# Why is chlorination less selective than bromination?

##### 1 Answer
Dec 24, 2014

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 ${E}_{\text{a}}$ are:

1. CH₃CH₂CH₂-H + ·Cl → CH₃CH₂CH₂· + H-Cl; ${E}_{\text{a}}$ = 17 kJ/mol
2. (CH₃)₂CH-H + ·Cl → (CH₃)₂CH· + H-Cl; ${E}_{\text{a}}$ = 13 kJ/mol
3. CH₃CH₂CH₂-H + ·Br → CH₃CH₂CH₂· + H-Br; ${E}_{\text{a}}$ = 67 kJ/mol
4. (CH₃)₂CH-H + ·Br → (CH₃)₂CH· + H-Br; ${E}_{\text{a}}$ = 54 kJ/mol

The Arrhenius equation predicts the rate constant for a reaction.

$k = A {e}^{\frac{- {E}_{\text{a}}}{R T}}$

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, $R T = {\text{8.314 J·K"^-1"mol"^-1 × "300 K" = "2494 J·mol}}^{-} 1$

For the chlorination reaction,

${k}_{2} / {k}_{1} = {e}^{\left({\text{-13 000 J·mol"^-1)/( "2494 J·mol"^-1))/ e^(("-17 000 J·mol"^-1)/( "2494 J·mol}}^{-} 1\right)} = \frac{{e}^{-} 5.212}{{e}^{-} 6.816} = {e}^{1.604}$

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,

${k}_{4} / {k}_{3} = \left({e}^{{\text{-54 000 J·mol"^-1)/( "2494 J·mol"^-1))/ e^(("-67 000 J·mol"^-1)/( "2494 J·mol}}^{-} 1}\right) = \frac{{e}^{-} 21.65}{{e}^{-} 26.86} = {e}^{5.21}$

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.