Total synthesis ITotal synthesis I

1

OH

OCl

OOfrom

Question 1

How would you make the molecule above (ibuprofen) from the three substrates above? You can use any other reagents but all the carbon atoms must come from the starting materials above.

2

OH

O

Cl

O O

Answer

Remember there is more than one right answer.

There are two possible approaches to answering such a question. The first is to map the starting materials onto the product. While the second involves a ‘proper’ retrosynthetic analysis.

With practice you will probably find yourself using the former approach. To start with you might need to work through the full analysis.

On the right you can see the most obvious mapping. Now we need to determine the order of reactions.

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Why don’t we add the ketone back into the molecule?

A ketone is meta directing and we would have the problem of attempting a chemoselective reduction.

If we approach this problem using retrosynthetic principles we need to remove the acid functionality. A guiding principle in retrosynthetic anaysis is removing reactive functionality as soon as possible (to avoid side reactions) and this has the added benefit of changing the oxidation state of the carbon to that of an epoxide (one of our starting materials).

In the forward direction this could be oxidation with Jones reagent.

OH

O

FGI OH

OH

O

O

4

OH C–COH

≡ ≡

O

The next disconnection removes the 3-carbon fragment. The forward reaction is Frieden-Crafts alkylation using an acid to activate the epoxide (secondary carbocation more stable than a primary carbocation).

Again we remove this fragment (right hand side) so that we remove the reactive functionality (the alcohol). Additionally, the left hand subunit is ortho/para directing.

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Friedel-Crafts alkylation would rearrange.

Meta directing

Another functional group interconversion prepares the molecule for the Friedel-Crafts acylation disconnection. We cannot directly add the alkyl unit via a Friedel-Crafts alkylation (see right) as it is liable to rearrange to the more stable cation.

We cannot add the epoxide unit to the ketone (bottom right) as this would give the wrong regiochemistry.

FGI

O

C–C

O

O

≡ ≡

Cl

Cl

O

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O

ClAlCl3

O

Et3SiH, H+or

Wolff-Kishner(NH2NH2 then NaOH/ROH)

i. epoxide, H+ii. Jones reagentOH

O

A potential synthesis of ibuprofen is shown here.

Friedel-Crafts acylation is followed by ketone reduction (hydrogenation would probably work as well as the Wolff-Kishner reaction or the silane-mediated reduction).

Friedel-Crafts alkylation with the epoxide followed by oxidation gives the desired compound.

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NC

MeO2CNH2

HO2C N CO2H

O

NNN

HN

Valsartan(diovan)

high blood pressure

Question 2

Propose a possible synthesis of diovan using the two starting materials shown and any other reagents you want.

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N CO2H

O

NNN

HN FGI N CO2Bn

O

NNN

HN

Answer

First step is to the protect the carboxylic acid as this reactive functionality could interfere with our synthesis. If, at the end of the planning we find it is not necessary then we will not add it!

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N CO2Bn

O

CNFGIN CO2Bn

O

NNN

HN

Next the tetrazole is removed. A number of reasons for this; firstly, I don’t like the cut of its gib (I have no idea how tetrazoles react). Secondly, I do know they are used as bioisosteres of carboxylic acids so I guess they behave in a similar manner so we want to get rid of reactive functionality early on.

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N CO2Bn

O

CN C–N NH

CO2BnCNCl

O

Remove the amide. C–X disconnections are obvious starting points to any retrosynthesis as we know so many reactions that allow their formation. The acyl chloride is simple enough to be considered commercially available.

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NH

CO2BnCN C–NBrCN

H2N CO2Bn

Another C–N disconnection. Amine alkylation can be problematic (over alkylation to give tertiary amines and quaternary ammonium salts is possible) but in this case the bulk of the amino acid and the fact that we need to start from valine make this disconnection very attractive.

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Finally, we have to make the ‘benzylic’ bromide from the ester starting material we are given. This is readily achieved from an alcohol, which in turn comes from the ester (in a forward sense this is simply reduction).

I do not think that amide formation from the ester directly is a good idea. To get to the desired product we would then have to achieve a selective reduction of the amide in the presence of an ester or acid. This could be challenging.

BrCNFGI

OHCNFGI OMeCN

O

NH

CN

OR

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OMeCN

Oi. DIBAL-H

ii. PBr3

BrCN

H2N CO2Bn

NH

CO2BnCN

EtNiPr2Cl

O

EtNiPr2N CO2Bn

O

CN

Here is one possible answer. I am sure there are many others.

Reduction is followed by bromide formation. Alkylation of valine gives the secondary amine.

Acylation forms the entire molecule bar the tetrazole.

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N CO2Bn

O

CNi. NaN3, ZnCl2ii. H2, Pd/C N CO2H

O

NNN

HN

Last two steps are tetrazole formation followed by hydrogenation to remove the protecting group.

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OH O

NH

OO

Question 3

How would you make this molecule from the two aldehydes given above?

Obviously this question combines our two favourite subjects … synthesis and asymmetric synthesis.

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OH O

NH

C–NH2N

OH O

OH

Answer

First, simplify the problem. Split the molecule into two subunits. The easy disconnection is the peptide link (C–N disconnection). This gives a simple homoallylic amine and the β-hydroxyacid.

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B

O

OBH

OH

HHO H

(–)-ipc2BCH2CH=CH2

≡

There are many routes to the homoallyic amine depending on how much background reading you have done (as you have of course being doing self-directed learning).

One route to the homoallylic amine using chemistry we have covered (or at least are in my notes and don’t require reading a book) would be the use of the Brown allylation chemistry.

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HO H

NH2

ON

NEtO2C CO2Et PPh3

H HN

ONaOH H NH2

This gives an alcohol to an amine so we need perform a stereospecific substitution reaction that inverts the stereocentre in a controlled manner. This can be achieved by the Mitsunobu reaction (I have used acetamide as the nucleophile … it is touch and go whether this would work. The Mitsunobu reaction requires the nucleophile to have a pKa of 15 or less and acetamide is 15.1). But let us pretend it does. Or …

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H

O O

NH

H2NH

NNHAc

NSi

OPh

Cl

H HNHNAc

SmI2H NH2

… we could use Leighton’s elegant silicon reagents (I covered one reacting with an aldehyde. It tur.s out they function very well with hydrazone as well).

For those of you that are interested a good paper to start with is J. Am. Chem. Soc. 2003, 125, 9596.

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O

B

(+)-ipc2Bcrotyl

OB H

H

HO

H

H≡

OH

We can make the second subunit using the Brown crotylation chemistry. We would need to the opposite enantiomer of pinene to the previous example and the E-crotyl reagent.

The pinene auxiliary controls the enantioselectivity while, as the reaction proceeds through a closed transition state, the geometry of the alkene controls the diastereochemistry.

Use the chair-like transition state to rationalise the stereochemical outcome.

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OH

i. TBSCl, imidazoleii. O3, PPh3iii. NaClO2

(Pinnck oxidation)TBSO O

OH

Cleaving the alkene can be achieved by ozonolysis (or osmium tetroxide and sodium periodate). This gives an aldehyde, which can be oxidised to the acid with the Pinnick oxidation.

Finally, peptide chemistry allows the two units to be coupled …

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TBSO O

OHH NH2

i. DCC, DMAP

ii. TBAFOH O

NH

… as shown here.

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