Question

Identify product 'B' in following series of reactions
2-methylpropan-2-ol \(\rightarrow _{363 \text{K}\,}^{20\% \text{H}_2\text{SO}_4\,}\) A \(\rightarrow _{\text{Conc. H}_2\text{SO}_4\,}^{\text{H}_2\text{O} \Delta \,}\) B

• A. n-Butylalcohol
• B. sec-Butylalcohol
• C. Iso-butylalcohol
• D. tert-Butylalcohol

Hint:

Dehydration gives 2-methylpropene; hydration follows Markovnikov rule to the tertiary alcohol.

Answer 1

The Correct Option is D

Step 1: Idea:
Dehydration followed by hydration of the same alkene brings back the alcohol whose carbocation is most stable.

Step 2: Track the carbons:
tert-Butyl alcohol loses water to give isobutene. In the second step, $\text{H}^+$ adds to the terminal $\text{CH}_2$ to give $(\text{CH}_3)_3\text{C}^+$, the tertiary carbocation.

Step 3: Finish:
Water attacks the carbocation and a proton is lost, giving $(\text{CH}_3)_3\text{COH}$.

Step 4: Option check:
n-Butyl, isobutyl and sec-butyl alcohols would require a primary or secondary carbocation or anti-Markovnikov addition, which is not favoured.

Final Answer:
B is tert-butyl alcohol, option (D). \[ \boxed{\text{tert-Butyl alcohol (D)}} \]