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UNIVERSITY OF MANITOBA DEPARTMENT OF CHEMISTRY
CHEM 3390 STRUCTURAL TRANSFORMATIONS IN ORGANIC CHEMISTRY
FINAL EXAMINATION
Dr. Phil Hultin
Saturday December 19, 2015 – 1:30 – 4:30 pm.
Section 1: Concepts and terminology (20 Marks)
Section 2: Stereochemistry and conformation (20 Marks)
Section 3: Reaction products and reagents (28 Marks)
Section 4: Mechanisms (20 Marks)
Section 5: Applications of spectroscopy (12 Marks)
TOTAL (100 Marks):
Put all answers in the space provided for each question. You may use the backs of the sheets if
necessary, but none of the answers requires a long written answer.
Clearly drawn chemical structures will greatly assist the marking process and will reduce the
likelihood that your answer will be misinterpreted.
You may use molecular models during the exam.
Note that a spectroscopic data sheet is not provided.
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1. (20 Marks Total)
CONCEPTS AND TERMINOLOGY.
a. (2 Marks)
Draw R-2-methylcyclohexanone and label the si face of the
carbonyl group.
b. (2 Marks)
Draw a specific molecule having 8 or fewer carbon atoms that
contains a pair of diastereotopic CH3 groups.
c.
(4 Marks)
Give 2 observations that indicate that anchimeric assistance may
be involved in a nucleophilic substitution reaction.
Molecule must contain a potential nucleophilic atom suitably located with
respect to a leaving group that it could promote displacement via a
reasonable cyclic pathway. The displacement occurs faster than would be
observed in the absence of the potential nucleophile.
If the reaction involves a stereogenic centre, the stereochemical outcome of
an SN2 type displacement will be net retention of configuration.
d. (4 Marks)
List 4 characteristics of a hard Lewis acid.
Small atomic or ionic radius.
Low polarizability
High oxidation state
(Relatively) High electronegativity
e. (4 Marks)
Briefly explain the difference between a stereospecific reaction
and a stereoselective reaction.
In a stereospecific reaction, the stereochemistry of the product is uniquely
determined by the stereochemistry of the starting material as a result of the
mechanism of the reaction. Stereoisomeric starting materials lead to specific
stereoisomeric products.
In a stereoselective reaction, two or more stereoisomeric products are
possible from a given starting material, but one is formed in preference to the
other(s).
f.
(4 Marks)
Briefly explain what transannular strain is.
In medium-sized rings, groups on one side of the ring cannot avoid close
contact with groups diametrically opposite them on the ring, while also
minimizing eclipsing (Pitzer) strain with their neighbours. This close contact
increases the energy of the ring conformation and is referred to as
transannular strain.
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2. (20 Marks Total)
STEREOCHEMISTRY AND CONFORMATION
a. (4 Marks)
Draw the structures of all cyclohexanetricarboxylic acid
configurational isomers that can be obtained (in principle) from the high-pressure
hydrogenation of 1,3,5-benzenetricarboxylic acid.
b. (6 Marks)
The acetal below could in principle adopt 3 different
conformations (A, B and C), but 13C NMR data suggest that only ONE is actually
present at equilibrium. Which of the conformers A, B or C is preferred and why?
Conformer A is preferred because of the Anomeric Effect. In all three
conformations both rings are chairs. The so-called spiro connection of the
rings means that with respect to one ring, either the CH 2 or O of the other
ring is axial, while the other is equatorial.
 In A both rings have a C-O bond axial, so there is a gauche
relationship CH2OCR2O for both rings and both benefit from
anomeric stabilization.
 In B only one of the rings has an axial C-O substituent, so it benefits
less from the anomeric effect.
 In C, the C-O bonds are equatorial with respect to both rings, so no
anomeric effect is possible.
NOTE: the distances between the two oxygens are more or less identical in
all three cases. There are no significant steric interactions between C-H
bonds of the two rings in any of the conformations, because the two rings
are offset from one another.
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c.
(4 Marks)
Draw wedge-and-dash structures for all distinct stereoisomers of
2,3,4-trihydroxypentane. Identify any that are related as enantiomers of one
another.
NOTE: the question asked for pentane, not cyclopentane!
d. (6 Marks)
Draw wedge-and-dash structures for all distinct acetal products
that are formed from each of the reactions below. Indicate whether the acetals
are chiral or not, and if more than one stereoisomeric acetal is possible indicate
how it relates to the other structures (i.e. as an enantiomer or as a diastereomer).
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3. (28 Marks Total)
REACTION PRODUCTS AND REAGENTS
(18 Marks) Provide the missing product or reagent(s)/solvent(s)/conditions for each of the
following. A standard aqueous workup can be assumed to follow all reactions unless
specifically noted. Show stereochemistry whenever it is appropriate.
This reaction is an
example on page
348 of the textbook.
a.
b.
(2)
(4)
The oxidative
cleavage of an αhydroxyketone is
shown in Scheme 525 on page 245 of
the textbook.
c.
(2)
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The WoodwardPrevost reaction
installs a cis diol on
the more-hindered
face of the alkene.
See Section 4.5.2.2
in the textbook.
d.
(2)
See Scheme 5-6,
page 212 in the
textbook for
elimination during
DMSO oxidations.
e.
The selective oxidation
of an allylic alcohol in
the presence of another
alcohol requires MnO2.
Other reagents would
oxidize both alcohols.
This question is almost
identical to the example
in Figure 5-6 on page
220 of the text.
f.
(2)
(2)
A Beckmann
rearrangement, the only
reaction we have seen
in CHEM 3390 that
starts from an oxime!
g.
(2)
An iodocyclization,
very similar to a
question from one of
the problem sets this
year. Also similar to
examples from Figure
4-13 on page 157 of
the textbook.
h.
(2)
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(10 Marks)
Complete the missing information in the following “road-map” problem.
4.
The use of
hydrogenation over
Pd(OH)2 catalyst to
reductively open an
epoxide is a bit
unusual. I
expected that the
starting material for
this step would be
derived from the
previous steps.
If you just hydrolyzed the
lactone without also
forming the epoxide from
the bromohydrin that
results, I was willing to
accept this provided that
you carried it forward
consistently. The next two
structures had to reflect
this choice.
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(20 Marks Total)
MECHANISMS
a. (5 Marks)
Aziridines are the nitrogen analogues of epoxides. The aziridinealcohol shown undergoes an intramolecular reaction to form a pyrrolidine when
treated with a strong base in a polar aprotic solvent like DMSO or DMF. Suggest
a mechanism for this reaction that is consistent with the stereochemistry shown.
NB: the solvent is not directly involved in the mechanism.
Recall that we saw nucleophilic attack on an alkyne was possible in one of the
problem sets this year, although it is not the most common process. The formation
of the epoxide ring is similar to other internal nucleophilic processes including the
formation of epoxides from bromohydrins under basic conditions.
Note that the question did not specify a particular base – just a “strong base”. This
means that the base just acts to deprotonate, it is not a nucleophile nor does it
participate in any other way.
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b. (5 Marks)
Halolactonization can be achieved by several reagents that form
the equivalent of an “X+” species. Iwata and coworkers devised an interesting
version of this process, sharing some features of the Swern mechanism.
Suggest a mechanism for the Iwata protocol shown below. Hint: CH3SCH3 is a
byproduct of this reaction.
It should be obvious from our discussion in class that most bromine-containing
molecules do not act as “bromonium ion donors”. The TMS-Br is no exception to this;
the bromine is polarized partially negative and the silicon partially positive as you
would expect from electronegativity.
The way the question is posed strongly implies that the conditions lead to the
formation of a “Br+” by some process that is related to the mechanism of the Swern
oxidation (which of course involves DMSO). The hint means that you should approach
the answer by remembering the Swern pathway and what happens to the DMSO in
this reaction. The formation of dimethylsulfinylbromide is the key part of this reaction
mechanism.
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c.
(5 Marks)
Aromatic rings bearing electron-withdrawing groups (so-called
Class II substrates) that also carry an alkoxy substituent para to the EWG
typically lose the alkoxy group during Birch reduction. Write out a mechanism
showing how this occurs.
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d. (5 Marks)
Rearrangements sometimes occur during the reactions of
electrophiles with strained alkenes. The product B is what we would expect from
the bromination of norbornene, but it is the minor product! Write mechanisms to
explain the formation of A and C.
The key observation that you should make is that specific stereoisomers of the
products are formed. This means that there is no way that SN1-like processes
can be involved.
This reaction illustrates a classic set of experiments. For many years there was
a debate about whether bridged cations of the kind depicted here were actually
real. The main protagonists in the debate were Saul Winstein (who advocated
for the delocalized bridged “non-classical carbonium ion” structure) and Herbert
C. Brown (the hydroboration guy, who argued that the cations were an
equilibrating set of conventional carbocation structures).
The debates were finally settled in favour of Winstein’s hypothesis by the
experimental work of George Olah, who was able to directly observe
carbocations in solution for the first time. Olah received the Nobel Prize in 1994
for this work.
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6. (12 MARKS TOTAL)
APPLICATIONS OF SPECTROSCOPY
a. (8 Marks)
When isobutyraldehyde (2-methylpropanal) is heated in the
presence of a small amount of sodium isobutoxide, a process known as the
Tischenko Reaction occurs.
The product of this reaction has the 1H and 13C NMR spectra shown below.
Note that we have not
discussed the
Tischenko Reaction,
but that you don’t
need to know the
mechanism to solve
the problem since the
spectra are pretty
clear. You can find
out about the
Tischenko Reaction
at
https://en.wikipedia.or
g/wiki/Tishchenko_re
action
Note that the product
has a carbonyl peak
at 177 ppm in the 13C
spectrum so it is
probably an ester, but
definitely NOT a
ketone.
NB: this signal
represents more than
one type of carbon
2H
d
1H
sept
1H
sept
6H 6H
d d
Draw the structure of the product in the box below. Using letters or numbers to
identify the C and H atoms in your structure, place each identifier next to the
appropriate chemical shift in the table.
13C
Structure
1H
NMR
NMR
177.17
C1
3.85
HA
70.36
C3
2.56
HC
34.15
C2
1.92
HB
27.83
C4
1.18
HE or HD
19.06
Both CH3 grps
0.94
HD or HE
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b. (4 Marks)
The Wieland-Miescher Ketone was treated with H2 (g) at 3 atm.
pressure, over a Pd/CaCO3 catalyst in pyridine solution, forming an 85:15 mixture
of products in 86% total isolated yield. The stereoisomeric products were
separated by chromatography.
Assuming that the 1H NMR signals of both products can be identified, suggest a
specific feature of the spectra that would be diagnostic of the stereochemistry of
each isomer.
In the trans decalin product (the minor product) the new C-H bond will be
axial with respect to both rings, since the ring junction substituents in a trans
decalin are always axial. In the major product, a cis decalin, the new C-H
bond will be axial with respect to one ring and equatorial with respect to the
other.
The Karplus relationship relates vicinal dihedral angles to vicinal coupling
constants. An antiperiplanar dihedral will be associated with a large 3J H-H
in the vicinity of 10-12 Hz, wheras a gauche dihedral is associated with a 3J
of around 4 Hz.
Only when a C-H bond is axial will it be part of an antiperiplanar alignment.
Thus, in the major product the new C-H will be a dddd with 1 large 3J and 3
very similar smaller 3J values. In the minor product this proton will give a
dddd with two large 3J and 2 smaller 3J values.
NOTE: When this exam was graded it became clear that many students did
not understand what this question was asking for. I decided to drop it from
the exam. I kept any marks that individual students may have earned on
this question, but ratioed the scores out of 96 points instead of 100,
effectively making this problem a “bonus”.