Questions: Free Radical Halogenation and Chain Reactions

Question 1

In the free radical chlorination of 2-methylbutane, which C–H bond is most likely to be abstracted by a chlorine radical?

• A. A primary C–H bond, because there are more of them and statistics favor abstraction there
• B. Any C–H bond with equal probability, since all C–H bonds have essentially the same bond energy
• C. The tertiary C–H bond, because the resulting tertiary radical is stabilized by hyperconjugation
• D. A secondary C–H bond, because secondary positions are most common in branched alkanes

Answer: (C)

Radical abstraction favors positions that produce the most stable carbon radical. Tertiary radicals are stabilized by hyperconjugation with adjacent C–H and C–C bonds, lowering the transition state energy for abstraction at that position. Although chlorine radical is not highly selective (selectivity ~5:4:1 tertiary:secondary:primary per hydrogen), it still preferentially abstracts the tertiary C–H. Statistics do favor primary positions by count, but the per-hydrogen selectivity still favors tertiary.

Question 2

A chemist wants to selectively functionalize only the tertiary C–H bond of isobutane. Should they use Cl₂ or Br₂ as the halogenating agent, and why?

• A. Cl₂, because its greater reactivity means it attacks more quickly, giving less time for competing reactions
• B. Br₂, because the bromine radical is less reactive and therefore more selective for the most stable (tertiary) radical
• C. Cl₂, because chlorine's selectivity ratio is higher in absolute terms
• D. Either, since both halogens show the same radical stability trend and produce identical selectivity ratios

Answer: (B)

Reactivity and selectivity are inversely related in radical reactions. Br• is far less reactive than Cl•, making its hydrogen abstraction step highly endothermic. By the Hammond postulate, a highly endothermic step has a late, product-like transition state, so the stability of the forming radical strongly influences the activation energy. This gives Br• a selectivity of roughly 1600:80:1 (tertiary:secondary:primary per hydrogen) compared to Cl•'s modest ~5:4:1. Choosing Br₂ gives predominantly the tertiary product.

Question 3

The propagation steps in radical halogenation regenerate a halogen radical, making the overall reaction a chain process where a single initiation event can drive the formation of thousands of product molecules.

• T. True
• F. False

Answer: True

The two propagation steps form a cycle: (1) X• + R–H → HX + R•, and (2) R• + X₂ → RX + X•. The halogen radical consumed in step 1 is regenerated in step 2, so the cycle can repeat thousands of times before termination. Only a small concentration of radicals is needed, and termination (radical coupling) is statistically rare compared to propagation because radicals exist at very low concentration.

Question 4

Because bromine radical is more reactive than chlorine radical, bromination of alkanes shows greater selectivity for tertiary C–H bonds.

• T. True
• F. False

Answer: False

This reverses the actual relationship. Bromine radical is LESS reactive than chlorine radical — its hydrogen abstraction step is more endothermic. The Hammond postulate predicts that a more endothermic step has a later, more product-like transition state, so the stability of the forming carbon radical matters more to the activation energy. It is precisely because Br• is less reactive (more selective about which C–H it abstracts) that bromination is far more regioselective. High reactivity and high selectivity are generally opposed: a very reactive radical attacks whichever C–H it encounters without discrimination.

Question 5

Why does the bromine radical show much greater selectivity for tertiary C–H bonds than the chlorine radical does? Explain using the Hammond postulate.

Model answer: The first propagation step — hydrogen abstraction from the alkane — is more endothermic for Br• than for Cl•, because the H–Br bond formed is weaker than H–Cl. By the Hammond postulate, an endothermic step has a late, product-like transition state: the transition state resembles the products (HBr + R•) more than the reactants. In a product-like transition state, the stability of the incipient carbon radical strongly influences the activation energy — a tertiary radical lowers the barrier significantly relative to a primary radical. For Cl•, the abstraction is less endothermic (earlier transition state, more reactant-like), so radical stability has a smaller influence on the barrier height, giving much lower selectivity.

This is a central application of the Hammond postulate: more endothermic (or more difficult) reactions are more selective because their transition states more closely resemble the product, making product stability a better predictor of relative rate.