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1
The Disconnection Approach
We start with a few simple problems to set you at ease with disconnections. Problem 1.1: Here is a two-step synthesis of the benzofuran 3. Draw out the retrosynthetic analysis for the synthesis of 2 from 1 showing the disconnections and the synthons.
Answer 1.1: As this is a simple SN2 reaction, the disconnection is of the C–O bond 2a and the synthons are nucleophilic phenolate anion 4, which happens to be an intermediate in the reaction, and the cation 5, which happens not be an intermediate in the reaction but is represented by the α-bromoketone 6.
Problem 1.2: Draw the mechanism of the cyclisation of 2 to 3. This is an unusual reaction and it helps to know what is going on before we analyse the synthesis. Answer 1.2: The first step is an acid-catalysed cyclisation of the aromatic ring onto the protonated ketone 7. Loss of a proton 8 completes the electrophilic aromatic substitution giving the alcohol 9.
Now protonation of the alcohol leads to loss of water 10 to give a stabilised cation that loses aproton 11 to give the new aromatic system 3. Problem 1.3: Now you should be in a position to draw the disconnections for this step.
Answer 1.3: We hope you might have drawn the intermediate alcohol 9. Changing 3 into 9 is not a disconnection but a Functional Group Interconversion (FGI) – changing one functional group into another. Now we can draw the disconnection revealing the synthons 12 represented in real life by 2.
A Synthesis of Multistriatin
In the textbook we gave one synthesis of multistriatin 17 and here is a shorter but inferior synthesis as the yields are lower and there is little control over stereochemistry.1 Problem 1.4: Which atoms in the final product 17 come from which starting material and which bonds are made in the synthesis? Hint: Arbitrarily number the atoms in multistriatin and try to trace each atom back through the intermediates. Do not be concerned over mechanistic details, especially of the step at 290°C.
Answer 1.4: However you numbered multistriatin, the ethyl group (7 and 8 in 17a) finds the same atoms in the last intermediate 16a and the rest falls into place. It then follows which atoms come from 14 and which from 15. Finally, you might have said that C-4 in our diagrams comes from formaldehyde.
So the disconnections also fall into place. Just one C–O bond was disconnected at first 17b then one C–O and one C–C 16b and finally the alkene was disconnected 14b in what you may recognise as an aldol reaction with formaldehyde. If you practise analysing published syntheses like this, you will increase your understanding of good bonds to disconnect.
References
1. W. E. Gore, G. T. Pearce and R. M. Silverstein, J. Org. Chem., 1975, 40, 1705.
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2
Basic Principles: Synthons and Reagents: Synthesis of Aromatic Compounds
This chapter concerns the synthesis of aromatic compounds by electrophilic and nucleophilic aromatic substitution. All the disconnections will therefore be of bonds joining the aromatic rings to the sidechains. We hope you will be thinking mechanistically, particularly when choosing which compounds can undergo nucleophilic aromatic substitution and the orientation of electrophilic aromatic substitution. Any textbook of organic chemistry1 will give you the help you need.
Problem 2.1: Compound 1 was needed2 for an exploration of the industrial uses of HF. Suggest how it might be made. Hint: consider which of the three substituents you would rather not add to the ring.
Answer 2.1: We can add the nitro group by nitration and the isopropyl group by Friedel-Crafts alkylation (as it is a secondary alkyl group) but we would rather not add the OMe group as there is no good reagent for MeO+. So we disconnect first the most deactivating group (nitro) 1a and then the isopropyl group 2.
Before writing out the synthesis, we should check that the orientation of the substitution will be what we want. The OMe group is ortho, para-directing so alkylation will go mainly para because of steric hindrance. Now we have a competition as isopropyl is also ortho, para-directing but, since OMe has a lone pair of electrons conjugated with the benzene ring, it will dominate so everything is fine. We therefore suggest:
Did you consider the alternative strategy? That is, disconnect the isopropyl group first 1b to give a new intermediate 4 and disconnect the nitro group second. The starting material, anisole 3, is the same in both routes.
Again we should check the orientation. Nitration of anisole will give a mixture of ortho 4 and para 5 products so much depends on the ratio and whether they can easily be separated. The Friedel-Crafts reaction will go ortho or para to the OMe group and meta to the nitro group so that is all right. However the deactivating nitro group might make the reaction difficult.
So what did the chemists prefer? One published synthesis2 used HF as a catalyst to alkylate ortho-nitro-anisole 4 with isopropanol. The yield was a respectable 84%. This made sense as they had a supply of 4. If anisole is nitrated with the usual HNO3 /H2 SO4 , a 31:67 ratio of ortho:para products is obtained. If the nitrating agent is an alkyl nitrite in MeCN, the ratio improves to 75:25. The best route nowadays is probably the nitration of available para-isopropyl phenol 6, probably quantitative, and methylation of the product 7 with, say, dimethyl sulfate.
Problem 2.2: Thes...
