Chapter 4: Alcohols and Alkyl Halides 
Selectivity
There are two components to understanding the selectivity of radical halogenations of alkanes:

RH 


Note how the bonds get weaker as we move down the table, so the R^{. }also gets easier to form, with 3^{o} being the easiest radical to form. 

CH_{3}H 




CH_{3}CH_{2}H 




(CH_{3})_{2}CHH 




(CH_{3})_{3}CH 


Halogen radical, X^{.}
The relative rates of reaction for X_{2} 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, R_{i}, and a statistical factor, nH_{i}.
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.
%P_{i }= % yield of product "i"
nH_{i }= number of H of type "i" R_{i }= reactivity factor for type "i" S_{i} = sum for all types 
Reactivity factors, R_{i}

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, CH_{3}CH_{2}CH_{3}
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).





Looking at the starting material alkane, i.e propane, we have two types of H:
(Don't make the mistake of looking at the number of types of H in the product that you are making, you need to look at the starting material)
Now for the calculations, so plugging the values into the equations we get (the reactivity factors R_{i }are in the table above):
% 1chloropropane = 100 x (6 x 1) / (6 x 1 + 2 x 3.9) = 100 x 6 / 13.8 = 43.5 % (experimental = 44 %)What about bromination of propane ?% 2chloropropane = 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 in the same, all we have to do is change the reactivty factors
% 1bromopropane = 100 x (6 x 1) / (6 x 1 + 2 x 82) = 100 x 6 / 170 = 3.5 % (experimental = 4 %)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 2^{o} radical, whereas in the chlorination the number of 1^{o} H dictates the regioslectivity.% 2bromoopropane = 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 sample problems.
© Dr. Ian Hunt, Department of Chemistry 