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Elimination Reactions Are Favored By Heat

A few posts back we saw how elimination reactions are often in competition with substitution reactions.

ow do we know when one reaction pathway is going to be preferred over another? As we’ll see, there are going to be several components to answering this question fully, but today we’ll talk about one simple rule of thumb going forward.

It’s this:

All else being equal, higher temperatures will lead to an increase in the amount of elimination products versus substitution products.

The rest of this post is about why.

Let’s say you have a reaction like this one. It’s possible for substitution or elimination products to be formed. [I’m keeping the identity of the base, substrate vague here since I don’t have a concrete experimental example to provide (although I’d greatly appreciate one)].


As temperature is increased, the relative amount of elimination products will increase relative to substitution products. You can imagine it looking like this.

Notice again how organic chemistry works. It’s not as if applying heat is an on/off switch that results in a reaction going from 100% substitution to 100% elimination. Instead, increasing temperature results in a gradual increase in elimination products relative to substitution. That’s because temperature is gradually leading to an increase in the rate constant for elimination versus rate constant for substitution.

So what’s going on here?

Here’s one thing we can say with confidence: at low temperatures, the activation energy for the substitution reaction is lower than that for the elimination reaction. Remember that the lower the activation energy, the higher the rate of the reaction. This might help to explain our product distribution: as we increase the temperature, more energy is available, so  so that the starting materials can ascend the activation barrier to provide elimination reactions also. This fits with what is observed.

However, there’s an even more fundamental reason why we might see more elimination products as heat is increased, and it has to do with some properties we know about thermodynamics that make rate constants (and activation energies) temperature dependent.

Let’s look again at the (generic) reactions:

What do we notice here? Notice that the substitution reaction we’re going from 2 species in the starting material to 2 species in the product. But in the elimination reaction, we’re going from 2 species in the starting material to 3 species in the product. An increase!

Since we’re birthing a new species in solution here, that’s going to result in an increase in entropy. And if you think waaay back to general chemistry, the Gibbs equation told us this relationship:

ΔG =  ΔH – T ΔS

Remember that the more negative ΔG is, the more favorable the reaction. As temperature incresaes, that TΔS term is going to start getting bigger and bigger; this will make Δ G more and more negative. At some point, as temperature is increased, the ΔG for elimination will become more negative than delta G for substitution. In other words, more favorable.

One thing to be careful about though! When we’re discussing ΔG, we really should be talking about the ΔG of the transition state, not that of the final product. [Why not? Because the stability of products isn’t related to reaction rate (if it was, our bodies would have combusted to CO2 and H2O a long time ago!)]. We give a special designation to thermodynamic terms of the transition state – we put a little double-dagger on them. Like this: ΔG‡  . This “Gibbs energy of activation” is how we define the activation energy of a reaction.

So you can see by analyzing this term that activation energy can change with temperature!

At low temperatures, the Gibbs energy of activation for substitution  (ΔG‡) is lower in energy (more negative) than that for elimination.  But at high temperatures, the Gibbs energy of activation ( ΔG‡ ) for elimination starts to be lower in energy than that for substitution reactions, and hence we get an increase in the amount of elimination product.

ΔG= ΔH‡–TΔS

Again, the bottom line is that, all else being equal,  heat will tend to favor elimination reactions.

Next Post: Two Types of Elimination Reactions

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