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.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Both Statement I and Statement II are correct.
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The Correct Option isB
Solution and Explanation
Step 1: Analysis of Statement I.
3-Phenylprop-1-ene is an alkene with a phenyl group attached to the carbon chain. When it reacts with HBr, the reaction follows Markovnikov's rule. The addition of HBr leads to the formation of a bromoalkane where the chiral center is created at the carbon where the bromine is attached. This statement is correct as the addition results in the formation of a chiral carbon.
Step 2: Analysis of Statement II.
The Gattermann reaction involves the formation of an aryl chloride using a benzene derivative, but it uses AlCl\(_3\) and HCl. The Sandmeyer reaction forms aryl halides and cyanides using copper halide/cyanide as the catalyst. Aryl cyanide is formed in Sandmeyer's reaction, but aryl chloride is typically formed in Gattermann's reaction. This statement is incorrect because both reactions don’t form both aryl chloride and aryl cyanide in all cases.
Step 3: Conclusion.
Since Statement I is correct and Statement II is incorrect, the correct answer is (B).
Final Answer: (B) Statement I is correct but Statement II is incorrect.
