Question One (7 marks)
i) Provide a mechanism for the following reaction.
Four marks for the mechanism.
ii) Explain the regiochemical outcome.
Protonation at the tertiary carbon yields a carbocation that is more effectively stabilized by the
lone pair on oxygen. The structure of the additional resonance structure was worth a mark.
While this does not follow Markovnikov’s Rule, it still follows the principle that the reaction
takes place via the most stable carbocation, which is the basis for the rule.
iii) What is the stereochemical outcome?
An achiral reactant give a product with two stereogenic carbons. Since the Lewis acid base
reaction between the carbocation and nucleophile can occur to either face, diastereomers are
produced, each racemic.
Class average on this question was 2.7. Many lost marks for not giving the additional
resonance structure. Arrow pushing was generally good.
Question Two (5 marks)
We did not look at the following reaction, nor is it in your textbook. Based on your
understanding of the addition reactions we have examined, speculate as to what the product of
the following reaction might be and offer a mechanism.
Br
Br
Br
Br
H3C
H
H3C
H
H3C
H
O
H
H
B:
H
Br
Br
H3C
H3C
O
H
H
H
H
B:
O
H
H
O
H
Br
H3C
H
H
O
H
B
B
This is a case of coming up with a mechanism to determine what product is formed, not a style
of question that I use often. Most people were able to get to the bromo enol and got three
marks. The reaction to this point is analogous to that of alkenes. Many people got the
structure of the final product, too, even if they didn’t get the whole mechanism. The
conversion of the enol to the ketone is based on the outcome of adding water to alkynes.
Class average was 3.6.
question.
I would consider this good performance on a fairly tough
Question Three (3 marks)
Rationalize the reactivity of the following three dienes. Their relative rate constants are
shown.
H3C
Cl
CH3
0.002
200
10-4
Cl is an electron withdrawing group that deactivates the diene, presumably by lowering the
HOMO energy. The methyl group in the second example does the opposite, hence the
reactivity. In the third case, the methyl group sterically destabilizes the cisoid conformation
necessary for reaction to occur.
Many people argued that the internal methyl interacts sterically with the incoming
dienophile. While true (for all terminal substituents, actually) such an effect is not as
significant as the impact on the cisoid conformation. Were it so, then cyclopentadiene would
not be such a reactive dienophile.
Class average was 2.0.
Question Four (6 marks)
In the addition of HBr to isoprene, there are, in principle, four possible products.
H
Br
Br
Br
Br
Br
Only two are actually formed. Circle those two, and provide a mechanism for their formation
that explains why these products are formed and not the others.
The first two are the products formed. The complication here is that protonation of the
diene can give two different allylic carbocations:
H
1A
Br
H
Br
Br
2B
2A
1B
Br
Br
Br
Br
Br
Br
Br
I have labeled the resonance structures. We know that even for dienes, this reaction
should follow Markovnikov’s Rule and that the most stable carbocation should lead to
the products.
This is how many people reasoned this out. The most stable of 1A, 1B, 2A and 2B
is 1A because it is 3° allylic making the corresponding bromide the major product. The
second product must be derived from the second best resonance structure 2A because it
is 2° while the others are 1° and therefore should not lead to product. This argument is
fundamentally incorrect, BECAUSE THERE ARE ONLY TWO CARBOCATIONS,
AND ONLY THE MOST STABLE ONE LEADS TO PRODUCTS.
The most stable carbocation is 1A/1B PERIOD. This cation leads to the two
products on the left by Markovnikov’s Rule because this cation is formed more quickly
than 2A/2B. To suggest that the 1° products cannot be formed is to fundamentally
misunderstand what resonance structures represent: 1A and 1B represent THE SAME
SPECIES AND THIS IS THE SPECIES THAT IS FORMED MORE QUICKLY.
There was a mark for identifying the correct products, four for the mechanism and 1 for
explaining that this reaction followed M’s rule.
Class average was 72%.
Question Five (2 marks)
The following reaction gives 100% trans product when the reactant is the meso-dibromide.
Br H
H
NaI, Acetone
CH3
CH3
H3C
HC
H Br
3
H
Is this reaction stereospecific? Explain how you know or what you would do to find out.
By definition, a stereospecific reaction is one in which different stereoisomeric reactants give
different stereoisomeric products. In this case, a meso-dibromide gives a trans-alkene product.
To determine if this is stereospecific, you would need to take the opposite diastereomeric
reactant (the RR, SS compound) and do the same reaction. If it gives only the cis-alkene
product, then the reaction is stereospecific.
Br H
H
CH3
H
H3C
H3C
H
Br
CH3
Many students argued that since the reverse process – addition of bromine to an alkeneis stereospecific, so must this reaction be. This makes no sense, however, since this reaction
involves iodide and so must have a different mechanism.
Question Six (10 marks)
i) Add arrows to the following two cycloaddition reactions to show electron movement.
ii) Of the two, which is/are symmetry allowed? Explain briefly.
Be brief explanation, I was hoping people would make reference to the number of pi electrons
involved in each process (6 and 8 respectively) and that since the first fits the Huckel rule, it
will be allowed. Most people did this. Some people gave full MO symmetry arguments for
both reactions which was, of course, perfectly acceptable.
One trap here is the triple bond. One of the pi bonds is not involved in this reaction and
so do not count in the pi elelctron count. Very few people got caught in this. I was pleased.
iii) For the second reaction, sketch either pair of Frontier MOs and indicate their symmetries.
We have a four atom system with 4 pi electrons and a three atom system with 4 pi electrons
The two possible FMO interactions are:
S
LUMO
oxide
A
LUMO
diene
HOMO
diene
HOMO
oxide
A
S
iv) This class of cycloaddition reaction either involves a HOMO diene/LUMO oxide or a
LUMO diene/HOMO oxide Frontier MO interaction. How would you determine which
interaction is involved in this reaction…
If you had a computer?
Calculate the HOMO and LUMO energies of each species and determine which
HOMO/LUMO combination has the smallest energy difference.
If you didn’t have a computer, but had an organic chemistry research lab?
Look at the impact of electron donating groups or electron withdrawing groups on the
rate of reaction.
Question Seven (6 marks)
Propose a synthesis of the following molecules. You may use any organic reactant of five
carbons or less and any inorganic reagents needed. You may need to use some reactions from
Chem 2500.
O
O
O
Br
O
O
O
(Br2)
Br
O
O
O
Br
CN
CN
(LDA, E2
CN
CN
(Br2)
Br
Class average was 69%. Three marks for the first one which most people got.
If you got the first reaction of the second, you got a mark. If you incorporated a leaving
group into your diene, you got additional credit.
Question Eight (5 marks)
i) For the following Diels-Alder reaction, draw the structure of the endo product.
H
CN
CN
Cyclopentadiene
ii) For the next Diels-Alder reaction, draw the structure of the exo product.
H
Furan
O
CN
CN
O
iii) Cyclopentadiene is an excellent Diels-Alder diene because it is locked in a cisoid
conformation. Furan shares this feature and it has an electron-donating O atom yet furan
reacts more slowly. Why?
Furan is an aromatic molecule and must lose its resonance energy when it reacts.
resonance energy is not nearly that of benzene, but it’s still a key factor.
It’s
iv) Reactions i and ii above are done for the same length of time. Cyclopentadiene gives
predominantly the endo product while furan give predominantly exo product. Is the reaction
with furan violating the endo rule? What may account for the differing outcomes?
In Diels-Alder reactions, the endo product is the kinetic product and the exo is the
thermodynamic product. Equilibrium is established by virtue of the fact that the DA reaction
is reversible. If, in given amount of time, equilibrium is achieved faster for the reaction with
furan, then it can form the exo product without violating the endo rule.
Aha, you say. That doesn’t make sense because the DA reaction with furan is slower
than with cyclopentadiene. But that’s not the point. It is the reverse reaction rate that will in
many cases dictate the rate at which equilibrium is achieved.
Class average was 33%.
The last part was a challenging question. I dropped these two marks from the total.
Question Nine (11 marks)
For each of the following reactions, fill in the missing reactants, products or reagents. Beware
of issues of stereochemistry and be sure to include appropriate stereochemistry when necessary
to distinguish between diastereomeric products. For reactions that give more than one product,
indicate which product is expected to be the major product.
1)
CHO
CHO
H3CO
H3CO
Major
H3CO
CHO
Minor
You got a mark if you at least got a cycloadduct out of this. Two marks if you had the correct
regiochemistry (ortho) and/or diastereoselectivity (endo). For three marks you needed to have
two products: the major product shown plus either the minor regioisomer or the minor
diastereomer.
2)
H
Ph
H
Ph
O
The conditions here are BH3 or a R2BH reagent followed by H2O2. One mark.
3)
Cl2, CH3OH H3CO
Cl
Two marks for correct (trans) stereochemistry and the correct regioselectivity.
4)
CN
CN
H
One mark for each correct reagent. I’m such a softy.
5)
Ph
H
Ph
Br2
H
Br
H
Br
H
Ph
Ph
Only the meso diastereomer shown is produced. Two marks.
6)
I
2 HI
H
I
CH3
H3CH2C
One mark.