Reaction (A):Ph-CH=CH-CH_3 xrightarrow[H_2O_2 | vedclass

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(D) Reaction $(A)$ is Hydroboration-Oxidation,which follows Anti-Markownikoff addition of water across the double bond.Reaction $(B)$ is Oxymercuratio

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In reaction $(B)$,the product is a racemic mixture.

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(D) Reaction $(A)$ is Hydroboration-Oxidation,which follows Anti-Markownikoff addition of water across the double bond.Reaction $(B)$ is Oxymercuration-Demercuration,which follows Markownikoff addition of water across the double bond.In reaction $(B)$,the starting material $CH_3-CH=CH-Ph$ is an unsymmetrical alkene. The addition of $H_2O$ follows the Markownikoff rule,resulting in the formation of a chiral center at the carbon atom attached to the phenyl group.Since the reaction proceeds through a cyclic mercurinium ion intermediate,the attack of water can occur from either side,leading to the formation of a racemic mixture of the product $CH_3-CH(OH)-CH_2-Ph$ (or $CH_3-CH_2-CH(OH)-Ph$ depending on regioselectivity,but the formation of a chiral center leads to racemization).

Reaction $(A)$:
$Ph-CH=CH-CH_3 \xrightarrow[H_2O_2 / OH^-]{B_2H_6 / THF}$ products.
Reaction $(B)$:
$CH_3-CH=CH-Ph \xrightarrow[NaBH_4]{Hg(OAc)_2 / H_2O}$ products.
Correct statement is:

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Reaction $(A)$:$Ph-CH=CH-CH_3 \xrightarrow[H_2O_2 / OH^-]{B_2H_6 / THF}$ products.Reaction $(B)$:$CH_3-CH=CH-Ph \xrightarrow[NaBH_4]{Hg(OAc)_2 / H_2O}$ products.Correct statement is: