ORGANIC CHEMISTRY
CUMULATIVE EXAM ANSWERS
October 29, 1999
[Doering, W. v. E.; Benkhoff, J.; Shao, L. J. Am. Chem. Soc. 1999, 121, 962.]
GENERAL INSTRUCTIONS: Feel free to use partial structures in writing mechanisms and structures.
I. (10 points) General
A. The senior author on this paper is William Doering. During a career that has spanned 55 years, he has made
significant contributions to a wide variety of research areas. For full credit, describe two distinctly different areas
in which he has made his mark; for partial credit, describe just one area. Write enough about each area so that
the grader(s) will know that you are not guessing at the answer. For example, no credit would be given if you
identified E. J. Corey simply as "a synthesis chemist" or Christopher Ingold simply as "a mechanisms chemist."
B. In 1967, R. B. Woodward wrote "Doering ... is a very playful fellow." I mention this because a recurring Latin
phrase, ad libitum, appears twice in the Abstract; once in the first paragraph of the text; twice on p. 963; and no
fewer than four times on p. 964. The dictionary translation is "in accordance with desire." In your own words,
explain the meaning of this phrase as it is applied in the article.
II. (35 points) Scheme 1
A. 1. Compound 1 is synthesized from á-tetralone (the structure without the methyl group), according to the
procedure given in the experimental section on p. 966. If Fe(CO)5 is omitted, the product is the result of an
aldol condensation followed by dehydration; show the structure of this aldol product and suggest what the
function of Fe(CO)5 is.
2. Speculate on why a much simpler procedure (á-tetralone + strong base + CH3I) might not have been
successful in this synthesis.
B. Write detailed mechanisms for 1 ÿ 2 and for 2 ÿ 3 under the conditions given in the experimental section.
C. Doering writes, p. 962, that the "Synthesis, although uncomplicated, contains a delightful interplay of kinetic and
thermodynamic control." Give at least one other example (from your course work, from your reading of the
literature, or from your research) that illustrates the idea of kinetic vs. thermodynamic control. Provide enough
information to show that you understand the meaning of this phrase.
D. The reaction of 3 with H2, Pd/C is said to give "a mixture of four stereoisomers"; what are they? [You may
describe them using structural drawings or words.]
E. Oxidation of the mp 112EC stereoisomer gives ketone trans-5 which, after treatment with PhMgBr and then
TsOH, gives a mixture of trans-6 and trans-7. Prolonged treatment with acid gave the equilibrium mixture of
96% trans-7 and 4% trans-6. Write a mechanism for this equilibration of 6 and 7; speculate on why one isomer
is so much more stable than the other.
III. (35 points) Scheme 2
A. Provide a mechanistic rationale for how trans-7 with strong base can give the anion called 6/8-.
B. In Chart 1, an equilibrium mixture of the two stereoisomers of both 6 and 8 is achieved by treating any one of the
four with KOC(CH3)3 in HOC(CH3)3. The ÄÄH and ÄÄS values are experimental; the SE and ÄSE are from
molecular mechanics calculations.
1. How was the composition of the mixture determined?
2. Why were so many temperatures (Table 1) used?
3. Why is it true that "the analytical error associated with trans-6 is inevitably larger" (p. 963, right-hand column,
3rd paragraph)?
4. Chart 1 shows some angles superimposed on the structures; what do these angles refer to? Be specific - for
example, answer this question for the two angles shown for cis-8.
C. The long paragraph beginning with "The steric energy of the global minimum ..." (p. 964, left-hand column) is
very confusing. Answer the following questions about it.
1. What is meant by "global minimum calculated by MM2"?
2. What is meant by" the steric energy" of this global minimum?
3. In your own words, explain why Doering concludes with the statement "These conclusions can scarcely both
be true."
2
IV. (20 points) Scheme 3
A. Clearly explain why the ÄH‡ is smaller for the E/Z transformation in 10 than for the one in 11. Be as clear and as
quantitative as possible.
B. Why is it true (p. 964, left-hand column, first paragraph) that McMurry reaction on racemic 3 gives four
stereoisomers of 9 vs. only two stereoisomers with enantiomerically pure 3.
C. Table 2 lists the rate constants and equilibrium constants at various temperatures for the interconversion of antiand syn-9. State clearly, without actually doing the calculations, how the values shown for Ea, log A, ÄH‡, and
ÄS‡ are obtained from the experimental data. (One doesn't have to reproduce all of the required equations; a
somewhat general answer will suffice.)
I.
A. There are numerous correct answers here. Some of the areas that Doering worked in (and often defined
completely) are: carbene additions and their stereochemistry; non-benzenoid aromatic compounds (tropylium,
heptafulvene, fulvalene, etc.); Cope rearrangement (chair vs. boat TS); fluxional molecules (e.g., bullvalene);
synthesis of quinine (with Woodward, 1944); vinylcyclopropane rearrangement (recall Baldwin's seminar on
October 7); the Doering-Zeiss mechanism for SN reactions; etc.
B. As used in the opening paragraph on p. 962 (twice) and in the final paragraph on p. 964, the phenyl group in
compounds 6 and 7 can freely rotate and can "assume whatever dihedral (torsional) angle optimizes interaction."
In contrast, the fused aromatic ring in 8 cannot vary its dihedral angle with the conjugated double bond. The
phrase ad libitum ("in accordance with desire") refers to the first type.
II.
CH2
A. 1. The aldol condensation product from á-tetralone and formaldehyde in KOH/EtOH is
the á,â-unsaturated ketone shown to the right. This suggests that Fe(CO)5, under
these conditions, serves as a reducing agent for the C=C bond.
O
2. Two problems might have been encountered; mentioning either one will receive full credit.
a.
Perhaps it was difficult to stop at the stage of
monoalkylation (as is often the case). Thus, 1 might
have been contaminated by significant (or even major)
quantities of the double C-alkylation product or the
product of C-alkylation followed by O-alkylation.
CH3
CH3
O
CH3
or
OCH3
H
b.
Perhaps the reaction did stop at monoalkylation of the enolate ion, but the
trapping occurred entirely at O.
OCH 3
B. The IUPAC names for DBN and MVK (Scheme 1) are given in the experimental section (p. 966). The first
reaction 1 ÿ 2 is merely a Michael addition.
N
N
+N
N
H
H
.. CH3
CH
O
H2C
C
CH3
O
CH3
O
H-A
2
CH3
CH
O
CH2 .. C
O
CH3
The conversion 2 ÿ 3 is an intramolecular aldol condensation with dehydration of the initial â-hydroxyketone
(which occurs stepwise via an E1cB mechanism).
3
CH3
O
CH3
CH2
O
H
O
..
CH2
O: -
:OH
-
O
-
:O:
H
H
..
-
O
H
HO:
-
O:
O
O
O:
H
H
:OH
O:
H
H
:OH
-
H
C. When a reaction can produce more than one product and when one product is formed faster but the other is
more stable, the proper choice of reaction conditions can favor formation of either the kinetic or thermodynamic
product. Shorter reaction times and lower temperatures favor production of the kinetic product; longer times and
higher temperatures favor accumulation of the thermodynamic product. Three examples, all of which are found
in most undergraduate texts, are the most commonly encountered cases of this phenomenon:
(a) addition of H-X to 1,3-butadiene, in which the 1,2-addition product is kinetic, the 1,4-product thermodynamic.
(b) Diels-Alder reaction (e.g., of cyclopentadiene with maleic anhydride), in which the endo product is kinetic, the
exo product thermodynamic.
(c) sulfonation of naphthalene, in which substitution at the á-position (C1) is kinetic, substitution at the â-position
(C2) thermodynamic.
D. They are the epimeric alcohols with a trans ring fusion (trans-4) and the epimeric alcohols with a cis ring fusion
(cis-4).
E. The equilibration is
accomplished (experimental
section, p. 967) by refluxing the
mixture in toluene using ptoluenesulfonic acid. The
equilibration occurs by addition
of a proton followed by
elimination of a proton from the
other side of the carbocation.
:OTs
-
H
H
H
+ H-OTs
+ Ph
Ph
6
H
H
Ph
H H
H
H
7
OTs
Me
One explanation for the greater stability of trans-7 comes from a similar
case that is discussed in Chem. 550. In the trans-fused 7, both bonds to
the left-hand ring are equatorial (designated e in the drawing) from the
Re
right-hand ring; in cis-fused 6, one of these bonds is equatorial but the
other is quasi-equatorial (designated eN). The result is that both rings are
H
unstrained in 7, but not in 6: either of the two rings (or both) has some
strain in it.
Me
e
7
e
R R e'R
H
6
NOTE: Ph omitted
for clarity reasons
III.
A.
CH3
H
H
CH3
H
..
H
H
Ph
H
:O-t-Bu
-
H
H
CH3
Ph
H
OTs
.. H
Ph
H
CH3
H
H
6/8
H
Ph
H
:O-t-Bu
-
B. 1. As stated in the text (p. 963, next-to-last paragraph), the sharp singlets of the bridgehead methyls were
clearly defined in the 1H-NMR spectrum (see, also, Chart 1).
4
2. The equilibrium constants were determined at many different temperatures so as to get ÄH and ÄS with the
greatest possible precision and accuracy. As stated on p. 963 (next-to-last paragraph), "a plot of the
experimental data ... as the natural logarithm of the ratios of the isomers (in %) against the reciprocal of the
temperature (K-1) provides slopes and intercepts, "ÄH" and "ÄS" ...." In other words, the standard equations
ÄG = -RT ln K and ÄG = ÄH - TÄS tell us that ln K = -ÄH/RT + ÄS/R. Thus, a plot of ln K vs. 1/T will give
slope of -ÄH/R and intercept ÄS/R.
3. Because trans- 6 is the least stable of the four isomers, its concentration is the smallest; thus the largest
experimental error is associated with its measurement. [Note from Table 1 that cis-8 predominates, that
trans-8 and cis-6 are present to the extent of about 50% of cis-8, but that trans-6 is present to a very small
extent (about1-2% of the major isomer).]
4. These are dihedral angles used as a measure of the degree of coplanarity of the aromatic ring and the C=C
bond. In cis-8, for example, there is a 21E angle between a C-C of the fused ring and the C=C bond; had the
angle been 0E, the system would have been exactly coplanar. The phenyl group makes a 49E angle to C3C4, which is of no concern here, but is of concern when it fails to achieve coplanarity in conjugated systems
like cis-6.
C. 1. MM2 is Molecular Mechanics Version 2, developed by Norman Allinger. It's equivalent to MMX, which is
available on the computers in Bu 654. (Allinger has a newer version, MM3. based on improved parameters
but it is not yet fully optimized). The phrase global minimum refers to the fact that with any molecular
modeling program it is possible to reach energy-minimized structures that may or may not be the most stable
of all structures possible. Often, the calculations lead to a local minimum, a structure that is more stable
than others "near it" (in bond distance, bond angle, dihedral angle, etc.). Further work must be done before
the "molecular modeler" is content to say that this is also the most stable of all possible structures (i.e., the
global minimum). As the authors write (p. 964, first paragraph), "painstaking care involving systematic
variation of initial conformations has been taken to avoid entrapment in local minima."
2. In molecular mechanics calculations, the steric energy (often confused with "strain energy") is the sum of
the individual energies that were minimized by the program. In MMX, the steric energy is the number called
MMXE which gives, in kcal/mol, the sum of the stretching, bending, van der Waals, etc. energies.
3. As the comparison among the isomers in Chart 1 shows, the conclusion seems to be that ð -electron
delocalization favors the ad libitum trans-6 by 1.44 kcal/mol but disfavors cis-6 by 0.46 kcal/mol (after steric
factors had been removed). .
IV.
A. The "reactions" shown in Scheme 3 are all rotations about the central C=C bond, normally a very high-energy
process. This sort of process can occur here because the diradical (after 90E rotation of the bond) is stabilized
by allylic resonance with the conjugated C=C bonds. Thus, 11 has a barrier of 40 kcal/mol, some 20-25 kcal/mol
below what would be expected for a simple double bond. Not only does one have doubly allylic resonance but
the carbons that bear the odd electron are bonded to alkyl groups that give additional stabilization. The barrier
for 10 (35 kcal/mol) is even lower because a phenyl (rather than an alkyl) stabilizes each allylic radical. As the
authors argue (p. 964, bottom right), this 5 kcal/mol difference "is an indication that an ad libitum phenyl group in
the 1-position of a cyclohexene ring is 2.5 kcal mol-1 more effective than an alkyl substituent in stabilizing an allyl
radical."
B. Compound 3 (in Scheme 1) is chiral and can have either R or S configuration at its one stereocenter. (Shown in
Scheme 1 is, in fact, the S enantiomer. McMurry coupling of two such enantiomerically pure ketones would give
9 with S configuration at both stereocenters (as shown in Scheme 3). Had the reaction been performed on the
racemic (50/50 R/S) mixture, the product would have been a mixture of R,R, R,S, S,R, and S,S materials; and,
of course, each of these could be either syn or anti at the central C=C. Because R,S and S,R are identical
(meso) for this system, one might have expected to get a mixture of six stereoisomers: anti,S,S; syn,S,S ;
anti,R,S; syn,R,S; anti,R,R; and syn,R,R. Why only "four stereoisomers can be expected" is not clear. But it is
clear that a mixture of stereoisomeric products will be formed. [When reference 14 is consulted, a privilege not
available to people taking this exam, one learns that what Doering means by "four stereoisomers" is the syn
racemic pair (i.e., R,R and S,S), the syn meso (R,S), the anti racemic pair (R,R and S,S) and the anti meso
(R,S).]
C. According to the Arrhenius equation, k = Ae-Ea/RT. Thus ln k = ln A - Ea/RT. A plot of ln k vs. 1/T will give a
straight line with slope -Ea/R and intercept ln A; from these, Ea and A can be derived.
As for the enthalpy and entropy of activation, there are equations that relate them to the Arrhenius terms; for full
credit, it's not necessary to give as much detail as int the following paragraph, but some mention must be made
that Ea and ÄH‡ are mathematically related, as are A and ÄS‡.
‡
‡
The absolute rate theory equation gives k = (kT/h)e-ÄH /RTeÄS /R (in which the k on the left is the rate constant
whereas that on the right is the Boltzmann constant). One "can show" that the relationship between this‡ equation
and the Arrhenius is such that Ea = ÄH‡ - RT and, so, ÄH‡ can be obtained from Ea; also A = kt/h e eÄS /R and
so ÄS‡ can be obtained from A.