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In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markonikoff's addition to alkenes because (1) both are highly ionic (2) one is oxidising and other is reducing (3) one of the steps is cndothermic in both the cases (4) all the steps are exothermic in both the cases

Short Answer

Expert verified
One of the steps is endothermic in both the cases.

Step by step solution

01

Identify Key Terms

Recognize and understand the key terms in the question: 'anti-Markovnikov's addition,' 'peroxide,' 'hydrogen chloride,' and 'hydrogen iodide.'
02

Understanding Anti-Markovnikov's Addition

Anti-Markovnikov's addition refers to the addition of HBr to alkenes in the presence of peroxides, leading to the formation of the product contrary to Markovnikov's rule.
03

Role of Peroxides

Hydroperoxides initiate a radical mechanism for the addition of HBr to alkenes, making the reaction follow an anti-Markovnikov pathway.
04

Analyzing Hydrogen Chloride and Hydrogen Iodide

Consider why HCl and HI don’t behave similarly to HBr. Analyze the energetic feasibility and the reaction mechanism for HCl and HI in the presence of peroxides.
05

Step Energetics

In the radical addition mechanism, certain steps of the reaction either absorb energy (endothermic) or release energy (exothermic).
06

Conclusion

For HCl and HI, one of the steps in the radical chain mechanism is endothermic, making the radical pathway unfavorable. Hence, anti-Markovnikov's addition is not observed with these acids.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radical Mechanism
When we talk about chemical reactions, a radical mechanism plays a crucial role, especially in the context of anti-Markovnikov's addition. In simple terms, a radical mechanism involves the formation of free radicals. These are highly reactive species that have unpaired electrons. The process starts with the initiation step, where a peroxide breaks down to form radicals. These radicals then interact with hydrogen halides, like HBr, to form new radicals and propagate the reaction. The chain of these steps drives the reaction forward and ultimately results in the addition of atoms to the carbon structure in an unconventional way, opposing Markovnikov's rule. This happens prominently with HBr in the presence of peroxides.
Hydroperoxides
Hydroperoxides are a specific type of peroxides, known for their role in initiating radical reactions. Peroxides break down upon heating or in the presence of light to form two radical species. In organic chemistry, these hydroperoxides are key players in anti-Markovnikov's addition. For instance, when HBr reacts with an alkene in the presence of hydroperoxides, it follows a radical mechanism. The energy provided helps break down the hydroperoxide, generating radicals that initiate the reaction. The radical formed from the peroxide adds to the alkene, creating a chain reaction that ultimately leads to the anti-Markovnikov product. However, this behavior is not observed with HCl or HI due to the differing energetics of the steps involved.
Energetic Feasibility
The energetic feasibility of a chemical reaction determines whether it can proceed or not. For anti-Markovnikov's addition using a radical mechanism, each step of the reaction must be energetically favorable. This means that the energy released has to be sufficient to keep the reaction going. With HBr, all the steps of the radical mechanism are exothermic, either releasing energy or requiring minimal energy input, thus making the reaction feasible. However, in the case of HCl and HI, one of the steps in their radical chain mechanism is endothermic, requiring more energy than the system can provide. This makes the radical pathway unfavored and thus, prevents the anti-Markovnikov addition for these hydrogen halides.
Hydrogen Halides
Hydrogen halides like HCl, HBr, and HI are commonly used in addition reactions with alkenes. Their behavior in the presence of peroxides, however, varies significantly. HBr is the most effective in facilitating anti-Markovnikov's addition through a radical mechanism, thanks to the favorable energy profile of its reaction steps. In contrast, HCl and HI don't engage in the same way. HCl is highly ionic and does not favor radical formation easily. On the other hand, HI forms radicals but the following steps demand more energy than what can be provided (endothermic steps), making the reaction energetically unviable. Therefore, unlike HBr, both HCl and HI fail to follow the anti-Markovnikov pathway in the presence of peroxides, sticking instead to the more expected Markovnikov addition pattern.

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Most popular questions from this chapter

Ozonolysis of 2,3 -dimethyl-1-butene followed by reduction with zinc and watcr gives (1) methanoic acid and 3 -methyl-2-butanone (2) methanal and 2 -methyl-2-butanone (3) methanal and 3-methyl-2-butanone (4) methanoic acid and 2 -methyl-2-butanone

In the presence of platinum catalyst, hydrocarbon (A) adds hydrogen to form \(\mathrm{n}\) -hexane. When hydrogen bromide is added to (A) instead of hydrogen, only a single bromo compound is formed. Which of the following is (A)? (1) \(\mathbf{C H}_{2}=\mathbf{C H}-\mathbf{C H}_{2} \mathrm{CH}_{2} \mathbf{C H}_{2} \mathrm{CH}_{3}\) (2) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (3) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{3}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}\)

The conversion of 3 -hexyne into trans- 3 -hexene can be effected by (1) \(\mathrm{Na} /\) liquid \(\mathrm{NH}_{3}\) (2) \(\mathrm{H}_{2} /\) Lindlar's catalyst (3) Clemmensen reduction (4) \(\mathrm{LiNH}_{2}\)

The reduction of 3 -hexyne with II \(_{2}\) / Lindlar's catalyst gives predominantly (1) \(\mathrm{n}\) -hexane (2) trans-3-hexene (3) cis-3-hexene (4) a mixture of cis and trans 3 -hexene

Among the alkencs which one produces tertiary butyl alcohol on acid hydration? (1) \(\mathrm{CII}_{3} \mathrm{CII}_{2} \mathrm{CII}=\mathrm{CII}_{2}\) (2) \(\mathrm{CII}_{3} \mathrm{CII}=\mathrm{CIICII}_{3}\) (3) \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{C}=\mathrm{CII}_{2}\) (4) \(\mathrm{CII}_{3} \mathrm{CII}=\mathrm{CII}_{2}\)

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