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Question: What would be the major product obtained from hydroboration-oxidation in \(2\)-methyl-\(2\)-butene?...

What would be the major product obtained from hydroboration-oxidation in 22-methyl-22-butene?

Explanation

Solution

Hint : The hydroboration–oxidation reaction is a two-step hydration reaction in organic chemistry that transforms an alkene to an alcohol. The anti-Markovnikov reaction for hydroboration–oxidation occurs when the hydroxyl group attaches to the less-substituted carbon.

Complete Step By Step Answer:
22-Methyl-22-butene, also known as 22-methylbut-22-ene, is an alkene hydrocarbon with the molecular formula C5H10{C_5}{H_{10}}. In trichloromethane and dichloromethane, it is used as a free radical scavenger. The English physician John Snow experimented with it as an anaesthetic in the 1840s, but discontinued use for unexplained reasons.
Hydroboration-Oxidation is a two-step process for producing alcohol. In the alkene double bond, the hydrogen (from BH3B{H_3} or BHR2BH{R_2}) bonds to the most substituted carbon and the boron binds to the least substituted carbon in a reaction that proceeds in an Anti-Markovnikov manner.
When 22-methyl-22-butene reacts with BH3B{H_3}in the presence of H2SO4{H_2}S{O_4}gives 33-methyl-22-butanol.
It must be noted that the product formed from hydroboration–oxidation of an alkene that is alcohol has the HH and OHOHgroups switched in comparison with the alcohol formed from the reaction with water in the presence of H2SO4{H_2}S{O_4}.
This takes place due to the fact that in hydroboration–oxidation BH3B{H_3}is the electrophile, not H+{H^ + }. This results with BH3B{H_3}being replaced by theOHOHgroup. The hydride ion H{H^ - }is a nucleophile.
C5H10C5H9OH{C_5}{H_{10}} \to {C_5}{H_9}OH, This happens in the presence of BH3B{H_3}and OHO{H^ - },H2O2{H_2}{O_2},H2O{H_2}O.
Thus, the major product obtained from hydroboration-oxidation in 22-methyl-22-butene is 33-methyl-22-butanol.

Note :
Since the left carbon contains a methyl and the right carbon contains two hydrogens, the hydrogen attaches to the side with fewer hydrogens rather than more, according to anti-Markovnikov insertion. This means that the hydroxide is added to the less substituted carbon at the end.