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(B) The reaction with $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ is Oxymercuration-Demercuration,which follows Markovnikov's rule to give product $A$ ($3$

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$A$ is Markovnikov product and $B$ is anti-Markovnikov product.

Vedclass · Answer

(B) The reaction with $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ is Oxymercuration-Demercuration,which follows Markovnikov's rule to give product $A$ ($3$,$3$-dimethylbutan$-2-$ol).The reaction with $(BH_3)_2$ followed by $H_2O_2/OH^-$ is Hydroboration-Oxidation,which follows anti-Markovnikov's rule to give product $B$ ($3$,$3$-dimethylbutan$-1-$ol).Therefore,$A$ is the Markovnikov product and $B$ is the anti-Markovnikov product.

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