Question

Identify the product 'B' in the following series of reactions:
$\text{Propan-1-ol} \xrightarrow{623\ \text{K},\ \text{Al}_2\text{O}_3} \text{A} \xrightarrow[\text{ii) }\text{H}_2\text{O}]{\text{i) conc. }\text{H}_2\text{SO}_4} \text{B}$

• A. Propanal
• B. Propan-2-ol
• C. Propene
• D. Propanone

Hint:

This entire sequence is a classic chemical strategy used to shift the functional position of a hydroxyl group from a terminal position (primary alcohol) to an internal position (secondary alcohol): $1^\circ\ \text{alcohol} \xrightarrow{\text{Dehydration}} \text{Alkene} \xrightarrow{\text{Markovnikov Hydration}} 2^\circ\ \text{alcohol}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the final organic product 'B' following a two-step transformation starting from the primary alcohol, propan-1-ol.

Step 2: Key Formula or Approach:
The first step ($\text{Al}_2\text{O}_3, 623\ \text{K}$) is an industrial dehydration reaction that converts an alcohol into an alkene. The second step (conc. $\text{H}_2\text{SO}_4$ followed by hydration) is the acid-catalyzed hydration of an alkene, which strictly obeys

Markovnikov's rule.

Step 3: Detailed Explanation:

Step 1: Formation of Product 'A' When propan-1-ol is heated over alumina ($\text{Al}_2\text{O}_3$) catalyst at $623\ \text{K}$, intramolecular dehydration takes place via elimination of a water molecule: $$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{Al}_2\text{O}_3,\ 623\ \text{K}} \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O}$$ Thus, product 'A' is

propene.

Step 2: Formation of Product 'B' Propene is then treated with cold concentrated sulfuric acid followed by heating with water. This results in the electrophilic addition of water across the double bond. According to Markovnikov's rule, the hydrogen ion ($\text{H}^+$) adds to the carbon with more hydrogen atoms, and the nucleophilic hydroxyl group ($\text{OH}^-$) attaches to the more substituted carbon atom: $$\text{CH}_3\text{CH=CH}_2 \xrightarrow{\text{conc. }\text{H}_2\text{SO}_4} \text{CH}_3\text{-CH(OSO}_3\text{H)-CH}_3 \xrightarrow{\text{H}_2\text{O}, \Delta} \text{CH}_3\text{-CH(OH)-CH}_3$$ The secondary alcohol formed is

propan-2-ol.

Step 4: Final Answer:
Product 'B' is propan-2-ol, which perfectly matches option (B).