AIIMS 2006 Chemistry Elimination Reactions MCQ Question

To analyze the reaction of the compound $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{X}$ with aqueous KOH, we need to understand the type of reaction that occurs under these conditions. Aqueous KOH is a strong base and typically promotes elimination reactions, specifically dehydrohalogenation when X is a halogen (like Cl, Br, or I).

Step-by-Step Explanation

1. Nature of the Reactant: The reactant is an allylic halide $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{X}$. It features a double bond between the first two carbon atoms which can be involved in elimination reactions.

2. Mechanism of Reaction: When treated with a strong base such as KOH, the base abstracts a proton (H) from a carbon adjacent to the carbon bearing the halogen (X). This results in the formation of a double bond and the elimination of the halide ion (X). The reaction can be summarized as: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{X} + \text{KOH} \rightarrow \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH} + \text{X}^-$

   Here, the hydroxyl group $\text{OH}$ replaces the halogen as the functional group.

3. Product Formation: In this scenario, the most likely site for elimination is between the carbon attached to the halogen (C4) and the adjacent carbon (C3). Thus, the major product formed will be: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}$

   This corresponds to option D.

Justifications for Answer Choices

• Correct Answer: D $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}$: This structure accurately shows the double bond formed between the first and second carbon atoms, while the hydroxyl group is on the terminal carbon.

• Option A $\text{CH}_3\text{CH}_2\text{CH}=\text{CH}-\text{CH}_2\text{OH}$: This structure suggests that a double bond has formed between C3 and C4 instead. This would not be favored due to sterics and the stability of the double bond formation, thus it is incorrect.

• Option B $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}$: This is identical to the correct answer but lacks clarity as it does not indicate the position of the hydroxyl group clearly compared to option D. It could potentially confuse if written incorrectly.

• Option C $\text{CH}_3-\text{CH}-\text{CH}_2|\text{| OH} \text{CH}_2$: This structure suggests branching and an improper placement of functional groups. It does not correspond to any plausible product from the given substrate and is therefore incorrect.

Conclusion

In summary, the major product formed from the reaction of $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{X}$ with aqueous KOH is correctly identified as option D: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}$. The elimination reaction leads to the formation of a double bond while replacing the halogen with a hydroxyl group.