Solveeit Logo
Login

Question

Question: Hydroboration-oxidation of \( C{H_3}CH = C{H_2} \) produces: (A) \( C{H_3}C{H_2}C{H_2}OH \) (B)...

Hydroboration-oxidation of CH3CH=CH2C{H_3}CH = C{H_2} produces:
(A) CH3CH2CH2OHC{H_3}C{H_2}C{H_2}OH
(B) CH3CH(OH)CH3C{H_3}CH(OH)C{H_3}
(C) CH3CH(OH)CH2OHC{H_3}CH(OH)C{H_2}OH
(D) CH3COCH3C{H_3}COC{H_3}

Explanation

Solution

Hydroboration is part of organic chemistry in which hydrogen boron is added to the carbon-carbon double bond of an organic compound. It can also be added to a carbon-carbon triple bond. Some organic compounds are made by hydroboration and it is very useful. Alkene is converted to alcohol.

Complete Step by step solution
Hydroboration is a reaction in the stream of organic chemistry, and it is basically a process of converting alkenes to neutral alcohols.
Hydroboration oxidation reaction is a process of converting an alkene, alkyne into alcohol. Hydroboration oxidation is a two-step reaction. This process creates a double or single bond between hydrogen and hydroxyl groups. Hydroboration oxidation does not follow markovnikov reaction, that is it is an anti-markovnikov reaction. By hydroboration oxidation we can add carbon and hydrogen atoms together. Hydroboration oxidation reaction is a process of converting an alkene, alkyne into alcohol. Hydroboration oxidation is a two-step reaction. This process creates a double or single bond between hydrogen and hydroxyl groups. Hydroboration oxidation does not follow markovnikov reaction, that is it is an anti-markovnikov reaction. By hydroboration oxidation we can add carbon and hydrogen atoms together.
Properties of hydroboration oxidation are:
It is anti-markovnikov in nature.
Hydroboration oxidation reaction is the best way to create an alcohol.
There is no formation of carbocation.
For example, propene when undergoes a process of hydroboration oxidation gives propan-1-ol as follows:
CH3CH=CH2→hydroborationB2H6/THF(CH3CH2CH2)3B→oxidationH2O2/OHCH3CH2CH2OHC{H_3}CH = C{H_2}\xrightarrow[{{\text{hydroboration}}}]{{{B_2}{H_6}/THF}}{(C{H_3}C{H_2}C{H_2})_3}B\xrightarrow[{{\text{oxidation}}}]{{{H_2}{O_2}/OH}}C{H_3}C{H_2}C{H_2}OH
Therefore, the correct answer is option A that is propan-1-ol (CH3CH2CH2OH)\left( {C{H_3}C{H_2}C{H_2}OH} \right) .

Note
When propene undergoes a process of hydroboration oxidation with propan-1-ol it also forms a by-product called Trialkyl Borane. Trialkyl borane is a weak Lewis acid, having a trigonal planar structure where boron is at centre.