Name:
Student No:
Page 1 of 13
CHEM 2220 Organic Chemistry II: Reactivity and Synthesis
Prof. P.G. Hultin, Dr. H. Luong
FINAL EXAM – Winter Session 2013R
Saturday April 13, 2013 1:30 pm – 4:30 pm
Frank Kennedy Gold Gym
Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN
notes they wish (both sides). Molecular model kits and calculators (no text or graphics memory!) are
permitted but no other aids may be used.
Question 1 – Reactions and Products
(30 Marks)
Question 2 – Synthesis Road-Map
(10 Marks)
Question 3 – Mechanism
(12 Marks)
Question 4 – Mechanism grab-bag
(32 Marks)
Question 5 – Laboratory
(10 Marks)
Question 6 – Spectroscopy
(6 Marks)
TOTAL:
(100 Marks)
CHEM 2220 Final Exam 2013R
Page 2 of 13
1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction
conditions to correctly complete the following reactions. Show stereochemistry when necessary.
(a)
(2 Marks)
(b)
(2 Marks)
(c)
(d)
(e)
(2 Marks)
(2 Marks)
(2 Marks)
(f)
(g)
(4 Marks)
(2 Marks)
CHEM 2220 Final Exam 2013R
(h)
(i)
(j)
(k)
(l)
Page 3 of 13
(2 Marks)
(2 Marks)
(2 Marks)
(4 Marks)
(4 Marks)
CHEM 2220 Final Exam 2013R
Page 4 of 13
2. (10 MARKS)
The polyether antibiotic monensin is produced by a particular strain of Streptomyces. Its
structure was determined in 1967, but the molecule was not successfully synthesized until the work of Y. Kishi
and his co-workers in 1979. The following sequence depicts the first steps in this successful effort. Fill in the
missing information in the boxes provided. Note: for this problem, R/S stereochemistry has been removed but
in fact the stereochemistry of monensin was a vital consideration in the actual synthesis.
CHEM 2220 Final Exam 2013R
Page 5 of 13
3. (12 MARKS) Mechanism. Nitrogen-containing rings are common structures in pharmaceuticals. One
convenient way to prepare 4-piperidinones is shown below. Fill in the structures of the missing intermediates,
and provide a detailed stepwise mechanism for each step.
CHEM 2220 Final Exam 2013R
4. (32 MARKS TOTAL)
(a) (5 MARKS)
Page 6 of 13
Mechanism grab-bag!
Give a mechanism for the formation of the cyclobutanone in the following reaction.
(b) (5 MARKS)
Acrolein (propenal) is made commercially from glycerol by heating in the presence of
sulfuric acid. Propose a stepwise mechanism to explain this reaction.
CHEM 2220 Final Exam 2013R
Page 7 of 13
(c) (5 MARKS)
Amides can be converted to nitriles by heating in thionyl chloride (SOCl 2). Propose a
stepwise mechanism for the formation of acetonitrile from acetamide.
(d) (5 MARKS)
The Vilsmeier Reagent offers an alternative method for forming acid chlorides from
carboxylic acids. Suggest a stepwise mechanism for this transformation.
CHEM 2220 Final Exam 2013R
Page 8 of 13
(e) (6 MARKS) The amino group in α-amino acids must be protected at certain stages of the synthesis of
peptides. The protecting group must be removed at other points, however, and the conditions for removal
must avoid extremes of acid or base. The so-called Alloc derivatives can be removed by warming with
diiodine in a mixture of acetonitrile and water as shown below. Provide a stepwise mechanism to explain
how this deprotection occurs.
(f) (6 MARKS) Here is another way to make piperidine rings, starting from a β-lactam such as 1. The
intermediate 2 from the first step is isolated before being submitted to the second step. Draw the structure
of 2 and provide detailed stepwise mechanisms for both steps in this process.
CHEM 2220 Final Exam 2013R
Page 9 of 13
5. Lab Questions (10 MARKS Total) Local physicist Dr. Sheldon Copper has decided to dress up as one of the
Twilight characters for Comic Con 2013. Of course the outfit is not complete unless there’s some fake blood.
The fake blood at the party store was pretty expensive, so to save money Sheldon decided he could make
tetraphenylcyclopentadienone (TPCD) from materials available in the organic chemistry lab to use as his fake
blood. This compound is a deep red colour because of its extended -conjugation.
TCPD can be synthesized by aldol condensation using diphenylacetone, benzil and sodium hydroxide, as
shown below. Sheldon has written out the following lab procedure intended to make 4 g of TPCD. Your
instincts developed from working in the CHEM 2220 laboratory tell you that although the reaction is possible,
Sheldon’s procedure will not work and modifications are necessary.
Note: The story, data and procedures in this question are fictitious. None of this should be used for experimental purposes.
Data
MW (g/mol)
MP (oC)
BP (oC)
Solubilities (g/mL)
Diphenylacetone
210
30-34
330
Ethanol – 1 g/mL
DCM – 0.5 g/L
Hexanes – 1 g/mL
Benzil
210
95
346
Ethanol – 1 g/mL
DCM – 2 g/mL
Hexanes – 0.2 g/L
TPCD
384
220
N/A
Ethanol – 1 g/mL
DCM – 0.1 g/L
Hexanes – 2 g/mL
Dissolve 50 g of diphenylacetone in 50 mL hexanes and sodium hydroxide (1 kg). Add 40 g of
benzil. Reflux the reaction for the length of time equivalent to one TV episode of Big Bang
Theory (about 30 minutes) and the product might have precipitated out. Filter off the product
and recrystallize the crude material with ethanol and dichloromethane.
(a) (7 Marks)
Critique Sheldon’s procedure and indicate how you would modify the procedure to make it
work better so that approximately 4 grams of product can be obtained. You will need a slight excess of one
of the reactants to push the equilibrium to favor the product. The goal is to obtain TPCD in the most pure
form possible. You are free to use any reagents and chemicals you wish.
Assume that your proposed reaction will produce a 50% yield and that there is residual starting
material (the one used in excess). State all assumptions. Of course there is some math involved
but we have chosen numbers that can be easily approximated.
CHEM 2220 Final Exam 2013R
Page 10 of 13
(b) (1.5 Marks)
Sheldon performed a TLC of his crude
product (shown at right). Unfortunately he did not label his
origins for starting material and product. Can you, once
again, help Sheldon figure out which is likely the product and
which is the starting material (benzil). Please justify your
reasoning.
(c) (1.5 Marks)
Let’s face it: Sheldon isn’t the ideal candidate to work in the synthetic chemistry lab. As the
dear friend that you are, you suggested that he consider being an advocate for green chemistry rather than
working in the lab. Using your proposed reaction and comparing it to Sheldon’s, name at least three
principles of green chemistry that are observed.
CHEM 2220 Final Exam 2013R
Page 11 of 13
6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H14O2. The
IR, 13C NMR and 1H NMR spectra of this compound are shown on the next page. Answer the following
questions about this compound.
(a) (0.5 MARK)
What is the unsaturation number for this compound?
(b) (1.5 MARK)
What functional group(s) does this compound contain? Indicate the specific evidence for
your conclusion.
13C
(c) (1 MARK)
What can you conclude from the number of
(d) (1 MARK)
What can you conclude from the 1H NMR signal at 2.3 ppm?
(e) (2 MARKS)
Draw the structure of this compound in the box below.
Structure for C8H14O2
NMR signals?
CHEM 2220 Final Exam 2013R
Page 12 of 13
Spectra for Question 6
IR
13C
NMR
NB: all signals are single lines.
1H
NMR
m
m
s
s
s
Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II
1H
NMR – Typical Chemical Shift Ranges
Chemical Shift (δ)
Type of Proton
C
0.7 – 1.3
CH3
Chemical Shift (δ)
Type of Proton
C
C
2.5 – 3.1
H
O
C
CH2
1.2 – 1.4
C
9.5 – 10.0
H
C
C
O
C
10.0 – 12.0
(solvent dependent)
1.4 – 1.7
H
OH
C
C
H
1.5 – 2.5
C
OH
1.0 – 6.0
(solvent dependent)
O
H
Aryl
C
H
2.1 – 2.6
O
C
H
3.3 – 4.0
2.2 – 2.7
Cl
C
H
3.0 – 4.0
4.5 – 6.5
Br
C
H
2.5 – 4.0
6.0 – 9.0
I
H
Aryl
H
Aromatic,
heteroaromatic
RCO2H
Y = O, NR, S
11
2.0 – 4.0
H
X–C–H
X = O, N, S, halide
Y
10
H
9
8
7
6
5
4
3
2

160
CR3-CH2-CR3
CHx-C=O
RCCR
CH3-CR3
RCN
140
120
100
80
60
40

13C
NMR – Typical Chemical Shift Ranges
IR – Typical Functional Group Absorption Bands
Group
C–H
C=C–H
C=C
C≡C–H
R–C≡C–R′
Aryl–H
Aryl C=C
Frequency
(cm-1)
2960 – 2850
3100 – 3020
1680 – 1620
3350 – 3300
2260 – 2100
3030 – 3000
1600, 1500
0
CHx-Y
Y = O, N
Aryl
Ketone, Aldehyde
Ester
Amide
Acid
180
1
“High Field”
Alkene
200
Y = O, NR, S
H
“Low Field”
220
R3C–H
Aliphatic, alicyclic
Y
H
12
C
Intensity
Medium
Medium
Medium
Strong
Medium (R ≠ R′)
Medium
Strong
Group
RO–H
C–O
C=O
R2N–H
C–N
C≡N
RNO2
Frequency
(cm-1)
3650 – 3400
1150 – 1050
1780 – 1640
3500 – 3300
1230, 1030
2260 – 2210
1540
Intensity
Strong, broad
Strong
Strong
Medium, broad
Medium
Medium
Strong
20
0
ANSWER KEY
Page 1 of 13
CHEM 2220 Organic Chemistry II: Reactivity and Synthesis
Prof. P.G. Hultin, Dr. H. Luong
FINAL EXAM – Winter Session 2013R
Saturday April 13, 2013 1:30 pm – 4:30 pm
Frank Kennedy Gold Gym
Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN
notes they wish (both sides). Molecular model kits and calculators (no text or graphics memory!) are
permitted but no other aids may be used.
Question 1 – Reactions and Products
(30 Marks)
Question 2 – Synthesis Road-Map
(10 Marks)
Question 3 – Mechanism
(12 Marks)
Question 4 – Mechanism grab-bag
(32 Marks)
Question 5 – Laboratory
(10 Marks)
Question 6 – Spectroscopy
(6 Marks)
TOTAL:
(100 Marks)
CHEM 2220 Final Exam Answers 2013R
Page 2 of 13
1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction
conditions to correctly complete the following reactions. Show stereochemistry when necessary.
(a)
(2 Marks)
(b)
(2 Marks)
(c)
(d)
(e)
(f)
(2 Marks)
(2 Marks)
(2 Marks)
(4 Marks)
CHEM 2220 Final Exam Answers 2013R
Page 3 of 13
(g)
(2 Marks)
(h)
(i)
(j)
(k)
(l)
(2 Marks)
(2 Marks)
(2 Marks)
(4 Marks)
(4 Marks)
CHEM 2220 Final Exam Answers 2013R
Page 4 of 13
2. (10 MARKS)
The polyether antibiotic monensin is produced by a particular strain of Streptomyces. Its
structure was determined in 1967, but the molecule was not successfully synthesized until the work of Y. Kishi
and his co-workers in 1979. The following sequence depicts the first steps in this successful effort. Fill in the
missing information in the boxes provided. Note: for this problem, R/S stereochemistry has been removed but
in fact the stereochemistry of monensin was a vital consideration in the actual synthesis.
CHEM 2220 Final Exam Answers 2013R
Page 5 of 13
3. (12 MARKS) Mechanism. Nitrogen-containing rings are common structures in pharmaceuticals. One
convenient way to prepare 4-piperidinones is shown below. Fill in the structures of the missing intermediates,
and provide a detailed stepwise mechanism for each step.
CHEM 2220 Final Exam Answers 2013R
4. (32 MARKS TOTAL)
(a) (5 MARKS)
Page 6 of 13
Mechanism grab-bag!
Give a mechanism for the formation of the cyclobutanone in the following reaction.
(b) (5 MARKS)
Acrolein (propenal) is made commercially from glycerol by heating in the presence of
sulfuric acid. Propose a stepwise mechanism to explain this reaction.
CHEM 2220 Final Exam Answers 2013R
Page 7 of 13
(c) (5 MARKS)
Amides can be converted to nitriles by heating in thionyl chloride (SOCl 2). Propose a
stepwise mechanism for the formation of acetonitrile from acetamide.
(d) (5 MARKS)
The Vilsmeier Reagent offers an alternative method for forming acid chlorides from
carboxylic acids. Suggest a stepwise mechanism for this transformation.
CHEM 2220 Final Exam Answers 2013R
Page 8 of 13
(e) (6 MARKS) The amino group in α-amino acids must be protected at certain stages of the synthesis of
peptides. The protecting group must be removed at other points, however, and the conditions for removal
must avoid extremes of acid or base. The so-called Alloc derivatives can be removed by warming with
diiodine in a mixture of acetonitrile and water as shown below. Provide a stepwise mechanism to explain
how this deprotection occurs.
(f) (6 MARKS) Here is another way to make piperidine rings, starting from a β-lactam such as 1. The
intermediate 2 from the first step is isolated before being submitted to the second step. Draw the structure
of 2 and provide detailed stepwise mechanisms for both steps in this process.
CHEM 2220 Final Exam Answers 2013R
Page 9 of 13
5. Lab Questions (10 MARKS Total) Local physicist Dr. Sheldon Copper has decided to dress up as one of the
Twilight characters for Comic Con 2013. Of course the outfit is not complete unless there’s some fake blood.
The fake blood at the party store was pretty expensive, so to save money Sheldon decided he could make
tetraphenylcyclopentadienone (TPCD) from materials available in the organic chemistry lab to use as his fake
blood. This compound is a deep red colour because of its extended -conjugation.
TCPD can be synthesized by aldol condensation using diphenylacetone, benzil and sodium hydroxide, as
shown below. Sheldon has written out the following lab procedure intended to make 4 g of TPCD. Your
instincts developed from working in the CHEM 2220 laboratory tell you that although the reaction is possible,
Sheldon’s procedure will not work and modifications are necessary.
Note: The story, data and procedures in this question are fictitious. None of this should be used for experimental purposes.
Data
MW (g/mol)
o
MP ( C)
o
BP ( C)
Solubilities (g/mL)
Diphenylacetone
210
30-34
330
Ethanol – 1 g/mL
DCM – 0.5 g/L
Hexanes – 1 g/mL
Benzil
210
95
346
Ethanol – 1 g/mL
DCM – 2 g/mL
Hexanes – 0.2 g/L
TPCD
384
220
N/A
Ethanol – 1 g/mL
DCM – 0.1 g/L
Hexanes – 2 g/mL
Dissolve 50 g of diphenylacetone in 50 mL hexanes and sodium hydroxide (1 kg). Add 40 g of
benzil. Reflux the reaction for the length of time equivalent to one TV episode of Big Bang
Theory (about 30 minutes) and the product might have precipitated out. Filter off the product
and recrystallize the crude material with ethanol and dichloromethane.
(a) (7 Marks)
Critique Sheldon’s procedure and indicate how you would modify the procedure to make it
work better so that approximately 4 grams of product can be obtained. You will need a slight excess of one
of the reactants to push the equilibrium to favor the product. The goal is to obtain TPCD in the most pure
form possible. You are free to use any reagents and chemicals you wish.
Assume that your proposed reaction will produce a 50% yield and that there is residual starting
material (the one used in excess). State all assumptions. Of course there is some math involved
but we have chosen numbers that can be easily approximated.
Critique:
Despite Dr. Copper’s intelligence, there are a number of issues with his proposed synthesis:
1. Amounts – To synthesize 4 g (about 0.01 mol) of TPCD, Sheldon would require a minimum of
4 g of each of benzil and diphenylacetone (keeping in mind the 50% yield). Choose benzil to
be in excess since it should be easier to remove at the end. Sodium hydroxide is a catalyst
and therefore should be used in much smaller amounts.
2. Solvent – The choice of solvent is critical to help the reaction proceed AND to help with the
workup. Sheldon chose hexanes as the solvent and this is inappropriate for two reasons: 1)
sodium hydroxide has limited solubility in hexanes and 2) benzil is insoluble in hexanes. Out of
the three solvents listed, the most appropriate is ethanol (at a minimum volume of 5 mL, adding
more as necessary) since it can dissolve all start materials and products. Dichloromethane
unfortunately cannot dissolve diphenylacetone and will likely not be able to dissolve sodium
hydroxide either.
3. Reaction time – It is unclear from the way the procedure is written whether there is certainty in
the yielding of product after refluxing for 30 minutes. The best thing to do is to monitor the
reaction by TLC (co-spot with starting materials and Sheldon should watch for the
disappearance of diphenylacetone).
4. Work up – Notice that the solvent chosen for the synthesis dissolves all starting material and
product. As well, in the monitoring of the reaction, all the diphenylacetone must be consumed
to allow for an ease of workup. The next step is to remove the solvent (and water byproduct)
by distillation (to concentrate the reaction down). Once the solvent has been removed, DCM is
added to the reaction to precipitate out the crude product.
5. Purification – no issues here!
CHEM 2220 Final Exam Answers 2013R
Page 10 of 13
(b) (1.5 Marks)
Sheldon performed a TLC of his crude
product (shown at right). Unfortunately he did not label his
origins for starting material and product. Can you, once
again, help Sheldon figure out which is likely the product and
which is the starting material (benzil). Please justify your
reasoning.
The spot on the left is more polar than the
spot on the right. Therefore, it’s likely that
the spot on the left is benzil while the spot
on the right is TCPD.
(c) (1.5 Marks)
Let’s face it: Sheldon isn’t the ideal candidate to work in the synthetic chemistry lab. As the
dear friend that you are, you suggested that he consider being an advocate for green chemistry rather than
working in the lab. Using your proposed reaction and comparing it to Sheldon’s, name at least three
principles of green chemistry that are observed.
Atom economy, catalysis, prevention of waste
CHEM 2220 Final Exam Answers 2013R
Page 11 of 13
6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H14O2. The
13
1
IR, C NMR and H NMR spectra of this compound are shown on the next page. Answer the following
questions about this compound.
(a) (0.5 MARK)
What is the unsaturation number for this compound?
2
(b) (1.5 MARK)
What functional group(s) does this compound contain? Indicate the specific evidence for
your conclusion.
-1
13
There is a carbonyl (IR ~1730 cm ) which is probably an ester ( C NMR ~173 ppm and NO OH in
IR).
13
There is also an alkene ( C NMR ~110 and ~148 ppm).
(c) (1 MARK)
What can you conclude from the number of
13
C NMR signals?
13
There are 7 C NMR signals but 8 carbons in the formula, therefore there is one pair of symmetryrelated carbons.
(d) (1 MARK)
1
What can you conclude from the H NMR signal at 2.3 ppm?
This is a 2-proton singlet. The chemical shift suggests it might be next to a double
bond. It could be a CH2C=O structure with no neighboring protons.
(e) (2 MARKS)
Draw the structure of this compound in the box below.
Structure for C8H14O2
Other relevant observations:
1
H NMR Integral ratio: 1:2:3:2:6.
1
3-Proton H Singlet at ~3.7 ppm is OCH3, from the ester.
1
Multiplet H NMR signals between 4.9 and 5.9 ppm are from alkene. There are three
protons in this region, so the alkene is monosubstituted R-CH=CH2.
6-Proton singlet at ~1 ppm is probably 2 identical CH 3 groups with no nearest neighbors.
13
Note that this is consistent with C NMR observation about a pair of symmetric groups.
CHEM 2220 Final Exam Answers 2013R
Page 12 of 13
Spectra for Question 6
IR
13
C NMR
NB: all signals are single lines.
1
H NMR
m
m
s
s
s
Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II
1
H NMR – Typical Chemical Shift Ranges
Chemical Shift (δ)
Type of Proton
C
0.7 – 1.3
CH3
Chemical Shift (δ)
Type of Proton
C
C
2.5 – 3.1
H
O
C
CH2
1.2 – 1.4
C
9.5 – 10.0
H
C
C
O
C
10.0 – 12.0
(solvent dependent)
1.4 – 1.7
H
OH
C
C
H
1.5 – 2.5
C
OH
1.0 – 6.0
(solvent dependent)
O
H
Aryl
C
H
2.1 – 2.6
O
C
H
3.3 – 4.0
2.2 – 2.7
Cl
C
H
3.0 – 4.0
4.5 – 6.5
Br
C
H
2.5 – 4.0
6.0 – 9.0
I
H
Aryl
H
Aromatic,
heteroaromatic
RCO2H
Y = O, NR, S
11
2.0 – 4.0
H
X–C–H
X = O, N, S, halide
Y
10
H
9
8
7
6
5
4
3
2

160
CR3-CH2-CR3
CHx-C=O
RCCR
CH3-CR3
RCN
140
120
100
80
60
40

13
C NMR – Typical Chemical Shift Ranges
IR – Typical Functional Group Absorption Bands
Group
C–H
C=C–H
C=C
C≡C–H
R–C≡C–R′
Aryl–H
Aryl C=C
Frequency
-1
(cm )
2960 – 2850
3100 – 3020
1680 – 1620
3350 – 3300
2260 – 2100
3030 – 3000
1600, 1500
0
CHx-Y
Y = O, N
Aryl
Ketone, Aldehyde
Ester
Amide
Acid
180
1
“High Field”
Alkene
200
Y = O, NR, S
H
“Low Field”
220
R3C–H
Aliphatic, alicyclic
Y
H
12
C
Intensity
Medium
Medium
Medium
Strong
Medium (R ≠ R′)
Medium
Strong
Group
RO–H
C–O
C=O
R2N–H
C–N
C≡N
RNO2
Frequency
-1
(cm )
3650 – 3400
1150 – 1050
1780 – 1640
3500 – 3300
1230, 1030
2260 – 2210
1540
Intensity
Strong, broad
Strong
Strong
Medium, broad
Medium
Medium
Strong
20
0