I am trying to calculate the energy barrier between the two possible enantiomeric components of a polymer with Gaussian. I wonder if I am able to locate a transition state between two of those structures with the keyword opt=QST2?

  • 1
    $\begingroup$ What kind of inversion would this be? $\endgroup$ – DSVA Jul 10 '17 at 23:04
  • $\begingroup$ The polymer has two crossed arms. With a shape similar to a cross. Depending on which arm is in front or behind, there will be a R-enantiomer and another one S-enantiomer. $\endgroup$ – S.G.V Jul 11 '17 at 15:39

Maybe, but more likely no. Depending on the complexity of your molecule the guess from QST2 is not good enough. Additionally you might run into some unforeseen coordinate system troubles.

As a small demonstration I'll use $\ce{CHFClBr}$ and the DF-M06L/def2-SVP level of theory and Gaussian 09 Rev. D.01.

I'll optimise one conformation and then just exchange the $y$ with the $z$ column.

#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)

title: optimisation

0 1
C        0.000000000      0.000000000      0.000000000
H       -0.530433945      0.045629756     -0.953288460
F       -0.905627118      0.105968719      1.035862034
Cl       1.161565217      1.332282245      0.093221601
Br       0.942449903     -1.690284863      0.135370837

That gives the following file for the transition state search. It is important, that the ordering in the two input blocks is identical, otherwise the program will fail horribly.

#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)
opt(QST2, maxcycle=50)

title: optimised conformation

0 1
 C     0.004249     0.005882     0.022885
 H    -0.510890     0.037500    -0.945427
 F    -0.880186     0.102687     1.016533
Cl     1.139127     1.361087     0.088671
Br     0.915653    -1.713561     0.128503

title: yz exchanged

0 1
 C     0.004249     0.022885     0.005882
 H    -0.510890    -0.945427     0.037500
 F    -0.880186     1.016533     0.102687
Cl     1.139127     0.088671     1.361087
Br     0.915653     0.128503    -1.713561

In this case the calculation will produce the following error:

 Berny optimization.
 Bend failed for angle     2 -     1 -     5
 Tors failed for dihedral     2 -     1 -     5 -     3
 Tors failed for dihedral     2 -     1 -     5 -     4
 FormBX had a problem.
 Error termination via Lnk1e in /path/to/g09/l103.exe at {a date}.

The problem is that in any case of switching stereo centres, you most likely have have an $\approx180^\circ$ angle at one point and then the automatic generation of redundant internal coordinates will fail.
You can (maybe) prevent this at a much higher computational cost by optimising in Cartesian coordinates. Alternatively you specify the redundant coordinates by yourself, but even then it is probably best to use QST3.
In some rare cases, such calculations can actually give you a starting structure for a real TS search.

In any case, a little chemical intuition is necessary and often a single step mechanism is much, much higher in energy than a step-wise path. Sometimes it is better to use your imagination and guess a transition state (which you can then optimise with very low level tools like force fields or pm6), and refine later on.

The example I chose is certainly an extreme one. In this case isomerisation will likely happen via $$\ce{^{$S$}CHFClBr <--> [CHFCl]+ + Br- <--> ^{$R$}CHFBrCl}$$ or similar, and a transition state won't be easy to find.

  • $\begingroup$ Thank you Martin - マーチン♦ . The transition state that I am trying to optimize is the one that involves the conversion between these enantiomers: R and S-2,2′-bis(2,2′-bithiophene-5-yl)-3,3′-bithianaphthene. $\endgroup$ – S.G.V Jul 15 '17 at 18:04
  • $\begingroup$ google.es/… $\endgroup$ – S.G.V Jul 15 '17 at 18:11
  • 1
    $\begingroup$ You may have luck with these complexes, but I'd rather recommend trying a series of relaxed/rigid scans and at least qst3 if not a straight TS search. PM6 does wonders for such things $\endgroup$ – Martin - マーチン Jul 15 '17 at 18:48
  • $\begingroup$ Thank you Martin - マーチン♦. I will check if my error is solved with PM6, because I can not optimize the TS by direct assumption due to the enormous steric impediment $\endgroup$ – S.G.V Jul 15 '17 at 19:36

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