Searching for Transition States

Today’s class revealed some of the practical difficulties of searching for transition states using a computer.

We demonstrated some of the difficulties by building eclipsed methanol and conducting a transition state search with (and without) symmetry. Neither search led to eclipsed methanol even though this structure is a transition state for internal rotation.

The subsequent lecture (download notes) showed that our pictorial view (“transition state = top of the energy hill”) is incomplete from the perspective of computerized calculations. For the purposes of computation, we need to treat a transition state as a saddle point where all first derivatives = 0, one second derivative > 0, and all other second derivatives < 0 (first derivative = gradient, second derivative = curvature).

It is hard to interpret these mathematical requirements on the curvatures when we operate in a Cartesian coordinate system so we use normal coordinates to describe the molecule’s internal geometry. From this point of view, a transition state can be described as a stationary point (all gradients = 0) at which one vibration frequency is imaginary (curvature with respect to reaction coordinate > 0) and all other vibration frequencies are positive (curvatures with respect to remaining normal coordinates < 0).

Finally, we performed a quick AM1 search for the Diels-Alder transition state which touched on some practical matters:

• Spartan can calculate normal coordinates, curvatures, and vibration frequencies (Setup: Calculation, check Compute: IR box)
• Spartan lists vibration frequencies sequentially (Display: Spectra) and normal modes can be inspected (adjust Amp and Steps). Imaginary frequencies are marked by the letter “i”
• Spartan’s default transition state geometry search take sthe normal coordinate associated with the first vibration in the list as the reaction coordinate (you can override this default, but I didn’t tell you how)
• Vibration frequency calculations are time-consuming and this makes transition state searches time-consuming because we need to calculate frequencies at least twice, once at the beginning and a second time at the end