Free-Radical Mechanism

Free-Radical Mechanism

It has generally been accepted that addition of halogen to alkenes, in the absence of light, is polar. Thus, for example, stewart et al. (1935) showed that the addition of chlorine to ethylene is accelerated by light. This suggests a free-radical mechanism :

Cl₂      hv     2Cl

CH₂=CH₂ + Cl  → CH₂Cl-CH₂     Cl₂     CH₂Cl-CH₂Cl + Cl  etc.

Poutsma (1965), however, has shown that chlorine adds, e.g., to but-2-ene at 25^oC in the absence of light, and proposed a spontaneous free-radical mechanism:

MeCH=CHMe + Cl₂   spont.  MeCHClCHMe   Cl₂   MeCHClCHClMe  +  Cl;  etc.

This mechanism is supported by the fact that the reaction was much slower in the presence of oxygen. Under these conditions, free-radical addition is inhibited and the reaction proceeds by the slower polar mechanism.

Instead of addition reactions with the halogens, the alkenes may undergo substitution provided the right conditions are used. Thus, when straight-chain alkenes are treated with chlorine at a high temperature, they form mainly monochlorides of the allyl type,i.e., in the chain –C-C-C=C-, it is the hydrogen of C that is substituted. This high temperature substitutions strongly presumptive of a free-radical mechanism, e.g.,

Cl₂      400 – 600^oC      2Cl

CH₃CH=CH₂ + Cl →  HCl + CH₂CH=CH₂   Cl₂   ClCH₂CH=CH₂ + Cl ;  etc.

The reason for allyl substitution occurring is not certain, but one possibility is that, of the three types of free radicals that may be formed, CH₃CH=CH, CH₃C=CH₂, and CH₂CH=CH₂, it is the allyl one that will have the greatest resonance stabilization (the contributing structures are identical):

CH₂CH=CH₂  ⟷  CH₂=CHCH₂

From the M.O. point of view, in the allyl free radical, each carbon atom is associated with a p_z electron. Thus an M.O. is formed covering all three carbon atoms, with consequent delocalization. Since the formation of the free radical with maximum stability means that the reaction path through this intermediate will be accompanied by the greatest decrease in free energy, the reaction will therefore proceed along this route.