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Question: Identify 'Z' in the following series of reaction: $Butan-2-ol \xrightarrow{PCl_5} X \xrightarrow{al...

Identify 'Z' in the following series of reaction:

Butan−2−ol→PCl5X→alc.Y→ii)H−OH/heatⅰ)H2SO4ZButan-2-ol \xrightarrow{PCl_5} X \xrightarrow{alc.} Y \xrightarrow[ii) H-OH/heat]{ⅰ) H_2SO_4} Z

A

Butan-2-ol

B

Butan-1-ol

C

2-Chlorobutane

D

But-2-ene

Answer

Butan-2-ol

Explanation

Solution

  1. Step 1:
    Butan-2-ol reacts with PCl₅ to form 2‑chlorobutane.

    Butan-2-ol→PCl52‑Chlorobutane (X)\text{Butan-2-ol} \xrightarrow{PCl_5} \text{2‑Chlorobutane (X)}

  2. Step 2:
    Under alcoholic conditions, 2‑chlorobutane undergoes dehydrohalogenation (elimination) following Zaitsev’s rule to give the more substituted alkene, but‑2‑ene (Y).

  3. Step 3:
    But‑2‑ene, when treated with H₂SO₄ (acid-catalyzed hydration) followed by H₂O/heat, undergoes Markovnikov addition of water to regenerate butan‑2‑ol (Z).

    But‑2‑ene→H2SO4,  H2O/heatButan-2-ol (Z)\text{But‑2‑ene} \xrightarrow{\text{H}_2\text{SO}_4, \; H_2\text{O/heat}} \text{Butan-2-ol (Z)}

Explanation (minimal):
Butan-2-ol → 2‑chlorobutane (with PCl₅) → but‑2‑ene (elimination under alcoholic conditions) → butan‑2‑ol (hydration with H₂SO₄/H₂O, heat) due to Markovnikov addition.