Reaction Mechanism

1. Understand Reaction Order and Collision Theory

- Reaction Order: The order of a reaction is the sum of the exponents of the concentration terms in the rate law. It reflects how the rate of the reaction changes with the concentrations of the reactants.

- Collision Theory: For a reaction to occur, reactant molecules must collide with sufficient energy (activation energy) and with the correct orientation.

2. Analyze the probability of simultaneous collisions
Higher-order reactions (order > 3) imply that more than three reactant molecules must simultaneously collide in the correct orientation for the reaction to proceed. The probability of such a simultaneous collision of multiple molecules with sufficient energy and proper orientation is extremely low.
For example, for a 4th-order reaction, you need four molecules to collide at the same time with enough energy and correct orientation for the reaction to happen.

3. Evaluate the given options

- (A) Shifting of equilibrium towards reactants due to elastic collisions: Elastic collisions do not directly affect equilibrium in a way that makes higher-order reactions rare.

- (B) Loss of active species on collision: While collisions can sometimes lead to the loss of active species, this is not the primary reason for the rarity of higher-order reactions.

- (C) Low probability of simultaneous collision of all reacting species: This is the most accurate explanation. The probability of multiple molecules colliding simultaneously with sufficient energy and proper orientation decreases drastically as the number of molecules increases.

- (D) Increase in entropy as more molecules are involved: While entropy changes are relevant to reactions, they don't directly explain why higher-order reactions are rare.

Final Answer:
(C) Low probability of simultaneous collision of all reacting species

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."

Identify A and B in the Reaction

Step 1: Elimination Reaction with Alcoholic KOH
When 1-bromopropane (CH₃CH₂CH₂Br) is heated with alcoholic KOH, it undergoes β-elimination to form propene (CH₃CH=CH₂).

Reaction:
CH₃CH₂CH₂Br + alc. KOH → CH₃CH=CH₂ + H₂O + KBr

So, A is: CH₃CH=CH₂ (Propene)

Step 2: Electrophilic Addition with Br₂
Propene reacts with bromine to give 1,2-dibromopropane.

Reaction:
CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br

So, B is: CH₃CHBrCH₂Br (1,2-Dibromopropane)

Final Answer:

• A = CH₃CH=CH₂ (Propene)
• B = CH₃CHBrCH₂Br (1,2-Dibromopropane)