Radical Substitution

• Substitution of R-H by -X provides the alkyl halide, R-X and HX.
• Alkane R-H relative reactivity order : 3^o > 2^o  > 1^o  > methyl
• Halogen reactivity F₂ > Cl₂ > Br₂ > I₂
• Only chlorination and bromination are useful in the laboratory.
• Reaction proceeds via a radical chain mechanism

This is because alkyl groups are weakly electron donating due to hyperconjugation and inductive effects.
Note that this is the same order as for carbocations. Resonance effects can further stabilize radicals when present.

Structure:
 

Alkyl radicals are sp2 hybridized, planar systems at the radical C center.  The p-orbital that is not utilized in the hybrids contains the single electron.

Reactivity:
As they have an incomplete octet, radicals are excellent electrophiles and react readily with nucleophiles.
Alternatively, loss of H^. can generate a p bond

Rearrangements:
Unlike carbocations, radicals do not tend to undergo rearrangements

Reactions involving radicals: Radical halogention of alkanes

Radical Substitution Mechanism

Unlike the large majority of reactions that you will see in your organic chemistry course, radical mechanism require that fishhook curly arrows that represent the motion of a single electron are used. These can be a little more confusing and more difficult to master. We suggest you get to grips with normal curved arrows first.

However, the mechanism for the bromination of methane is shown below, but the mechanism for chlorination or higher alkanes in the same. Note that it contains three distinct type of steps, depending on the net change in the number of radicals that are present.
 

RADICAL CHAIN MECHANISM FOR REACTION OF METHANE WITH Br2
Step 1 (Initiation) Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.
Step 2 (Propagation) (a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then (b) The methyl radical abstracts a bromine atom from another molecule of Br2 to form the methyl bromide product and another bromine radical,  which can then itself undergo reaction 2(a) creating a cycle that can repeat.
Step 3 (Termination) Various reactions between the possible pairs of radicals allow for the formation of ethane, Br2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

 

Selectivity

There are two components to understanding the selectivity of radical halogenations of alkanes:

• reactivity of R-H system
• reactivity of X^.

Halogen radical, X^.

• Bromine radicals are less reactive than chlorine radicals
• Br^.  tends to be more selective in its reactions, and prefers to react with the weaker R-H bonds.
• The more reactive chlorine radical is less discriminating in what it reacts with.

The selectivity of the radical reactions can be predicted mathematically based on a combination of an experimentally determined reactivity factor, R_i, and a statistical factor, nH_i. In order to use the equation shown below we need to look at our original alkane and look at each H in turn to see what product it would give if it were to be susbtituted. This is an exercise in recognizing different types of hydrogen, something that will be important later.
 

%Pi = % yield of product "i" nHi = number of H of type "i" Ri = reactivity factor for type "i" Si = sum for all types

Reactivity factors, Ri Br Cl 1o 1 1 2o 82 3.9 3o 1640 5.2

What do the reactivity factors indicate ? Well as an example of the conclusions we could make:

• Bromination is 1640 times more likely to occur at a 3^o position than 1^o
• Chlorination is 5.2 times more likely to occur at a 3^o position than 1^o

Now for the calculations, so plugging the values into the equations we get (the reactivity factors R_i are in the table above):

% 1-chloropropane = 100 x (6 x 1) / (6 x 1 + 2 x 3.9) = 100 x 6 / 13.8 = 43.5 %  (experimental = 44 %)

% 2-chloropropane = 100 x (2 x 3.9) / (6 x 1 + 2 x 3.9) = 100 x 7.8 / 13.8 = 56.5 % (experimental = 56 %)

Most of the process is the same, all we have to do is change the reactivty factors.

% 1-bromopropane = 100 x (6 x 1) / (6 x 1 + 2 x 82) = 100 x 6 / 170 = 3.5 %  (experimental = 4 %)

% 2-bromoopropane = 100 x (2 x 82) / (6 x 1 + 2 x 82) = 100 x 164 / 170 = 96.5 % (experimental = 96 %)

There are other examples in the Self Assessment.

Radical Halogenation of Alkanes

Summary:

• When treated with Br₂ or Cl₂, radical substitution of R-H generates the alkyl halide and HX.
• Alkane R-H relative reactivity order : tertiary > secondary > primary > methyl.
• Halogen reactivity F₂ > Cl₂ > Br₂ > I₂
• Only chlorination and bromination are useful in the laboratory.
• Bromination is selective for the R-H that gives the most stable radical.
• Chlorination is less selective
• Reaction proceeds via an radical chain mechanism which involves radical intermediates.
• The termination steps are of low probability due to the low concentration of the radical species meaning that the chances of them colliding is very low.

RADICAL CHAIN MECHANISM FOR REACTION OF METHANE WITH Br2
Step 1 (Initiation) Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.
Step 2 (Propagation) (a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then  (b) The methyl radical abstracts a bromine atom from another molecule of Br2 to form the methyl bromide product and another bromine radical,  which can then itself undergo reaction 2(a) creating a cycle that can repeat.
Step 3 (Termination) Various reactions between the possible pairs of radicals allow for the formation of ethane, Br2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

More highly brominated by-products are possible if methyl bromide reacts with a bromine radical in the same fashion as methane does. Can you draw the cycle that leads to the formation of dibromomethane ?