Here’s a question that comes up a lot: What’s a transition state?
In the Harry Potter series, Remus Lupin changed to his werewolf form when the moon was full. But he didn’t go from human to werewolf just by snapping his fingers: the fangs and hair had to grow, the spine curled, the hands became more pawlike and the fingernails lengthened into claws. If you saw the movies, the business of transforming from human to werewolf didn’t look like a whole lot of fun! You can imagine that the extent to which Remus disliked transforming into a werewolf was proportional to the pain involved. That point of maximum pain – where Lupin is half-man, half-werewolf – is a lot like a transition state.
Like the most painful part of transforming from a human to a werewolf, a transition state is at a local energy maximum: it has partial bonds and cannot be isolated as an individual molecule.
Let’s try to demonstrate what that means.
Under certain conditions, when you take a primary alkyl halide and add a good nucleophile, you end up forming a new carbon-nucleophile bond and breaking the carbon-halide bond. It’s a substitution reaction, since we’re exchanging one functional group for another. Here’s an example: the reaction between NaOH and CH3Br.
You might recognize this as an SN2 reaction (nucleophilic substitution, bimolecular). Try to imagine how this is happening. You’re breaking the C-halide bond while at the same time forming a new carbon-nucleophile bond. Since you can’t have a stable carbon with 5 full bonds (it would have 10 electrons and thus violate the octet rule) here has to be some point where you have partial bonds.
[Imagine runners in a relay race. There is a brief moment in each race where one runner is passing the baton to the other, where neither has complete possession of the baton (electrons, in our analogy) but it is shared between the two. That’s the analogy here.]
Drawing out the reaction, it looks a little like this (using Nu as a substitute for OH(–) and M as a substitute for Na(+) to make it more general). That’s the transition state. Again, it is the point of highest energy (most unstable), since you have two partial bonds at their maximum (weakest) lengths. Furthermore, the geometry has gone far from the ideal (tetrahedral) to one with more steric interactions (trigonal bipyramidal)
Since this is a pretty painful and unhappy state of affairs for both molecules involved, the lifetime of the transition state is extremely short – on the order of femtoseconds – and rapidly relaxes to the product.
We have a name for this point of “maximum pain”: the difference between the energy of the starting material and the transition state (the point of “maximum pain”) is called the “Activation energy“.
So given a diagram for the progress of a reaction, you can figure out the activation energy by subtracting the energy of the transition state from the energy of the product.
This graph shows the reaction of CH3Br (A) with a nucleophile (Nu) that goes through a transition state (B) to yield a product, CH3Nu (C) [in addition to Br(–), not shown]
- The activation energy would be E of the TS (B) minus the E of the starting material (A).
- Notice another thing for this reaction: the product (CH3Nu) is lower in energy than the starting material (CH3Br). We can figure out the overall energy for the process by subtracting the energy of the product from the energy of the starting material:Energy given off by the reaction = Energy of product (C) minus Energy of starting material (A).
- You can even figure out the activation energy for the reverse reaction: that would be the energy of the transition state (B) minus the energy of the product (C); in this case the activation energy will be higher since the product C is more thermodynamically stable:
- Activation energy for reverse reaction = energy of transition state (B) – energy of product (C).
Key things to remember here:
1) you can’t isolate a transition state – it has partial bonds
2) you can calculate the energies of a reaction (qualitatively) by examining a reaction energy diagram.