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Selectivity

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

R-H
Within the series of sp3 C-H bonds, the strength of the C-H bonds varies slightly depending on whether the H is 1o, 2o or 3o. The following table shows the bond dissociation energy, that is the energy required to break the bond in a homolytic fashion, generating R and H ..
 
Type
R-H
kJ/mol
kcal/mol
Note how the bonds get weaker as we move down the table, so the Ralso gets easier to form, with 3o  being the easiest radical to form.
CH3-H
435
104
1o
CH3CH2-H
410
98
2o
(CH3)2CH-H
397
95
3o
(CH3)3C-H
380
91

Halogen radical, X.

The relative rates of reaction for X2 relative to chlorine are : F =108, Cl = 1, Br = 7 x 10-11 and I = 2 x 10-22 i.e. relative to chlorination, F reacts rapidly, Br very slowly and I very, very, very slowly.

For a given set of reaction conditions, the selectivity of the radical reactions can be predicted mathematically based on a combination of an experimentally determined reactivity factor, Ri, and a statistical factor, nHi.

Don't be intimidated by this equation, it's not as bad as you fear! In order to use the equation 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 substituted. This is an exercise in recognising different types of hydrogen, something that will be important later.

predicting outcome of radical halogenations %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 F
1o 1 1 1
2o 82 3.9 1.2
3o 1640 5.2 1.4
(note that Ri depends on the reaction conditions)

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

Let's look at a worked example, say chlorination of propane, CH3CH2CH3

How many different monochlorides can be produced by radical chlorination ? ANSWER

This means there are two types of H atom in propane (use the JSMOL diagrams below to highlight this if you are unsure).




propane
1-chloropropane
2-chloropropane

Looking at the starting material alkane, i.e propane, we have two types of H:

Now for the calculations, so plugging the values into the equations we get (the reactivity factors Ri 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 %)

What about bromination of propane ?

Most of the process in 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 %)

Note that the results match well with experimental values (under the same conditions) and that they illustrate the high regioselectivity of the bromination reaction for the 2o radical, whereas in the chlorination the number of 1o H dictates the regioslectivity.

There are other examples in the sample problems.


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organic chemistry © Dr. Ian Hunt, Department of Chemistry University of Calgary