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123.312 Advanced Organic : Retrosynthesis Tutorial

Question 1. Propose a retrosynthetic analysis of the following two compounds. Your answer should include both the , showing your thinking, and the reagents that would be employed in the actual synthesis.

Compound A

O

Answer:

FGI OH C–C OH

O dehydration O aldol 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 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 ! FGI O O HO 1 2 OH HO O 3 5 reduction 4

C–C

O O

HO ! !

O O O EtO OEt

The first disconnection should be relatively simple, break the C–O bond to give the and . 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 . 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; NaBH4 does not reduce .

+ O O H , O O O O base NaBH4 EtO C H2O EtO2C 2 O O EtO OEt CO Et 2

Question 2. Give the retrosynthetic analysis for the following three compounds. Pay special attention to the relationship between the functional groups.

CO2H CO2H CO2H

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 …I’m not convinced it helps but if it allows you to rationalise what is going on more readily then use it!

FGI C=C CO2H CO2Et CO2Et hydrolysis ! !

PPh3 CHO CO2Et

The second is probably the hardest; there is no simple enolate disconnections so we have to look slightly further a field. Whilst we can go via an , the best route probably involves FGI to a nitrile and then simple C–C bond formation by a substitution with a anion.

FGI C–C CO H N 2 N hydrolysis ! !

Br NaCN

Alkylation of an enolate offers the most rapid approach to the third structure. Not much needs to be said about this one.

FGI C–C CO H CO Et 2 2 CO2Et hydrolysis ! !

CO2Et Br CO2Et

Question 3. How would you make these compounds?

H OH CO2H NH2 N

Answers

The first is simply a case of reduction amination. We cannot form an amide so it has to proceed by the imine.

H FGI N N C=N O H2N

reduction

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.

CO2H FGI EtO2C CO2Et C–C Br Br EtO2C CO2Et decarboxylation

The final compound is a primary amine. This could either be prepared by reductive amination of the appropriate (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.

NH2 FGI NH O C=N

reduction

NH2 FGI N3 OTs OH C–N C–S

reduction

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 NEt2

O NH2

Answer

O O O OH O OH NEt2 FGI C–O

O O reduction O

NH2 remove the NH2 NO2 (with the reactive functionality) the amine can be problematic we can now remove the . so we convert it to the less Attepts to do this earlier reactive nitro group. This also C–O would have met with failure prepares the way for its due to alkylation of the amine eventual disconnection O OH O OH O OH FGI C–N

diazonium + – OH N2 BF4 OH NO2 I would stop here as I don't know how much the phenol group is ortho, para aromatic chemistry you have done. But if you directing but should favour the FGI have done enough then we can take the leaser hindered position (and diazonium synthesis all the way back to the acid we might be able to argue about H-bonding)

O OH O OH O OH FGI C–N

reduction NH2 NO2

remember, the acid is electron withdrawing so is meta directing

Question 5. (a) How would you synthesise

From

Answer:

i. BH3 ii. H2O2 / NaOH PBr3 OH Br

Remember, we need to get anti-Markovnikof addition of the hydroxyl group so we use hydroboration / oxidation.

(b) How would you synthesise

OH

From

Answer:

i. BH3 Br2, ii. H2O2 / t-BuOK hv Br NaOH OH The key to this one is functionalisation of the hydrocarbon. This is achieved by radical bromination, then its plain sailing.

(c) How would you synthesise

OH

From

Br

Answer:

Mg OH Br MgBr Et2O CH3CHO

Lets be honest, if you can’t do this one then you’re in trouble!

(d) How would you synthesise

HO

From

HO

Answer:

PCC EtMgBr HO O HO Likewise, this one is not that taxing but hopefully gets you thinking about interconversions.

(e) How would you synthesise

OH

From

Br

Answer:

t-BuOK mCPBA EtMgBr OH Br O

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:

HBr OH acid ROOR dehydration 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

NH2 O

From

O

Answer:

HBr PPh3 Br N3 NH2 O ROOR NaN3 H2O O O O

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

Ph N H

From

O

OEt

Answer:

O O O O + LiAlH Ph H3O SOCl2 PhNH2 Ph 4 N OEt OH Cl N H H

Hopefully, this one doesn’t cause too many problems.

(i) How would you synthesise

OH Et Et

From

EtO2C CO2Et

Answer:

+ i. base i. base i. H3O OH ii. Et–I EtO2C CO2Et ii. Et–I EtO2C CO2Et ii. heat Et CO2H LiAlH4 EtO2C CO2Et Et Et Et Et Et Et This one is all about recognising which functional groups are required and how they can be interconverted.

(j) How would you synthesise

OH OH

From

O

Answer:

i. base O OH OH OH O ii. EtCHO NaBH4

Don’t get fooled by how similar may appear. Count the , look for real relationships. In this case we have a 1,3-diol derived from a ketone; start thinking about an aldol reaction right away.

(k) How would you synthesise

Br

From

OH

Answer:

HBr OH PCC O MeMgBr acid ROOR OH Br