Question

Which of the reagents shown below would accomplish the following transformations?

	(A)	(B)
	${{H}_{3}}{{O}^{+}}$	$B{{H}_{3}}THF;{{H}_{2}}{{O}_{2}}/NaOH$
	$NaOH$	$B{{H}_{3}}THF;{{H}_{2}}{{O}_{2}}/NaOH$
	$HBr$ in ether	$Hg{{(OAc)}_{2}}/{{H}_{2}}O;NaB{{H}_{4}}$
	$NaN{{H}_{2}}$	$Hg{{(OAc)}_{2}}/{{H}_{2}}{{O}_{2}};NaB{{H}_{4}}$

Answer 1

Hint: We know that the Oxymercuration-demarcation reaction is an alternative method of hydration of alkene. Given alkene is an asymmetric alkene. Additions take place according to Markovnikov’s rule. The first step of the reaction is the formation of cyclic mercurinium ions. The second step is the opening of mercurinium ions by the nucleophilic attack of water. Finally, Demercuration takes place by reduction with \[NaB{{H}_{4}}\] the reagent.

Complete answer: In Oxymercuration – Demercuration reaction alkene is converted to alcohol. In this reaction the reagent used is mercury $\left( II \right)$ acetate in tetrahydrofuran which is used as the solvent. During the reaction, reduction is also taking place with the help of reducing agents. The oxymercuration-demercuration reaction is a type of addition reaction which is used to convert the alkene into alcohol.
In this reaction, the alkene first reacts with mercury $\left( II \right)$ acetate \[\left[ Hg{{\left( OAc \right)}_{2}} \right]\] which is followed by the reduction reaction done by sodium borohydride \[\left( NaB{{H}_{4}} \right).\] In this reaction propene is reacted with mercury $\left( II \right)$ acetate in presence of tetrahydrofuran followed by reduction using sodium borohydride to form isopropanol. The mechanism of this reaction follows Markovnikov’s rule of regioselectivity where the hydroxyl group is joined to the most substituted carbon atom and the hydrogen atom is joined to the least substituted carbon atom.
In the given reaction, the first elimination reaction of the compound takes place with \[NaN{{H}_{2}}\] to form propene. This propene is then reacted with \[Hg{{\left( OAc \right)}_{2}}/{{H}_{2}}O;NaB{{H}_{4}}.\]It undergoes oxymercuration-demercuration reaction to form isopropyl alcohol as a major product.

Therefore, the reagent A and B here are $NaN{{H}_{2}}$ and $Hg{{(OAc)}_{2}}/{{H}_{2}}{{O}_{2}};NaOH$ respectively.

Note:
Remember that in addition to reactions of mixed unsaturated compounds, Markownikoff’s rule is followed which tells us about the major product form in the reaction. Oxymercuration-Demercuration reaction is an anti-addiction reaction and converts alkenes to neutral alcohols.