2.222 – introductory organic chemistry ii – term test...
TRANSCRIPT
NAME: STUDENT NUMBER: Page 1 of 7
University of Manitoba Department of Chemistry
2.222 – Introductory Organic Chemistry II – Term Test 2 Thursday March 20, 2003
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 table of spectroscopic data is attached at the end of the exam.
1. (6 MARKS) Provide a detailed stepwise mechanism for the following reaction.
O
COOEt
O
O K
OO
EtOOC
THF
(H3O+ workup)
Page 2 of 7
2. (10 MARKS) Write a stepwise mechanism explaining both steps involved in the formation of a peptide (amide) bond between a carboxylic acid and an amine using the reagent dicyclohexylcarbodiimide, as shown below:
C NN C6H11C6H11
CH2Cl2R OH
OR'H2N
R O
O
NH
NC6H11
C6H11
R NH
R'
O
NH
NH
O
C6H11C6H11
+
Page 3 of 7
3. (5 MARKS) The Meerwein-Ponndorf-Verley Reaction is closely related to one of the reactions in Chapter 12 of our textbook. In this process, a ketone is reduced to a secondary alcohol using an excess of aluminum tri(isopropoxide). Suggest a mechanism for this reaction.
O
+ Al
O
O O
H3C CH3
CH3
CH3
H3C
CH3
H
H
HHO H
+O
(after aqueous acid workup)
Page 4 of 7
4. (10 MARKS) The following multi-step sequence is missing some information. Fill in the blanks with the necessary structure or reagent/solvent to correctly complete the synthesis.
O
O
OHH
H
H2 (g)Pd/CEtOH
p-toluenesulfonic
acid (cat.)
toluene, 80 oC
NaOHH2O
C6H12O
reagent, solvent, workup
Page 5 of 7
5. (14 MARKS) Provide the necessary product, reagent/solvent or starting material to correctly complete each of the following reactions. You may assume that reactions are followed by an aqueous workup procedure if necessary. Mechanisms are NOT required.
a.
O
H
1) NH3Cl, KCN EtOH, reflux
2) aq. H2SO4, reflux
b.
O O
c.
Br
1) Ph3P, EtOH
2) n-BuLi, THF then add
O
H
d.
O
O NaOEtEtOH, reflux
e.
O O
f.
COOH
1)
2)
Page 6 of 7
6. (5 MARKS) You have been presented with a sample of a liquid for structural analysis. It has a pepperminty odour, and is slightly water soluble. Mass spectrometry suggests a molecular weight of 138. The compound has the IR, 13C and 1H NMR spectra shown below. You also perform some chemical tests: • It slowly decolourizes bromine water; • It gives a negative iodoform test; • After sodium fusion, it tests negative with Prussian Blue and also with silver nitrate.
What is the structure of the unknown material?
4 0 0 0 3 0 0 0 2 0 0 0 1 5 0 0 1 0 0 0 5 0 0
NB: This signal is actually 2 very close singlets!
Structure of unknown sample is:
Spectroscopy “Crib Sheet” for 2.222 – 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
←δ
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
←δ
CH3-CR3 CHx-C=O
CR3-CH2-CR3
CHx-Y Y = O, N Alkene
Aryl
Amide Ester Ketone, Aldehyde
Acid
ANSWER KEY Page 1 of 7
University of Manitoba Department of Chemistry
2.222 – Introductory Organic Chemistry II – Term Test 2 Thursday March 20, 2003
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 table of spectroscopic data is attached at the end of the exam.
1. (6 MARKS) Provide a detailed stepwise mechanism for the following reaction.
O
COOEt
O
O K
OO
EtOOC
THF
(H3O+ workup)
O
COOEt
O
H OtBu
O
COOEt
O
COOEt
O
O
12
345
O
OEt
O
O
H3O+duringworkup
OO
EtOOC H
C
BC
This is an aldol followed by a retro-aldol. Notice that the collapse of the alkoxide expels an enolate. The cleavage of a C-C bond is possiblebecause the enolate is a relatively stable anion. This problem is Groutas # 61.
A
BA
1 2
34
5
ANSWER KEY Page 2 of 7
2. (10 MARKS) Write a stepwise mechanism explaining both steps involved in the formation of a peptide (amide) bond between a carboxylic acid and an amine using the reagent dicyclohexylcarbodiimide, as shown below:
3.
C NN C6H11C6H11
CH2Cl2R OH
OR'H2N
R O
O
NH
NC6H11
C6H11
R NH
R'
O
NH
NH
O
C6H11C6H11
+
R OH
O
C
N
N
C6H11
C6H11
R O
O
CN
N
C6H11
C6H11
HR O
O
CNH
N
C6H11
C6H11
R'H2N
RO
O
CNH
N
C6H11
C6H11
NR'H H
RO
OH
CNH
N
C6H11
C6H11
NR'H
R
OH
NR'
HO
CNH
N
C6H11
C6H11
R
O
NR'
HO
CNH
HN
C6H11
C6H11
Several variations on this basic mechanism are possible. A particularly good variation is to use the carboxylic acid to protonate the carbodiimide in the very first step, and then attackwith the resulting carboxylate. This can avoid separation of charge.
This reaction is on page 18-20. Note that the intermediate structure is actually similar to an acid anhydride in many respects, and thus creates a good leaving group for displacement by the amine.
ANSWER KEY Page 3 of 7
(5 MARKS) The Meerwein-Ponndorf-Verley Reaction is closely related to one of the reactions in Chapter 12 of our textbook. In this process, a ketone is reduced to a secondary alcohol using an excess of aluminum tri(isopropoxide). Suggest a mechanism for this reaction.
O
+ Al
O
O O
H3C CH3
CH3
CH3
H3C
CH3
H
H
HHO H
+O
(after aqueous acid workup)
O
Al
O
O O
H3C CH3
R
R
H
+
OAl
OO
OCH3
CH3
RR
H
OAl
OO
O
CH3H3C
RR
H
Acidworkup
HO H
+O
( plus Al2O3 etc)
This reaction is similar to the Cannizzaro Reaction (pp 607-8) and also tothe material in Exercise 12.27 (p 639).
ANSWER KEY Page 4 of 7
4. (10 MARKS) The following multi-step sequence is missing some information. Fill in the blanks with the necessary structure or reagent/solvent to correctly complete the synthesis.
O
O
OHH
H
H2 (g)Pd/CEtOH
p-toluenesulfonic
acid (cat.)
toluene, 80 oC
NaOHH2O
C8H12O
reagent, solvent, workup
O3, CH2Cl2thenZn/HOAc
O
12
34
56
78
12
34
5
678
OH
H
CH3MgBr, Etherthenaq. NH4Cl
H
IntramolecularAldol - from
Exercise 13.7 b
Catalytic Hydrogenation gives syn product
Ozonolysis with
reductive workup,
or KMnO4, H3O+, heat
Grignard addition to ketone
Acid-catalyzedelimination
H
H CH3
This structure (or the ortho isomer), arising from Friedel-Crafts reaction with toluene, is mechanistically reasonable and will also receive full marks in the last step.
ANSWER KEY Page 5 of 7
5. (14 MARKS) Provide the necessary product, reagent/solvent or starting material to correctly complete each of the following reactions. You may assume that reactions are followed by an aqueous workup procedure if necessary. Mechanisms are NOT required.
a.
O
H
1) NH4Cl, KCN EtOH, reflux
2) aq. H2SO4, reflux
NH2
COOH
b.
O OH2Pd or Pt catalystMeOH or other alcohol solvent
c.
Br
1) Ph3P, EtOH
2) n-BuLi, THF then add
O
H
H
d.
O
O NaOEtEtOH, reflux
O
e.
O O
LDA, THF, -78 oC
then Br
f.
COOH
1) AcC l, A lC l3, CH2C l2
2) KM nO 4, heat
NB: You could also have used 1) B r2, FeBr3; 2) Mg/Ether then CO 2
In step 1) you could have alky lated w ith any alkyl halide, sincethese would be oxidized to the ac id as well. The acylation is the better choice because it deactivates the ring towards further reaction, but e ither w ill rece ive fu ll m arks here.
ANSWER KEY Page 6 of 7
6. (5 MARKS) You have been presented with a sample of a liquid for structural analysis. It has a pepperminty odour, and is slightly water soluble. Mass spectrometry suggests a molecular weight of 138. The compound has the IR, 13C and 1H NMR spectra shown below. You also perform some chemical tests: • It slowly decolourizes bromine water; • It gives a negative iodoform test; • After sodium fusion, it tests negative with Prussian Blue and also with silver nitrate.
What is the structure of the unknown material?
4 0 0 0 3 0 0 0 2 0 0 0 1 5 0 0 1 0 0 0 5 0 0
Structure of unknown sample is:
NB: This signal is actually 2 very close singlets!
O
Isophorone
C9H14OMol. Wt.: 138.21
O O
A BAlthough structures A and B are not correct, they
are both similar enough to the actual structure
that they will also be given full marks. Please
note, however that structure B would NOT
produce a 1H NMR spectrum that was entirely
singlets since the vinylic H would be coupled to
the adjacent CH2.
Spectroscopy “Crib Sheet” for 2.222 – 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
←δ
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
←δ
CH3-CR3 CHx-C=O
CR3-CH2-CR3
CHx-Y Y = O, N Alkene
Aryl
Amide Ester Ketone, Aldehyde
Acid