Chem 459/559—Medicinal Chemistry Problem Set 3

Chem 459/559—Medicinal Chemistry
Problem Set 3
Fall 2008
Name________________________ 20 points
In the concentration-response plots below
% Response
% Response
125
100
75
50
25
0
-25
125
100
75
50
25
0
-25
-10.0 -7.5 -5.0 -2.5 0.0
Log[Drug]
-10.0 -7.5 -5.0 -2.5 0.0
Log[Drug]
A
B
C
A
A + 10 uM D
A + 100 uM D
Identify compounds A-D as agonist, competetive antagonist, noncompetetive
antagonist, inverse agonist, partial agonist, or inactive compound. (8 points).
A is a full agonist
B is a partial agonist
C is inactive or could be an antagonist – not enough data to say for sure
D is a noncompetetive antagonist

Draw acetylcholine and indicate all possible modes of interaction with its receptors (5
points).
H-bond
acceptor
π-bond
O van-Der
Waals
CH3 can do
charge transfer
H3C O
N
+
H-bond acceptor
CH 3
CH 3
ion-ion or ion-dipole
All indicated dipoles can participate in
dipole-dipole and ion-dipole interactions
(the bonds participate, not the individual atoms)
Given that the carbonyl oxygen is required for interaction with nicotinic receptors and the
ester oxygen is required for interaction with muscarinic receptors, suggest a potentially
selective bioisosteric analog for nicotinic receptors and an analogous bioisostere that
would be muscarinic selective (4 points).
may be cyclic
X CH3 X = S, CH2 O
Muscarinic
N
CH 3
CH 3
may be cyclic
X = O, N
X CH 3
Nicotinic
Although acetylcholine has no chiral centers, draw a structure indicating how it might
interact in a stereoselective fashion with a receptor based on conformations (3 points).
O H CH3 O
H
H 3C O
N
CH 3
CH 3
H 3C O
These are two examples. Any conformation in which the
acetate or ammonium are not in the same plane (eclipsed syn or
anti) are chiral conformations. All gauche conformers are
chiral if conformation is restricted by binding to the receptor.
H 3C
N
CH 3
H
H
N
CH 3
CH 3
CH 3

Modeling/conformational analysis exercise (20 points extra credit)
To perform this exercise, you will need a model kit and will need to use the computers in
room S51J, on which are installed ChemDraw and Chem3D.
Using ChemDraw, sketch nicotine as shown at right (without numbering).
Open Chem3D. Copy the structure from ChemDraw and paste it
into Chem3D. Check to be sure the methyl group is oriented
correctly according to the structure above. Rotate the molecule
using the sliders on the right hand and bottom edges of the screen
until you are confident you have the correct structure (Do not
select any atoms or bonds any bonds yet.).
Using a model kit, build a handheld model of nicotine. Attach hydrogens to all atoms
using the 20 mm bonds (shorter grey).
Move the cursor over an atom and leave it for a moment, a pop-up box will give a
description of the atom and measurements associated with it. Likewise having the cursor
over a bond will show the bond characteristics.
Note the initial dihedral angle between the the aromatic ring and the CN bond of the
pyrrolidine (you can do this by holding down the shift key and highlighting the four
atoms indicated in the structure above – be careful not to select the lone pair on the
nitrogen). From the Object menu select Set Dihedral Angle. A message box will open at
the side showing the dihedral angle for the selected atoms (Do not enter a value – it will
distort the structure). The initial angle should be near 0 o . If it is not, click and highlight
the bond between the two rings (2 and 3 as shown above). Rotate the bond with the
slider on the left until the dihedral angle in the window is nearly 0 o . Rotate the bond on
your handheld model to match.
From the MM2 menu, select Compute Properties and select Steric Energy Summary in the
dialog box and click OK. A message will appear at the bottom of the screen with the
Steric Energy. Write this number down here. 23.709 kcal/mol
From the MM2 menu, select Minimize Energy and set the RMS gradient to 0.05 and click on
Run to execute the minimization. This will relax the molecule into a low energy state for
the pyrrolidine ring and the torsional energy about the bond connecting the two rings.
Once you have done so, readjust the dihedral angle so that the angle is again very near 0 o .
Write this number down in the left hand column of the table on the next page.
(Note: initial energy 7.7-8.1 kcal/mol is at a dihedral angle of -28 o )
Select and rotate the bond connecting the two rings by 30 o increments and calculate the
associated energies using Compute Properties and plot them in the graph on the right.
Each time, move your handheld model to the same position and observe the two in
comparison.
N
1
2
3
N 4
CH 3

φ( o ) dieq nomin diax Cis
0 8.5 23.7 9.2 210
30 16.9 46.3 9.6 24.6
60 25.7 51.9 9.1 9.7
90 19.0 36.0 9.5 9.8
120 8.5 21.9 15.4 25.7
150 8.1 21.2 12.9 72.0
180 9.6 23.6 10.2 179
210 23.2 45.3 11.4 22.7
240 23.5 52.1 9.8 9.4
270 14.4 37.1 10.8 9.4
300 8.4 22.2 23.8 24.9
330 7.8 21.1 16.8 89.8
M1 7.5 7.5 8.2 8.60
M2 7.5 7.5 8.3 8.56
Dieq - Values for trans diequatorial Nomin - Dieq Values w/o minimization
Diax - Values for trans diaxial Cis - Values for cis
Rotate the bond between the rings such that the dihedral angle is around 60 o . and
minimize the structure. In the table, write down the dihedral angle (M1) and steric
energy from the minimized structure.
From the minimized structure, rotate the bond between the rings to -60 o and repeat the
minimization. Record the dihedral angle (M2) and energy in the table.
Based on your experiment, answer the following questions.
Steric energy kcal/mol
Compare your 0 o energy in the table with the value you wrote on the preceeding page. Is
there any difference and if so, what would account for it (5 points)?
The energy before minimization will be higher because the bond angles are
arbitrary and have not been optimized.
60
50
40
30
20
10
0
0
0 30 60 90 120 150 180 210 240 270 300 330 360
Dihedral angle (degrees)
Trans-nomin
Trans-dieq
Trans-diax
Cis
How well do the minimized structures agree with the graph you constructed? Are the
dihedral angles and energies close to what you expected (5 points)?
The minimized structure will have angles of about -55 o or +129 o for the backward
and forward rotations respectively for the correct (trans-diequatorial) structure.
This correlates nicely with the graph above, but the energies are better due to a full
optimization of structure. The angles for the diaxial are (-97 o , +64 o ) and the cis (-
103 o , +66 o ) will be different of course.
From examination of your handheld model and/or the computer generated model, can
you suggest a structural reason why the minimum energy conformations found should be
preferred over the higher energy conformations seen (Be specific, 5 points)?
In those conformations there is less interaction between the two rings due to
coplanarity of the N-methyl and/or the C3 methylene of the pyrrolidine ring.
250
200
150
100
50
Steric energy (Cis only)