addition of diborane to the double bond takes place. This results in the addition of hydrogen atom to the carbon which is adjacent to the carbon attached to the boron atom to form alkyl borane.
STEP-2
Oxidation step
In the oxidation step, alkyl borane is treated with any base or water and hydrogen peroxide. This results in the replacement of boron-carbon bond with carbon-OH bond. In this hydrogen peroxide act as an nucleophile
The given reaction is shown below.
$C{H_3}CH = C{H_2}\xrightarrow[{{H_2}{O_2}}]{{{B_2}{H_6},DMF}}C{H_3}C{H_2}C{H_2}OH$
In the above reaction, prop-1-ene reacts with diborane and hydrogen peroxide to give propan-1-ol. In this reaction dimethylformamide is used as a solvent.
Therefore, the correct option is C.


Note:
In the hydroboration-oxidation reaction, the boronating agent keeps adding itself to the alkene until all the hydrogen atom is lost to the alkene and no more hydrogen atom is available in the boronating agent.