In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markonikoff ’s addition to alkenes because

one of the steps is endothermic in both the cases

The correct answer is option 3. one of the steps is endothermic in both the cases.

Let us look at why option (3) is the correct answer regarding the addition of hydrogen halides (\(HCl\) and \(HI\)) to alkenes in the presence of peroxides:

Mechanism:

Initiation (Peroxide Decomposition): Peroxides (such as benzoyl peroxide, \( \text{ROOR} \)) decompose under heat or light to form free radicals:

\(\text{ROOR} \rightarrow 2 RO^•\)

These radicals (\( \cdot \text{RO} \cdot \)) initiate the reaction.

Propagation Steps:

The alkene reacts with a peroxide-derived radical to form an alkyl radical:

\(\text{CH}_2=CH-\text{CH}_3 +  \text{RO}^•  \rightarrow CH_2-C^•H_2 + \text{R}-\text{OH}\)

This alkyl radical can then react with \(HCl\) or \(HI\).

Reaction with Hydrogen Halides:

The alkyl radical formed reacts with \(HCl\) or \(HI\) to form an alkyl halide (halogenated alkane):

For example, with \(HCl\):

\(CH_2-C^•H_2 + \text{HCl} \rightarrow \text{CH}_3-\text{CH}_2-\text{Cl}\)

Endothermic Step:

Energy Considerations:

The key reason why anti-Markovnikov addition occurs in the presence of peroxides is because one of the steps in this radical mechanism is endothermic.

Specifically, the abstraction of a hydrogen atom from the alkene to form the initial alkyl radical \((CH_2-C^•H_2)\) is endothermic. This means it requires energy input rather than releasing energy (exothermic).

This endothermic nature of the initial step destabilizes the intermediate formed, favoring the formation of the more stable tertiary or secondary radicals over primary radicals.

Conclusion: Therefore, option (3) is correct because one of the steps in the radical mechanism initiated by peroxides (the abstraction of a hydrogen atom from the alkene) is endothermic. This endothermic step is crucial in explaining why hydrogen halides do not follow Markovnikov's rule under these conditions, leading instead to anti-Markovnikov addition products.