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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.

d. (4 Marks) List 4 characteristics of a hard Lewis acid.

e. (4 Marks) Briefly explain the difference between a stereospecific reaction and a stereoselective reaction.

f. (4 Marks) Briefly explain what transannular strain is.

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.

Small atomic or ionic radius. Low polarizability High oxidation state (Relatively) High electronegativity

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).

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 CH2 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.

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).

NOTE: the question asked for pentane, not cyclopentane!

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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.

a. (2)

b. (4)

c. (2)

This reaction is an example on page 348 of the textbook.

The oxidative cleavage of an α-hydroxyketone is shown in Scheme 5-25 on page 245 of the textbook.

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d. (2)

e. (2)

f. (2)

g. (2)

h. (2)

See Scheme 5-6, page 212 in the textbook for elimination during DMSO oxidations.

The Woodward-Prevost reaction installs a cis diol on the more-hindered face of the alkene. See Section 4.5.2.2 in the textbook.

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.

A Beckmann rearrangement, the only reaction we have seen in CHEM 3390 that starts from an oxime!

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.

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(10 Marks) Complete the missing information in the following “road-map” problem.

4.

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.

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.

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(20 Marks Total) MECHANISMS

a. (5 Marks) Aziridines are the nitrogen analogues of epoxides. The aziridine-alcohol 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.

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.

Structure 13C NMR 1H 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

2H d

1H sept

1H sept

6H d

6H d

NB: this signal represents more than one type of carbon

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.org/wiki/Tishchenko_reaction 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.

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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”.