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UNIVERSITY OF MANITOBA DEPARTMENT OF CHEMISTRY

CHEM 3390 STRUCTURAL TRANSFORMATIONS IN ORGANIC CHEMISTRY FINAL EXAMINATION

Dr. Phil Hultin Friday December 10, 2010 – 9:00 am.

Section 1: Concepts and terminology (20 Marks)

Section 2: Stereochemistry and conformation (20 Marks)

Section 3: Reaction products and reagents (25 Marks)

Section 4: Mechanisms and experimental explanations (25 Marks)

Section 5: Applications of spectroscopy (10 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.

Because there was an error in

the third mechanism problem

when the exam was written,

this 10-mark question was

dropped and the exam was

scored out of 90 points instead.


1. (20 Marks Total) CONCEPTS AND TERMINOLOGY.

a. (4 Marks) Define the difference between a reaction that is stereospecific and one that is 100% stereoselective.

b. (2 Marks) Draw a specific example of a molecule that is chiral but which does not contain any stereogenic centres. Do not use generic “R-groups” or “X-groups”.

c. (2 Marks) Draw an example of a molecule that contains a diastereotopic pair of atoms, and indicate which of these atoms is pro-R.

d. (2 Marks) Draw a specific example of a meso compound. Do not use generic “R-groups” or “X-groups”.

e. (2 Marks) Define the difference between a hydrogenation reaction and a hydrogenolysis reaction.

A stereospecific reaction is mechanistically constrained such that stereoisomeric

starting materials lead exclusively to stereoisomeric products. A reaction that is

100% stereoselective is one that can form stereoisomeric products but which

produces only one of the possible products.

A hydrogenation is an addition reaction in which two hydrogen

atoms are added to the substrate molecule but only pi bonds are

cleaved.

A hydrogenolysis is a reaction in which a sigma bond is cleaved in

the process of adding two hydrogen atoms.


f. (3 Marks) List three characteristics typical of a hard Lewis acid.

g. (2 Marks) Draw a specific example of a compound containing a threo 1,2-diamine.

h. (3 Marks) Briefly explain why rings of between 7 and 12 members are more strained than cyclohexane, whereas rings larger than 12 members are essentially free of strain.

1) Small (or localized positive charge)

2) Low polarizability

3) High oxidation level (or high positive charge)

In a ring of between 7-12 members, there is no possible conformation

in which all dihedrals can be staggered without bringing groups on

opposite sides of the ring into close contact (transannular

interactions). In larger rings, there is sufficient distance and

flexibility that relaxed and uncrowded geometries with staggered

dihedrals similar to those in acyclic structures are possible.


2. (20 Marks Total) STEREOCHEMISTRY AND CONFORMATION

a. (2 Marks) The compound shown is highly symmetrical, yet the 1H NMR

signals from the CH2 groups of the esters are surprisingly complicated. Briefly explain why.

DaSilva, J.A.; Barría, C.S.; Jullian, C.; Navarrete, P.; Vergara, L.N.; Squella, J.A. J. Braz. Chem. Soc., 2005, 16, 112-115.

b. (8 Marks) Compounds 1 and 2 shown at right favour different conformations in solution. One prefers a chair, while the other primarily adopts a twist-boat geometry. Draw perspective views showing the preferred conformations of 1 and 2, and briefly explain why one of them adopts the twist-boat geometry. You do not have to show the conformation of the C6H11 sidechains.

1 2

Rychnovsky, S.D.; Yang, G.; Powers, J.P. J. Org. Chem., 1993, 58, 5251–5255.

The cyano group has a small A-value, because of its essentially cylindrical structure. In compound 1 it thus

does not suffer from severe 1,3-synaxial interactions with the methyls at C2. Notice that putting the cyano

axial also may benefit from the anomeric effect, which is not available to the methyl in compound 2.

The methyl in compound 2 spreads out substantially more, and suffers from large synaxial conflicts. Recall

that because C-O bonds are shorter than C-C bonds, these conflicts are greater here than in cyclohexane.


c. (10 Marks Total) The enantiomeric purity of chiral secondary alcohols can be determined through 31P NMR integrations as shown below. In this method, the chiral alcohol is reacted with PCl3 and pyridine in CDCl3. A phosphonate diester forms because of traces of water present in the solvent.

The resulting CDCl3 solution of phosphonate diesters is then analyzed by 31P NMR. The 31P NMR of the phosphonate diester derived from racemic 2-butanol contains three signals in a ratio of 2:1:1 as shown below.

i. (6 Marks) Draw the structures of all stereoisomeric phosphonate

diesters that are formed when racemic 2-butanol is submitted to this reaction. Label the stereogenic centres in each structure as R or S and identify any enantiomeric pairs or meso compounds among them.

ii. (2 Marks) Why are there three signals in the 31P NMR spectrum derived from the racemic alcohol?

iii. (2 Marks) Sketch the 31P NMR spectrum you would expect to observe if this procedure was applied to a pure sample of R-2-butanol.

(Feringa, B.L.; Smaardijk, A.; Wynberg, H. J. Am. Chem. Soc. 1985, 107, 4798-9)

1.000

2.000

0.997

31P NMR (80.988 MHz; proton decoupled)

Chemical shifts relative to H3PO4 = 0.0 ppm

P HO

(R)

O

(R)

O

P HO

(S)

O

(S)

O

P HO

(R)

O

(S)

O

PH O

(R)

O

(S)

O

Enantiomers

Meso Meso

but note these are not the same!

The RR and SS enantiomers give identical 31P NMR signals, but each of the two meso forms gives a distinct signal because they are diastereomers of one another (as well as being diastereomers of the chiral forms).


3. (25 Marks Total) REACTION PRODUCTS AND REAGENTS

a. (2)

b. (4)

c. (3)

d. (2)

e. (2)

This question also

appeared on the 2010

Midterm.

The key here is H-

bonding by the allylic

alcohol, which directs

epoxidation to the “top”

face of the more

electron-rich alkene. See

pages 203-206 of the

textbook.

The formation of the

unsaturated ketone is

problem 6-7 in the textbook.

Elimination during Swern

oxidations is noted in the

text on page 223.

This process is analogous

to chemistry discussed on

page 184 of the textbook,

and also in class. Amide

(and ester) carbonyls can

participate in

intramolecular attack on

activated alkenes. In these

reactions, the amide

nitrogen substituent is lost.

Epoxidation in step 1

followed by an aqueous

strong acid in step 2 was

also acceptable, since

aq. acid would open an

epoxide to a diol.

Pinacol rearrangement

gives the two products.


f. (4)

g. (3)

h. (3)

i. (2)

This is the example of post-

Baeyer-Villiger

rearrangement given in the

text and discussed in class.

This example of

deoxygenation appears

on page 260 of the

textbook, and its

mechanism is explained

on pages 258-259.


4. (25 Marks Total) MECHANISMS AND EXPERIMENTAL EXPLANATIONS

a. (8 Marks) Beckmann rearrangement of the oxime 1 with the very electrophilic reagent SOCl2 led to nitrile 2. Draw a stepwise mechanism to explain the formation of 2.

NOTE: other researchers have shown that the oxime 3 rearranges normally to give the expected amide under a variety of Beckmann conditions.

Note that since you are told that 3 does not open up in this way, your

mechanism must reflect how sulfur participates in forming 2.


b. (7 Marks) Dissolving-metal reduction of tri-O-methyl gallic acid (3,4,5-trimethoxybenzoic acid) leads to loss of the 4-methoxyl group. Draw a mechanism that explains how this group in particular is lost.

It is especially important in reactions like this to keep track of the valency of the reacting

centres. Draw in the hydrogens, for example, so that you can account for all the electrons in

the valence shell. Be sure to keep careful count of the electrons and the charges. It is very easy

to lose track of what the charge is when fragments are being lost.


c. (10 Marks) While the following oxidative ring enlargement is not particularly high-yielding, it is mechanistically interesting. Provide a detailed stepwise mechanism to explain the formation of the two observed products.

When this exam was

written, the structure of the

product ketone in the

reaction was incorrect,

which made the question

very hard to answer!

The question has been

corrected here, and the

answer given is the expected

mechanism for the correct

process.

Students should note that this is the same chemistry featured in 2010 Problem Set 3,

question 1. In that problem set, a different molecule was used, but both reactions

come from the same paper and have the same mechanism.


5. (10 MARKS TOTAL) APPLICATIONS OF SPECTROSCOPY

a. (6 Marks) When 2 mmol of biacetyl monooxime and 1 mmol of ethylenediamine are stirred together in methanol, a white crystalline material crystallizes from the solution after 45 minutes. This compound is insoluble in dichloromethane or chloroform, but does dissolve in DMSO. Its

1H NMR

spectrum in DMSO solution is very simple, and the data are summarized below:

Chemical Shift (ppm) Multiplicity and integral

11.43 s, 2H

3.71 s, 4H

2.03 s, 6H

1.90 s, 6H

What is the structure of this new compound? Assign the proton NMR signals to appropriate atoms in your proposed structure.

(Roy, A.S.; Weyhermüller, T.; Ghosh, P. Inorg. Chem. Commun., 2008, 11, 167.)

b. (4 Marks) The 1H NMR spectrum of one isomer of hexenol (C6H12O) is

shown below. Draw its structure.

Notice that there are three vinylic protons,

so the alkene is at one end of the chain.

Also notice there are no CH3 groups!

Therefore there is no branching, and the

OH group must be at the other end of the

chain. The compound is thus hex-5-en-1-ol

The key here is a symmetrical

structure. You can draw either E or Z

oximes as long as symmetry is

maintained. Also, as long as your

signal assignment shows the correct

symmetry, you can assign the CH3

groups to either value.

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