Chapter 4: Alcohols and Alkyl Halides

Selectivity

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

• reactivity of R-H system
• reactivity of X.
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 R.  also gets easier to form, with 3o  being the easiest. CH3-H 435 104 1o CH3CH2-H 410 98 2o (CH3)2CH-H 397 95 3o (CH3)3C-H 380 91

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 fast, Br very slow and I very, very, very slowly.

• Bromine radicals, Br ., are less reactive than chlorine radicals, Cl . (because Br is less electronegative than Cl)
• Br tends to be more selective in its reactions, and prefers to react with the weaker R-H bonds.
• The more reactive Cl . 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, Ri, and a statistical factor, nHi. 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 recognising 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 3o position than 1o
• Chlorination is 5.2 times more likely to occur at a 3o position than 1o
• Bromination is more selective than chlorination

Let's work an 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).

 Highlight primary H Highlight secondary H propane 1-chloropropane 2-chloropropane

Looking at the starting material, propane, we have two types of H:

• 6 x 1o in the two -CH3 so nH1 = 6 (replacing any of these H gives 1-chloropropane)
• 2 x 2o in-CH2- so nH2 = 2 (replacing either of these H gives 2-chloropropane)
• (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 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 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.

 © Dr. Ian Hunt, Department of Chemistry