Question

Statement-I: 3-Phenylprop-1-ene will react with HBr and give alkyl halide as the major product having 1 chiral carbon atom. Statement-II: Aryl chloride and Aryl cyanide both can be formed by Gattermann and Sandmeyer reactions.

• A. Both Statement I and Statement II are incorrect.
• B. Statement I is correct but Statement II is incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are correct.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question evaluates an addition reaction on an alkene and the scope of Gattermann/Sandmeyer reactions.
Step 2: Detailed Explanation:
Statement-I: 3-Phenylprop-1-ene (\(Ph-CH_2-CH=CH_2\)) reacts with HBr.
- Addition of \(H^+\) gives a \(2^\circ\) carbocation: \(Ph-CH_2-CH^+-CH_3\).
- A hydride shift occurs to form a more stable benzylic carbocation: \(Ph-CH^+-CH_2-CH_3\).
- Bromide ion attacks this carbocation to give 1-bromo-1-phenylpropane: \(Ph-CH(Br)-CH_2-CH_3\).
- This carbon (\(C_1\)) is attached to four different groups (\(H, Br, Ph, Et\)), making it chiral. Thus, Statement-I is correct.

Statement-II: Sandmeyer and Gattermann reactions are used to convert benzene diazonium salts into aryl halides (\(Cl, Br, I\)) or aryl cyanides (\(CN\)). However, Gattermann reaction specifically refers to the use of Copper powder and HX to form aryl halides. Aryl cyanide is typically formed using Sandmeyer conditions (\(CuCN/KCN\)). More importantly, Statement-II is often deemed incorrect in competitive exams because Sandmeyer is used for \(Cl, Br, CN\) while Gattermann is primarily for \(Cl, Br\). Specifically, Gattermann reaction is not used for cyanide formation. Thus, Statement-II is incorrect.
Step 3: Final Answer:
Statement I is correct and Statement II is incorrect.