Why are tertiary alkyl halides obtained in higher yield than secondary or primary during free radical halogenation of alkanes?

Question

From my textbook:

Halogens ($\ce{Cl_2}$ and $\ce{Br_2}$ ) react with alkanes in the presence of ultraviolet light to form haloalkanes.This reaction is free radical substitution and gives a mixture of mono, di, or polysubstituted haloalkanes which are difficult to separate into pure components. Moreover, the yield of any one compound is low because of the formation of other compounds. For example, in case of butane, a mixture of 1-chlorobutane (28%) and 2-chlorobutane (72%) is obtained, where the secondary alkyl halide is obtained as the major product rather than the primary one. $$\ce{CH_3CH_2CH_2CH_3->[\ce{Cl_2,~UV~light}]CH_3CH_2CH_2CH_2Cl + CH3CH_2CHClCH_3}$$
In general, the ease of substitution of various hydrogen follows the sequence:
$${\rm tertiary > secondary >primary}$$

After reading through all the above statements in my book, I now have the following questions:

• Why are tertiary alkyl halides obtained as major products instead of secondary or primary ones? It seems that a hydrogen from a tertiary carbon can be more easily replaced by a halogen than that of a secondary or primary carbon. If it is so, why is hydrogen from a tertiary carbon more easily replaced than a hydrogen from a secondary or primary carbon?

• Why are fluorine and iodine not considered in case of free radical halogenation (the first two lines of the above passage extracted from the book mentions only chlorine and bromine)?

• Why is a tertiary alkyl halide not produced in the above example? Even if we think it might be produced in a lesser amount, there should have been more of it than the primary or secondary alkyl halides. It seems the reason for tertiary alkyl halide not being in the equation is because it might not have formed or it might have formed in a lesser amount than primary or secondary alkyl halides. If it is either any case, it would violate the order mentioned in the book.

Answer 1

The reason for the ordering is that tertiary radicals have a lower energy (and are thus easier to form) than secondary radicals, which are in turn easier to form than primary radicals. This is because of a phenomenon called sigma conjugation - the weak overlap between the half filled orbital and the adjacent side group orbitals gives a slight lowering of energy. This means that it is easier to remove a hydrogen to give a tertiary radical intermediate than to give a secondary radical intermediate and so on. This is similar to Markovnikov's rule.

The reason a tertiary product is not formed is that for butane there is no tertiary centre to begin with - nowhere that a Cl radical can react to give a tertiary butyl radical. Forming a tertiary product would require a rearrangement of the carbon chain.

Answer 2

Why are tertiary alkyl halides obtained as major than secondary or primary? It seems that hydrogen from tertiary carbon can be easily replaced by halogen than from secondary or primary carbon. If it is so, why is hydrogen from tertiary carbon can be easily replaced than from secondary or primary?

The mechanism involve: $$\ce{X_2->[\Delta][\text{or UV-light}]2X.}\\\ce{X.+RH->HX +R.}\\\ce{R.+X2->RX +X.}$$ where last two steps keep on repeating cyclically. The difficult step is step is as follows which decide the rate of the mechanism:

The relative rates of primary, secondary, tertiary alkyl halides depends on two factors: probability and Activation Energy.Probability factor for a typical case of propane is $3^\circ:2^\circ:1^\circ=0:6:2$(directly equal to number of hyrogens of that type.) The activation energy of $\ce{R-H + X.->R. +H-X}$ in KCAL/MOL are: $$\begin{array}{c|c|c}\ce{R&X=Cl&X=Br\\CH3&4&18\\$1^\circ$&1&13\\$2^\circ$&0.5&10\\$3^\circ$&0.1&7.5}\end{array}$$ The more stable the radical, more easily it is formed.Factors that stabilize the free radical tend to stabilize the incipent free radical in the transition state: $$\ce{-C-H+X. -> [C^{\delta.}\bond{...}H\bond{...}X^{\delta.}] -> -C.+H-X}$$ A certain factor (delocalization of the odd electron) causes the energy difference between radicals. Stability of free radicals:$3^\circ>2^\circ>1^\circ>\ce{CH3}$ Ease of formation of free radicals:$3^\circ>2^\circ>1^\circ>\ce{CH3}$

Why is fluorine and iodide not considered in case of free radical halogenation (as you can see, the first two lines of the above passage extracted from the book, mentions only chlorine and bromine)?

Iodination is reversible but it may be carried out in the presence of an oxidising agent such as $\ce{HIO3,HNO3,HgO,}$etc., which destroys the hydrogen iodide as it is formed and so drives the reaction to the right: $$\ce{CH4 +I2<=>CH3I +HI\\5HI +HIO3->3I2 +3H2O}$$ Iodines are more conviniently prepared by treating the chloro- or bromo- derivatives with $\ce{NaI}$ in methanol or acetone solution: $$\ce{RCl+NaI->[\text{acetone}]RI+NaCl}$$ This reaction is possible because $\ce{NaI}$ is soluble in methanol or acetone, whereas sodium chloride and bromide are not. This reaction is known as Finkelstein or Conant-Finkelstein reaction.

Direct Fluorination is usually explosive; special conditions are necessary for the preparation of the fluorine derivatives of the alkanes.One such example is Swarts reaction.

Why is tertiary alkyl halide not produced in the above example? Even if we think it might be produced in less amount, it should have been greater than primary alkyl or secondary alkyl halide, is n't (because of the order)? It seems the reason for tertiary alkyl halide not being in the equation is because of either, it might not have formed or it might have formed in lesser amount than primary or secondary. If it is either any case, it would violate the order mentioned in the book.

See carefully there's no tertiary alkylhalide possible, because no tertiary hydrogen is available. As radicals do not rearrange, the structure never changes!

Answer 3

To address your second question, fluorine and iodine are not considered for this reaction because the first is too reactive while the latter is too unreactive.

Fluorine will react harshly with alkanes, producing an unselective mixture of fluorocarbons. The chain propagation steps of the reaction are so exothermic that it proceeds explosively, and special measures are necessary to control the reaction (e.g. low reactant concentrations).

In contrast, iodine will react very slowly, if at all. Although the activation energy for the reaction (+151 kJ/mol) is the lowest of all halogens, the first chain propagation step (+142 kJ/mol) and the overall reaction enthalpy (+54 kJ/mol) are very endothermic, which disfavors radical iodination of alkanes (reference).