Questions: Free Radical Chain Reactions: Halogenation of Alkanes

Question 1

2-methylbutane is treated with Br₂ under UV light. Which C-H bond is preferentially brominated, and why?

• A. The primary C-H bonds, because there are more of them and statistical abundance dominates
• B. The secondary C-H bonds, because they represent the best balance of abundance and stability
• C. The tertiary C-H bond, because the resulting tertiary radical is most stable
• D. All C-H bonds equally, because radicals are non-selective

Answer: (C)

The selectivity-determining step is H-abstraction by the bromine radical, and the transition state closely resembles the product alkyl radical (Hammond's postulate — the step is endothermic). Therefore, the stability of the radical formed directly lowers activation energy. A tertiary radical is stabilized by hyperconjugation with three adjacent alkyl groups, making the tertiary C-H bond by far the preferred site for bromination. Statistical abundance of primary C-H bonds does not overcome the large stability difference for Br, which is why option A is wrong — that reasoning applies better to the less selective Cl₂.

Question 2

Chlorination of propane gives both 1-chloropropane and 2-chloropropane in roughly 45:55 ratio despite there being 6 primary H's and only 2 secondary H's. What does this tell you about chlorine radical selectivity?

• A. Chlorine is highly selective, strongly favoring secondary C-H bonds over primary
• B. Chlorine shows modest selectivity — secondary C-H bonds react faster per hydrogen, but the difference is small enough that primary abundance still gives significant primary product
• C. Chlorine is entirely non-selective, and the 55% secondary product is explained solely by statistical abundance
• D. Chlorine selectively attacks secondary C-H bonds via an ionic mechanism

Answer: (B)

If chlorine were perfectly non-selective, the 6:2 primary:secondary ratio of H's would give 75% primary product. Instead, ~55% is secondary, meaning secondary positions react at roughly 3.7× the rate per H. This modest per-hydrogen selectivity (compare Br₂ where selectivity is ~1600× for tertiary) reflects a relatively exothermic H-abstraction step with an early, reactant-like transition state that doesn't fully express radical stability differences. The key contrast with Br₂ — which is endothermic in the H-abstraction step and therefore has a late, product-like TS that strongly responds to radical stability — is explained by Hammond's postulate.

Question 3

Once a free radical chain reaction is initiated, it requires continuous UV light exposure to keep producing product.

• T. True
• F. False

Answer: False

Initiation only needs to generate the first few radicals. After that, the propagation steps are self-sustaining: each propagation cycle regenerates a halogen radical that feeds directly back into the next cycle. A single initiation event can produce thousands of product molecules before termination (radical-radical recombination) ends the chain. Continuous UV light is not required — once the chain is running, it sustains itself. This is the defining characteristic of a chain reaction.

Question 4

Termination steps in free radical halogenation reduce the overall yield of the halogenated product because they consume radicals without producing HX.

• T. True
• F. False

Answer: True

True, and the reasoning is important. Termination (e.g., R· + X· → R–X, or R· + R· → R–R) removes chain-carrying radicals from the pool, ending those particular chain sequences. Termination via R· + R· coupling actually forms a byproduct (dimeric alkane) rather than the desired alkyl halide, directly reducing yield. R· + X· → R–X does produce an alkyl halide, but it still terminates the chain, preventing the thousands of additional product molecules that propagation would have generated. Minimizing termination (by keeping radical concentrations low) maximizes chain length and yield.

Question 5

Explain why bromination of alkanes is highly regioselective while chlorination typically gives product mixtures, even though both proceed through the same chain mechanism.

Model answer: The H-abstraction step in the propagation cycle is more endothermic for Br· than for Cl·. By Hammond's postulate, an endothermic step has a late, product-like transition state that closely resembles the alkyl radical being formed. This means the TS energy is strongly influenced by radical stability differences (3° vs. 2° vs. 1°), so bromination is dramatically selective. For chlorination, H-abstraction is exothermic, giving an early, reactant-like TS that barely 'feels' the stability of the incipient radical — hence modest selectivity and mixed products.

The key is the energetics of the selectivity-determining step, not the mechanism itself (which is identical for Cl₂ and Br₂). Hammond's postulate connects TS structure to reaction thermodynamics: exothermic steps have early TSs that look like reactants, while endothermic steps have late TSs that look like products. Because radical stability differences are expressed more fully in the late TS of Br H-abstraction, bromination achieves ~1600:80:1 selectivity (3°:2°:1° per H), while chlorination achieves only ~5:4:1.