Techniques of Molecular Modelling: Molecular Mechanics

Case Study 4: Deciding Stable Ring Conformations.

A modelling problem often comprises a complex series of molecules, their properties and their reactions. The first stage would be to break down the problem into smaller problems, each of which might require a separate technique to study them. Take for example the following reactions. When compound 12 is photolysed in heptane, a solution of "x" is formed, consisting of an equilibrium mixture of 13 and 14. If the solution is heated to 98C, "x" then reacts to form a new compound 15 in which the two vinyl groups are specifically trans. The transformation of 12 to 13, and of 13 to 14 are classical pericyclic reactions where a method appropriate for breaking bonds and studying transitions states should be used. Molecular mechanics is clearly NOT appropriate for these steps.

But what about the transformation of 13 to 15? The original authors make no comment on the origins of the stereoselectivity, but it is clearly tempting to conclude that this reaction must be a transition state controlled reaction. But it might also depend on the conformation of the rather unusual ten membered ring, containing two trans alkene units. Molecular mechanics is in fact an excellent and very rapid probe of the conformation of such a ring. Its worth finding out how much we can learn about the system in this manner.

The first step in modelling this reaction would be to define the connectivity using the molecule editor, and then minimise the structure using a MM force field suitable for alkenes such as MM2. When 13 is so minimised, one should be aware that in several conformations might be possible in a 10-membered ring. Two such conformations are readily located differing in energy by 4.6 kcal/mol. The lowest has a trans relationship at the bond formed in 15,

stretch         =    0.490        angle           =    1.353
stretch bend    =    0.044        dihedral        =    8.366
improp torsion  =    0.254        van der Waals   =    6.036
electrostatics  =    0.133        hydrogen bond   =    0.000

The energy of the final structure is         16.6773 kcal/mol.

The higher has a cis relationship, the difference between the two isomers arising arising purely from the angle term in the MM2 force field.

stretch         =    0.497        angle           =    5.453
stretch bend    =    0.076        dihedral        =    8.415
improp torsion  =    0.292        van der Waals   =    6.406
electrostatics  =    0.146        hydrogen bond   =    0.000

The energy of the final structure is         21.2843 kcal/mol.

Thus the stereospecificity of this reaction is readily explained in terms of angle strain and not Woodward Hoffmann rules (which by the way would allow both cis and trans products to form for 15).

Literature Citation. S. W. Staley and T. J. Henry, J. Am. Chem. Soc., 1970, 92, 7613

Case Study 5: Including Metal Ions

Here, one can see that a variety of bond types have to be defined in the mechanics force field, including values for various Mg...O and Mg...C weak bonds. At best these are going to be guestimates, but this does allow some exploration of the steric environment of the reacting site to be made. The original MM2 force field does not allow such bonds to be calculated, but some more recent modelling programs such as CAChe do allow such estimates. One interesting conclusion emerging from the calcuations is that the nature of the 4th coordination site on the Mg (THF is shown below) does affect the outcome of the calculations, suggesting the reaction may in fact be solvent sensitive. This in turn might suggest further experiments to be carried out.

stretch         =   14.962        angle           =   20.117
stretch bend    =    0.023        dihedral        =   25.369
improp torsion  =    0.307        van der Waals   =   16.707
electrostatics  =   -3.453        hydrogen bond   =    0.000

The energy of the final structure is         74.03 kcal/mol.

stretch         =   14.734        angle           =   19.834
stretch bend    =   -0.058        dihedral        =   27.585
improp torsion  =    0.315        van der Waals   =   16.633
electrostatics  =   -3.313        hydrogen bond   =   -0.529

The energy of the final structure is         75.20 kcal/mol.

Literature Citation. A. G. Shultz, L. Flood and J. P. Springer, J. Org. Chemistry, 1986, 51, 838.

Case Study 6: The Stability of Carbonium Ions

Organic chemists are taught that carbonium ion stability is predominantly due to electronic factors such as whether they are primary, secondary, tertiary, allylic, benzylic etc. What is mentioned much less is that their relative stability is also very sensitive to their geometries, and in particular the angles of the three substituents at the carbon. So what does Molecular Mechanics make of the following carbonium ions? They are arranged in the expected order of increasing stability. Notice that the MM2 method gradually decreases the predicted angle energies. Notice also however how it predicts that the tertiary carbonium ion is actually LESS stable than the preceeding primary ion, which is surprising to say the least. Clearly, electronic factors must be playing an important role as well. Suffice to say that even without ANY consideration of electronic factors, the MM2 method does not do too badly in its prediction of relative stabilities. This re-inforces the conclusion that bond angles in carbonium ions are just as important as substitution!

To make the next jump in realism in our modelling technique, it is clearly time to move on to the Quantum Mechanics methods, where electronic effects are explictly calculated. A preview of the results obtained with such a method is shown here for comparison with the MM results.

stretch         =    0.868        angle           =   19.173
stretch bend    =   -0.672        dihedral        =    6.497
improp torsion  =    0.155        van der Waals   =    2.663
electrostatics  =   -1.341        hydrogen bond   =    0.000

The energy of the final structure is 27.342 kcal/mol. (Final QM heat of formation = 249.11 Kcal/mol)

stretch         =    0.447        angle           =   18.758
stretch bend    =   -0.078        dihedral        =    6.916
improp torsion  =    1.124        van der Waals   =    1.622
electrostatics  =    0.000        hydrogen bond   =    0.000

The energy of the final structure is 28.788 kcal/mol. (Final QM heat of formation = 232.62 Kcal/mol)

stretch         =    0.571        angle           =    5.694
stretch bend    =   -0.027        dihedral        =    7.179
improp torsion  =    0.009        van der Waals   =    5.678
electrostatics  =   -0.545        hydrogen bond   =    0.000

The energy of the final structure is 18.559 kcal/mol. (Final QM heat of formation = 211.67 Kcal/mol)

stretch         =    0.512        angle           =    3.439
stretch bend    =   -0.153        dihedral        =    6.662
improp torsion  =    0.002        van der Waals   =    3.872
electrostatics  =    0.683        hydrogen bond   =    0.000

The energy of the final structure is 15.017 kcal/mol. (Final QM heat of formation = 195.79 Kcal/mol)

Literature Citation. 2nd Year Organic Problem Sheet, Imperial College, 1991. Set 3.

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