Waals or dispersion forces, or as electrostatic attractions? Is any one of these
approaches to be preferred in principle?
No, none is to be preferred “in principle”, meaning on grounds of theoretical
appropriateness. This is because MM is severely practical, in the sense that the
forcefield need only satisfactorily and swiftly reproduce molecular properties,
mainly geometries. The method makes no apologies for ad hoc additions which
improve results. An example of this is seen in the inclusion of a special term to force
the oxygen of cyclobutanone to lie in the ring plane [1]. Identifying the terms in a
forcefield with distinct theoretical concepts like force constants and van der Waals
forces is at best an approximation.
Hydrogen bonding can be dealt with in principle in any way that works. A
weak covalent bond would be simulatedby a small bond stretch constant
(roughly, a force constant), a strong van der Waals force could be modelled by
adjusting the two constants in the Lennard-Jones expression, and electrostatic
attraction by a Coulomb’s law inverse distance expression. These are only sim-
ple examples of how these methods might beimplemented; a brief discussion is
given by Leach [2]. The choice of method to be implemented is determined
by speed and accuracy. Treating stronghydrogenbondsbyMMhasbeendis-
cussed [3].
References
- Leach AR (2001) Molecular modelling, 2nd edn. Prentice Hall, New York; section 4.6
- Leach AR (2001) Molecular modelling, 2nd edn. Prentice Hall, New York; Section 4.13
- Vasil’ev VV, Voityuk AA (1992) J Mol Struct 265(1–2):179
Chapter 3, Harder Questions, Answers
Q6
Replacing small groups by “pseudoatoms” in a forcefield (e.g. CH 3 by an “atom”
about as big) obviously speeds up calculations. What disadvantages might accom-
pany this simplification?
The obvious disadvantage is that one loses the directional nature of the group
and thus loses any possibility of simulating conformational effects, as far as that
group is concerned. Rotation around a C–CH 3 bond alters bond lengths and
energies, albeit relatively slightly, but if we pretend that the CH 3 group is spherical
or ellipsoidal, then clearly it cannot engender a torsional energy/dihedral angle
curve.
The loss of the conformational dimension could be a significant defect for a polar
group like OH, where rotation about a (say) C–OH bond could in reality lead to
Answers 605