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(A) Step $1$: Dehydration of $Propan-1-ol$ with $Al_2O_3$ at $623 \ K$ gives propene $(A)$ as the product.$CH_3-CH_2-CH_2-OH \xrightarrow[623 \ K]{Al_

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Propan$-2-$ol

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(A) Step $1$: Dehydration of $Propan-1-ol$ with $Al_2O_3$ at $623 \ K$ gives propene $(A)$ as the product.$CH_3-CH_2-CH_2-OH \xrightarrow[623 \ K]{Al_2O_3} CH_3-CH=CH_2 (A)$Step $2$: Addition of $conc. H_2SO_4$ to propene follows $Markovnikov's$ rule to form isopropyl hydrogen sulphate,which upon hydrolysis with water $(H_2O, \Delta)$ yields $Propan-2-ol$ $(B)$.$CH_3-CH=CH_2$ $\xrightarrow{conc. H_2SO_4} CH_3-CH(OSO_3H)-CH_3$ $\xrightarrow{H_2O, \Delta} CH_3-CH(OH)-CH_3 (B)$Thus,the final product $B$ is $Propan-2-ol$.

Identify the product $B$ in the following sequence of reactions:
$CH_3-CH_2-CH_2-OH$ $\xrightarrow[623 \ K]{Al_2O_3} A$ $\xrightarrow[(ii) H_2O, \Delta]{(i) conc. H_2SO_4} B$

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Identify the product $B$ in the following sequence of reactions:$CH_3-CH_2-CH_2-OH$ $\xrightarrow[623 \ K]{Al_2O_3} A$ $\xrightarrow[(ii) H_2O, \Delta]{(i) conc. H_2SO_4} B$