chemistry 2.222 organic chemistry ii: reactivity and ...hultin/chem2220/archive/2002/... ·...
TRANSCRIPT
Practice Final – Not for Marks Page 1 of 12
Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
PRACTICE “FINAL EXAM” – Winter 2002 This is a sample of what to expect on the Final Exam for the course. The format of the final will be EXACTLY the same as this one. The distribution of the questions will be similar, and their difficulty will be the same or easier. Give yourself 3 hours, using only your “cheat sheet” and lots of scrap paper.
PART I: Question 1 (20 Marks)
Question 2 (4 Marks)
Question 3 (10 Marks)
Question 4 (12 Marks)
Question 5 (8 Marks)
Question 6 (8 Marks)
PART II: (8 Marks)
TOTAL: (70 Marks)
Practice Final – Not for Marks Page 2 of 12
PART I: DO ALL QUESTIONS – THERE IS CHOICE IN QUESTION 1 ONLY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions
to correctly complete TEN (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked.
(a)
O
O
H
O
H2 (g), Pd/CEthanol
(b)
OH
heat
(c)
O
NaBH3CN, EtOH
NH2
(d)
O
NH2 O
OEtH2N
toluene (solvent)heat
C13H16N2O
(e)
O
OCH3
O
H
1) CH3OH, H2SO4 (cat.)
2) LiAlH4, THF(H2O workup)
(f)
Br(C6H5)3P
Ethanol
n-BuLiTHF
then addO
H
Practice Final – Not for Marks Page 3 of 12
(g)
N Br
AlCl3, CH2Cl2
(h)
O
O
NaOEt, EtOHheat
C10H8O
(i)
NH
O
NH2
O
OCF3CO2H
(j)
OCH3
CH3CO2H, H2O2H2SO4 (cat.)
(k)
HOOCBr2, CH2Cl2
(l)
CH3
O
CH3
O
CH3
(m)
OO
H3CO OCH3Br Br
2 NaOCH3CH3OH
H2SO4, H2Oheat
(n)
CHO
CHO(Two Steps)
Practice Final – Not for Marks Page 4 of 12
(o)
+ heatCO2CH3
CO2CH3
(p)
Hg(OAc)2H2O, THF
followed byNaBH4
(q)
NO
CH3LiAlH4THF
followed byH2O workup
(r)
COOH COOH
Cl
2. (4 MARKS) The final four steps of the synthesis of the alkaloid retronecine are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks.
N
O
O
CO2EtN
O
CO2EtHO
NOH
CO2EtHO
N
HO OH
NaOEt, EtOH
retronecineC10H15NO3
Practice Final – Not for Marks Page 5 of 12
3. (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation.
OHO CN
CO2CH3
HCl, CH3OH
heat
KCN, H3O+
Practice Final – Not for Marks Page 6 of 12
4. (a) (2 MARKS) Tollens’ Test for aldehydes is carried out under alkaline conditions. The reagent is usually formed by first adding a small amount of aqueous NaOH to a solution of AgNO3, and then treating the mixture with aqueous NH4OH until a clear solution is obtained. Provide chemical equations showing the processes that occur at each step in the preparation of Tollens’ Reagent. Explain why ammonium hydroxide is used instead of simply adding more sodium hydroxide.
(b) (1 MARK) When distilling a liquid and attempting to record its boiling temperature, the bulb of the thermometer is placed in the neck of the distillation apparatus and not immersed in the liquid. Explain why this is important.
(c) (1 MARK) When carrying out a distillation as described in part (b), an accurate boiling temperature will only be recorded when both vapour and condensing liquid are simultaneously present on the bulb of the thermometer. Explain why this is so.
(d) (2 MARKS) The catalyst (AlCl3 or FeCl3) used in Friedel-Crafts reactions must be dry and pure. Why is this important for the success of the reaction?
Practice Final – Not for Marks Page 7 of 12
(e) (2 MARKS) When synthesizing an azodye from a phenol and a primary aromatic amine, a cold, acidic solution of sodium nitrite and the amine is poured into a basic solution of the phenol. Why is it advantageous to use a basic solution of the phenol?
(f) (4 MARKS) In the Hinsberg test for amines, the following reaction is performed and the solubility of the product in aqueous NaOH is observed.
Amine + S
O
O
ClNaOH (aq.)
Product
• Primary amines: The product is soluble in aqueous NaOH. • Secondary amines: The product is insoluble in aqueous NaOH. • Tertiary amines: No new product is formed.
Draw the structures of the different products formed when the Hinsberg test is performed on ethylamine, diethylamine, and triethylamine, and use these structures to explain the solubility patterns noted above.
Practice Final – Not for Marks Page 8 of 12
5. (8 MARKS) Simple alkenes are readily converted to epoxides using peroxycarboxylic acids, but are unaffected by alkaline hydrogen peroxide. In contrast, the C=C bonds of α,β-unsaturated ketones (enones) are epoxidized relatively slowly by peroxycarboxylic acids but are rapidly converted to epoxides by hydrogen peroxide under basic conditions.
Briefly explain this pattern of reactivity, using your knowledge of organic structures and reaction mechanisms. (HINT: the pKa of H2O2 is 11.62).
RR
R'CO3Hfast RR
O
R O
R
R'CO3HRO
R
Overy slow
H2O2, NaOH (aq.)
H2O2, NaOH (aq.)
fastR O
R
O
no reaction!
Practice Final – Not for Marks Page 9 of 12
6. (8 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C6H13NO are shown below.
4000 3000 2000 1500 1000 500
tr s m m tr
(b) Draw the structure of the compound in the
box at right (4 Marks).
(c) Clearly indicate which H atom(s) in your structure is(are) responsible for each of the signals in the 1H NMR spectrum (4 Marks).
Practice Final – Not for Marks Page 10 of 12
PART II: CHALLENGE PROBLEM. DO ONE PROBLEM ONLY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks.
A. SYNTHESIS Propose a multi-step synthetic route to the compound shown at right. You may use benzene, diethyl malonate, and any other organic molecule having 4 carbons or fewer as starting compounds. For each step in your synthesis, provide all needed reagents, solvents and reaction conditions. You can employ any valid reagents and reactions in your synthesis. Mechanisms and discussion are not required.
B. MECHANISM The Hantsch Reaction provides a route to substituted dihydropyridine and pyridine compounds. It has been used in preparing a variety of pharmaceutical products. It is based entirely on processes and mechanistic steps that we have discussed in considerable detail during this course. Provide a detailed stepwise mechanism for the Hantsch reaction shown below.
OEt
O O
2 + H
O
NH3 (excess)EtOH
N
H
O
OEt
O
EtO
C. SPECTROSCOPY The RCMP have confiscated a shipment of chemicals destined for an organic chemist at the U of Manitoba, because they were tipped off that the package included a substantial amount of methamphetamine (“Speed” or “Crystal Meth”). You have been given the suspicious white powder to analyze. Unfortunately, your automated mass spectrometer with its computerized database of spectra of illicit drugs is broken this week, so you turn instead to chemical tests, IR, 1H and 13C NMR spectra. Your elemental analysis suggests that the powder probably has the right molecular formula to be methamphetamine. The spectra are displayed on the next page.
1. Is the suspicious compound actually methamphetamine? Briefly explain what evidence from the spectra supports your answer.
2. If it is not methamphetamine, what is it? Briefly explain how the spectra fit your proposed structure.
C18H20
CH3
NH CH3
methamphetamine
Practice Final – Not for Marks Page 11 of 12
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
AmideEster
Ketone, Aldehyde
Carbox. Acid
Practice Final – Not for Marks ANSWER KEY Page 1 of 17
Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
PRACTICE “FINAL EXAM” – Winter 2002 This is a sample of what to expect on the Final Exam for the course. The format of the final will be EXACTLY the same as this one. The distribution of the questions will be similar, and their difficulty will be the same or easier. Give yourself 3 hours, using only your “cheat sheet” and lots of scrap paper.
PART I: Question 1 (20 Marks)
Question 2 (4 Marks)
Question 3 (10 Marks)
Question 4 (12 Marks)
Question 5 (8 Marks)
Question 6 (8 Marks)
PART II: (8 Marks)
TOTAL: (70 Marks)
Practice Final – Not for Marks ANSWER KEY Page 2 of 17
PART I: DO ALL QUESTIONS – THERE IS CHOICE IN QUESTION 1 ONLY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions
to correctly complete TEN (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked.
(a)
O
O
H
O
H2 (g), Pd/CEthanol O
O
H
HH
(b)
OH
heat
O
(c)
O
NaBH3CN, EtOH
NH2 NH
(d)
O
NH2 O
OEtH2N
toluene (solvent)heat
C13H16N2O
N
N
H O
(e)
O
OCH3
O
H
1) CH3OH, H2SO4 (cat.)
2) LiAlH4, THF(H2O workup)
OH
O
H
O
CH3
H3C
(f)
Br(C6H5)3P
Ethanol
n-BuLiTHF
then addO
H
H
Practice Final – Not for Marks ANSWER KEY Page 3 of 17
(g)
N Br
AlCl3, CH2Cl2
NN
and/or
(h)
O
O
NaOEt, EtOHheat
C10H8O
O
(i)
NH
O
NH2
O
OCF3CO2H
NH2
O
NH2
NH3
O
NH2
O
O
CF3
or
(j)
OCH3
CH3CO2H, H2O2H2SO4 (cat.)
O OCH3
(k)
HOOCBr2, CH2Cl2
HOOC
Br
Br
plus enantiomer
(l)
CH3
O
CH3
O
CH3
(CH3CH2)2CuLiEther or THFor CH3CH2MgBr/CuIEther or THFthen H3O+ workup
(m)
OO
H3CO OCH3Br Br
2 NaOCH3CH3OH
H2SO4, H2Oheat
COOH
(n)
CHO
CHO(Two Steps)
O3, CH2Cl2thenZn/HOAc
Practice Final – Not for Marks ANSWER KEY Page 4 of 17
(o)
+ heatCO2CH3
CO2CH3
CO2CH3
H3CO2C
(p)
Hg(OAc)2H2O, THF
followed byNaBH4
OH
(q)
NO
CH3LiAlH4THF
followed byH2O workup
N
CH3
(r)
COOH COOH
Cl
Cl2, AlCl3 or FeCl3CH2Cl2
2. (4 MARKS) The final four steps of the synthesis of the alkaloid retronecine are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks.
N
O
O
CO2EtN
O
CO2EtHO
NOH
CO2EtHO
N
HO OH
NaOEt, EtOH
retronecineC10H15NO3
NaOEt/EtOHNaBH4, H2O or CH3CH2OH
N
CO2EtHOLiAlH4, THFthenH2O w/u
Practice Final – Not for Marks ANSWER KEY Page 5 of 17
3. (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation.
O HO CNCO2CH3
HCl, CH3OH
heat
KCN, H3O+
O
H3O
OH
CNHO CN
NB: KCN + H3O+ will create an equilibrium mixture containing a small proportion of CN
HO CN
HCl
H2O CNH
( + H )
( - H )
( - H2O )
C N
H
Note that the second reaction occurs in methanol, and that there is initially NO water present.
C N H
C N H
CH3OH
CN HO
H3C
H
( - H )
CN HO
H3C
H
CN
HO
H3C
H
H2O
CN
HO
CH3 H
OH
H( - H )
CN
HO
CH3 H
O
HH
CN
HO
CH3 H
O
H
H
CO
CH3O
H
( - NH3 )
CO
CH3O
( - H )
Practice Final – Not for Marks ANSWER KEY Page 6 of 17
4. (a) (2 MARKS) Tollens’ Test for aldehydes is carried out under alkaline conditions. The reagent is usually formed by first adding a small amount of aqueous NaOH to a solution of AgNO3, and then treating the mixture with aqueous NH4OH until a clear solution is obtained. Provide chemical equations showing the processes that occur at each step in the preparation of Tollens’ Reagent. Explain why ammonium hydroxide is used instead of simply adding more sodium hydroxide.
Silver hydroxide is insoluble in water, as are most metal hydroxides. The Tollens test requires alkaline conditions, however. In order to obtain a soluble source of silver ion, it is necessary to form the soluble di-amine complex of Ag+, using ammonium hydroxide (aqueous ammonia).
(b) (1 MARK) When distilling a liquid and attempting to record its boiling temperature, the bulb of the thermometer is placed in the neck of the distillation apparatus and not immersed in the liquid. Explain why this is important.
In a distillation, the idea is to separate compounds on the basis of their differing boiling points. The vapour composition and temperature will not in general match the liquid composition and temperature except in the case of a pure substance. The liquid temperature will reflect any non-ideality in the mixture being distilled. Thus, to obtain the boiling temperature of the material actually being collected at the condenser outlet, it is important to measure the vapour temperature.
(c) (1 MARK) When carrying out a distillation as described in part (b), an accurate boiling temperature will only be recorded when both vapour and condensing liquid are simultaneously present on the bulb of the thermometer. Explain why this is so.
The boiling point is by definition the point at which both liquid and vapour phases coexist at equilibrium. Unless both phases are present at the stillhead, it is impossible to determine whether one is actually at the boiling point.
(d) (2 MARKS) The catalyst (AlCl3 or FeCl3) used in Friedel-Crafts reactions must be dry and pure. Why is this important for the success of the reaction?
Water present in a Friedel-Crafts reaction will react with the carbocations produced, forming alcohol by-products.
2 AgNO3 (aq.) + 2 NaOH (aq.) 2 Ag(OH) (s) + 2 NaNO3 (aq.)
Ag(OH) (s) + 2 NH4OH [Ag(NH3)2 OH] (aq.) + 2 H2O
Practice Final – Not for Marks ANSWER KEY Page 7 of 17
(e) (2 MARKS) When synthesizing an azodye from a phenol and a primary aromatic amine, a cold, acidic solution of sodium nitrite and the amine is poured into a basic solution of the phenol. Why is it advantageous to use a basic solution of the phenol?
Most phenols are insoluble or only slightly soluble in water, whereas their anionic conjugate bases are quite soluble. The phenol will react quickly with the aryl diazonium salt formed in the acidic nitrite solution ONLY if it is in solution at the outset.
(f) (4 MARKS) In the Hinsberg test for amines, the following reaction is performed and the solubility of the product in aqueous NaOH is observed.
Amine + S
O
O
ClNaOH (aq.)
Product
• Primary amines: The product is soluble in aqueous NaOH. • Secondary amines: The product is insoluble in aqueous NaOH. • Tertiary amines: No new product is formed.
Draw the structures of the different products formed when the Hinsberg test is performed on ethylamine, diethylamine, and triethylamine, and use these structures to explain the solubility patterns noted above.
CH3CH2 NH2 S
O
O
HN
H3CH2C
CH3CH2NH S
O
O
N
H3CH2CH3CH2C
H3CH2C
CH3CH2N S
O
O
N
H3CH2CH3CH2C
H3CH2C
CH2CH3H3CH2C
H2O, NaOH
S
O
O
N
H3CH2C
H3CH2CH3CH2C O
NaThe sulfonamide formed from a primary amine and benzenesulfonyl chloride has a relatively acidic N-H bond, which is deprotonated in basic solution to form a soluble salt. Secondary amines form secondary sulfonamides that lack this N-H bond and that cannot be deprotonated. They are insoluble in aqueous media. Tertiary amines cannot form stable covalent compounds with sulfonyl chlorides – the product of nucleophilic addition to the sulfonyl chloride is readily hydrolysed in water.
Practice Final – Not for Marks ANSWER KEY Page 8 of 17
5. (8 MARKS) Simple alkenes are readily converted to epoxides using peroxycarboxylic acids, but are unaffected by alkaline hydrogen peroxide. In contrast, the C=C bonds of α,β-unsaturated ketones (enones) are epoxidized relatively slowly by peroxycarboxylic acids but are rapidly converted to epoxides by hydrogen peroxide under basic conditions.
Briefly explain this pattern of reactivity, using your knowledge of organic structures and reaction mechanisms. (HINT: the pKa of H2O2 is 11.62).
RR
R'CO3Hfast RR
O
R O
R
R'CO3HRO
R
Overy slow
H2O2, NaOH (aq.)
H2O2, NaOH (aq.)
fastR O
R
O
no reaction!
Peroxycarboxylic acids require relatively electron-rich alkenes in order to form epoxides. The pi cloud of the alkene must interact with the somewhat electron-poor terminal oxygen of the peroxyacid in order to set up the transition structure for the concerted epoxidation reaction. Alkaline hydrogen peroxide solutions contain the highly nucleophilic HOO– anion. This is because hydrogen peroxide is almost 2.5 pK units more acidic than water, and thus will be deprotonated by hydroxide ion. Isolated alkenes are highly unreactive towards nucleophilic reagents of this kind, and therefore will not be affected by alkaline hydrogen peroxide solutions. The C=C bond of an enone is quite electron-poor, because of the electron-withdrawing character of the ketone carbonyl group. This will make epoxidation of enones by electrophilic reagents like peroxyacids rather slow. On the other hand, nucleophiles can add in a 1,4 fashion to enones. This provides a pathway for HOO– to react.
(A detailed mechanism of the kind shown above would NOT be necessary for full marks. Your answer would have to identify the key fact that epoxidation of an enone by a nucleophilic
reagent must proceed by some kind of nucleophilic reaction.)
R
O
OOHR
O
O
OH
R
O
O
+ HO
Intermediate enolate attacksoxygen, displacine hydroxide
Practice Final – Not for Marks ANSWER KEY Page 9 of 17
6. (8 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C6H13NO are shown below.
4000 3000 2000 1500 1000 500
tr s m m tr
(b) Draw the structure of the compound in the
box at right (4 Marks).
(c) Clearly indicate which H atom(s) in your structure is(are) responsible for each of the signals in the 1H NMR spectrum (4 Marks).
O
NH
Practice Final – Not for Marks ANSWER KEY Page 10 of 17
PART II: CHALLENGE PROBLEM. DO ONE PROBLEM ONLY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks.
A. SYNTHESIS Propose a multi-step synthetic route to the compound shown at right. You may use benzene, diethyl malonate, and any other organic molecule having 4 carbons or fewer as starting compounds. For each step in your synthesis, provide all needed reagents, solvents and reaction conditions. You can employ any valid reagents and reactions in your synthesis. Mechanisms and discussion are not required.
B. MECHANISM The Hantsch Reaction provides a route to substituted dihydropyridine and pyridine compounds. It has been used in preparing a variety of pharmaceutical products. It is based entirely on processes and mechanistic steps that we have discussed in considerable detail during this course. Provide a detailed stepwise mechanism for the Hantsch reaction shown below.
OEt
O O
2 + H
O
NH3 (excess)EtOH
N
H
O
OEt
O
EtO
C. SPECTROSCOPY The RCMP have confiscated a shipment of chemicals destined for an organic chemist at the U of Manitoba, because they were tipped off that the package included a substantial amount of methamphetamine (“Speed” or “Crystal Meth”). You have been given the suspicious white powder to analyze. Unfortunately, your automated mass spectrometer with its computerized database of spectra of illicit drugs is broken this week, so you turn instead to chemical tests, IR, 1H and 13C NMR spectra. Your elemental analysis suggests that the powder probably has the right molecular formula to be methamphetamine. The spectra are displayed on the next page.
1. Is the suspicious compound actually methamphetamine? Briefly explain what evidence from the spectra supports your answer.
2. If it is not methamphetamine, what is it? Briefly explain how the spectra fit your proposed structure.
C18H20
CH3
NH CH3
methamphetamine
Practice Final – Not for Marks ANSWER KEY Page 11 of 17
Practice Final – Not for Marks ANSWER KEY Page 12 of 17
Answers to “Challenge Problems”
Synthesis This synthesis problem is actually a bit longer than I had intended, so it definitely is harder than what you will face on the “real exam”. There are several ways you could break this target down retrosynthetically, but perhaps the simplest is the convergent scheme shown below. This scheme uses only a few types of reactions:
The preparation of benzophenone first requires the syntheses of bromobenzene and benzaldehyde.
These are then assembled in another Grignard reaction, followed by oxidation:
The ketone is easily made by a Grignard reaction between isopropylmagnesium bromide and acetaldehyde. This synthesis is directly out of Chapter 15 in the text.
The final assembly of the target involves the Aldol condensation between the KINETIC enolate of our ketone and the benzophenone, followed by deoxygenation of the ketone.
O
O
O
(kinetic enolate)
+
MgBr
+O
H
MgBr
+
CHO
benzene
BrMg, ether
then add CH3CHO(H3O+ w/u)
OH CrO3pyridine
O
OLDATHF, -78 oC
OLi O
O
H2NNH2KOHheat
Br2, FeBr3
CH2Cl2
Br
Mg, Etherthen CH2O
(H3O+ w/u)
OH
CrO3pyridine
OH
Br
Mg, Ether
then add benzaldehyde
(H3O+ w/u)
OH O
CrO3pyridine
Practice Final – Not for Marks ANSWER KEY Page 13 of 17
Mechanism The Hantsch reaction is an application of the Acetoacetic ester synthesis, combined with the chemistry of imines and enamines. Since ammonia is a basic reagent, the processes are all occurring under basic conditions, although protonated ammonia (NH4+) will be available as a proton source when needed, as will the alcohol solvent. If you look at the product, you will see that two molecules of acetoacetate form the dihydropyridine ring, with the phenyl group and one carbon of the ring coming from benzaldehyde. Nitrogen has been introduced at the site of the ketone group in the two acetoacetate fragments. In part one of the mechanism we see attack of an acetoacetate nucleophile on the benzaldehyde
carbonyl, with elimination to form the conjugated ketoester. This process is known as the Knoevenagel condensation. A second enolate then carries out a 1,4 addition (the Michael reaction) leading to the product shown at the bottom of this scheme.
OEt
O O
2 + H
O
NH3 (excess)EtOH
N
H
O
OEt
O
EtO
OEt
OO
H H
NH3
OEt
OO
HO
H
OEt
OO
HO
NH4
OEt
OO
HOH
NH3OEt
OO
EtO
O O
H
OEt
OO
EtOO
OH
(NH4)
OEt
OO
EtOO
OH
Go to next page!
Practice Final – Not for Marks ANSWER KEY Page 14 of 17
In the next steps, we have to introduce the nitrogen. There are a couple of ways we could formulate this process, and either would be acceptable. The first way is based entirely on imine chemistry. While this is fairly straightforward, it does involve a lot of steps.
EtO
O O
EtOO
OH
NH3 EtO
O O
EtOO
OH
N
HH
H
EtO
O HO
EtOO
OH
N
H
H
EtO
O
EtO
O
O
H
NHH
EtO
O
OEt
O
O
H
NH2
HNH3EtO
O
OEt
O
OH
N
H
H
( - H / + H )
EtO
O
OEt
O
OHH
NH
NH3 EtO
O
OEtO
NH
Practice Final – Not for Marks ANSWER KEY Page 15 of 17
We could also write a mechanism based on conjugate additions to enols and enamines. In fact, this is probably a better formulation of this process, because beta ketoesters exist with a very high proportion of enol due to the conjugation that occurs when a C=C group is present. As you can see, this way of viewing the mechanism is also shorter.
EtO
O OH
EtOO
OH
NH3 EtO
O HO
EtOO
OH
N
HH
H
EtO
O HO
EtOO
OH
N
H
H
OEt
O
OEtO
OH
NH2EtO
O
OEtO
OH
N
H
H
EtO
O
OEtO
NH
Base
EtO
O
OEtO
OH
NH
Practice Final – Not for Marks ANSWER KEY Page 16 of 17
Spectroscopy The first question is: “Is Dr. Hultin busted?” The spectra indicate that the answer is NO – the mystery powder is not methamphetamine. The strongest evidence of this is in the 1H NMR. If it were methamphetamine, there would be 5 different aliphatic signals, in the ratio of 1:1:2:3:3. There would be a 3H singlet at around 3 ppm for the N-CH3 group, and a 3H doublet at about 1.1 ppm for the C-CH3 group, which has one nearest-neighbor hydrogen. The benzylic CH2 group would be a doublet at somewhere around 2.5 ppm, and the C-H grouping would appear as a complex multiplet signal near 3 ppm. The 1H spectrum we have here is far too simple to be methamphetamine. So, the second question is: “What is the mystery powder?” If we accept that it is an isomer of methamphetamine, as indicated in the question, we have a good starting point. The 13C NMR shows 4 aromatic C atoms and 3 different aliphatic C atoms, two of which are considerably downfield of the other. The 1H NMR shows 4 different kinds of protons, in a ratio of 5:2:2:6. Clearly the 5H signal at about 7.3 ppm is due to the aromatic protons, indicating a monosubstituted benzene. The 6H singlet at about 1.1 ppm is good evidence of two identical CH3 groups, with NO nearest neighbour hydrogens. The other 2H singlet, at about 2.8 ppm is probably a CH2 group with no nearest neighbour hydrogens that can cause coupling, but this grouping must be attached to a polar atom because of its chemical shift. There is only one polar atom in this molecule – nitrogen. The IR clearly shows a strong N-H band, and there remains a 2H signal in the 1H NMR spectrum. This suggests the presence of an NH2 group. So, we have the following fragments, indicating the structure shown.
R H3C
C
CH3
R R
CH2
NH2C
R
R
R
NH2
H3C CH3
But, there are only 3 different kinds of aliphatic carbons, according to the 13C NMR spectrum.
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
AmideEster
Ketone, Aldehyde
Carbox. Acid