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CHEM 2220 Final Exam 2012R Page 2 of 14
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) (2 Marks)
(d) (2 Marks)
(e) (2 Marks)
(f)
OHPBr3 Ph3P
Product is ionic - show bothcation and anion parts!
nBuLi, THFthen add
H
O
(6 Marks)
CHEM 2220 Final Exam 2012R Page 3 of 14
(g) (2 Marks)
(h) (2 Marks)
(i) (2 Marks)
(j) (2 Marks)
(k) (6 Marks)
CHEM 2220 Final Exam 2012R Page 4 of 14
2. (a) (7 MARKS) Synthesis. 1-(Hydroxymethyl)cyclohexene can be prepared in four steps starting from diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary reactions, as well as the intermediate products formed at each step.
OO
OEtEtO
OH
Reaction 1
Reaction 2
Reaction 3
Reaction 4
(b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required starting materials can be obtained, briefly explain why or why not.
CHEM 2220 Final Exam 2012R Page 5 of 14
3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that this process occurs at an unusually low temperature for this type of reaction.
CH2Cl2/MeOH25 oC
H
H
OH
OHHO OH(racemic)
CHEM 2220 Final Exam 2012R Page 6 of 14
4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes.
This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17).
(a) (5 MARKS) Johnson’s team studied the following reaction as a model process. Suggest a stepwise mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)
CHEM 2220 Final Exam 2012R Page 7 of 14
(b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of them. Write a mechanism for the formation of the two products shown.
CHEM 2220 Final Exam 2012R Page 8 of 14
5. (20 MARKS TOTAL) Mechanism grab-bag!
(a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CHEM 2220. Write a mechanism for this reaction.
(b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction. Write a stepwise mechanism to explain this transformation.
CHEM 2220 Final Exam 2012R Page 9 of 14
(c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure.
(d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).
CHEM 2220 Final Exam 2012R Page 10 of 14
6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAlH4) is a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville’s proposed synthetic procedure is shown below.
Most data and procedures in this question are fictitious and should not be used for experimental purposes.
Acid A Alcohol B LiAlH4 Diethyl ether MW (g/mol) 136.15 122.14 38.39 74.12 MP (oC) 52 20 N/A -116 BP (oC) 120 195 N/A 35 Solubilities (g/mL)
Water – 0.3 Ethanol - 6 Diethyl ether - 3
Water – 0.05 Ethanol - 10 Diethyl ether - 20
Water – N/A Ethanol – N/A Diethyl ether – 10
Water - 0.03 Ethanol – miscible
Densities 1.76 0.89 N/A 0.88 Neville’s Proposed Procedure
Dissolve Acid A (10 g) in diethyl ether (100 mL). Cool the solution on an ice bath. Slowly add LiAlH4 (8 g). After 30 minutes of stirring, slowly add ice cold water (50 mL) to the solution. Remove and discard the top (aqueous) layer. The organic layer is extracted three times with 50 mL aliquots of 1 M HCl. The solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from the distillation flask (round bottom flask).
(a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with Neville’s procedure as you can, and indicate how you would modify the procedure to make it work. Your goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield. State all assumptions.
CHEM 2220 Final Exam 2012R Page 11 of 14
(b) (3 MARKS) After implementing your recommendations, Neville collected his product and then brought you the TLC plate shown in Figure 1. Unfortunately Neville had used up all the starting material so he could not use it as a reference spot. However, he found a sample of pure Alcohol B and used that as the reference. Help Neville interpret this TLC plate by explaining whether the product has been made and comment on the purity of the product.
(c) (3 MARKS) Neville performed the reaction again and examined his product by TLC on silica gel. The plate is shown on the left in Figure 2.
Using the blank TLC plate template on the right in Figure 2, draw a diagram of how the plate should look if the plate substrate (absorbent) was modified to the LESS POLAR cellulose but using the same eluent. The spots from left to right are starting material, co-spot, and product.
Eluent: ethyl acetate:hexanes (1:9)Silica plate
Solvent front
Origin
Alcohol Bcospot
Productfrom the reaction
Figure 1: TLC of Neville's first product.
Figure 2: TLC for Neville's second attempt.
CHEM 2220 Final Exam 2012R Page 12 of 14
7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H16O. The IR, 1H NMR and 13C NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (1 MARK) What is the unsaturation number for this compound?
(b) (1 MARK) What functional group(s) is(are) present? What specific spectroscopic evidence supports your answer?
(c) (1 MARK) Briefly explain the signals in the 1H NMR spectrum between 0.80 – 0.88 ppm.
(d) (2 MARKS) Draw the structure of this compound in the box below.
Structure for C7H16O
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Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II
1H NMR – Typical Chemical Shift Ranges
Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)
C CH3 0.7 – 1.3 C C H 2.5 – 3.1
C CH2 C 1.2 – 1.4 O
H
9.5 – 10.0
C H
C
C
C
1.4 – 1.7 O
OH
10.0 – 12.0 (solvent dependent)
C H
1.5 – 2.5 C OH
1.0 – 6.0 (solvent dependent)
O
H
2.1 – 2.6 CO H 3.3 – 4.0
Aryl C H 2.2 – 2.7 CCl H 3.0 – 4.0
H 4.5 – 6.5 CBr H
2.5 – 4.0
Aryl H 6.0 – 9.0 CI H 2.0 – 4.0
13C NMR – Typical Chemical Shift Ranges
IR – Typical Functional Group Absorption Bands
Group Frequency
(cm-1) Intensity Group
Frequency (cm-1)
Intensity
C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium Aryl C=C 1600, 1500 Strong RNO2 1540 Strong
12 11 10 9 8 7 6 5 4 3 2 1 0
R3C–H Aliphatic, alicyclic
X–C–H X = O, N, S, halide
Y
HH
Aromatic, heteroaromatic
Y
H
RCO2H
Y = O, NR, S Y = O, NR, S
“Low Field” “High Field”
220 200 180 160 140 120 100 80 60 40 20 0
CH3-CR3 CHx-C=O
CR3-CH2-CR3
CHx-Y Y = O, N Alkene
Aryl
Amide Ester
Ketone, Aldehyde
Acid RCNRCCR
ANSWER KEY Page 1 of 15
CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong
FINAL EXAM – Winter Session 2012R
Friday April 20, 2012 1:30 pm – 4:30 pm Frank Kennedy Brown 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 (10 Marks)
Question 3 – Mechanism (8 Marks)
Question 4 – Mechanism (12 Marks)
Question 5 – Mechanism grab-bag (20 Marks)
Question 6 – Laboratory (15 Marks)
Question 7 – Spectroscopy (5 Marks)
TOTAL: (100 Marks)
CHEM 2220 Final Exam 2012R ANSWER KEY Page 2 of 15
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) (2 Marks)
(d) (2 Marks)
(e) (2 Marks)
(f) (6 Marks)
CHEM 2220 Final Exam 2012R ANSWER KEY Page 3 of 15
(g) (2 Marks)
(h) (2 Marks)
(i) (2 Marks)
(j) (2 Marks)
(k) (6 Marks)
CHEM 2220 Final Exam 2012R ANSWER KEY Page 4 of 15
2. (a) (7 MARKS) Synthesis. 1-(Hydroxymethyl)cyclohexene can be prepared in four steps starting from
diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary
reactions, as well as the intermediate products formed at each step.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 5 of 15
(b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the
Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required
starting materials can be obtained, briefly explain why or why not.
The Diels-Alder reaction generally requires an electron-withdrawing group on
the dienophile. The cyclohexene product will have the double bond between
the carbons that were C2 and C3 of the diene. The diene and dienophile
needed to make 1-(hydroxymethyl)cyclohexene in the simplest way do not fit
these requirements, so the DA reaction is probably not appropriate.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 6 of 15
3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following
reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the
separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that
this process occurs at an unusually low temperature for this type of reaction.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 7 of 15
4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from
acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes.
This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the
laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17).
(a) (5 MARKS) Johnson’s team studied the following reaction as a model process. Suggest a stepwise
mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)
CHEM 2220 Final Exam 2012R ANSWER KEY Page 8 of 15
(b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of
reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of
them. Write a mechanism for the formation of the two products shown.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 9 of 15
5. (20 MARKS TOTAL) Mechanism grab-bag!
(a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CHEM 2220. Write a
mechanism for this reaction.
(b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction.
Write a stepwise mechanism to explain this transformation.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 10 of 15
(c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic
hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure.
(d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient
access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of
this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous
workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).
CHEM 2220 Final Exam 2012R ANSWER KEY Page 11 of 15
6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g
of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAlH4) is
a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville’s
proposed synthetic procedure is shown below.
Most data and procedures in this question are fictitious and should not be used for experimental purposes.
Acid A Alcohol B LiAlH4 Diethyl ether
MW (g/mol) 136.15 122.14 38.39 74.12
MP (oC) 52 20 N/A -116
BP (oC) 120 195 N/A 35
Solubilities (g/mL)
Water – 0.3 Ethanol - 6 Diethyl ether - 3
Water – 0.05 Ethanol - 10 Diethyl ether - 20
Water – N/A Ethanol – N/A Diethyl ether – 10
Water - 0.03 Ethanol – miscible
Densities 1.76 0.89 N/A 0.88
Neville’s Proposed Procedure
Dissolve Acid A (10 g) in diethyl ether (100 mL). Cool the solution on an ice bath. Slowly add LiAlH4 (8
g). After 30 minutes of stirring, slowly add ice cold water (50 mL) to the solution. Remove and discard
the top (aqueous) layer. The organic layer is extracted three times with 50 mL aliquots of 1 M HCl. The
solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from
the distillation flask (round bottom flask).
(a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with
Neville’s procedure as you can, and indicate how you would modify the procedure to make it work. Your
goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You
are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield.
State all assumptions.
Here are some key points that could be part of your answer:
1) The amount of Acid A is incorrect. Since we want 122 g of Alcohol A (1 mole) and we know that
the reaction produces product at 50% yield, then we require 2 moles of starting material, or 272 g
of Acid A.
2) To dissolve Acid A, notice that the solubility in diethyl ether is 3 g/mL, we’ll require a minimum of
90 mL to dissolve all of Acid A.
3) Reduction of an acid requires 3 hydride equivalents: 1 to neutralize the acidic proton plus 2 more
to add to the carbonyl. A minimum of 28.8 g of LiAlH4 is required (0.75 moles) assuming the
reagent delivers all four hydrides. In practice one would probably add at least 1 mole, and
probably more just to be sure.
4) As mentioned by Dr.Hultin in class, the LiAlH4 reduction reaction will have to be refluxed (note
that THF is usually used as the solvent since it has a higher boiling point than diethyl ether) and
the reaction should be monitored by TLC to determine its completion.
5) After water is added to quench the reaction, the ether will be the top layer and not the bottom.
6) A WASHING should be done and not an EXTRACTION using aliquots of 1 M NaOH to remove
the residual acid. Note that the product is in the ether layer.
7) Dry the ether layer with a drying agent potassium carbonate would be a good choice to remove
any final traces of acid.
8) The product can be isolated by distilling off the solvent and saving the liquid in the round bottom
flask.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 12 of 15
(b) (3 MARKS) After implementing your recommendations, Neville
collected his product and then brought you the TLC plate shown in
Figure 1. Unfortunately Neville had used up all the starting
material so he could not use it as a reference spot. However, he
found a sample of pure Alcohol B and used that as the reference.
Help Neville interpret this TLC plate by explaining whether the
product has been made and comment on the purity of the product.
(c) (3 MARKS) Neville performed the reaction again
and examined his product by TLC on silica gel. The
plate is shown on the left in Figure 2.
Using the blank TLC plate template on the right in
Figure 2, draw a diagram of how the plate should
look if the plate substrate (absorbent) was modified
to the LESS POLAR cellulose but using the same
eluent. The spots from left to right are starting
material, co-spot, and product.
Figure 1: TLC of Neville's first product.
Figure 2: TLC for Neville's second attempt.
There are two spots in the product lane. The one with
the higher Rf corresponds to the product while the
bigger spot with the lower Rf corresponds to the
starting material. Therefore the TLC suggests that very
little product was made in the reaction. The product, be
it the acid or alcohol is considered impure since there
are two spots in the TLC.
The samples will travel to higher Rf values since the plate is less
polar. Alcohol A will still have a higher Rf than Acid A.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 13 of 15
7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H16O. The
IR, 1H NMR and
13C NMR spectra of this compound are shown on the next page. Answer the following
questions about this compound.
(a) (1 MARK) What is the unsaturation number for this compound?
(b) (1 MARK) What functional group(s) is(are) present? What specific spectroscopic evidence supports
your answer?
(c) (1 MARK) Briefly explain the signals in the 1H NMR spectrum between 0.80 – 0.88 ppm.
(d) (2 MARKS) Draw the structure of this compound in the box below.
Structure for C7H16O
Unsaturation number is ZERO.
There is an alcohol because the IR shows a broad band at around 3400 cm-1
.
There are 9 protons in this region. One signal is a 3-proton singlet suggesting a CH3 group with no
neighbour protons. The other signal integrates for 6 protons and at first might appear to be a triplet.
However, the peaks of a triplet must be in a 1:2:1 ratio and this is not the case. It looks more like this
is actually two different signals on top of one another, perhaps a 3-proton triplet with a 3-proton
singlet on top.
The 1-proton quartet at about 3.5 ppm tells us that the CH next to the OH group is also
next to a CH3 and nothing else. We see that CH3 as the doublet at 1.1 ppm.
Our analysis for part c above also tells us that there must be two methyls with no
neighbours, and one methyl with a CH2 next to it.
CHEM 2220 Final Exam 2012R ANSWER KEY Page 14 of 15
Spectra for Question 7
IR
1H NMR
13C NMR
NB: all signals are single lines.
Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II
1H NMR – Typical Chemical Shift Ranges
Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)
C CH3 0.7 – 1.3 C C H 2.5 – 3.1
C CH2 C 1.2 – 1.4
O
H
9.5 – 10.0
C H
C
C
C
1.4 – 1.7
O
OH
10.0 – 12.0
(solvent dependent)
C H
1.5 – 2.5 C OH
1.0 – 6.0
(solvent dependent)
O
H
2.1 – 2.6 CO H
3.3 – 4.0
Aryl C H 2.2 – 2.7 CCl H
3.0 – 4.0
H
4.5 – 6.5 CBr H
2.5 – 4.0
Aryl H 6.0 – 9.0 CI H
2.0 – 4.0
13C NMR – Typical Chemical Shift Ranges
IR – Typical Functional Group Absorption Bands
Group Frequency
(cm-1
) Intensity Group
Frequency (cm
-1)
Intensity
C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad
C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong
C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong
C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad
R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium
Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium
Aryl C=C 1600, 1500 Strong RNO2 1540 Strong
12 11 10 9 8 7 6 5 4 3 2 1 0
R3C–H
Aliphatic, alicyclic X–C–H
X = O, N, S, halide
Y
H H
Aromatic,
heteroaromatic
Y
H
RCO2H
Y = O, NR, S Y = O, NR, S
“Low Field” “High Field”
220 200 180 160 140 120 100 80 60 40 20 0
CH3-CR3
CHx-C=O
CR3-CH2-CR3
CHx-Y
Y = O, N Alkene
Aryl
Amide
Ester
Ketone, Aldehyde
Acid RCN
RCCR