Synthetic Organic Chemistry Final Exam (6KM33)
Tuesday, 8th April, 2014, 9:00–12:00 (3 h)
This is an open-book written exam. The course textbooks (Warren & Wyatt, 2nd Ed.)
with personal notes and worked exercises, one good general undergraduate textbook
(e.g. Organic Chemistry, 7th Ed. Bruice) and a copy of the literature assignment with
personal notes are all permitted. Additional books and material are also allowed (but
not necessary) EXCEPT access to the internet is STRICTLY FORBIDDEN. There are
four questions in total including one bonus question. The first three questions are
compulsory. The bonus question is optional.
Maximum no. points per question, and recommended time are highlighted in yellow.
Maximum no. points for exam: 130 points
Explain your answers clearly.
You may write in English or in Dutch.
GOOD LUCK!
Question 1.(48 points, 1 h) In this course you have synthesized numerous different alcohol derivatives. a) Give examples of five different reactions, which result in the formation of an alcohol functional group (e.g. see Question 2, part b). For each reaction, you do not need to write mechanisms; simply write clearly the structures of the starting material(s), product(s) and any other necessary reagents.(15) ‐ 1 point starting material. ‐ 1 point reagent. ‐ 1 point product. ‐ e.g (but not exclusively): i) BH3
ii) H2O2/KOH(aq)
EtMgBr
OH
O
OH
r ac
NaBH4
OH
O
OsO4
H 2O
N2
OH
OH
r ac
OH
Alcohols are extremely versatile building blocks. b) For the above four syntheses, each starting from alcohol 1, draw the structure of the intermediates after each step, including the most likely product(s) A and C, and suggested, B and D. Where applicable correctly assign all stereo‐ and geometric isomers.[Note: For B and D, more than one synthetic step may or may not be necessary.](15) ‐
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‐
‐
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Approximate marking scheme (at examiner’s discretion): 1 point per correct intermediate and final structure (A & C). 1 point per reagent, 1 point per correct intermediate (B & D). Max. 15 points. e.g (but not exclusively): i) (COCl)2,
DMSO, NEt3
OH
ii) PhMgBr
chiral racemic
O
OH
A
1
B
i) TsCl, Py
ii) NaN3
N3
OTs
OH
1
Br
ii) PPh3
i) PBr3
Br
OH
PPh3
1
O
Br
Z-isomer
iii) KOtBu
PPh3
PPh3
C
D
i) NaCN
ii) Pd/C, H2
OTs
NH2
N
iii)
O
H
N
NH2
then NaBH4
Likewise, simple and inexpensive aromatic building blocks, such as benzene, methoxybenzene (anisole) and toluene are quickly elaborated into a diverse range of richly functionalized benzene AND cyclohexane derivatives. e.g. compound 2, an experimental analgesic, can be disconnected to 3 and 4. [Note: for parts 1c‐e, write clearly all intermediate structures, and suggested reagents. You do not need to provide mechanisms.] c) Propose a synthesis of methoxybenzene starting from benzene?(5) ‐
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‐
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1 point per correct intermediate. 1 point per reagent. Max. 5 points. e.g (but not exclusively): d) Propose a synthesis of 3 and 4 both starting from either benzene or toluene.(10) ‐
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‐
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1 point per correct intermediate. 1 point per reagent. 5 points per molecule, 3 and 4. Max. 10 points. e.g (but not exclusively): e) How would you synthesize 2 from 3 and 4?(3) ‐
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‐
‐
‐
‐
‐
See above. 1 point per correct intermediate. 1 point per reagent. Any form of carboxylic acid activation would do (HATU, DCC etc.) Protection of alcohol is not necessary; reaction can go chemoselective for the amine. Max. 3 points. e.g (but not exclusively): Question 2.(50 points, 1 h) α,β‐unsaturated carbonyls are “chemical chameleons”, which change their reactivity according to their chemical environment. a) Give examples of five different ways to make an α,β‐unsaturated carbonyl functional group in no more than 2 steps. For each reaction sequence you do not need to write mechanisms; simply write clearly the structures of the starting material(s), product(s) and any other necessary reagents.(15) ‐ 1 point starting material. ‐ 1 point reagent. ‐ 1 point product. ‐ e.g (but not exclusively): α,β‐unsaturated ketone (enone) 5 reacts with Grignard reagents at the carbonyl carbon e.g. 1,2‐addition of n‐BuMgBr to form alcohol 6, or with organolithium cuprates at the β‐carbon, e.g. 1,4‐Michael addition of methylcuprate en route to ketone 7. b) Provide a detailed curly arrow mechanism for the synthesis of 7?(3) ‐ 1,4‐Michael addition of methyl cuprate to 5.(1 point) ‐ Formation of thermodynamic enolate intermediate stabilized by lithium ion.(1 point) ‐ “Quench” enolate with BnBr, leading to alkylation α to carbonyl AND to methyl group.(1 point) c) How could you selectively synthesize products 8‐10 starting from enone 5? Suggest suitable reagents A‐C.[Note: Multiple synthetic steps may be required. Write clearly all intermediate structures, and suggested reagents. You do not need to provide mechanisms.](12) ‐ General markign scheme: ‐ 1 point per correct intermediate. ‐ 1 point per reagent. ‐ Max. 12 points distributed evenly across all three molecules. ‐ e.g (but not exclusively): α,β‐unsaturated carbonyls also behave as dienophiles in Diels‐Alder reactions. d) Perform Diels‐Alder disconnections on the following five compounds, 11‐
15, draw the corresponding diene and dienophile reagents.(10) ‐
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1 point per correct diene. 1 point per correct dienophile. Max. 2 points per molecule. 10 points in total. e) Propose a synthesis of 16 starting from valerolactone. You do not need to provide reaction mechanisms; simply write clearly the structures of the intermediates formed and reagents used.(10) ‐
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In general: 1 point per correct reagent. 1 point per correct intermediate. Max. 10 points per molecule. 10 points in total. Clearly, lots of ways to go about synthesizing this molecule. Use flexible marking scheme. One possible route could be: Question 3. (based on literature assignment).(32 points, 45 mins) [Citation: Breitler & Carreira, Angew. Chem. Int. Ed., 2013, 52, 11168.] NOTE: for this question the compound numbers are identical to those used in the citation! With reference to Scheme 3: a) Compounds 6 and 7 are not commercially available. Propose a synthesis of 6 and 7 from A and B, respectively?(7) ‐
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In general: 1 point per correct reagent. 1 point per correct intermediate. Max. 2 points for 6 and 5 points for 7. Max. points = 7. Literature routes used are: b) Draw a detailed curly arrow mechanism for the formation of 8 from 6 and 7 over two steps.(5) ‐ Step a ‐ Deprotonation of 6 α to ester group (1 point) and 1,4‐Michael addition at enone 7.(1 point) ‐ Deprotonation of α to ketone and 1,2‐addition at aldehyde to form the six‐membered ring.(1 point) ‐ E1cb elimination of water.(1 point) ‐ Step b ‐
Krapcho decarboxylation of tert‐Butyl ester functional group.(1 point) c) What would be the expected product be if compound 8 were reacted under catalytic hydrogenation conditions (e.g.Pd/C/H2)? Explain therefore why the authors chose to use Na/NH3(l)/EtOH instead.(2) ‐
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Catalytic hydrogenation conditions would reduce all unsaturated bonds.(1 point) Birch reduction conditions (Na/NH3(l)/EtOH) are therefore more chemoselective than catalytic hydrogenation.(1 point) d) Draw the structure of the compound formed after step e, and provide a detailed curly arrow mechanism for its formation.(3) ‐
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Formation of thermodynamic trisubtituted enol form of ketone functional group.(1 point) Nucleophilic attach of enol 1,2 at the aldehyde functional group to form the bicyclic ring system.(1 point) Correct structure above.(1 point) e) Briefly discuss the different roles played by EtOH in steps c and g.(2) ‐
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In step c, EtOH functions as a proton source for the Birch reduction of the enone to the ketone.(1 point) In step g, EtOH functions as a co‐solvent to increase the solubility of 10 in water/EtOH thus facilitating enzymatic formation of (‐)‐5.(1 point) With reference to Scheme 3, step h: f) Briefly explain the stereochemical outcome for this reaction.(2) ‐ Kinetic explanation: axial attack is slower than equatorial due to steric encumbrance of the axial hydroxyl group.(1 point) ‐ Thermodynamic explanation: ‘Equatorial’ product is thermodynamically more stable than axial.(1 point) ‐ Grigard is a hard nucleophile, therefore kinetic explanation is likely to be more applicable in this case. g) Why do you think that an early attempt at this reaction using analog C (below) instead of (‐)‐5 did not lead to the anticipated product D?(1) ‐
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One possible explanation could be that the Grigard reagent prefers to function as a base and deprotonate α to the ketone instead of reacting 1,2. The bulky TBS group prevents coordination build‐up of negative charge as a consequence of 1,2‐Grignard addition, which would normally be stabilized through coordination of the magnesium cation (which is more favored in the presence of the alcohol, but less favored in the presence of the OTBS group). This or any other similarly well‐reasoned answer will score 1 point (at examiner’s discretion). h) Why was it necessary to use 2,2 equivalents of 2‐isopropenylMgBr for this reaction?(1) ‐
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1st equivalent is needed to first deprotonate the alcohol. 2nd equivalent then reacted 1,2 at the carbonyl group. i) Briefly explain how adding lanthanide salt LaCl3.2LiCl to the reaction improved the yield of 11.(2) ‐
Trans‐metallation with lanthanum results in a less basic carbon‐centred nucleophile (1 point) and therefore increased selectivity for 1,2‐
addition over α‐deprotonation.(1 point) With reference to Scheme 5: j) Explain with as much mechanistic detail as possible what is happening in step c, which results in the formation of the natural product (+)‐
Crotogoudin from 19.(7) ‐ Protection of lactone as triisopropylsilyl ketene acetal using TIPSOTf and NEt3.(1 point) ‐ Increase chemoselectivity. Lactone was found in earlier studies to react with electrophiles in preference to the ketone.(1 point) ‐ α‐deprotonation of ketone with LiHMDS (1 point) and quench with Eschenmoser’s salt (CH2=NMe2+I‐).(1 point) ‐ Hofmann elimination (MeI, then basic Al2O3).(1 point) ‐ Two additional points for curly arrow mechanism to illustrate the ‐ Once again, any well‐reasoned answer will score points Bonus Question. Total synthesis of Spongistatin 1. Spongistatin 1 is a marine macrolide natural product, which has generated considerable interest in the field due to its potent antitumour properties and outstanding structural beauty. In one reported synthesis of 1, the CD spiroketal fragment 2 was synthesized from chiral enantiomerically pure epoxides 3 and 4. Propose a synthesis of 2 from 3 and 4 ignoring the stereochemistry at the C23 anomeric carbon. ‐ This is optional. ‐ Any attempts at a sensible answer will score points.