In the reaction CH_3CH_2CH_2Br xrightarrow{aq | vedclass

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(A) Step $1$: $CH_3CH_2CH_2Br + aq. NaOH ightarrow CH_3CH_2CH_2OH$ ($X$ is propan$-1-$ol).Step $2$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, 	ext{Heat

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$CH_3-CH(OH)-CH_2Cl$

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(A) Step $1$: $CH_3CH_2CH_2Br + aq. NaOH ightarrow CH_3CH_2CH_2OH$ ($X$ is propan$-1-$ol).Step $2$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, 	ext{Heat}} CH_3CH=CH_2$ ($Y$ is propene).Step $3$: $CH_3CH=CH_2 + HOCl ightarrow CH_3CH(OH)CH_2Cl$ (major product) and $CH_3CH(Cl)CH_2OH$ (minor product).According to Markovnikov's rule,the electrophile $Cl^+$ attacks the double bond to form the more stable carbocation,followed by the attack of $OH^-$. The major product is $CH_3CH(OH)CH_2Cl$.

In the reaction $CH_3CH_2CH_2Br$ $\xrightarrow{aq. NaOH} (X)$ $\xrightarrow{Al_2O_3, \text{Heat}} (Y)$ $\xrightarrow{HOCl} (Z)$,what is $Z$?

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