Question

What would be the major product obtained from hydroboration-oxidation in $2$-methyl-$2$-butene?

Answer 1

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:
$2$-Methyl-$2$-butene, also known as $2$-methylbut-$2$-ene, is an alkene hydrocarbon with the molecular formula ${C_5}{H_{10}}$. 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 $B{H_3}$ or $BH{R_2}$) 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 $2$-methyl-$2$-butene reacts with $B{H_3}$in the presence of ${H_2}S{O_4}$gives $3$-methyl-$2$-butanol.
It must be noted that the product formed from hydroboration–oxidation of an alkene that is alcohol has the $H$ and $OH$groups switched in comparison with the alcohol formed from the reaction with water in the presence of ${H_2}S{O_4}$.
This takes place due to the fact that in hydroboration–oxidation $B{H_3}$is the electrophile, not ${H^ + }$. This results with $B{H_3}$being replaced by the$OH$group. The hydride ion ${H^ - }$is a nucleophile.
${C_5}{H_{10}} \to {C_5}{H_9}OH$, This happens in the presence of $B{H_3}$and $O{H^ - }$,${H_2}{O_2}$,${H_2}O$.
Thus, the major product obtained from hydroboration-oxidation in $2$-methyl-$2$-butene is $3$-methyl-$2$-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.