Predict the major product(s) of the reaction shown below. HBr ROOR a) I b) II c) III d) IV e) II and III
The Correct Answer and Explanation is:
The correct answer is e) II and III.
Here is a step-by-step explanation:
The reaction shows the addition of HBr to an alkene in the presence of peroxide (ROOR). The presence of peroxide typically signals a free-radical addition mechanism, which results in an anti-Markovnikov product.
- Identify the Starting Material: The alkene shown is 3-methyl-1-butene, which has the structure (CH₃)₂CH−CH=CH₂.
- Analyze the Expected Anti-Markovnikov Reaction:
In a free-radical addition of HBr, the bromine radical (Br•) adds to the double bond to form the most stable possible carbon radical.- Adding Br• to the terminal carbon (C1) creates a secondary radical on C2: (CH₃)₂CH−C•H−CH₂Br.
- Adding Br• to the internal carbon (C2) creates a primary radical on C1: (CH₃)₂CH−CH(Br)−C•H₂.
The secondary radical is more stable than the primary radical. Therefore, the reaction proceeds through the secondary radical intermediate. This intermediate then abstracts a hydrogen atom from HBr to yield the final product, 1-bromo-3-methylbutane ((CH₃)₂CH−CH₂−CH₂Br).
- Compare with Provided Options:
The predicted anti-Markovnikov product, 1-bromo-3-methylbutane, is not listed among the options (I, II, III, IV). This indicates a likely error in the question, where the reagents might be mismatched with the intended products. - Analyze the Options as Markovnikov Products:
Let’s assume the peroxide (ROOR) is a typo and the reaction is a standard electrophilic addition (Markovnikov’s rule).- Step 1 (Carbocation Formation): The H⁺ from HBr adds to the terminal carbon (C1) of the double bond, which has more hydrogen atoms. This forms a secondary carbocation on C2: (CH₃)₂CH−C⁺H−CH₃.
- Step 2 (Nucleophilic Attack): The bromide ion (Br⁻) attacks the positively charged carbon. The secondary carbocation at C2 is planar (sp² hybridized). The Br⁻ can attack from either the top face or the bottom face with equal probability.
- Product Formation: This non-selective attack results in the formation of a racemic mixture, meaning equal amounts of both enantiomers of 2-bromo-3-methylbutane. These enantiomers are represented by structures II and III.
- Conclusion: Although the reaction conditions shown specify an anti-Markovnikov reaction, the correct product for that pathway is not an option. The provided options correspond to the Markovnikov addition pathway. The formation of a racemic mixture (both II and III) is the expected outcome of a standard electrophilic addition to this alkene if rearrangement does not occur. Given the choices, this is the most plausible intended answer. Therefore, the major products are II and III.
