Question

Which of the following hydrocarbon will give same product with \(\mathrm{HBr}\) at \(-80^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\) ? (A) (B) (C) (D)

Short answer

In order to provide a short answer, please include the structures of the hydrocarbons (A, B, C, and D). Based on the provided information, the answer would involve comparing the products formed during the electrophilic addition reaction at low (Markovnikov's rule) and high temperatures (anti-Markovnikov addition) and identifying the hydrocarbon that yields the same product in both cases.

Step-by-step solution

Analyze each hydrocarbon

For each hydrocarbon (A, B, C, D), we need to identify the most stable carbocation formed in the reaction (for low-temperature, Markovnikov addition) and predict the resulting product.

Predict products for high-temperature reaction

For each hydrocarbon, we then need to predict the products formed during the high-temperature reaction (anti-Markovnikov addition) due to the free radical mechanism.

Compare products in both cases

Finally, we look for the hydrocarbon that gives the same product at both low and high temperatures. This will be the hydrocarbon that satisfies the given condition in the exercise. Note: Without the structures of the given hydrocarbons (A, B, C, and D), it's impossible to provide a definitive solution to this problem. Please provide the structures of the hydrocarbons in order to get a complete step-by-step solution.

Key concepts

Markovnikov addition

In organic chemistry, the concept of Markovnikov addition is crucial when predicting the outcome of reactions involving hydrocarbons and strong acids like HBr. Markovnikov's rule states that when a protic acid (HX) adds to an unsymmetrical alkene, the hydrogen atom bonds to the carbon with the greater number of hydrogen atoms already attached. This occurs because the process involves the formation of a carbocation, which is more stable when the positive charge is located on a more substituted carbon.
This rule is guided by the idea that more substituted carbocations, involving carbons bonded to more alkyl groups, are generally more stable, making this the favored path.

• The hydrogen from HBr attaches to the less substituted carbon.
• The halide ion (Br⁻) then adds to the more substituted carbon.
• This results in the formation of the Markovnikov product.

Understanding this mechanism helps predict not only the final product but also the yield and efficiency of the reaction.

anti-Markovnikov addition

Anti-Markovnikov addition, on the other hand, takes place under certain conditions where the Markovnikov rule doesn't apply, such as in the presence of peroxides with HBr. This type of addition to alkenes results in the hydrogen atom bonding to the more substituted carbon, the opposite of the normal Markovnikov preference.
This phenomenon primarily arises due to the formation of free radicals, which are different from carbocations and involve a completely different mechanism.

• Peroxides aid in generating free radicals by breaking down into reactive species.
• The free radical process involves the homolytic cleavage of a bond, leading to radicals that initiate a chain reaction.
• Radicals stabilize differently and prefer the least hindered paths, leading to anti-Markovnikov products.

This process is significant for synthetic chemistry, enabling products that wouldn't normally form under standard conditions.

carbocation stability

Carbocation stability is a central concept in many organic reactions, including the Markovnikov addition. A carbocation is a positively charged carbon atom with three covalent bonds instead of the usual four. It can play a critical role in determining the stability and the path that a reaction will take.
In these processes, the most stable carbocation is generally the one that is more substituted.
This is because:

• Alkyl groups donate electron density through inductive effects, stabilizing the positive charge.
• More substituted carbocations allow for hyperconjugation, an interaction that further stabilizes the cation.
• The presence of electron-donating groups enhances stability.

By analyzing the stability of possible carbocations, chemists can predict the predominant reaction pathway and product.

free radical mechanism

The free radical mechanism describes a unique reaction pathway that can override traditional carbocation mechanisms, particularly in the presence of peroxides like in anti-Markovnikov addition reactions. Free radicals are atoms or molecules that contain an unpaired electron, making them highly reactive.
The mechanism generally consists of three stages:

• Initiation: Peroxides decompose to form radicals in the presence of heat or light.
• Propagation: Radicals react with non-radical species to produce new radicals, continuing the chain reaction.
• Termination: Two radicals combine to form stable, non-radical products, ending the reaction.

This chain reaction approach allows for the addition of atoms in an anti-intuitive order compared to carbocation pathways. Understanding free radical mechanisms expands the possibilities for synthesis and product formation.