Question

Peroxide effect is observed with the addition of HBr but not with the addition of HI to unsymmetrical alkene because

• A. H-I bond is stronger that H-Br and is not cleaved by the free radical
• B. H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules
• C. Bond strength of HI and HBr are same but free radicals are formed in HBr
• D. All of these

Answer 1

The Correct Option is B

To solve this problem, we need to understand the peroxide effect in the addition of hydrogen halides (HBr and HI) to unsymmetrical alkenes and how the free radical mechanism influences the reaction.

1. The Peroxide Effect:
The peroxide effect refers to the phenomenon where the addition of hydrogen halides like HBr to an unsymmetrical alkene, in the presence of peroxides, results in the anti-Markovnikov product (the halide attaches to the less substituted carbon). This occurs through a free radical mechanism rather than the ionic mechanism typical of halide additions.

2. Free Radical Mechanism with HBr:
In the case of HBr, peroxides generate free radicals that initiate the reaction. The free radicals formed cleave the H-Br bond, leading to the formation of a bromine radical (Br•) which reacts with the alkene. This results in the anti-Markovnikov product. This process occurs because the bond dissociation energy of H-Br is lower than that of H-I, allowing the free radical formation in HBr.

3. In the Case of HI:
On the other hand, HI does not exhibit the peroxide effect. The reason is that the I-H bond is much weaker than the H-Br bond, making it more prone to bond cleavage. However, iodine radicals (I•) tend to recombine with each other to form molecular iodine (I₂), rather than participating in the addition to the alkene. Therefore, free radicals are not effectively formed in the reaction with HI in the presence of peroxides.

4. Analyzing the Options:
- Option (A) "H-I bond is stronger than H-Br and is not cleaved by the free radical" is incorrect because the H-I bond is weaker than the H-Br bond, making it easier for the H-I bond to break. 

- Option (B) "H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules" is correct. The weak H-I bond allows the formation of iodine radicals, but these radicals tend to recombine to form I₂, so the peroxide effect does not occur. 

- Option (C) "Bond strength of HI and HBr are the same but free radicals are formed in HBr" is incorrect because the bond strength of H-I is weaker than H-Br, making free radicals easier to form in HBr. 

- Option (D) "All of these" is incorrect because option (A) and (C) are incorrect.

Final Answer:
The correct answer is (B) "H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules."

Answer 2

Let's evaluate each option:

(A) H-I bond is stronger than H-Br and is not cleaved by the free radical: This is incorrect. The H-I bond is weaker than the H-Br bond and can be easily cleaved by free radicals.

(B) H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules: This is correct. The H-I bond is weaker, so iodine free radicals are formed and tend to combine to form iodine molecules, preventing the peroxide effect with HI.

(C) Bond strength of HI and HBr are the same but free radicals are formed in HBr: This is incorrect. The bond strength of H-I is weaker than H-Br, and it is this weaker bond that makes free radical formation easier in HBr.

(D) All of these: This is not correct as Option A and Option C are incorrect.
 

The correct answer is (B) : H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules.