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123. 312 Advanced Organic Chemistry: Retrosynthesis
Tutorial
Question 1.
Propose a retrosynthetic analysis of the following two compounds. Your answer should include both the synthons, showing your thinking, and the reagents that would be employed in the actual synthesis. Compound A O Answer: O FGI dehydration O OH C–C aldol OH O
O O Remember that a conjugated double bond can easily be prepared by dehydration, thus we can perform an FGI to give the aldol product. The 1, 3 - diO relationship should make spotting the disconnection very easy. Of course, in the forward direction the reaction is not quite that simple; we have two carbonyl groups so we must selectively form the correct enolate but this should be possible by low temperature lithium enolate formation prior to the addition of cyclohexanone. The aldol condensation is such a common reaction that it is perfectly acceptable to do the following disconnection: O (^) O
O O Compound B
O O Answer O O O O
OH O HO FGI reduction O O HO O O HO C–C
O EtO O OEt O 1 2 3 4 5 The first disconnection should be relatively simple, break the C–O bond to give the acid and alcohol. The next stage might be slightly tougher…your best bet is to look at the relationship between the two functional groups; it is 1 ,5. This can be formed via a conjugate addition of an enolate. To do this we need two carbonyl groups so next move is a FGI to form the dicarbonyl. Two possible disconnections are now possible depending on which enolate we add to which activated alkene. The one I have drawn is simpler, diethyl malonate is commercially available as is the enone (or it can be prepared by the self- condensation of acetone). Additionally, conjugate addition of malonates prefers 1 , 4 to 1, 2 addition, which can be an issue with simple carbonyls. Chemoselectivity in the reduction step is not an issue; NaBH 4 does not reduce esters. O EtO O OEt O (^) base O EtO 2 C CO 2 Et NaBH 4 O O EtO 2 C H+, H 2 O O O
Question 2.
Give the retrosynthetic analysis for the following three compounds. Pay special attention to the relationship between the functional groups. CO 2 H CO 2 H CO 2 H Answers: The first is the easiest; it is an !,"-unsaturated compound so we are looking at either aldol condensation or a simple Wittig reaction. Sometimes you will see double bond disconnections drawn with a double charge synthon…I’m not convinced it helps but if it allows you to rationalise what is going on more readily then use it!
H N FGI reduction N (^) C=N O H 2 N The next isn’t much harder…we have an alcohol, this should yell Grignard addition to a carbonyl and hence the disconnections are: OH O C–C BrMg^ Br This one is potentially a little harder…but not much. The best route to the acid is via alkylation of diethyl malonate. The latter is easily enolised, will only undergo two additions, is fairly robust yet will readily undergo decarboxylation. CO 2 H (^) FGI decarboxylation EtO 2 C CO 2 Et C–C Br Br EtO 2 C CO 2 Et The final compound is a primary amine. This could either be prepared by reductive amination of the appropriate ketone (made from oxidation of the secondary alcohol made earlier) or by substitution of an appropriately derivatised secondary alcohol (tosylation of the secondary amine) with azide followed by reduction. NH (^2) FGI reduction NH (^) O C=N NH (^2) FGI N 3 OTs OH reduction C–N C–S
Question 4.
Perform the retrosynthetic analysis of the following compound. Remember, your planned synthesis must be synthetically possible and shouldn’t suffer from regio- or chemoselectivity issues. O O NEt 2 O NH 2 Answer
O O NEt 2 O NH 2 O OH O NH 2 C–O FGI reduction O OH O NO 2 C–O O OH OH NO 2 O OH OH FGI C–N diazonium O OH N 2 +BF 4 – FGI diazonium O OH NH 2 FGI reduction O OH NO 2 C–N O OH remove the ester (with the reactive functionality) the^ amine^ can^ be^ problematic so we convert it to the less reactive nitro group. This also prepares the way for its eventual disconnection we can now remove the ether. Attepts to do this earlier would have met with failure due to alkylation of the amine the phenol group is ortho, para directing but should favour the leaser hindered position (and we might be able to argue about H-bonding) I would stop here as I don't know how much aromatic chemistry you have done. But if you have done enough then we can take the synthesis all the way back to the acid remember, the acid is electron withdrawing so is meta directing
Question 5.
(a) How would you synthesise From Answer: i. BH 3 ii. H 2 O 2 / NaOH OH PBr 3 Br Remember, we need to get anti-Markovnikof addition of the hydroxyl group so we use hydroboration / oxidation.
HO Answer: HO PCC O EtMgBr HO Likewise, this one is not that taxing but hopefully gets you thinking about functional group interconversions. (e) How would you synthesise OH From Br Answer: Br t-BuOK mCPBA^ O EtMgBr^ OH This one is quite hard. But again, it is all about FGI and recognising where the original carbons are. I recommend number you carbons and then trying to identify relationships between functional groups and this numbering. (f) How would you synthesise Br From OH Answer: OH acid dehydration HBr ROOR Br
Quick, but not necessarily straightforward; the more reactions you know the easier this becomes. Here we require anti - Markovnikov addition of the HBr. Therefore, we add peroxide to allow a radical reaction. (g) How would you synthesise O NH 2 From O Answer: O O Br O HBr N 3 ROOR NaN 3 PPh 3 H 2 O O NH 2 Again, need anti - Markovnikov so use radical bromination. The add nitrogen via the azide. Of course, there are other answers (hydroboration , oxidation and reductive amination??) (h) How would you synthesise N H Ph From OEt O Answer: OEt O H 3 O+ OH O SOCl 2 Cl O PhNH 2 N H Ph O LiAlH (^4) N H Ph Hopefully, this one doesn’t cause too many problems. (i) How would you synthesise
OH Answer: OH (^) PCC O (^) MeMgBr OH acid HBr ROOR Br
