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: Understanding the Reaction Sequence:
2-Methylpropan-2-ol is tert-butyl alcohol. Step 1 uses 20% \(\text{H}_2\text{SO}_4\) at 363 K, which dehydrates tertiary alcohols by a mild route. Step 2 hydrates the product.

Step 2: Step A:
Dehydration gives the alkene 2-methylpropene (isobutene):
\[ (\text{CH}_3)_3\text{C-OH} \xrightarrow{20\%\ \text{H}_2\text{SO}_4,\ 363\ \text{K}} (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O} \]

Step 3: Step B:
Acid-catalysed hydration follows Markovnikov addition. The proton adds to the \(\text{CH}_2\) carbon, giving the stable tertiary carbocation, then water adds to it:
\[ (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} (\text{CH}_3)_3\text{C-OH} \]
So B is tert-butyl alcohol.

Step 4: Why the other options are wrong.
Anti-Markovnikov hydration would be needed for n-butyl alcohol, isobutyl alcohol or sec-butyl alcohol, and acid hydration does not do that. A tertiary carbocation is the most stable, so only tert-butyl alcohol forms.

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