Activated complex theory, tells us that due to the collision between the molecules of the reactants, they form a transition specie before the product is formed, which is called active complex. On the other hand we have consecutive reactions on which is also formed a intermediate product before forming the actual product we're interested on. My question is :

Where is the difference between the activated complex and the intermediate product, since they are both formed before the actual product ?


1 Answer 1


The key difference is that transition states occur at a maximum of the potential energy curve for the reaction whereas intermediates occur at a local minimum. Take this example of the reaction profile for an $\ce{S_{N}1}$ reaction:

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You will see that the products and reactants occur at minima on the curve and the intermediate also occurs at a minimum, albeit a higher energy one. In between the minima are located the maxima where you find the transition states. Unless the activation energy is zero (see this question for rare examples) there will always be a transition state located between any two minima.

Transition states are usually represented using dashed bonds to show bonds in the process of being broken or formed as opposed to being fully formed as in intermediates or products. Additionally the Hammond postulate says that the transition state will most resemble the stable species closest to it in energy, in this case the carbocation, and this can be used to help predict the structures of transition states.

Transition states are very short lived because they immediately 'roll downhill' on the potenetial energy curve to reach an intermediate or product. By contrast some intermediates are actually quite stable and can be isolated.

  • $\begingroup$ Thanks, so I assume there isn't any way we can know if some molecule is a transition state or a intermediate without seeing the diagram or having any energy value of them ? $\endgroup$ May 4, 2015 at 20:50
  • $\begingroup$ @Ndrina Limani Edited my answer to give more information. $\endgroup$
    – bon
    May 4, 2015 at 20:56

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