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(C) $(1)$ Acid-catalyzed hydration of propene follows Markovnikov's rule,where the electrophile $H^+$ adds to the terminal carbon to form the more sta

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$A: CH_3CH(OH)CH_3, B: CH_3CH_2CH_2-OH$

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(C) $(1)$ Acid-catalyzed hydration of propene follows Markovnikov's rule,where the electrophile $H^+$ adds to the terminal carbon to form the more stable secondary carbocation $(CH_3-CH^+-CH_3)$,which is then attacked by $H_2O$ to form propan-$2$-ol as the major product $A$.$(2)$ Hydroboration-oxidation of propene follows anti-Markovnikov's rule,where the boron atom adds to the terminal carbon,and subsequent oxidation with $H_2O_2/OH^-$ replaces boron with an $-OH$ group,resulting in propan-$1$-ol as the major product $B$.

Identify the major product $A$ and major product $B$ in the following set of reactions:
$CH_3-CH=CH_2 \xrightarrow{H_2O, H^+} \text{Major product } A$
$CH_3-CH=CH_2 \xrightarrow{(BH_3)_2, H_2O_2, OH^-} \text{Major product } B$

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Identify the major product $A$ and major product $B$ in the following set of reactions:$CH_3-CH=CH_2 \xrightarrow{H_2O, H^+} \text{Major product } A$$CH_3-CH=CH_2 \xrightarrow{(BH_3)_2, H_2O_2, OH^-} \text{Major product } B$