I've been reading an explanation about transition states in Smith's Organic Synthesis:

A transition state is an energy maximum along the reaction coordinate. However, a chemical reaction where the reactant contains N atoms actually involves movement over an energy surface with 3N-6 dimensions, and a transition state is actually an energy minimum in each of the remaining (3N-7) dimensions. Liken a chemical reaction to a drive across a mountain range. The “goal” is to go from one valley to another, just as the goal of a chemical reaction is to go from one minimum on the energy surface (the reactants) to another minimum (the product). The road moves about a two-dimensional surface whereas the energy surface for a chemical reaction may involve dozens or even hundreds of dimensions. There is no need to climb to the top of a mountain (a maximum in both dimensions) to go from one valley to another. Rather, is sufficient to go through a pass between mountains (a maximum in one dimension but a minimum in the other). Similarly, it is only necessary to pass through an energy maximum in one coordinate (the reaction coordinate) for a chemical reaction. Because a transition state is an energy maximum (albeit only in one of many dimensions) it cannot function to “trap” a molecule, let alone collect a population of “trapped” molecules. This means that a transition state cannot even be observed experimentally let alone isolated or characterized. Quite simply, a transition state does not exist in the sense that a molecule exists.

This multidimensional explanation completely confounds me. Could someone explain this in a more understandable language? What is this surface with 3N-6 dimensions? How did the author use it to conclude that TS can't be observed experimentally?

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    $\begingroup$ related: chemistry.stackexchange.com/questions/51903/… You may get some results if you search for "potential energy surface", since that is exactly what this paragraph is describing. We can only depict them in 3D, but if you get the hang of the idea, then it is not to difficult to convince yourself that it can be generalised to 4D, etc. $\endgroup$ Commented Oct 2, 2016 at 9:21
  • $\begingroup$ Those formidable math formulas don't help me much. Is there an intuitive explanation as I don't understand the one in the quoted text? $\endgroup$
    – EJC
    Commented Oct 2, 2016 at 9:45

2 Answers 2


The picture below shows a simple potential energy surface with a transition state at the maximum point on the reaction path, which is at the 'saddle point'.
The reaction path is the solid black line. (Oscillations in this show that the reactant and product are vibrating). The contours are at different energy with large negative being lowest. This picture is based on a calculation using known experimental parameters.

Along the reaction path the energy is as small as it can be, even though it increases as we go along, because by moving to the right or left (perpendicular to the general direction of the path) the energy is even higher. So the saddle point is just like a pass between high mountain peaks. At the highest point on the reaction path the only way to lower the energy is to go down, either to return on the original path or to continue to products. The time it takes to cross the transition state is definitely sub picosecond, < $10^{-12}$ s and possibly up to a 100 times faster.

It is not possible to visualise a multidimensional reaction path, so we have to make do with 2D plots, as below, of one reaction coordinate vs. the other with contours at different energy. In principle many similar plots could be needed for a complex reaction but very often most of the atoms in the molecule are bystanders and it possible then to only look at the separation and energy of the atoms in bonds being broken and made.


Trajectory of $\ce {Cl + H2}$ in a gas phase reaction. The H atom approaches the chlorine molecule at the right. The yellow circle shows the transition state.

  • $\begingroup$ Could you elaborate a bit? What is the curved black line? Are the negative numbers, energies? What does the green dashed line represent? Why is the TS situated at that single position? $\endgroup$
    – EJC
    Commented Oct 2, 2016 at 17:34
  • $\begingroup$ And why are there 3N-6 dimensions? $\endgroup$
    – EJC
    Commented Oct 2, 2016 at 19:00
  • $\begingroup$ The black line is the trajectory, and the oscillations are the effect of vibrations at the molecule moves. The numbers are energies. The Cl approaches the $\ce{H2}$ in the deep valley on the lower right of the picture and moves leftwards. After reaction the HCl is vibrating more than the $\ce{H2 }$ was. Note that the plot is of separation $\ce{Cl + H2}$ on the lower axis and of $\ce{H + HCl}$ on the other. $\endgroup$
    – porphyrin
    Commented Oct 3, 2016 at 6:40
  • $\begingroup$ The $3N-6$ are the number of internal degrees of freedom in a non-linear molecule of N atoms. The 6 is to remove whole molecule translational motion plus whole molecule rotations along/around each of x, y & z axes. $\endgroup$
    – porphyrin
    Commented Oct 3, 2016 at 6:45

First, the question may pose itself why $3N-6$ at all? And for that, we should take a step back and ask ourselves: Why $3N$?

These are the degrees of freedom of a molecule. If you break a molecule down into its atoms, each atom would have three degrees of freedom: $x, y$ and $z$. In a molecule, these degrees of freedom are generally retained; however, you must modify them. It is commonplace to say that three degrees of freedom of the entire molecule represent translation (into $x, y$ and $z$ directions) while another three (of a non-linear molecule) represent rotations ($R_x, R_y$ and $R_z$). That leaves us with $3N-6$ degrees of freedom which are typically called vibrational. The key is that the $6$ nonvibrational ones do not affect the interior of the molecule while the other degrees do move the atoms relatively to each other.

As an example, let’s assume $N = 3$, thus $3N-6 = 3$. This allows us to keep an analogy of a three-dimensional something; you would just need to expand the thought experiment to more dimensions. Furthermore, I’m just going to say that one of the three atoms is a central one and the movement of the other atoms can be expressed as relative location to that central one. (I’m effectively ignoring a degree of freedom here, which I would like to define as vibration of the central atom towards its partners so that nothing productive can happen. It is a strech, I know.)

You can now imagine a 3D-plot where the $x$ axis is one atom’s position in arbitrary units, $y$ is the second one’s and $z$ is energy. Different arbitrary $x$ and $y$ positions, so positions of atom $1$ and $2$ lead to a different energy. Two spots of this $x,y$ grid must be minima, though: they correspond to the reactant and the product. To react, the atoms must somehow take a path across these arbitrary units from one minimum to the other.

This is now where the mountain range analogy comes in. As somebody from close to the alps, I have no problems imagining mountain ranges, valleys, summits and passes. I don’t know if people from areas as flat as pancakes would, though.

Across that path I just mentioned, there must be a transition state. This is the highest altitude you reach on your way from basin 1 (or valley 1; the reactant) to basin 2 (or valley 2; the product). Of course you need to go up and back down but is there anything else you can say about the transition state? Let’s go back to the dimensions, $x$ and $y$. Let’s assume the ‘best’ path is following a straight line abiding to $x+2y =0$. Of course, neither dimension $x$ nor $y$ tell us much about the transition state — it is a maximum along both dimensions —, however, we can mathematically transform $x$ and $y$ until one of the transformed coordinates, $p$, lines up nicely with the reaction direction and the other one, $q$, is orthogonal to it. If we express the path as functions of our new coordinates (= dimensions) $p,q$ then:

  • the transition state is still a maximum in $p$ direction; but

  • it is a minimum in $q$ direction!

Why is it a minimum in $q$? Well, the transition state is a pass. The lowest possible spot to cross a ridge. So if you dissect the mountain range along said ridge, you can find a minimum on that ridge — this corresponds to your pass or your transition state.

What we just found out for $p$ and $q$ holds true for all $3N-6$ dimensions: when defined correctly, there is always one in which the transition state is a maximum; call it the reaction coordinate. And for all the other $3N-7$ dimensions it must be a minimum (if they are defined correctly). If it doesn’t work out of the box, ‘transform dimensions’.

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    $\begingroup$ Well, of course it's a stretch. It sounds like the symmetric stretch to me... $\endgroup$
    – hBy2Py
    Commented Oct 3, 2016 at 2:39

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