<Li> We can also look at the bond dissociation energies (BDEs) to understand the selectivity of bromination . The BDE of a bond is the energy required to break it by homolytic cleavage, and these values can be used to determine if a reaction or step in a reaction is exothermic or endothermic . In a chain reaction step of a Br radical reacting with a hydrogen on a secondary carbon to cleave the H − C bond requires 397 kJ / mol and an H − Br is formed . We can look at the BDE of H − Br (= 366 kJ / mol) and subtract this value from 397 kJ / mol to get + 31 kJ / mol . This positive value tells us that this step in the reaction requires energy (endothermic) and the reactants are more stable than the products . Comparing this with the same situation of a Cl radical we see that this step is exothermic (397 − 431 kJ / mol = − 20 kJ / mol). From these values we can conclude that in bromination it is more important that the most stable radical (tertiary) be formed during this step and thus it is more selective than chlorination . This is because less energy is required to form the H--Br and tertiary radical (380 − 366 kJ / mol = + 14 kJ / mol). This value is 17 kJ / mol less than the secondary radical formation . Iodine does not even participate in free - radical halogenation because the entire reaction is endothermic . </Li> <Li> It is possible to predict the product distribution of different monochloro derivatives resulting from the chlorination of an alkane with non-equivalent hydrogens . From experimentation it has been determined that the relative rates of chlorination to primary, secondary, and tertiary positions are 1, 3.8, and 5 respectively (this ratio is used in the example following this paragraph). This corresponds with alkyl radical stability: tertiary radical species are more stable than secondary radical species, and secondary radical species are more stable than primary radical species--thus any single chlorination will favor substitution at the most substituted carbon . Because of this trend, the percentages of each product formed from the parent radical can be estimated with relatively high accuracy . For example, 2 - methyl butane ((CH) CHCH CH) exhibits the following results: </Li> <Dl> <Dd> For clarity, the unique hydrogens will be labeled as follows: <Dl> <Dd> a = (CH) (primary), b = CH (tertiary), c = CH (secondary), d = CH (also primary) </Dd> </Dl> </Dd> <Dd> Applying the aforementioned substitution ratios in the formula: ((number of hydrogens) × (ratio factor)) / ((primary hydrogens × 1) + (secondary hydrogens × 3.8) + (tertiary hydrogens × 5)) <Dl> <Dd> a: 6 × 1 = 6 a = 6 / 21.6 = 28% </Dd> <Dd> b: 1 × 5 = 5 b = 5 / 21.6 = 23% </Dd> <Dd> c: 2 × 3.8 = 7.6 c = 7.6 / 21.6 = 35% </Dd> <Dd> d: 3 × 1 = 3 d = 3 / 21.6 = 14% </Dd> </Dl> </Dd> <Dd> Denominator total for this species = 6 + 5 + 7.6 + 3 = 21.6 </Dd> </Dl> <Dd> For clarity, the unique hydrogens will be labeled as follows: <Dl> <Dd> a = (CH) (primary), b = CH (tertiary), c = CH (secondary), d = CH (also primary) </Dd> </Dl> </Dd>

Explain why free-radical halogenation usually gives mixtures of products