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Medicinal Chemistry - Exam 1 Solved Questions | CHEM 4170, Exams of Chemistry

Material Type: Exam; Professor: Gates; Class: Medicinal Chemistry; Subject: Chemistry; University: University of Missouri - Columbia; Term: Unknown 2007;

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Name______________________________________CHEM 4170 Exam 1 all answers must fit in sp ace provid ed
1
Matching (22 pts). Place the Letter of the Best
Answer in the Blank on the Left
1. Pharmacokinetics __N___
2. Pharmacodynamics __J___
3. Pharmacophore __L___
4. Autocoid ___C___
5. X-ray crystallography __F___
6. Covalent Bond ___H__
7. van der Waals interaction __D___
8. Hydrogen Bond __R___
9. Ionic Bond __G___
10. QSAR __P___
11. KB __A___
A. = 1/KD
B. Any compound used to cure, mitigate, prevent, or diagnose
a disease or physical malady
C. Endogenous ligand for a receptor
D. Strength varies as 1/r6 (r = distance)
E. = [D][R]/[D•R]
F. Can provide a three-dimensional picture of a
macromolecular drug target
G. Strength varies as 1/r2 (r = distance)
H. Stable bond at room temp. Worth 85 kcal/mol
I. Stable bond at room temp. Worth 8.5 kcal/mol
J. Interaction of drug with its biological target
K. Quantitative system to assay receptors
L. Core set of functional groups required for biological
activity within a structural group of molecules
M. Often thermoneutral a vaccum
N. Absorbtion, distribution, metabolism, excretion, and
toxicity
O. Favored by electron withdrawing groups
P. Quantitative structure-activity relationship
Q. Qualitative stereochemistry assessment ratio
R. Strongest when bond angle is 180˚
12. (5 pts). Why were almost all early, successful medicines derived from plants?
Answer: See class notes regarding the early history of medicinal chemistry
13. (6 pts). List one advantage and two disadvantages of an in vivo bioassay for lead discovery.
i. Answer: See class notes regarding bioassays.
ii.
iii.
14. (5 pts). Addition of an ionic interaction worth –6.8 kcal/mol will alter the binding constant of a
drug by a factor of about (circle the best answer and briefly show how you calculated your answer):
Answer: 100,000. Must show equation that you used to calculate. See handout on Keq/G
0.01 0.1 3.4 10 100 1000 10,000 100,000 1,000,000
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Matching (22 pts). Place the Letter of the Best

Answer in the Blank on the Left

1. Pharmacokinetics N_

2. Pharmacodynamics J_

3. Pharmacophore L_

4. Autocoid C

5. X-ray crystallography F_

6. Covalent Bond _H

7. van der Waals interaction D_

8. Hydrogen Bond R_

9. Ionic Bond G_

10. QSAR P_

11. KB A_

A. = 1/K (^) D B. Any compound used to cure, mitigate, prevent, or diagnose a disease or physical malady C. Endogenous ligand for a receptor D. Strength varies as 1/r 6 (r = distance) E. = [D][R]/[D•R] F. Can provide a three-dimensional picture of a macromolecular drug target G. Strength varies as 1/r 2 (r = distance) H. Stable bond at room temp. Worth 85 kcal/mol I. Stable bond at room temp. Worth 8.5 kcal/mol J. Interaction of drug with its biological target K. Quantitative system to assay receptors L. Core set of functional groups required for biological activity within a structural group of molecules M. Often thermoneutral a vaccum N. Absorbtion, distribution, metabolism, excretion, and toxicity O. Favored by electron withdrawing groups P. Quantitative structure-activity relationship Q. Qualitative stereochemistry assessment ratio R. Strongest when bond angle is 180˚

12. (5 pts). Why were almost all early, successful medicines derived from plants?

Answer: See class notes regarding the early history of medicinal chemistry

13. (6 pts). List one advantage and two disadvantages of an in vivo bioassay for lead discovery.

i. Answer: See class notes regarding bioassays.

ii.

iii.

14. (5 pts). Addition of an ionic interaction worth –6.8 kcal/mol will alter the binding constant of a

drug by a factor of about (circle the best answer and briefly show how you calculated your answer):

Answer: 100,000. Must show equation that you used to calculate. See handout on Keq/G

15. (5 pts). How are molecules like the one shown below used in diversity-oriented synthesis

(combinatorial chemistry)? A complete answer will show a chemical reaction along with a short

written explanation.

O O

Cl Cl Cl CHN 2 O

Answer: See class notes on combinatorial chemistry.

16. (10 pts). A combinatorial library of tripeptides containing Gly, Phe, Leu, Glu, Trp and Gln was

constructed using Merrifield's resin and Still's tagging method for encoding.

( a ) How many peptides are possible in this library? (Show the equation you used to calculate this

number, please.)

N = b x^ (b = # of building blocks, x = # of cycles)

The active bead was isolated, tags removed by photolysis, and identified by electron capture gas

chromatography to give the following result: Tags 1, 2, 5, and 8 were detected.

( b ) Use your knowledge of Still's encoding method to determine the structure of the active peptide on

the polymer bead (please indicate the carboxy and amino ends on your answer).

Assume the following “code”:

100 = Gly 010 = Phe 001 = Leu 110 = Glu Trp = 101 111 = Gln

Answer: See assigned problems in textbook. Also see classnotes and posted exam keys

for previous years.

17. (5 pts). Calculate Log P for the molecule shown below.

N H

HN

O

O

CN

18. (5 pts). Why is a Log P > 5 detrimental to the pharmacokinetic properties of a drug candidate?

Answer: see class notes regarding Log P and Lipinski’s Rule of Fives

22. (5 pts). Regarding lead optimization: Provide an argument based upon thermodynamic

consideration of drug-target interactions that explains why a chainring modification of a lead

compound might yield a more active compound. Your answer must fit in the space below (compose

your answer before you begin writing).

Rigid compounds, if designed properly, are “preorganized” for binding to the target site on

the biological macromolecule. Thus, they do not pay an entropic penalty upon binding to the

target. This is reflected in a “better” delta G of binding and a “better” equilibrium binding

constant K.

23. (5 pts). Draw a stable analog of the compound shown below that contains one bioisosteric

substitution. Please propose only one analog.

HN N

NH (^2)

Many possible. See handouts on isosteric substitutions. See class

notes on the development of cimetidine.

24. (8 pts). The biological activities of three drug candidates are shown below.

NC NH

I NH

H 3 C NH

-0.4 0.2 0.

Activity = 1.1 Activity = 2.1 Activity = 3.

-0.2 0.0 0.

Axes… Log activity (vertical) and sigma (horizontal).

a) Sketch a Hammett plot on the graph above. Label the axes and show the data points. The points

do not need to be highly accurate. Is the Hammett rho value () positive or negative?

Negative

b) Use your Hammett plot to suggest an analog that is likely to have higher activity than any of those

shown above. Draw the complete structure of your analog below and briefly explain your choice.

NH2, OH, OCH3, CO2-, t-Bu, Et Any of these should be better.

Log P Values From Leo, A.; Hansch, C.; Elkins, D. Chem Rev. 1971 , 71 , 525. Table courtesy of Prof. Richard B. Silverman Compound log Poct Compound log Poct Compound log Poct CH3OH -0.66 CH2=CHCOOH 0. O CH (^3)

CH3NH2 -0.57 CH 3 CH2CN 0.16 CH2=CH-OCH2CH3 1.

CCl3COOH 1. H 3 C C CH (^3)

O -0.24 CH3CH2CH2COOH 0.

BrCH2COOH 0.41 CH2=CHCH2OH 0.17 CH3CH2CH2CH2OH 0. ClCH2COOH 0.47 CH3CH2CHO 0.38 CH3CH2OCH2CH3 0. FCH2COOH -0.12 CH3CO2Me 0.18 CH3CH2OCH2CH2OH -0. ICH2COOH 0.87 CH3CH2COOH 0.33 CH3CH2NHCH2CH3 0. CH3CN -0.34 CH3OCH2COOH -0.

N H

CH3CHO 0.43 CH3CH2CH2Br 2.10 CH3CH2CH2CH2CH2F 2. CH3COOH -0.17 CH3CH2CH2NO2 0.65 PhCH2OH 1. HOCH2COOH -1.11 CH3OCH2OCH3 0.00 PhCH2NH 1. CH3CH2Br 1.74 CH3OCH2CH2OH -0.

NH

O

O

CH3CH2Cl 1.54 Me3N 0.27 PhCH2COOH 1. CH3CH2I 2.00 CH3I 1.69 PhOCH2COOH 1. CH3CONH2 -1.46 CH3NO2 -0.33 2.

CH3CH2NO2 0.

HN O NH

O -1.07 OH

C CH

CH3CH2OH -0.32 HOOCCH=CHCOOH 0.28 O 0.

Me2NH -0.

NH

O

O

-1.21 O O 1.

CH3CH2NH2 -0.19 CH2=CH-O-CH=CH2 1.81 O

O

HOCH2CH2NH2 -1.31 CH3CH=CHCOOH 0.72 N^ 2.

HC CCO 2 H 0.46 HOOCCH2CH2COOH -0.59 3.

CH2=CHCN -0.92 CH2=CHCH2OCH3 0.

N H

-CH 2 - (x ) 0.50 -CH 3 (x ) 0. Branching -0.20 -H 2 C=CH-CH=CH 2 - (x ) "two thirds of benzene"

H 2 C=CH- (x ) "one third of benzene"