Why is chlorination less selective than bromination?

Answer 1

Because there are less differences in the activation energy for attack at the 1°, 2°, and 3° positions, chlorination is less selective than bromination.

Think about propane's halogenation at the 1° and 2° positions.

(CH₃)₂CHX and CH₃CH₂CH₃ + X₃ → CH₃CH₂CH₂X

During the chain propagation stages, the various halopropanes are formed.

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

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

The rate constant of a reaction is predicted by the Arrhenius equation.

#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#

Regarding the chlorination process,

#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.

Regarding the reaction of bromination,

#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.

However, we observe that selectivity increases as the difference in activation energies increases.

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Answer 2

Chlorination is less selective than bromination because chlorine atoms are smaller than bromine atoms, allowing them to react more readily with a wider range of substrates. Additionally, the weaker C-Cl bond compared to the C-Br bond in the intermediate species makes the reaction less selective. This means that chlorination reactions are more likely to result in multiple substitution products, whereas bromination reactions tend to yield predominantly the desired mono-substituted product.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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