The addition of HI in the presence of peroxide catalyst does not follow anti-Markovnikov's rule because :

The question asks why the addition of HI to an alkene in the presence of peroxide does not follow anti-Markovnikov's rule. First, recall that anti-Markovnikov addition occurs for HBr with peroxides due to a free radical mechanism. Peroxides generate radicals that initiate a chain reaction: the bromine radical adds to the less substituted carbon of the alkene, forming a more stable carbon radical, which then abstracts a hydrogen from HBr, giving the anti-Markovnikov product.

Now, consider HI. The options must be evaluated:

Option A suggests HI is a strong reducing agent. While HI can act as a reducing agent, this property does not directly prevent anti-Markovnikov addition. The reducing ability might affect other reactions but is not the primary reason here.

Option B claims the H-I bond is too strong to break homolytically. However, the bond dissociation energy (BDE) of H-I is approximately 299 kJ/mol, which is weaker than H-Br (366 kJ/mol) and H-Cl (431 kJ/mol). Thus, the H-I bond breaks more easily, not less, so this option is incorrect.

Option C states that the iodine atom combines with a hydrogen atom to reform HI. In the free radical mechanism, the iodine radical (I•) adds to the alkene to form a carbon radical, which then abstracts a hydrogen from HI to produce the alkyl iodide and regenerate I•. This propagation step is essential for the chain reaction and does not inherently prevent anti-Markovnikov addition. The reversibility of the addition step is key, not reformation of HI.

Option D states that the iodine atom is not reactive enough to add across a double bond. This is correct. The iodine radical has low reactivity due to its large size and low electronegativity. The carbon-iodine bond formed is weak (BDE ≈ 234 kJ/mol), making the addition to the alkene reversible. For anti-Markovnikov addition to occur, the radical addition must be irreversible and fast enough to sustain the chain reaction. However, the iodine radical's addition is slow and reversible, allowing the ionic mechanism (which follows Markovnikov's rule) to dominate. The ionic mechanism is faster for HI because I⁻ is a good nucleophile, and the reaction proceeds via carbocation intermediates.

Hence, the correct answer is Option D.