Ch4 : Selectivity

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 1^o, 2^o or 3^o. 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 3^o being the easiest.
	CH₃-H	435	104
1^o	CH₃CH₂-H	410	98
2^o	(CH₃)₂CH-H	397	95
3^o	(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 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, 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 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

	Br	Cl
1^o	1	1
2^o	82	3.9
3^o	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
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 1^o in the two -CH₃ so nH₁ = 6 (replacing any of these H gives 1-chloropropane)
2 x 2^o 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 R_iare 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 2^o radical, whereas in the chlorination the number of 1^o H dictates the regioslectivity.

There are other examples in the sample problems.