Selectivity in Free Radical Reactions: Bromine vs. Chlorine

by James

in Organic Chemistry 1

In last blog post on radicals we saw this data that compares the chlorination of propane vs. the bromination of propane:
1-propane1Bromination:
5-bromination

We also saw that when you take statistics into account, the selectivity of these reactions are different.
For chlorination, the reaction is selective for secondary C-H over primary C-H by a factor of 55/(45/3) = 3.6 to 1
For bromination, the reaction is selective for secondary C-H over primary C-H by a factor of 97/(3/3) = 97 to 1.

The question that this post hopes to answer is  “Why is bromine more “selective” for the secondary carbon than chlorine?”. 

It’s kind of a long answer. This post goes through the data and makes the scientific argument. In the next post I’ll put forward a simple analogy that simplifies this idea for many students.

The first thing to note is formation of the different chloropropanes happens during the chain propagation step (that is, after initiation). So for our purposes here we are only going to analyze the two propagation steps and assume that initiation has already occurred.  In other words, this step:

1-select

1. This Might Seem Irrelevant… But Trust Me, It Is Not

Before addressing the topic of selectivity directly, let’s first begin by talking about activation energy. You might recall that in order for a reaction to occur, the reactants must come into contact with each other with sufficient energy to overcome the repulsions between them (electron clouds). They also must collide in such a way that allows for transfer of electrons (aka “orbital overlap”)

In other words, the rate is equal to:

[concentration of molecules with sufficient energy] *[probability they collide in the “right way”]

This is dealt with by the Arrhenius equation:

arrhenius

Here, the “probability they collide in the right way” is dealt with by the pre-exponential factor A and is unique for each reaction.  The exponential factor e-Ea/RT is what’s known as the “distribution function” and this is how we calculate the % of molecules in solution that have sufficient energy to react (heretofore known as the “activation energy”. )

Note the fact that this is temperature dependent. Why might that be? Well, as a collection of molecules is heated, the average speed of those molecules will increase. This is described by a related function known as the Boltzmann distribution, shown right here at three different temperatures [a=1, a=2, and a=5]. The y-axis shows the # of molecules (“distribution”) and the x-axis shows energy  [thank you Wikipedia]

Heat Increases The Average Velocity Of Molecules

Maxwell-Boltzmann_distribution_pdf.svg

This shows the “distribution” (aka “number”) of molecules at specific energies. The hump in the middle is the “most probable” energy, and notice how it resembles a bell curve with tails at both ends. At the far right of the scale are the most energetic molecules. Note what happens when temperature is increased: on average, molecules have greater energy, and also the right-hand “tail” extends further out.

Now imagine we have a reaction with activation energy Ea. At very cold temperatures, very few molecules have sufficient energy to react. But as the reaction mixture is heated, the proportion of those molecules increases. Ergo, the rate of the reaction increases with heat. Here is an example of how it works. Note how in this case below, the reaction doesn’t occur at 300K, but starts to occur at a reasonable rate at 330K!

Ea

Okay, that covers changing the reaction temperature. But what does this have to do with the selectivity of halogenation? 

Here, we’re keeping reaction temperature constant, but the activation energy for each reaction is slightly different.

We have four reactions in total to think about (two different halogens and two different C-H bonds).  The activation energy for each reaction has been experimentally measured (and here, we’re talking about the activation of C-H bond breaking [i.e. propagation] not initiation.  Here’s what they look like.

3-act-en-hal

This is actually all the information we need to be able to make a rough estimate of selectivities.
For a reaction at 300K,  we can calculate RT using the gas constant (1.987 cal/ K mol) and plug in the activation energy for each reaction. By dividing the two equations by each other, the pre-exponential factor A will roughly cancel out and we can obtain estimates for selectivities.
The bottom line here is that due to the nature of the Arrhenius equation, the greater the difference between activation energies, the larger the selectivity. The effects can be dramatic, even when going from a difference of 1kcal/mol (for chlorination) to 3 kcal/mol (for bromination)

small diff

All this is fine – it’s based on experimental data for activation energies. But it just opens up another question.

WHY is the difference for activation energies greater for bromination (3 kcal/mol) than for chlorination (1 kcal/mol). Great question!

Understanding this point begins with understanding the energy profile of these two different reactions (chlorination and bromination).

We’ll do the math in a second, but the key difference is that in chlorination, the key propagation step is exothermic and in bromination, the key propagation step is endothermic. This is because chlorination forms a strong H-Cl bond (103 kcal/mol) and bromination forms a much weaker H-Br bond (87 kcal/mol).

The two reaction coordinates roughly look like this:

Look closely where the transition state is for each reaction.
In chlorination, the reaction is exothermic, and the transition state resembles the reactants. According to Hammond’s postulate, we could say that this transition state is “early”.
In bromination, the reaction is endothermic, and the transition state resembles the products. According to Hammond’s postulate we say that this transition state is “late”.

What this means is that for chlorination, the difference in activation energies between the two radical pathways (i.e. “secondary” and “primary”) will most closely resemble the reactants (which are identical in energy). So there will be a very small difference in activation energies between the two.

1-chlor-prim-sec-rxn-coord

 

The difference in activation energies is small (1 kcal/mol) because of the early transition state. Note that even though there is a fairly large difference in energy between the products (3kcal/mol) it doesn’t affect the activation energies.

We can do the same analysis for bromination. Here, it’s a “late” transition state, so the difference in activation energies between primary and secondary will closely resemble the differences in energy between the two. So we would expect the activation energy difference to more closely resemble the difference between the energies of the products. And that is the case!

brom prim vs sec

So the bottom line for today’s post is a few things:

1) going from an activation energy difference of 1kcal/mol to about 3 kcal/mol can mean the difference between a reaction with a selectivity of 3.5:1 and a reaction with a selectivity of 97:1. Wow!!

2) We compared a reaction with an “early” transition state and a “late” transition state and saw that the reaction with the “late” transition state was more selective. This is to be expected when we’re starting with identical reactants and there are significant differences in the energies of the products.

In the next post I’ll use a simple analogy to drive home the point in a way that has helped students to understand this point on a more intuitive level.

Next Post: Halogenation At Tiffany’s


 

PS Math: calculations for the energies of each reaction. Chlorination:

1-chlor

Bromination:

1bromo vs 2 bromo

 

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{ 9 comments… read them below or add one }

Josh

This is awesome from a physical organic chemistry perspective. How on earth are you not a professor at some school. You actually know your chemistry. I wish you were a professor at my school. This is a great write-up!

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Rinku kushwaha

Nice way of explaining the concept of physical organic chemistry

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muskan

Great job!
Seriously I loved reading the concepts explained so nicely!

Reply

Sophie

Wow, really detailed explanations. I really appreciate your website.

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Stefan

Very well explained topic! There’s only one thing I don’t get (although it’s not important for the regioselective step):

Why is the formation of the bond between the halogen and carbon energetically identical for the primary and secondary carbons? (83 kcal/mol in case of Cl, 70 kcal/mol in case of Br?) I would assume that there’s also a slight difference? Unfortunately I couldn’t find any thermodynamical data.

I would appreciate your short answer.

Reply

Vinay Deep Punetha

Ohhh my my my !!!! Hats off to you…..I am speechless…..its too detailed and lucid to understand.

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Idan

Hello,
Just pointing out R=1.987 cal/ K mol (not kilo) written at the “Calculating the selectivities based on differences in activation energy” image and the paragraph above it.
Also, could you please explain “[…] in chlorination, the key propagation step is exothermic and in bromination, the key propagation step is endothermic.”
What makes the first propagation step the “key” step? Why not look at the net energy difference for both as you did in the image at the bottom?

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James

The first propagation step is the key step because it results in formation of a radical at the alkane – this is the slow step (breakage of C-H). The second step is fast because a considerably weaker bond is being broken (Cl-Cl) .

Reply

James

Thanks for the correction on cal vs kcal. Fixed!

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