Question

Why is chlorination less selective than bromination?

Answer 1

Chlorination is less selective than bromination because chlorination has smaller differences in activation energy for attack at 1°, 2°, and 3° positions.

Consider the halogenation of propane at the 1° and 2° positions.

CH₃CH₂CH₃ + X₂ → CH₃CH₂CH₂X and (CH₃)₂CHX

Formation of the different halopropanes occurs during the chain propagation steps.

CH₃CH₂CH₂-H + ·X → CH₃CH₂CH₂· + H-X
(CH₃)₂CH-H + ·X → (CH₃)₂CH· + H-X

The rates depend on the activation energies. The approximate values of `E_"a"` are:

1. CH₃CH₂CH₂-H + ·Cl → CH₃CH₂CH₂· + H-Cl; `E_"a"` = 17 kJ/mol
2. (CH₃)₂CH-H + ·Cl → (CH₃)₂CH· + H-Cl; `E_"a"` = 13 kJ/mol
3. CH₃CH₂CH₂-H + ·Br → CH₃CH₂CH₂· + H-Br; `E_"a"` = 67 kJ/mol
4. (CH₃)₂CH-H + ·Br → (CH₃)₂CH· + H-Br; `E_"a"` = 54 kJ/mol

The Arrhenius equation predicts the rate constant for a reaction.

`k = Ae^((-E_"a")/(RT))`

If the `A` values are the same for each reaction and the temperature `T` is constant, we can calculate the relative rates of reactions with different activation energies.

`k_2/k_1 = e^((-E_"a")/(RT_2))/ (e^((-E_"a")/(RT_1)))`

At 300 K, `RT = "8.314 J·K"^-1"mol"^-1 × "300 K" = "2494 J·mol"^-1`

For the chlorination reaction,

`k_2/k_1 = e^(("-13 000 J·mol"^-1)/( "2494 J·mol"^-1))/ e^(("-17 000 J·mol"^-1)/( "2494 J·mol"^-1)) = (e^-5.212)/ (e^-6.816) = e^1.604`

So `k_(2°)/k_(1°) = 4.97`

This assumes that the `A` values are the same. This is not quite true, but it's a reasonable estimate. The actual value of `k_(2°)/k_(1°)` is 3.9.

For the bromination reaction,

`k_4/k_3 = (e^("-54 000 J·mol"^-1)/( "2494 J·mol"^-1))/ e^(("-67 000 J·mol"^-1)/( "2494 J·mol"^-1)) = (e^-21.65)/ (e^-26.86) = e^5.21`

So `k_(2°)/k_(1°) = 183`

This is again a bit high, because the `A` values are not quite the same. The actual value of `k_(2°)/k_(1°)` is 82.

But we see that as the difference in activation energies increases, the selectivity increases.