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Patrick: An Introduction to Medicinal Chemistry 5e MOLECULAR MODELLING EXERCISES Chapter 20 Exercise 20.1 Nevirapine Introduction Neviripine (Fig. 1) is a non-nucleoside reverse transcriptase inhibitor that is used in the treatment of HIV as part of a multi-drug therapy (see also An Introduction to Medicinal Chemistry, section 20.7.3.2).

Figure 1 Nevirapine The Tasks Nevirapine is a chiral molecule although it is not apparent from the structure shown. In this exercise you will carry out the following tasks 1. Build the structure of nevirapine 2. Identify the possible conformations using molecular mechanics 3. Describe the shape of the structure. 4. Explain why it is chiral. 5. Create the opposite enantiomer 6. Create diagrams that show clearly why the two enantiomers are not superimposable. A detailed set of procedures is given below, but you may wish to tackle the exercise yourself before following them.

© Oxford University Press, 2013. All rights reserved.

Patrick: An Introduction to Medicinal Chemistry 5e PROCEDURES The following procedures are written for the Apple Mac version of Spartan 10 (version 1.0.1). Mouse operations will vary for Windows users. Modifications of the procedures may also be required for other versions of Spartan. You are advised to familiarize yourself with basic Spartan procedures described in the file ‘Spartan Introduction’ before attempting any of the exercises. 1. Open a new Spartan file and build the structure of nevirapine (Fig. 1). Energy minimize the structure and describe the resulting structure (Fig. 2).

Figure 2 3D-Model of nevirapine from different perspectives. The energy minimized structure shows that the tricyclic ring system is folded in a ‘butterfly’ shape with the cyclopropane ring pointing in the opposite direction. 2. Choose Set Torsions from the Geometry Menu. This shows that the program identifies one bond (indiciated by a yellow cylinder) that could be rotated to generate different conformations, as well as an invertible nitrogen (indicated by a yellow circle). (Fig. 3).

Figure 3.. Rotatable bond and invertible nitrogen identified by Spartan. Choose Set up from the Calculations menu and Calculate Conformer Distribution at Ground state with Molecular Mechanics MMFF. Click Submit. Save the file as nevirapine and wait a second or two for a message stating that the calculation is complete. Click OK.

© Oxford University Press, 2013. All rights reserved.

Patrick: An Introduction to Medicinal Chemistry 5e Three conformations are generated in a separate file. Study the three conformations using the slider bar at the bottom left of the window. The three conformations are due to rotation of the bond linking the cyclopropyl group to the tricyclic ring system. The tricyclic ring system remains constant for all three conformations. 3. Reveal the window showing the first of the three conformations. Copy the structure and paste it into a new file. Choose New Molecule from the File menu and paste the structure into the new window. Double click on the structure with both the shift and command keys depressed. This instruction will generate the opposite enantiomer of the compound. The two enantiomers can be viewed as in figure 4.

Figure 4. Both enantiomers of nivaripine viewed to emphasise the butterfly shape. By studying these structures, it should become evident that the two structures are indeed enantiomers. They are non-superimposable mirror images of each other. If you rotated the right hand structure such that the tricyclic ring system was orientated the same way, then the methyl substituent and the amide group are in different positions compared to the left hand structure (Fig. 5).

Figure 5 Comparison of both enantiomers having rotated the right hand structure shown in figure 4. Nevirapine is an example of a structure that is chiral, but has no asymmetric (chiral) centres. Chirality is a result of the overall asymmetry of the molecule, resulting from a) the butterfly shape, and b) the presence of the methyl substituent and the amide group. If the tricyclic ring system had been planar, it would not have been chiral. If the methyl substituent and the amide group had not been present, the molecule would not have been chiral.

© Oxford University Press, 2013. All rights reserved.