Download Principles of Drug Action I, Spring 2004

Principles of Drug Action I, Spring 2004
Sample Exam Questions
1. What is the likely bonding arrangement (neutral state) between chlorine (atomic number 17) an
atom with an atomic number of 12? An atom with an atomic number of 12 will have 2
electrons in its valence shell (3 shell). Such an atom will most likely "give up" those two
electrons in ionic bonding to achieve and octet (in it's 2 shell)
A.
B.
C.
D.
E.
1 covalent bond
2 covalent bonds
6 covalent bonds
1 ionic bond
2 ionic bonds #
2. What is the formal charge on an atom with an atomic number of 5 when it forms four covalent
bonds with hydrogen? This atom has 3 electrons in its valance shell. Thus in the neutral form
it will form 3 bonds. If it forms an additional bond it will be sharing an additional electron
and have a -1 charge. A better way to solve such problems is using the formal charge
equation: FC = Valence - [1/2BEs + NBEs] which for this atom is: 3- [1/2(4) + 0] = -1
A.
B.
C.
D.
E.
0
+1
+2
–1 #
–2
3. The amine compound below is not optically active because: it is sp3 hybridized, tetrahedral and
contains a chiral atom (N)
A.
B.
C.
D.
there is no sp3 hybridized atom in this compound
there is no chiral atom in this compound
there is rapid inversion about the nitrogen atom #
the compound is not a tetrahydral
N
H
CH3
1
4. What is the stereochemical relationship between the two compounds shown below: Same
configurations about the double bond (cis) and chiral carbon (S). Only difference is
rotational position about a C-C single bond
HO
CH3
CH3
C
CH3
H
A.
B.
C.
D.
E.
H3C
CH2CH3
and
H HO
C
CH3
C
CH2CH3
C
H3C
CH3
CH3
Positional isomers
Conformational isomers #
Geometric isomers
Optical isomers
Tautomers
5. Which order (A-E) correctly ranks the relative water solubility (from highest to lowest) of
compounds I-V below? A. Generally diols > alcohols > ethers > hydrocarbons for water
solubility of comparable structures. Also, secondary alcohols (branched) are more water
soluble than their corresponding primary alcohols!
A.
B.
C.
D.
E.
V > II > III > IV > I
V > III > II > IV > I
V > IV > II > III > I
II > III > V > IV > I
IV > V > III > II > I
I
III
II OH
IV
OCH3
OH
V
OH
OH
2
6. Which compound shown below (A-E) would be the major product formed from the electrophilic
addition reaction shown below? B. The reaction involves addition across an alkene bond and
will proceed by addition of H+ to give the most stable intermediate carbocation (tertiary) to
which Br- will add.
CH2
HBr
Product A, B, C, D or E?
CH3
CH2
Br
CH3
CH3
Br
Br
CH3
CH2
CH3
B
A
C
CH2
CH2Br
Br
CH3
D
E
7. How many different (regioisomeric and stereoisomeric) electrophilic addition products could form
in the following reaction? Two regioisomers are possible by addition of OH- and Cl+ across
the double bond. During addition a chiral center is created at each addition site (for each
regioisomer). Thus two chiral centers are generated for each regioisomer, meaning there can
be four optical isomers for each regioisomer, or a total of 8 products. Note that the chiral
center present in the starting material is not involved in the reaction and does not change (it
is fixed!).
A.
B.
C.
D.
E.
2
4
6
8#
12
H
H
Cl
CH2CH3
HOCl
Number of Products?
H
3
8. The nucleophilic substitution reaction shown below would proceed most rapidly with which
electrophile (variable X group)? I is a better leaving group than Cl (larger, more polarizable
halogen) and halogens in general are better leaving groups than amines and alcohols.
Alkenes (and other hydrocarbons) do not function as leaving groups typically (too unstable)
A.
B.
C.
D.
E.
Cl
I #
NHCH3
OH
CH=CH2
X
H2N
+
Nucleophilic
Substitution
Product
9. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V
below? C. Phenols are more acidic than ethers, so II is least acidic. Phenols with electron
withdrawing groups in conjugation with the site of ionization (phenol O) enhance acidity to a
greater extent by stabilizing the negative charge of the conjugate base.
OH
OCH3
I
II
OH
O
CF3
OH
H3C
III
A.
B.
C.
D.
E.
OH
CH3
IV
OCH3
V
III > IV > I > V > II
V > III > I > IV > II
III > I > IV > V > II
I > III > IV > V > II
III > IV > V > I > II
10. How many electrophilic mono-substitution products are possible in the following reaction (include
even those less likely to form)? Did this in the review session. There are four possible
positions for substitution and each one yields a different regioisomeric product!
A.
B.
C.
D.
E.
1
2
3
4 #
5
OH
CH3Cl
AlCl3
Total Products Possible?
Cl
4
11. Which compound below (A-E) would be the primary elimination product formed in the following
reaction? B. The reaction by definition involves loss of halogen (Br-) by removal of a proton
on the carbon atom adjacent to that carbon bound to Br. In this case there are two such
carbons, but the "interior" one will lose a proton faster because it gives the more stable
alkene (more substituted), and generates and alkene where the C=C is in conjugation with
the aromatic ring!!!!
CH3
HO
-
Primary Elimination Product?
Br
CH3
CH2
CH3
OH
C
B
A
CH3
HO
CH3
Br
OH
D
E
12. What is the stereochemical relationship between the two compounds shown below? Note the
asymmetric carbons at the site of restricted rotation (C=C) and that the priority groups are
oriented in different directions!
A.
B.
C.
D.
E.
Positional isomers
Conformational isomers
Geometric isomers #
Optical isomers
Tautomers
H
H
Br
and
H
H
CH3
H
H
CH3
H
CH3
H
CH3
H
H
Br
5
13. Based on the mechanism shown below, which X substituent listed below (A-E) would this enhance
this reaction to the greatest extent? This reaction proceeds by formation of a carbocation, so
any functional group that stabilizes + charge, especially by resonance donation (+R) will
enhance this reaction to the greatest extent. Methoxy is the only +r group shown. Methyl is
+I (so it helps, but weakly), H is "neutral" and F and NO2 are withdrawers
CH3
CH3CH2SH
Br
X
A.
B.
C.
D.
E.
CH3
CH3
X
X
Br-
SCH2CH3
X=H
X = OCH3 #
X = NO2
X=F
X = CH3
14. How many optical isomers are possible for the compound shown below? C. This compound has
two chiral centers, but is symmetrical. Thus while RR, SS, RS and SR isomers are possible,
the SR and RS isomers are the same compound!!!! (meso)
A.
B.
C.
D.
E.
0
2
3
6
8
CH3
CH3
6
15. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V
below? Relative acidity trends: Acids > phenols > alcohols > alkynes > alkenes/alkanes.
Furthermore, phenols with electron withdrawing groups (V) are more acidic than those with
donating groups (IV) since they stabilize negative charge better!
Br
C CH
Br
CH2OH
II
I
CH3O
OH
III
Br
OH
IV
A.
B.
C.
D.
E.
Br
COOH
V
II > V > IV > III > I #
V > IV > II > III > I
II > IV > V > I > III
II > III > V > IV > I
III > II > V > IV > I
16. Which compound below (A-E) will undergo electrophilic aromatic substitution most readily? C.
That aromatic ring with the best electron donating group will be most reactive toward
electrophilic substitution. In this case only one compound has an electron donor (C), and it is
a weak donor by +I. If an electron donor by resonance (such as OH) was included in this
probloem, it would be the correct choice!
O
NO2
C N
CH3
C CH3
N(CH3)3
A
B
C
D
E
7
17. Which description below (A-E) accurately describes the regiochemical or stereochemical
relationship between compounds I and II below? Note the asymmetric carbons at the site of
restricted rotation (C=C) and that the priority groups are oriented in different directions!
A.
B.
C.
D.
E.
Tautomers
Regioisomers
Enantiomers
Geometric isomers #
Conformers
H3C
H H
CH3
H
H
H CH3
H
H3C
I
II
18. Which order (A-E) correctly ranks the relative ease of oxidation with CrO3 (from most to least) of
compounds I-V below? Should be III > I > II (no correct choice present…sorry)
A.
B.
C.
D.
E.
I > II > III
II > I > III
III > II > I
II > III > I
I > III > II
OH
I
H3C
CH2OH
OH
III
II
19. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V
below? A. Sulfonic acids are more acidic than COOH which > thiphenols > phenols >
alcohols. See class discussion notes
A.
B.
C.
D.
E.
V > IV > III > I > II
IV > V > I > III > II
V > IV > I > III > II
V > IV > II > III > I
V > IV > III > II > I
OH
CH2OH
SH
COOH
SO 3H
NO 2
NO 2
II
NO 2
III
NO 2
IV
NO 2
V
I
8