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Name: __________________________ Student Number: _____________________
University of Manitoba - Department of Chemistry
CHEM 2220 - Introductory Organic Chemistry II - Term Test 1
Thursday, February 13, 2014; 7-9 PM
This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions.
Put all answers in the spaces provided. If more space is required you may use the backs of the
exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached
at the end of the exam.
QUESTION MARKS
1. Mechanism
(8 Marks)
2. Mechanism
(4 Marks)
3. Mechanism
(4 Marks)
4. Mechanism
(8 Marks)
5. Reactions and Products
(20 Marks)
6. Spectra and Structures
(6 Marks)
TOTAL (50 Marks)
CHEM 2220 Test #1 Page 2 of 9 Feb 13, 2014
1. (8 MARKS) An example of the Ritter Reaction, a general process for the formation of amides
by the acid-catalyzed reaction of nitriles with alkenes, is shown below. Write a stepwise
mechanism for this reaction.
CHEM 2220 Test #1 Page 3 of 9 Feb 13, 2014
2. (4 MARKS) 1-Bromobicyclo[2.2.2]octane (A) does not undergo an E2 elimination when
treated with a strong base. Briefly explain why not.
3. (4 MARKS) Write a stepwise mechanism to explain the following rearrangement.
CHEM 2220 Test #1 Page 4 of 9 Feb 13, 2014
4. (8 MARKS) When compound A was treated with HBr in CH2Cl2 and with complete exclusion
of light, bromocyclohexane product B was formed. However, when the reaction was conducted in
the presence of peroxides, product C was obtained instead. Write out stepwise mechanisms for
these two products and briefly summarize why the different results are obtained.
CHEM 2220 Test #1 Page 5 of 9 Feb 13, 2014
5. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to
correctly complete the following reactions. Mechanisms are NOT required. Show relative
product stereochemistry (wedge and dash bonds) where appropriate – if a racemic product is
formed, simply indicate “+/−” or “racemic”.
a. (2 Marks)
b. (2 Marks)
c. (2 Marks)
d. (2 Marks)
CHEM 2220 Test #1 Page 6 of 9 Feb 13, 2014
e. (4 Marks)
f. (2 Marks)
g. (2 Marks)
h. (4 Marks)
CHEM 2220 Test #1 Page 7 of 9 Feb 13, 2014
6. (6 MARKS Total) The IR and NMR spectra of an unknown organic compound A having the
formula C6H13NO are shown on the next page. Based on these data, answer the following
questions about compound A.
a. (1 Mark) What is the degree of unsaturation in compound A?
b. (1 Mark) What functional group is present in A?
c. (1 Mark) What is the significance of the 3-proton singlet at δ1.98 ppm in the 1H NMR
spectrum?
d. (3 Marks) What is the structure of compound A?
CHEM 2220 Test #1 Page 8 of 9 Feb 13, 2014
IR
C6H13NO
13C NMR
C6H13NO
1H NMR
C6H13NO
tr,
2H
tr,
3H m,
2H
m,
2H
s,
3H s,
1H
Page 9 of 9
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
ANSWER KEY
University of Manitoba - Department of Chemistry
CHEM 2220 - Introductory Organic Chemistry II - Term Test 1
Thursday, February 13, 2014; 7-9 PM
This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions.
Put all answers in the spaces provided. If more space is required you may use the backs of the
exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached
at the end of the exam.
QUESTION MARKS
1. Mechanism
(8 Marks)
2. Mechanism
(4 Marks)
3. Mechanism
(4 Marks)
4. Mechanism
(8 Marks)
5. Reactions and Products
(20 Marks)
6. Spectra and Structures
(6 Marks)
TOTAL (50 Marks)
CHEM 2220 Test #1 ANSWERS Page 2 of 9 Feb 13, 2014
1. (8 MARKS) An example of the Ritter Reaction, a general process for the formation of amides
by the acid-catalyzed reaction of nitriles with alkenes, is shown below. Write a stepwise
mechanism for this reaction.
You did not have to show all the resonance structures in your answer. A correct mechanism can be written using only one of the possible canonical forms.
CHEM 2220 Test #1 ANSWERS Page 3 of 9 Feb 13, 2014
2. (4 MARKS) 1-Bromobicyclo[2.2.2]octane (A) does not undergo an E2 elimination when
treated with a strong base. Briefly explain why not.
3. (4 MARKS) Write a stepwise mechanism to explain the following rearrangement.
E2 elimination requires that the C-H and C-X bonds adopt an
antiperiplanar arrangement, which permits orbital overlap in
the transition state. The ring bonds in this molecule cannot
rotate, and the C-Br bond is nearly orthogonal to the adjacent
C-H bonds. Thus, the necessary geometry is impossible, and
E2 elimination cannot occur.
It can also be noted that the product would be impossibly
strained, since if such an alkene could form the two ends would
be rotated almost 90° from one another!
Note the image on
the right, which
shows a view
similar to a
Newman projection.
The C-Br is in the
background, and the
CH2 is in the
foreground. Clearly
the C-H and C-Br
bonds are not
antiperiplanar.
This is Groutas #21.
This is Klein problem 8.73.
CHEM 2220 Test #1 ANSWERS Page 4 of 9 Feb 13, 2014
4. (8 MARKS) When compound A was treated with HBr in CH2Cl2 and with complete exclusion
of light, bromocyclohexane product B was formed. However, when the reaction was conducted in
the presence of peroxides, product C was obtained instead. Write out stepwise mechanisms for
these two products and briefly summarize why the different results are obtained.
The cationic cyclization is a straightforward extension of electrophilic addition to alkenes. Similar reactions are found in Groutas, for example #29.
The conditions clearly indicate a radical mechanism. By analogy with the cationic mechanism, the initial radical can attack the remaining alkene. This mechanism is a chain in which the propagation requires a few steps. It is written here simply as a linear sequence but obviously the final step produces bromine radical to re-enter the first step.
CHEM 2220 Test #1 ANSWERS Page 5 of 9 Feb 13, 2014
5. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to
correctly complete the following reactions. Mechanisms are NOT required. Show relative
product stereochemistry (wedge and dash bonds) where appropriate – if a racemic product is
formed, simply indicate “+/−” or “racemic”.
a. (2 Marks)
b. (2 Marks)
c. (2 Marks)
d. (2 Marks)
Any hydroboration reagent could be used here.
This reaction cannot form the Zaitsev product, since the tertiary C-H cannot align antiperiplanar to the C-OTs. See Klein pp. 359-360, and SkillBuilder 8.7.
CHEM 2220 Test #1 ANSWERS Page 6 of 9 Feb 13, 2014
e. (4 Marks)
f. (2 Marks)
g. (2 Marks)
h. (4 Marks)
Klein suggests NaOH as the base. You could also have used the non-nucleophilic NaH in THF solution, or several other relatively strong bases like NaOMe or NaOEt. This is very similar to problem 3 from Sapling Homework #4.
Recall Klein Figure 8.25 on page 375. Sulfur compounds RS− are nucleophiles but not bases, so an SN2 reaction is the only possibility here (Klein section 8.13).
Note that there is no water present during the mercuration, so the only nucleophile available is the alcohol group.
Alkylation of terminal alkynes was first presented in CHEM 2210, see Klein section 10.3. This reaction has featured in a few Sapling problems this year. You could also have suggested forming the ketone by simple acid hydration but this would require heating as well as aqueous acid. Remember that this process is slow due to a termolecular rate-limiting step.
CHEM 2220 Test #1 ANSWERS Page 7 of 9 Feb 13, 2014
6. (6 MARKS Total) The IR and NMR spectra of an unknown organic compound A having the
formula C6H13NO are shown on the next page. Based on these data, answer the following
questions about compound A.
a. (1 Mark) What is the degree of unsaturation in compound A?
b. (1 Mark) What functional group is present in A?
c. (1 Mark) What is the significance of the 3-proton singlet at δ1.98 ppm in the 1H NMR
spectrum?
d. (3 Marks) What is the structure of compound A?
Unsaturation = 1
2×6 + 2 = 14; 14 – 13 = 1; 1 + 1 (for N) = 2; 2/2 = 1.
Strong IR bands at ~1675 and ~3280 cm-1
, plus 13
C NMR at
~176 ppm, plus N in the formula = AMIDE
A 3-proton singlet strongly suggests a methyl with no nearest neighbors;
the chemical shift of 1.98 ppm says it is next to the carbonyl of the amide
(i.e. we have CH3C(O)-N).
If you just put
“carbonyl”, ½ Mark.
You were not required to assign the signals to specific parts of your
structure. For future reference:
13
C NMR: 170.54 2 39.47 3 31.73 4 23.06 1 20.19 5 13.73 6 1H NMR:
7.05 (1H, broad s, NH) 3.21 (2H, tr, CH2-3) 1.98 (3H, s, CH3-1) 1.49 (2H, m, CH2-4) 1.35 (2H, m, CH2-5) 0.92 (3H, tr, CH3-6) IR: (notice that detailed analysis can get complicated!) 3289 (amide N-H stretch) 3088 (related to an overtone of 1558) 2960, 2934, 2874 (C-H stretches) 1654 (amide I – mostly C=O stretch) 1558 (amide II – mostly N-H bend)
CHEM 2220 Test #1 ANSWERS Page 8 of 9 Feb 13, 2014
IR
C6H13NO
13C NMR
C6H13NO
1H NMR
C6H13NO
tr,
2H
tr,
3H m,
2H
m,
2H
s,
3H s,
1H
Page 9 of 9
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
