Consider the following reaction sequence:
$$ ext{Ethanol} \xrightarrow{PBr_3} X \xrightarrow{ ext{Alc. KOH}}

Question

Consider the following reaction sequence:
$$	ext{Ethanol} \xrightarrow{PBr_3} X \xrightarrow{	ext{Alc. KOH}} Y \xrightarrow[(ii) H_2O, \Delta]{(i) H_2SO_4, 	ext{room temp.}} Z$$
The product Z is:

Accepted Answer

1.  **Step 1 (Formation of Alkyl Halide):** Ethanol ($CH_3CH_2OH$) reacts with Phosphorus Tribromide ($PBr_3$) to undergo nucleophilic substitution, forming Ethyl bromide ($X$).
    *   Reaction: $3CH_3CH_2OH + PBr_3 \rightarrow 3CH_3CH_2Br + H_3PO_3$ .
2.  **Step 2 (Elimination):** Ethyl bromide ($X$) reacts with Alcoholic KOH. This is a dehydrohalogenation reaction (elimination) which yields Ethene ($Y$).
    *   Reaction: $CH_3CH_2Br + KOH(alc) \rightarrow CH_2=CH_2 + KBr + H_2O$ .
3.  **Step 3 (Hydration):** Ethene ($Y$) reacts with concentrated $H_2SO_4$ at room temperature to form ethyl hydrogen sulphate ($CH_3CH_2OSO_3H$). Upon subsequent hydrolysis with water and heat, it regenerates Ethanol ($Z$). This is the standard laboratory method for converting alkenes to alcohols.
    *   Reaction: $CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$.
4.  **Conclusion:** The reaction sequence converts ethanol to ethene and then back to ethanol. Thus, product Z is Ethanol ($CH_3CH_2OH$).

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