Question

Identify the incorrect statement.

• A. When \( \text{(–)–2-bromooctane} \) is reacted with aqueous KOH, the substitution follows \( S_N2 \) mechanism and product formed is \( \text{(–)} \)-Octan-2-ol.
• B. Benzylic halides show higher reactivity towards \( S_N1 \) reaction.
• C. The reaction of \( \text{C}_6\text{H}_5\text{CH}_2 \text{– CHCl} \) with alc. KOH on heating yields two products.
• D. Optically active \( \text{(–)} \)-2-Methylbutan-1-ol on reaction with HCl yields a product \( (+) \)-1-Chloro-2-methylbutane.

Hint:

In \( S_N2 \) reactions, the nucleophile attacks from the opposite side of the leaving group, leading to inversion of configuration. In \( S_N1 \) reactions, the carbocation formed is planar, allowing for a mix of products with retention or inversion of configuration.

Answer 1

The Correct Option is A

Step 1: Analyze option (A).
The reaction of \( \text{(–)–2-bromooctane} \) with aqueous KOH undergoes an \( S_N2 \) mechanism, meaning the reaction proceeds with backside attack, resulting in inversion of configuration. The product formed in this case should be \( \text{(–)} \)-Octan-2-ol, which is optically active. However, the statement that the product formed is \( \text{(–)} \)-Octan-2-ol is incorrect since, in the \( S_N2 \) mechanism, the product should be an inverted configuration, leading to a different product. Hence, this is the incorrect statement.

Step 2: Analyze option (B).
Benzylic halides (such as benzyl chloride) indeed show higher reactivity towards \( S_N1 \) reactions due to the stability of the benzylic carbocation formed during the reaction. Therefore, this statement is correct.

Step 3: Analyze option (C).
The reaction of \( \text{C}_6\text{H}_5\text{CH}_2 \text{– CHCl} \) with alcoholic KOH leads to the elimination of HCl, forming a product with a double bond. This reaction follows the \( E2 \) mechanism. The statement is correct since two products (cis and trans isomers) can be formed from this reaction.

Step 4: Analyze option (D).
\( \text{(–)} \)-2-Methylbutan-1-ol is optically active, and when reacted with HCl, it undergoes an \( S_N2 \) reaction, leading to inversion of configuration and forming \( (+) \)-1-Chloro-2-methylbutane. This statement is correct.

Step 5: Conclusion.
The incorrect statement is option (A). The product formed in the \( S_N2 \) reaction should have an inverted configuration, not the same configuration. Therefore, the correct answer is option (A).