Organic Chemistry II (CHE325)
REVIEW PROBLEMS
Key
1. Draw a complete orbital picture for the molecule shown below. Is this molecule chiral? Explain.
H
H
C C C
CH3
CH3
p orbital
sp2 orbital
H
H
H
H
H
s orbital
C
C
C
C
H
C
sp3 orbital
H
H
H
mo lec ul e is c hir al - it ha s a non-s upe ri mp o s able mi r ro r im a ge
2. For each of the following compounds, draw the important resonance structures.
a.
(CH3)2N CH
CH CH NH2
(CH3)2N CH
CH CH NH2
(CH3)2N CH
CH CH NH2
(CH3)2N CH
CH CH NH2
b.
O
CH2 CH N
O
O
CH2 CH N
O
CH2 CH N
O
O
c.
O
O
O
O
3. Indicate the direction of polarity for each of the bonds indicated:
O
C
O
CH3
OCH3
CH3 MgBr
4. Give IUPAC names for each of the following compounds:
CH3
CH3
CH2 CHCH3
C C C CH2CH3
CH3
Cl
4-et hyl -4,6- di met hy l-2- h ept yn e
I
(E)-2-c hl o ro- 4-i o do -2- bute ne
H
(S)-2 -m et hyl- 3-i s op ro py lcyc lo hex en e
Br
1-br om o- 3- met hyl pe nt an e
NO2
o- nit ro p he nyl be nz ene
O
OH
Cl
CH2CH3
4-chlo r o-3 -et hyl benz oic ac id
5. Identify the functional groups in the following molecules.
amine
alkene
alkene
alcohol
NH2
OH
arene
N
halide
(chloride)
Cl
O
H
H
O
ether
O
O
CH3
alkyne ester
amide
6. Using Newman projections, show all possible conformations of 1-bromobutane. Identify the lowest
and highest energy conformations.
H
CH2Br
H
H
H
CH3
CH3
CH2Br
H
H
H
H CH2Br
H
H
H
CH3
CH3CH2Br
H
H
H
H
H
lowest
energy
highest
energy
7. Draw the two possible chair conformations of the following molecules. In each case, identify the
lowest energy conformation, and explain your reasoning.
co mp o und
po s si bl e c o nfo rm ati o ns
CH3
CH3
CH3
CH3
CH3
CH3
CH3 CH3
CH3
A
B
A is the low energy conformation - two equatorial Me groups in A whereas in
conformation B there are two axial methyl groups
CH3
CH3
OH
CH3
CH3
OH
OH
CH3
CH3
CH3
CH3
A
CH3
B
A is the low energy conformation - since the OH group is sterically smaller
than a methyl group, the steric interaction between axial OH & CH3 groups in
A is less severe than that in B (between two axial CH3 groups)
8. define the following terms, and give an example of each:
a. nucleophile: can donate electron density to an electron deficient species
examples:
:NH3, PhS-, C=C, etc.
b. electrophile: can accept electron density from an electron rich species
examples:
Br2, R3C-X, H+ , etc.
9. Show all possible stereoisomers of the following compound. Note that there are 3 stereogenic
centers. Identify the relationship of each of these isomers with one another, e.g. enantiomers,
diastereomers, etc.
H NH2
H
H
enantiomers:
H NH2
H
H NH2
H
H NH2
H
H NH2
H
H
H
H
H
A
C
E
G
H NH2
H
H NH2
H
H NH2
H
H NH2
H
H
H
H
H
B
D
F
H
A/B , C /D, E/F, G/H
all other pairs are diastereomers
10. Assign the absolute configuration of each stereogenic center present in the molecule shown below.
CH3 OH
S
S
R CO2H
OCH3
11. Write a complete mechanism to explain the following observation:
CH3
CH3
C CH CH2
CH3
HCl
CH3
Cl CH3
C CH CH3
CH3
+
CH3
major product
minor product
a
CH3
CH3 C CH CH2
CH3
HCl
CH3 Cl
C
CH CH3
CH3
Cl
CH3
H
CH3 C CH CH2
CH3
2° cation
methyl
shift
(path a)
b
CH3 H
CH3 C CH CH2
CH3
3° cation
(more stable)
Cl
trap by
chloride
(path b)
CH3
CH3 Cl
C CH CH3
CH3
CH3
Cl CH3
C CH CH3
CH3
major product
minor product
12. Predict all products from the following reactions. Label major and minor products.
Cl Br
CH CH2
KOH
ethanol
(1 equiv.)
A + B
NaNH2
NH3
NaNH2
C
1. BH3
2. H2O2
F
D
CH3CH2Br
E
H2
Lindlar Pd
G
Cl
C
major
CH2
CH
A + B
Br
CH
C
CH
C
C
minor
C CH2CH3
H
H
C
CH2 CH
E
Na
D
O
C
C
C
CH2CH3
F
G
13. For each of the following transformations (A - D), the major product is shown. In each case, explain
why the product formed is the major one. It may help to consider what minor products may be formed.
CH3
CH3
CH3
H 2O
H3O+
Br2
H2SO4
heat
CCl4
CH3
A
Br
OH
CH3
CH3
B
CH3
C
CH3
KOH
ethanol
Br
CH3
D
CH3
A: Protonation of the double bond in A leads to a tertiary carbocation which is then trapped by
water to give the more highly substituted alcohol B (Markovnikov addition)
B: Dehydration of the tertiary alcohol leads to formation of the more highly substituted double
bond via an E1 elimination (Zaitsev's rule).
C: Anti addition of bromine across the carbon-carbon double bond (e.g. formation of a bromonium
ion intermediate, followed by backside displacement to give the anti product.
D: E2 elimination requires anti orientation of proton and leaving group (trans-diaxial orientation of
Br and H). In this case, anti elimination of Bra proceeds so that the more hichly substituted
double bond in formed. In the elimination of Brb, anti elimination leads to formation of the least
highly subtituted double bond.
Br
Hb
CH3
Hc
Br
H
anti elimination if not possible Hc is equatorial
CH3
anti elimination of Br and Ha leads to more
highly substituted double bond
Ha
anti elimination of Br and Hb leads to less
highly substituted double bond
14. An unknown compound had the following MS:
a) identify the base peak and the parent ion in this spectrum
base peak:
m/z = 43
parent ion:
m/z = 72
b) what is the formula weight of this compound?
FW = 72
15. Match a structure from the list below to the following IR spectra. Clearly identify and label
diagnostic absorbances in each IR.
O
O
OH
OH
A
a. Compound:
sp3
B
D
C-H
C=O
b. Compound:
B
OH
sp2
C-H
C CH
C
D