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(B) The first reaction is Hydroboration-Oxidation of propene. This reaction follows anti-Markovnikov addition of water,resulting in the formation of $

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Positional isomers

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(B) The first reaction is Hydroboration-Oxidation of propene. This reaction follows anti-Markovnikov addition of water,resulting in the formation of $A = CH_3-CH_2-CH_2OH$ (propan$-1-$ol).The second reaction is Oxymercuration-Demercuration of propene. This reaction follows Markovnikov addition of water,resulting in the formation of $B = CH_3-CH(OH)-CH_3$ (propan$-2-$ol).Comparing $A$ and $B$,both are alcohols with the same molecular formula $C_3H_8O$,but the position of the hydroxyl $(-OH)$ group is different. Therefore,they are positional isomers.

Identify the relationship between $A$ and $B$ in the following reactions:
$CH_3-CH=CH_2 \xrightarrow[(2)H_2O_2, OH^-]{(1)BH_3-THF} A$
$CH_3-CH=CH_2 \xrightarrow[(2)NaBH_4]{(1)(AcO)_2Hg, H_2O} B$

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Identify the relationship between $A$ and $B$ in the following reactions:$CH_3-CH=CH_2 \xrightarrow[(2)H_2O_2, OH^-]{(1)BH_3-THF} A$$CH_3-CH=CH_2 \xrightarrow[(2)NaBH_4]{(1)(AcO)_2Hg, H_2O} B$