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Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
FINAL EXAM – Winter 2003 Paper Number 546
Thursday April 24, 2003 6:00 – 9:00 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 are also permitted but no other aids may be used. The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam paper itself.
An answer to ONE of the “Challenge Questions” in PART II may be written in the exam booklet for EXTRA CREDIT.
PART I: Question 1 (20 Marks)
Question 2 (10 Marks)
Question 3 (15 Marks)
Question 4 (15 Marks)
Question 5 (10 Marks)
SUB-TOTAL: (70 Marks)
PART II (EXTRA CREDIT) (7 Marks)
TOTAL:
Page 2 of 13
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 of the following reactions. All reactions do in fact lead to products. Clearly indicate which TEN responses you want marked!
(a)
Br
1) Br2, AlBr3 CH2Cl22) Sn, HCl
(b)
H
H
HH
(c)
O
Br
LDA, THF, -78 oC then
(d)
O
O
O
(e)
Hg(OAc)2 (cat.)H2SO4 (aq.)
Page 3 of 13
(f)
O
OBr2, AlBr3
CH2Cl2
(g)
NH
O
LiAlH4, THF
(H3O+ workup)
(h)
O O
OMeMeO
NaOMeMeOH
(i)
OO
O
OMeOO
OH
reagent, solvent, workup
(j)
OHMeO
HBr, CH2Cl2
Page 4 of 13
(k)
O O
(l)
N
O
H
C6H5NH2, Ether
(m)
O
O OHCOOH
(n)
NH2 1) NaNO2, HCl (aq.)
2) H2SO4 (aq.) heat
(o)
O
O
LDA, THF-78 oC
then heat(H3O+ workup)
Page 5 of 13
2. (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from ethyl acetoacetate (B). You may use any additional organic starting materials having three or fewer carbons, as well as any reagents or solvents you require. This transformation can be accomplished in fewer than 5 steps. Remember that although a retrosynthetic analysis may be useful to you in solving this problem, the answer I want is the “forward synthesis” complete with reagents and products.
O OO
OEt
A B
starting fromMake
Page 6 of 13
3. (7 MARKS) Provide a detailed stepwise mechanism for the following transformation.
O
O
OH
O
O
O
OO
NaOCH3CH3OH
+
Page 7 of 13
(b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide a detailed stepwise mechanism for this reaction.
CH3
OOO
O
O
CH3
+NaOH
H2O, 95 oC
Page 8 of 13
4. (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'-di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in CH2Cl2, using FeCl3 as a catalyst. Explain how A could be formed under these conditions.
(b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation reactions with acetone (C). The reaction of B under these conditions is much faster than that of A. Briefly explain why this is so, taking into account electronic effects.
+ ClFeCl3
CH2Cl2+
CH H
A - trace impurity
major product
H3CO
O
HO2N
O
H H3C CH3
O
A B C
Page 9 of 13
(c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2 advantages of using reflux conditions in a chemical reaction.
Page 10 of 13
5. (10 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C7H15N are shown below. NB: In the 1H NMR spectrum, m=multiplet, d=doublet, tr=triplet, q=quartet.
4 0 0 0 3 0 0 0 2 0 0 0 1 5 0 0 1 0 0 0 5 0 0
m m q d tr
Draw the structure of the compound in the box below.
Page 11 of 13
PART II: CHALLENGE PROBLEM (EXTRA CREDIT).
DO ONE PROBLEM ONLY. Part II is worth up to XX additional marks. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Insert the booklet into Part I when you hand it in.
A. SYNTHESIS There is something seriously wrong with the following proposed synthesis of the drug terfenadine. Identify the problem, and propose specific steps needed to make this route workable.
N
OH
Br
H
O N
OHOH
Mg
Ether
terfenadine
B. MECHANISM When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism to explain this transformation.
C. SPECTROSCOPY The unusual compound diketene has the molecular formula C4H4O2. Its 1H and 13C NMR spectra are shown on the next page. Heating diketene in the presence of cyclopentadiene leads to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene?
O
Diketene + heat
O OHH3O+
100 oC
A B
Page 12 of 13
PPM 150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0
PPM 4.90 4.80 4.70 4.60 4.50 4.40 4.30 4.20 4.10 4.00 3.90 3.80
PPM 4.900 4.896 4.892 4.888 4.884 4.880 4.876 4.872 4.868 4.864 4.860 4.856 4.852 PPM 4.504 4.502 4.500 4.498 4.496 4.494 4.492 4.490 4.488 4.486 4.484
PPM 3.906 3.904 3.902 3.900 3.898 3.896 3.894 3.892 3.890 3.888 3.886
Diketene 13C NMR Four signals, δ 165.18, 147.66, 87.06 and 42.40 ppm.
Diketene 1H NMR Three signals, δ 4.877, 4.494 and 3.897 ppm.
Double triplet Double triplet
Double doublet
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
ANSWER KEY Page 1 of 15
Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
FINAL EXAM – Winter 2003
Paper Number 546
Thursday April 24, 2003 6:00 – 9:00 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 are also permitted but no other aids may be used. The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam paper itself.
An answer to ONE of the “Challenge Questions” in PART II may be written in the exam booklet for EXTRA CREDIT.
PART I: Question 1 (20 Marks)
Question 2 (10 Marks)
Question 3 (15 Marks)
Question 4 (15 Marks)
Question 5 (10 Marks)
SUB-TOTAL: (70 Marks)
PART II (EXTRA CREDIT) (7 Marks)
TOTAL:
Page 2 of 15
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 of the following reactions. All reactions do in fact lead to products. Clearly indicate which TEN responses you want marked!
(a)
Br
O 1) Br2, AlBr3 CH2Cl22) Sn, HCl
(b)
H
H
HHCH2I2, Zn, CH2Cl2
(I-CH2-Zn-I)
(c)
O
Br
OLDA, THF, -78 oC then
(d)
O
O
OmCPBA, CH2Cl2
or any other specific peroxyacid reagent
(e)
OHg(OAc)2 (cat.)H2SO4 (aq.)
Page 3 of 15
(f)
O
OBr2, AlBr3
CH2Cl2 O
OBr
ortho isomer is also acceptable, although this would be the minor product
(g)
NH
O
NHLiAlH4, THF
(H3O+ workup)
(h)
O O
OMeMeO
OO
OMe
NaOMeMeOH
(i)
OO
O
OMeOO
OH
reagent, solvent, workup
2 PhMgBr, Etherw/u with aq. NH4Cl
(j)
OHMeO MeO
Br
HBr, CH2Cl2
Page 4 of 15
(k)
O O
(CH3CH2)2CuLiTHF, -78 oC(acid w/u)
You could also use CH3CH2MgBr/CuI (cat.)/Ether/Acid workup
(l)
N
O
H
C6H5NH2, Ether
O
Cl
O
O
Oor
(m)
O
O OHCOOH
NaOHH2O
(n)
NH2 OH1) NaNO2, HCl (aq.)
2) H2SO4 (aq.) heat
(o)
O
O
O
OH
LDA, THF-78 oC
then heat(H3O+ workup)
Page 5 of 15
2. (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from ethyl acetoacetate (B). You may use any additional organic starting materials having three or fewer carbons, as well as any reagents or solvents you require. This transformation can be accomplished in fewer than 5 steps. Remember that although a retrosynthetic analysis may be useful to you in solving this problem, the answer I want is the “forward synthesis” complete with reagents and products.
O OO
OEt
OO
OEt
OO
OEt
OO
OEt
OO
OEt
O
A B
starting fromMake
NaOEtEtOH
CH3CH2Br
NaOEt
EtOHCH3Br
NaOEtEtOHCH3CH2Br
orLDA, THF -78 oCCH3CH2Br
1) NaOH (aq)2) H3O+
orH3O+, heat
( and CO2 and EtOH)
( The Acetoacetic Ester synthesis of ketones)
Page 6 of 15
3. (7 MARKS) Provide a detailed stepwise mechanism for the following transformation.
O
O
OH
O
O
O
OO
O
O
OH
O
O
O
OH
O OMe
OH
O
O
OMe O
OH
O
O
OMe
O
O
O
OMeO
O
O
OMe
O
O
O
HMeO
NaOCH3CH3OH
+
+
OMe
OMe--
-
--
-
Page 7 of 15
(b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide a detailed stepwise mechanism for this reaction.
CH3
OOO
O
O
CH3
CH3
OOH CH3
OO
OO
OO
O
OO
H
O
OO
O
O
CH3
O
O
O
CH3
OHH
O
O
CH3
+NaOH
H2O, 95 oC
-OH-
-H2O
HO-
-
-
H2O
HO -
Page 8 of 15
4. (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'-di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in CH2Cl2, using FeCl3 as a catalyst. Explain how A could be formed under these conditions.
(b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation reactions with acetone (C). The reaction of B under these conditions is much faster than that of A. Briefly explain why this is so, taking into account electronic effects.
+ ClFeCl3
CH2Cl2+
CH H
A - trace impurity
major product
The CH2 grouping comes from the solvent, CH2Cl2. Dichloromethane is not very prone to reaction of this kind, but it may undergo reaction with Lewis acids to a slight extent, especially if the reaction is left for an extended period of time. This is probably what happened here.
CH
H
Cl
ClFeCl3
CH
H
Cl
Cl FeCl3δ
δ Cl CH2 FeCl4
Cl CH2Cl
H2CH
ClH2C
and then do the Friedel-Crafts alkylation over again with the other phenyl group to form the product
H3CO
O
HO2N
O
H H3C CH3
O
A B C
These condensation reactions are Aldol processes, in which the enolate of acetone attacks the aldehyde. The rates of the two processes will depend on the degree of electrophilicity of the two aldehydes. Since B issubstituted with a strong electron-withdrawing group, whereas A carries a strong electron-donating group, the carbonyl group of B will have a greater degree of partial positive charge. Thus, reaction of B will be faster.
O2NO
HO2N
O
HO2N
O
H
O2NO
HO2N
O
H
These resonance forms show that the nitro group can directly withdraw electron density from the carbonyl.
Page 9 of 15
(c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2 advantages of using reflux conditions in a chemical reaction.
Water hose IN
Water hose OUT
Conden
ser
Flask
Boiling chipsLiquid
Clamp attached to stand
Heat source
Top of condenser is OPEN!
Advantages: 1) Reflux apparatus keeps solvent (and possibly also reagents and
products) from boiling away by recondensing it (them) and returning the condensate to the flask.
2) Refluxing maintains a steady reaction temperature, which is the boiling temperature of the solvent used.
Page 10 of 15
5. (10 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C7H15N are shown below. NB: In the 1H NMR spectrum, m=multiplet, d=doublet, tr=triplet, q=quartet.
4 0 0 0 3 0 0 0 2 0 0 0 1 5 0 0 1 0 0 0 5 0 0
m m q d tr
Draw the structure of the compound in the box below.
NCH2
H3C
H3C
Page 11 of 15
PART II: CHALLENGE PROBLEM (EXTRA CREDIT).
DO ONE PROBLEM ONLY. Part II is worth up to 7 additional marks. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Insert the booklet into Part I when you hand it in.
A. SYNTHESIS There is something seriously wrong with the following proposed synthesis of the drug terfenadine. Identify the problem, and propose specific steps needed to make this route workable.
N
OH
Br
H
O N
OHOH
Mg
Ether
terfenadine Answer: You cannot form a Grignard reagent from a halide that also contains an alcohol group. The solution to the problem is to protect the alcohol with a group that is unaffected by Grignard formation or reaction. This group must be removed from the final product. Protection/deprotection is not necessarily straightforward, however. A benzyl ether could be easily installed in the starting material, which would permit Grignard formation. This answer (with appropriate reagents specified) would be acceptable for 6 out of 7 marks. (NB: making an ester would NOT be suitable, because you can’t make a Grignard that contains a carbonyl group either).
N
OH
BrBr
NaHTHF
N
O
Br
OH group incompatible withGrignard formation
Benzyl ether protects OH, will notinterfere with Grignard formation
However, in fact the removal of the benzyl group from the final product would likely damage the product. Hydrogenation will cleave benzylic C-O bonds, but there are actually two such groups in this molecule. We do not want to remove the existing hydroxyl group!
N
OOH
N
OH
H H
H2 (g)Pd cat.
MeOH
You can also cleave benzylic ethers using very strong acids, particularly HBr. However, this would certainly lead to elimination reactions.
N
OOH
NHBr
Either or both of the hydroxylgroups would eliminate
The 7th bonus mark would be given for commenting on the problem of deprotection. I don’t require you to have a solution to this problem. You could get around the difficulty by completely re-designing the synthesis. That is actually not what the problem asked for, but I will evaluate any proposed alternates on their merits.
Page 12 of 15
B. MECHANISM When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism to explain this transformation.
O OH
O
H3O+
O
H
O
H
H
O
H3O+
OH
H
H
OH
H
O
H
O
H
O
H3O+
100 oC
A B
+
+
There are actually a couple of ways you could have written this process. It requires onlya series of protonation/deprotonation steps to move the double bonds around.
+
( + H+ ) ( - H+ )
( + H+ )
( - H+ )
(1,2-shift)
+
+
The 1,2-shift step could also be written as an elimination/re-protonation sequence.
Page 13 of 15
C. SPECTROSCOPY The unusual compound diketene has the molecular formula C4H4O2. Its 1H and 13C NMR spectra are shown on the next page. Heating diketene in the presence of cyclopentadiene leads to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene?
O
O
O
H H
H
H
O
O
H H
H
H
OC
H H
Diketene + heat
Diketene has the structure:
As its name implies, diketene is a dimer of the molecule ketene, andthey are interconverted by an electrocyclic reaction.
2
It is the monomer that undergoes a Diels-Alder reaction with cyclopentadiene. You could see that if you worked backwards from the product in the reactionshown.
The two vinylic hydrogens are double triplets. This is because each isdistinct, splitting the other into a doublet, which is then split by the twoother hydrogens (which are both equivalent). The CH2 group is adouble doublet because the signals are split by the two different vinylichydrogens.
The 13C NMR is harder to interpret. The signal at 165 ppm is the carbonylcarbon, and the one at 42 ppm is the CH2 group. The alkene portion of thestructure is a bit strange. The carbon in the ring is at 147 ppm, but the terminalcarbon is upfield at 87 ppm. This spectrum really just told you that all 4 carbonswere different, and that one was probably a carbonyl
Page 14 of 15
PPM
PPM
PPM 4.900
Diketene 13C NMR Four signals, δ 165.18, 147.66, 87.06 and 42.40 ppm.
150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0
Diketene 1H NMR Three signals, δ 4.877, 4.494 and 3.897 ppm.
4
4.896
PPM 3.906 3.904 3.902 3.900 3.898 3.896 3.894 3.892 3.890 3.888 3.886
Double triplet.90 4.80 4.70 4.60
4.892 4.888 4.884 4.880 4.876 4.872 4.868 4.864 4.860 4.856 4.852 PPM 4.504
Double triplet
4.50 4.40 4.30 4.20 4.10
4.502 4.500 4.498 4.496 4.494 4.492 4.490 4.488 4.486 4.484
Double doublet
4.00 3.90 3.80
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
