CHM 222: Organic Chemistry III
Review 2 (Chapt 18-19)
1. Attempted oxidation of 1,4-butanediol to butanedioic acid results in significant yields of the -butyrolactone.
Explain this mechanistically.
H
OH
HO
O
O
[O]
OH
HO
OH
O
O H
O
OH
O + H2O
O
2. SN2 reactions of simple carboxylate ions with haloalkanes in aqueous solution do not give good yields of
esters. Explain why this is so.
In aqueous solution, the charged carboxylate ion will hydrogen bond to the water molecules
which decreases the nucleophilicity of the carboxylate ion. If the reaction proceeds, then the
product formed would have less extensive hydrogen bonding with the water solvent.
3. Reaction of 1-iodobutane with sodium acetate gives an excellent yield of ester if acetic acid is used as the
solvent. Why is acetic acid a better solvent for this process than water?
Acetic acid is better able to solvate both polar and non-polar molecules. The product ester is
predominantly non-polar and its formation is disfavored in water since water cannot easily solvate
non-polar molecules. The acetic acid can solvate the non-polar ester product with its methyl
group.
4. Benzene rings of many compounds in nature are prepared biosynthetically by a pathway similar to that
operating in fatty acid synthesis. Acetyl units are coupled, but the ketone functions are not reduced. The
result is a polyketide thioester, which forms rings by intramolecular aldol condensation. Draw out a
mechanism for the formation of o-orsellinic acid from the thioester shown below. Hydrolysis of the thioester to
give the free carboxylic acid is the last step.
HO
O
O
O
COOH
CH3
O
S protein
OH
o-Orsellinic acid
Polyketide thioester
enol
formation
enol
formation
O
O
O
OH
S protein
HO
O
CH3 O
H
O
CH3 O
S-protein
H
-H2O
O
S-protein
O
H
5. Propose reasonable mechanisms for the following two reactions.
O
O
Br2, CH2Cl2
O
ONa
Br
Br
Br
Br
CH3
O
O
Br
ONa
Br
O
O
ONa
H
O
CH3
Br
Br
2
(Note stereochemistry)
Br
O
H+, H2O
OH
OH
O
OH
OH
H+, H2O
O
(Cis and trans isomers)
OH
HO
O
HO
OH
H
HO
OH
O
OH
H
OH
OH
OH
O
O
OH
-H
O
OH
O
OH
O
OH
O
O
OH
O
H
O
O
CH3
H3C
O
O
O
O
CH3
H3C
O
O
H
O
O
CH3
H3C
O
OH
OH
H
6. One way to determine the number of acidic hydrogens in a molecule is to treat the compound with NaOD in
D2O, isolate the product, and determine its new molecular weight by mass spectrometry. For example, if
cyclohexanone is treated with NaOD in D2O, the product has a new MW = 102. Explain how this method
works.
The hydrogens  to the carbonyl group of a ketone are slightly acidic. In a solution of NaOD and
D2O, these acidic hydrogens will be exchanged with deuterium which has a mass of 2 instead of
hydrogen’s mass of 1. Once all of the acidic hydrogens have been exchanged with the deuterium
isotope of hydrogen, the new molecular weight will be one mass unit higher for each exchanged
hydrogen. Cyclohexanone has a MW = 98 and has 4 acidic  hydrogens. When these
hydrogens are exchanged with deuterium, the new MW = 102.
7. Treatment of 1-phenyl-2-propenone with a strong base such as LDA does not yield an anion, even though it
contains a hydrogen on the carbon atom next to the carbonyl group. Explain.
O
LDA
No Rxn
H
The molecule is an  unsaturated ketone and the conjugated -system will want to remain coplanar so delocalization of the -electrons can occur. In this conformation, the C-H bond is
perpendicular to the carbonyl -bond and deprotonation will not occur because there is no way to
push the electrons through the system to make an enolate anion. The C-H bond broken during
this deprotonation must overlap with the adjacent -bond to form the electron sink that makes
these  hydrogens slightly acidic.
8. Predict the product(s) of the following reactions:
CO2H
CO2H

+ CO2 + H
CO2H
O
O
1. NaOEt
CH3
O
O
2. CH3I
O
O
Br2, PBr3
O
H2O
Br
OH
Br
O
OH
A
Br
B
O
NaOH, H2O
OH
+ CHI3
I2
9. Base promoted chlorination and bromination of esters at the -position occur at the same rate. Explain why.
The reaction involves the attack of an enolate anion from the ester on an activated halogen
(either from the molecular halogen or a reagent such as N-bromo or N-chlorosuccinimide). The
rate limiting step is the formation of the enolate anion and not the subsequent attack on the
halogen. Therefore, the rate of the two reactions will be the same since the rate limiting step for
each reaction is the formation of the enolate anion.
10. When an optically active carboxylic acid such as (R)-2-phenylpropanoic acid is brominated under HellVolhard-Zelinskii conditions, will the product be optically active or racemic? Explain.
The product will be racemic since it is the chiral carbon that gets deprotonated when the enolate
for the acid bromide is formed. The chiral carbon gets deprotonated to make the achiral enolate
which can then attack bromine from either face of the enolate -system. This will result in loss of
optical activity and the product will be racemic. (See problem 8 for an example of the HellVolhard-Zelenski reaction.)
11. Fill in the reagents (a-c) that are missing from the following reaction scheme.
O
O
O
O
CO2CH3
CH3
CO2CH3
CH3
B = H2O, H+, 
A = CH3I
H3C
CH3
C = 1.1 eq. LDA
CH3I
12. An interesting consequence of the base catalyzed isomerization of unsaturated ketones is that 2-substituted
2-cyclopentenones can be interconverted with 5-substituted 2-cyclopentenones. Propose a mechanism for
this isomerization.
O
O
NaOH
CH3
HO
CH3
H
O
O
HO
H
H OH
O
CH3
CH3
H OH
13. The Favorskii reaction involves treatment of an -bromocycloketone with base to yield a ring-contracted
product. For example, reaction of 2-bromocyclohexanone with aqueous NaOH yields cyclopentanecarboxylic
acid. Propose a mechanism.
O
HO
H
Br
COOH
1. NaOH
2. H3O+
HO
H
O
HO
Br
O
HO
O
O