Question

Identify the product B in the following sequence of reactions?

• A. Propan-2-ol
• B. Propan-1-ol
• C. Isopropyl hydrogen sulphate
• D. Propene

Hint:

In acid-catalyzed hydration of alkenes, the product is always the more substituted alcohol (Markovnikov addition). The sequence via alkyl hydrogen sulfate is a two-step industrial method for alcohol synthesis.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We start from propan-1-ol (1-propanol) and treat it with alumina at 623 K to form A. Then A is treated with concentrated sulfuric acid followed by concentrated water with heating to yield B. We must identify B.

Step 2: Key Formula or Approach:
Alumina at high temperature (623 K) acts as a dehydrating agent for primary alcohols, producing alkenes via elimination. Then alkenes react with concentrated sulfuric acid to form alkyl hydrogen sulfates (Markovnikov addition). Finally, hydrolysis of the alkyl hydrogen sulfate gives the corresponding alcohol (also Markovnikov).

Step 3: Detailed Explanation:
- Step 1: Propan-1-ol (\( CH_3CH_2CH_2OH \)) on heating with \( Al_2O_3 \) at 623 K undergoes dehydration to form propene (\( CH_3CH=CH_2 \)). So A = propene. - Step 2: Propene reacts with concentrated \( H_2SO_4 \). The addition follows Markovnikov’s rule: H adds to the less substituted carbon (terminal carbon), and \( HSO_4^- \) adds to the more substituted carbon (central carbon), giving isopropyl hydrogen sulphate (\( (CH_3)_2CHOSO_3H \)). - Step 3: Isopropyl hydrogen sulphate is hydrolyzed with hot concentrated water to yield propan-2-ol (isopropanol, \( (CH_3)_2CHOH \)). Thus B = propan-2-ol.

Step 4: Final Answer:
The final product B is propan-2-ol, which matches option (A).