Free Radical Reactions
Halogenation At Tiffany’s
Last updated: December 7th, 2022 |
Thinking Through The Selectivity of Bromination vs Chlorination: An Intuitive Analogy
As we discussed in the last post on radicals, bromine radicals are considerably more selective than chlorine radicals in the halogenation of alkanes.
Why? As we saw, the detailed answer involves looking at reaction energy diagrams, the Hammond postulate, and the Arrhenius equation.
When I tutor, I have found that this explanation goes over many students’ heads. So I try to explain it with an analogy. This may or may not work for you.
Table of Contents
- Bromination Is Highly Selective For Tertiary C–H Bonds. Chlorination Is Not
- Tertiary C–H Bonds Are Significantly Weaker Than Secondary and Primary C–H Bonds
- “Halogenation At Tiffany’s” : An Analogy
- The More Unstable Radical (Cl•) Is Less Sensitive To Bond Strength Than The More Stable Radical (Br•), Which Is Only Reactive Enough To Break The Weakest C–H Bond
- Just So You Know: Low Reactivity Does Not Always Mean High Selectivity (or Vice Versa)
1. Bromination Is Highly Selective For Tertiary C–H Bonds. Chlorination Is Not.
As shown in this chart, bromination is much more selective for tertiary C–H bonds than chlorination is.
To be exact, bromination is 19,400 times more selective for tertiary C–H bonds vs primary C–H bonds. Chlorination is a mere 6 times more selective for tertiary C–H bonds.
For example, a hydrocarbon exposed to Br2 in the presence of heat or light (for initiation) can selectively be brominated at the tertiary C–H bond, resulting in one major product:
In contrast, free-radical chlorination of alkanes can result in a mixture of products, stemming from halogenation at primary, secondary, and tertiary C–H bonds.
2. Tertiary C–H Bonds Are Significantly Weaker Than Secondary and Primary C–H Bonds
As you may recall from the previous post, tertiary C-H bonds also happen to be weaker [about 93 kcal/mol] than both secondary [96 kcal/mol] and primary [100 kcal/mol] C-H bonds. (See article: Selectivity in Free-Radical Reactions: Bromination vs Chlorination)
In the last post, I gave the proper (and correct) explanation for the higher selectivity of bromine over chlorine in the free radical halogenation of alkanes: chlorination is exergonic, bromination is endergonic, and applying Hammond’s Postulate leads us to the conclusion that bromination, due to the late [“product-like”] transition state, should have a greater difference in activation energies for the carbon radical-forming step than chlorination [early transition state, more “reactant like”], and hence should have higher selectivity [selectivity is proportional to the difference between activation energies].
On occasion (actually quite frequently) in the course of my line of tutoring work this first explanation has been met with silence, a blank stare, or some other reaction that is generally consistent with incomprehension. At this point I have a choice: given the limited time available, do I persevere with this explanation, albeit with a slightly different tack, or cut bait and try something completely different?
The calculation is: how vital is it for the student’s needs to fully comprehend the Hammond’s Postulate explanation? Depending on the situation, I might decide to break out my backup plan for explaining this phenomenon, which will fall into the category of convenient lies told by chemistry instructors.
I call this analogy “Halogenation At Tiffany’s”. Here it goes (caveats afterward)
3. “Halogenation At Tiffany’s”: An Analogy
Imagine you work at a jewelry shop – Tiffany’s, let’s say. One day you’re standing behind the counter and a man in an immaculate, expensive suit barges in. Clearly he’s in a rush. His pockets are bulging with money. “I’ll have a diamond!” he says.
Almost before you can speak, his eyes are already fixated on a particular jewel in the display case – one that happens to be one of the largest and most expensive in the store. “That one looks nice!” he says. “I’ll take it!”. And as he pulls out his wads of cash, you take the diamond out of the locked display case and make the transaction. “Thanks!” he says, running out the door.
If only every sale were that simple, you think.
As if to prove your point, the next person to walk through the door looks to be in his early twenties, wearing baggy jeans and a baseball cap. He walks up and down the display cases, slowly looking at each one. After a few minutes, he stops to ask: “could you direct me to some of your least expensive diamonds, please?”.
Even when standing in front of your cheapest stones, he looks hesitant. Finally, after many delays, he finally settles on buying your absolute cheapest diamond in the store. As he pays, and you hand him the diamond, he smiles. “This is the only one I can actually afford! ” he says.
4. The More Unstable Radical (Cl•) Is Less Sensitive To Bond Strength Than The More Stable Radical (Br•), Which Is Only Reactive Enough To Break The Weakest C–H Bond
Get the analogy? The guy in the suit, in a hurry, with wads of cash, is like the chlorine radical. In a hurry (unstable) and not very sensitive to price (“bond strength”). Expensive primary C-H, secondary C-H, “cheap” tertiary C-H ; it will react with them all.
The cheap student, slow and methodical, is like the bromine radical. Slower (less unstable) and very sensitive to price. His “selectivity” is high due to the fact that he can’t “afford” to buy the more “expensive” primary and secondary C-H bonds, and can only afford the “cheapest” tertiary C–H bond.
5. This Is Good Enough For Our Purposes, But Low Reactivity Does Not Always Mean High Selectivity (or Vice Versa)
This analogy “works” with almost everyone I use it with because it taps into a story everyone can understand. At the end of the day the student will understand two things: chlorine is more reactive than bromine, and tertiary C–H is weaker than secondary or primary C–H.
So what’s the flaw? Well, just because a species is reactive, doesn’t mean it’s unselective; it depends on the nature of the competing transition states. The “Reactivity Selectivity Principle” has been called an “imperishable myth” of organic chemistry for good reason.
Is this analogy a “good enough” explanation? I think it depends on the student. For many students, who will never encounter organic chemistry again, I suppose it is. Is a student taking an online organic chemistry course at UNE en route to a nursing career really being cheated if he/she doesn’t learn the 100% proper explanation as to why tertiary C-H bonds are selectively brominated? I don’t think so. Would a chemistry major at an R1 research school be cheated by this explanation? Absolutely.
As an instructor I have to choose the oversimplifications I can live with, given that my students’ time, attention, and motivation are finite resources. C’est la vie.
Next post: let’s talk about allylic bromination.
Next Post: Allylic Bromination
4 thoughts on “Halogenation At Tiffany’s”
You make organic chemistry so easy to understand. This website is a life-saver. Thank you SO MUCH.
Thank you so much!! I was struggling with a question and this analogy just made it so simple! You’re great!
You are absolutely awesome
What a great analogy! Makes much more sense now and easy to remember, thanks!