retrosynthesis: 123.312

...analysis

...Retrosynthetic

123.312 gareth j rowlands: massey university©gareth j rowlands

Retrosynthetic analysis

This file does not contain my lecture notes.It does not contain all the information in my lectures.It is not intended to be printed.It is not intended to be printed (so no complaints that it is 400 pages long).

This file does not contain my lecture notes.It does not contain all the information in my lectures.

©gareth j rowlands


OH

The idea of this file is to allow you to look at the retrosynthesis of a variety of molecules at your own pace and for me to experiment with methods of communicating the material. If you click on the arrow you’ll get the general idea...

terminology guidelines aromatics aliphatics two group patterns C–C bonds


OH

I have attempted to make various bonds (in this case adjacent to the alcohol) interactive thus allowing you to see potential disconnections. It may be helpful, it may not...we will have to wait and see.Funnily enough, I like to experiment...



All the information you need can be found from these sources.Of course, to actually pass the course, you need to understand the material...

www.massey.ac.nz/~gjrowlan/teaching.htmlgeneral specific


http://www.massey.ac.nz/~gjrowlan/teaching.html

http://www.massey.ac.nz/~gjrowlan/teaching.html

http://www.amazon.com/Organic-Chemistry-Jonathan-Clayden/dp/0198503466/ref=sr_1_1?ie=UTF8&s=books&qid=1289343075&sr=8-1

http://www.amazon.com/Organic-Chemistry-Jonathan-Clayden/dp/0198503466/ref=sr_1_1?ie=UTF8&s=books&qid=1289343075&sr=8-1

http://encore.massey.ac.nz/iii/encore/record/C%7CRb2119655%7CSstuart+warren%7COrightresult%7CX4?lang=eng&suite=pearl

http://encore.massey.ac.nz/iii/encore/record/C%7CRb2119655%7CSstuart+warren%7COrightresult%7CX4?lang=eng&suite=pearl

synthon synthetic equivalent

functional group interconversion

target molecule retrosynthetic analysis

disconnectionreverse step

Terminology






Terminology

OH2N

The target molecule (TM) is the goal, the target, the molecule you are trying to make...






Terminology

Retrosynthetic analysis (or retrosynthesis) is the idea of working backwards, one step at a time, to simplify a molecule. It is the logical approach to planning a synthesis. Each precursor becomes the target for further analysis. -EXAMPLE-

target molecule precursor 1 precursor 2 starting

material






TerminologyA logical backwards steps. This arrow effectively means “can be made from.”

To be of any value, there must be a real reaction that corresponds to the forward reaction






Terminology

A retrosynthetic (reverse) step involving the breaking of a bond to form two (or more) synthons.

The more reactions you know the more possibilities you can invoke.

X YX Y X Y

X YX• Y•






TerminologyA synthon is an idealised fragment.

It does not have to exist. It aids thought/retrosynthesis. It should have a synthetic equivalent to be of any use.

-EXAMPLE-

X YX YX• Y•






Terminology

The synthetic equivalent is a real compound that corresponds to the synthon. Ideally, a commercially available reagent (or the next target in your retrosynthesis)

O

≡

≡

H

Cl

O

(& AlCl3)






Terminology

The imaginary conversion of one functional group into another in order to aid simplification, help planning or uncover a disconnection. There must be a good ‘forward’ (real) reaction. -EXAMPLE-

OH2N FGI

OO2N


...analysis: planning

...Retrosynthetic

©Tom Coates@Flickr

➎ consider FGI ➏ repeat

➊ identify functional groups ➋ identify patterns

➍ identify problems➌ examine disconnections

Guidelines

Where do we start when we plan a synthesis?

Below are a set of guidelines to help you logically approach retrosynthesis or the planning stage. They are not rules, the only rule is that you want to simplify the problem whilst using chemically allowable transformations!

OH2N


OH2Namine ketone




.....identify functional groups

Functional groups are the signposts to retrosynthesis. Without functionality, we have a very limited range of reactions at our disposal. Frequently, they control where we can apply disconnections.

➊


OH2Namine ketone

ortho/para directing

metadirecting




.....identify patterns

The pattern or connections between functional groups often reveal which reactions you can employ. Learning to recognise patterns of functional groups is very important for retrosynthesis. The pattern of functional groups frequently indicates the order reactions should be approached.

➋





.....examine disconnections

To begin with, you need to examine all possible disconnections. With practice you will learn that some can readily be ignored. You must also remember not to look at backwards just one step but to go further back. Shortsightedness has ruined many a retrosynthesis.

➌

OH2N

aC–N

H2N

OH2N

ObC–Cba





.....identify problems

Are all the disconnections chemically allowable? Will the reaction proceed with the correct regio-, stereo- or chemoselectivity? Does the disconnection simplify the problem? Try and answer these questions before you proceed.

➍route a

H2N

no synthetic equivalent

route b

H2No,p-directing,

NOT meta-directing





.....consider FGI

Functional group interconversions do not simplify a structure, but they do overcome problems and/or allow disconnections that will simply the target

➎O

H2N FGI

reduction

OO2N

(note: I have written the forward reaction under the arrow; this shows the FGI is possible & highlights potential problems)





.....repeat 1-5 (until you have simple SM)

Just keep repeating the steps until you have a commercially available starting material. Approached logically, and with a good working k n o w l e d g e o f r e a c t i o n s , retrosynthesis can be both fun & easy.

➏O

H2N FGI

reduction

OO2N

C–C

O2N O

-synthetic equivalents-


OH2N FGI

reduction

OO2N

C–C

O2N O

Cl




.....repeat 1-5 (until you have simple SM)➏

-finish retrosynthesis-


OH2N FGI

reduction

OO2N

C–C

O2N O

Cl

C–NO2N

-Synthesis-




.....repeat 1-5 (until you have simple SM)➏


-Synthesis-




.....repeat 1-5 (until you have simple SM)➏O

H2N FGI

reduction

OO2N

C–C

O2N O

Cl

C–N

HNO3H2SO4

AlCl3

SnHCl


...analysis: examples

...Retrosynthetic©lennox_mcdough@Flickr

.Retrosynthesis of a benzene derivative

The synthesis of aromatic compounds is relatively simple; we have a limited number of reliable reactions and a well-defined set of guiding principles.

Therefore, we will start here...


.Simple retrosynthesis

NH2

BrHow could you make this compound?Consider our guidelines...



NH2

Br

amine

bromide

o,p-directing

o,p-directing(just)

identify FG & patterns connecting them (guidelines 1 & 2)

Which bond C–N or C–Br would you

disconnect?



NH2

BrChoose a bond! I’m not doing all the work for you...



NH2

BrAre there any other approaches to this molecule?


NH2

Br

FGI

reduction

NO2

Br


Change order of events & perform FGI first.Do we disconnect C–N or C–Br next?


NH2

Br

FGI

reduction

NO2

Br


Now you are just being lazy. Choose a bond!



Which route is best?Really depends on your

definition of best...

-next example-

NH2

Br

©carbonNYC@Flickr


.Simple aromatic retrosynthesis

How could you make this compound?Consider our guidelines...

Br

OH



Br

OHsecondary alcohol

bromide

o,p-directingo,p-directing

alkyl

identify FG & patterns connecting them (guidelines 1 & 2).Which bond, C–Br or C–C would you disconnect first?



FGI introduces a ketone. This aids simplification by permitting standard Friedel Crafts chemistry.

Which bond, C–Br or C–C would you disconnect first?

Br

OHFGI

reductionBr

O



Br

OHFGI

reductionBr

O

C–C

Br

C–Br

bromination Cl

O

and the synthesis



Br

OHFGI

reductionBr

O

C–C

Br

C–Br

bromination Cl

O

Br2/Fe

AlCl3NaBH4


.Retrosynthesis of a benzene derivatives

Aromatic chemistry limits your choices (but allows some very reliable reactions). It was a good place to start but now lets turn our attention to more complex systems...


... aliphatic examples...Retrosynthetic analysis:

©Horia Varian@Flickr

.Simple retrosynthesisHow would you make chlorbenside (anti-tick/mite)?Consider our guidelines...

©graftedno1@Flickr

Cl

S

Cl



Identify FG & patterns connecting them (guidelines 1 & 2)C–heteroatom bonds are easy to identify & wide range of reactions available to form them. These disconnections are our starting point...

Cl

S

Clsulfide

chloride

reactive benzylic position

chloride


Cl

S

Cl

a

b c

d


Which C–heteroatom bond would you disconnect first?Remember we want to simplify the problem & use reliable reactions.



Cl Br

ClSH

NaOEt

Cl

S

Cl



©Peter Keyngnaert@Flickr

OHN

O

ONH

OOMe

OH

ICI-D7114(anti-obesity drug)


.Simple retrosynthesisHow would you make this intermediate from the synthesis of ICI-D7114?Consider our guidelines...

PhHN

O

O Ph



PhHN

O

O Ph

etheramine

C–X disconnections

no simple aromatic

disconnections

Identify FG & patterns connecting them (guidelines 1 & 2)C–heteroatom (C–X) bonds are easy to identify.These disconnections are our starting point...


PhHN

O

O Pha

bcd




.Remove reactive functionality

Removing reactive functionality early limits side reactions & increases the chance of selectivity.

The rest of the retrosynthesis... ©Alexandra Polido@Flickr


.Simple retrosynthesis continued

Which C–heteroatom bond should we disconnect next?Remember we want to simplify the problem & use reliable reactions.

O Ph

OBr

e

f



A simple understanding of basic reactions (& the principles behind them) allows a rapid synthesis of this precursor.

HO

OHBnCl

excess

base

HO

OBn

BrBr

base

O

OBnBr

excess

BnNH2

O

OBnNHBn


.Retrosynthesis

©alancleaver_2000@Flickr

F3C

HN

fenfluramineneuroactive drug

appetite surpressantterminology guidelines aromatics aliphatics two group patterns C–C bonds

.Retrosynthesis

F3C

HN

amine

C–X disconnections

Identify FG & patterns connecting them (guidelines 1 & 2)C–heteroatom (C–X) bonds are easy to identify.These disconnections are our starting point...


.Retrosynthesis

The obvious disconnection does NOT work.Why?Think about the chemistry/reactivity of primary vs. secondary amines...

F3C

HN C–N

F3C

H2N

≡ ≡

F3C

NH2Br


.Problem

Over alkylation can be a serious problem.The solution to which is...

F3C

NH2 Br

F3C

HN

Br

F3C

NBr

F3C

NBr

more reactive

MORE reactive


.Solution

...Functional Group InterconversionAn amide is deactivated compared to an amine, so get single addition.The synthesis is...

F3C

HN

F3C

HNFGI

amide reduction

O

C–Namide

F3C

NH2

Cl

O


F3C

HN

F3C

HNFGI

amide reduction

O

C–Namide

F3C

NH2

Cl

O

baseLiAlH4

.Synthesis

...Functional Group InterconversionAn amide is deactivated compared to an amine, so get single addition.The synthesis is...


There are many other FGI for the formation of amines.A common solution is given on the next page...

.Solution

F3C

HN

F3C

NFGI

imine reduction

C–Nimine

F3C

OH2N


F3C

HN

F3C

NFGI

imine reduction

C–Nimine

F3C

OH2N

NaBH4 or NaBH3CN

H+

And now for an example...

.Synthesis


.Retrosynthesis

N

N

Ph

OOMe

F

Ocfentanilpainkiller


.Retrosynthesis

amine C–N amide

Identify FG & patterns connecting them (guidelines 1 & 2)C–N bonds are easy to identify.These disconnections are our starting point...

N

N

Ph

OOMe

F

amide ether

C–N amine


N

N

Ph

OOMe

F

a

b

c

.Retrosynthesis



.Retrosynthesis

N

N

Ph

OOMe

F

C–N

N

HN

Ph

F

OOMe

Cl

N

N

Ph

Famine

condensation

N

O

Ph H2N

F

amide

FGIimine

reduction

C–N

...and the synthesis...


N

N

Ph

OOMe

F

C–N

N

HN

Ph

F

OOMe

Cl

N

N

Ph

Famine

condensation

N

O

Ph H2N

F

amide

FGIimine

reduction

C–N

base

NaBH3CNH+

.Retrosynthesis


...disconnections

...Two group

.Two group disconnections

OPh

OH

How could you make this compound?Consider our guidelines...



OPh

OH{ allyl group ether

alcohol

C–Xat centre of

molecule

Identify FG & patterns connecting them (guidelines 1 & 2)C–X bonds are easy to identify.These disconnections are our starting point...


OPh

OHa b


Which would be the better disconnection?



Epoxides are relatively stable.Epoxides are easy to prepare (and control stereochemistry)Therefore...

Ph

OH≡

Ph

OHBr ≡

PhO


.1,2-diX

Forwards - 1 functional group gives 2 new ones.Backwards - look for two functional groups next to each other & we know we can make them from a single functional group.

So the synthesis is...

RX1

R2

X2

X1 ≠ (or =) X2 = O, N, S

1,2-diX

R2X2

1

2


.Retrosynthesis

lets look at an example...

OPh

OHb C–O

O

Ph

OH

≡

≡

OH

PhO

NaH


.Retrosynthesis

©non-partizan@Flickr

OOH

NH

propranololbeta blockerstress relief


.Retrosynthesisamine

C–N amine

Identify FG & patterns connecting them (guidelines 1 & 2)C–X bonds are easy to form & our new 1,2-diX pattern is visible twice.These disconnections are our starting point...

alcoholether

C–O phenolic

ether OOH

NH

1 12

1,2-diX


.Retrosynthesis

]

Which C–X bond would you disconnect first?

OOH

NH

a b


.Retrosynthesis

Looks good but can you see a potential problem?

O

H2N

OOOH

NH C–N

C–O

OO

≡

≡OH

OCl


.There is a problem!

©tibchris@Flickr

stereochemistry!Don’t care? Go to next -retrosynthesis-Want to know what is going on? -here-


.Retrosynthesis

©limowreck666@Flickr

ON

O

H2N

N O

intermediate towards moxnidazole (anti-parasitic)


.Retrosynthesis

amine

Identify FG & patterns connecting them (guidelines 1 & 2)C–X bonds & two possible 1,2-diX disconnections.These disconnections are our starting point...

carbamate

hydrazone

2 x C–X carbamate

1 12

2 x 1,2-diX

ON

O

H2N

N O


ON

O

H2N

N O

a b c

d

.Retrosynthesis

Where would you start?

1 12


OHNHH2N

N Oe f

.Retrosynthesis

Which order to we add the ‘amines’? (both are 1,2-diX disconnections)

1 12


.Retrosynthesis

Here is the complete retrosynthesis and here is the synthesis

ON

O

H2N

N O

b c2 x C-X OHNHH2N

N Oe

1,2-diX

N ONH2H2N

Of

C–XCl

OHN

O


ON

O

H2N

N O

b c2 x C-X OHNHH2N

N Oe

1,2-diX

N ONH2H2N

Of

C–XCl

OHN

O

Basehydrazine

MeO OMe

O

.Retrosynthesis

Now lets look at another useful pattern to identify


.Retrosynthesis

©shellgreenier@Flickr

O

Ph

N

atropine mimic


.Retrosynthesis

amine

Identify FG & patterns connecting them (guidelines 1 & 2)New pattern has heteroatoms three carbons apart (1,3-diX).Why is this a useful pattern?

ketone

3 12

1,3-diX

O

Ph

N


.1,3-diX

Conjugate addition is a reliable reaction. So the pattern is...

O

Ph

N C–XO

Ph

N

≡

O

Ph

Br

≡

O

Ph 123

too reactive


.1,3-diX

Conjugate addition or Michael addition or 1,4-addition.Good disconnection as it is a reliable forward reaction and there are many methods for the formation of the enone.

so the synthesis would be...

OO OX

12

3 12

3

≡


.1,3-diX

Note: the proton could be replace by other electrophiles to make more complex compoundsNote: in this case the mechanism is probably more complicated (look up iminium ion activation)

Remember, it’s not just ketones that can activate alkenes...

O

PhNH

O

Ph

N

HOO

O

Ph

N


.1,3-diX

Conjugate addition or Michael addition or 1,4-addition.alkene can be activated by carbonyl group, sulfones, nitriles, nitro groups or any strongly electron-withdrawing group.

Lets look at an example

.X.X .XX

12

3 12

3

≡


.Retrosynthesis

How would you make this compound?Follow our normal thought process...

Look for patterns

O NH2


.Retrosynthesis

amine

Identify FG & patterns connecting them (guidelines 1 & 2)Two heteroatoms (1,3-diX) but neither are electron-withdrawing groups.

Do we know any FGI that could convert one into a EWG?

ether

3

1

2

1,3-diX

O NH2


.Retrosynthesis

FGI allows amine to be converted to nitrile (reduction in a forward reaction). Nitrile strong electron-withdrawing group & sets up 1,3-diX.

-Synthesis-

O NH2FGI

reductionO

N

1,3-diX

ON

≡

OHN


O NH2FGI

reductionO

N

1,3-diX

ON

≡

OHN

base

LiAlH4

.Retrosynthesis

1,3-diX disconnection is very useful. It allows molecules to be rapidly divided.

-Now the biggy-terminology guidelines aromatics aliphatics two group patterns C–C bonds

.C–C Bond Formation

©the albino@Flickr


.Retrosynthesis

©Somalia ya swan@Flickr

carnation perfume intermediate


.Retrosynthesisalkyne

Identify FG & patterns connecting them (guidelines 1 & 2)Only FG is alkyne.C–C bonds next to functional groups are good starting points.

-Retrosynthesis-

C–Cnext to FG


.Retrosynthesis

Alkynes can be deprotonated with strong base and make good nucleophiles.

-Synthesis-

C–C

≡ ≡

H

H

Br


C–C

≡ ≡

H

H

Br

i. NaNH2ii. R–Br

.Retrosynthesis

Alkynes can be deprotonated with strong base and make good nucleophiles.

-Lets try something harder-


.Retrosynthesis

©{ pranav }@Flickr

OH

violet oil component


.Retrosynthesis

alkene

Identify FG & patterns connecting them (guidelines 1 & 2)Two functional groups makes life a little easier but still some potential problems...

-Retrosynthesis-

C–Cnext to FG

OH

alcohol


OHab

FGI

.Retrosynthesis

Where would you start this retrosynthesis?


.Retrosynthesis

-another example-

OH FGI OH C–C

OH

≡

H

C–CBr

H

H

NaNH2

i. NaNH2ii. oxirane

H2Lindlar's catalyst


.Retrosynthesis

this is not a pea moth ©Dell’s Pics@Flickr

O

O

pea moth pheromone


.Retrosynthesis

alkene

C–Cnext to FG

ester

O

O

reactive functionality

Identify FG & patterns connecting them (guidelines 1 & 2)

-Retrosynthesis-


O

O

FGI

a b

.Retrosynthesis



.Retrosynthesis

In the actual synthesis the THP protecting group was used.Make sure you understand each step (and know all the mechanisms)Next a -key disconnection-

Br O O

NaOTHP( )7

i. NaNH2ii. MeI

OTHP( )7

Na(s), NH3(l)

OTHP( )7

i. HOii. AcCl

O

O


.How would you make?

©vitroid@Flickr

PhOH


.Retrosynthesis

C–Cnext to FG

alcohol

Identify FG & patterns connecting them (guidelines 1 & 2)New pattern is called 1,1-C–C

-1,1-C–C-

PhOH


.1,1-C–C

If you see an alcohol, the first disconnection you should think about is the addition of a nucleophile to a C=O bond (but not the only disconnection).

-Retrosynthesis-

R R'

OH 1,1-C–C

R

OHR'

≡ ≡

R

OBrMg R'


PhOH

a

b

c

.Retrosynthesis



.Retrosynthesis

©jeffreyklassen.com

NPh

OH

Ph

fenpiprane precursor(anti-histamine)


.Retrosynthesis

1,1-C–Cnext to FG

amine

Identify FG & patterns connecting them (guidelines 1 & 2)The alcohol group allows 1,1-C–C disconnections.two heteroatoms indicates that we should look to set up 1,3-diX (in other words this is going to influence the order of steps) -retrosynthesis-

NPh

OH

Ph

alcohol

1,3-diX(if we form C=O)

312


So, which way around should we perform the disconnections?

NPh

OH

Ph

a b

.Retrosynthesis


.Retrosynthesis

Now that was easy. Want another example?

NPh

OH

Ph

2 x1,1-C–C

N

OHPhPh

N

O

≡

MeO

Ph MgBr

1,3-diX

O

MeO

N≡O

MeO

HN

mix together

2 x


.Retrosynthesis

©tranchis@Flickr

OCl

precursor to chlophedianol

(cough suppressant)


.Retrosynthesis

1,1-C–Cnext to FG

ketone

Identify FG & patterns connecting them (guidelines 1 & 2)Looks straightforward? No way I could be trying to trick you?-retrosynthesis-

aryl ringOCl

aryl ring


.Retrosynthesis

1,1-C–C disconnection removes aryl ring.Can you see why this will not work?

OCl1,1-C–C

OCl

≡ ≡

OClBrMg

OEt


.Problem

Multiple addition will occur (ketones normally more reactive than esters)

Solution? Functional group interconversion

OClBrMg

OEt

Cl OH

PhPh


.Retrosynthesis

FGI gives the precursor for a single addition (or 1,1-C–C disconnection)-synthesis-

OClFGI

oxidation

OHCl

1,1-C–C

OHCl

≡

Cl OMgCl


.Retrosynthesis

How else could we make this compound (with reactions you have been taught)?

Cl OMgCl

OHCl

Jones reagent

OCl


.Retrosynthesis

Think about aromatic substitution and the Friedel-Crafts reaction.

Which side would you disconnect?

OCla b


.Retrosynthesis

Synthesis is a simple Friedel-Crafts reaction.

Now a more complex example involving the carbonyl group...

Cl O

ClFeCl3

OCl


.Retrosynthesis

How would we make this compound?

-retrosynthesis-

O O


.Retrosynthesis

1,1-C–Cnext to FG

(acid?)

Identify FG & patterns connecting them (guidelines 1 & 2)Technically only a lactone but this could be derived from an alcohol & an acid. Such a disconnection permits 1,1-C–C to be used.-retrosynthesis-

(alcohol?)

lactone

O O1,1-C–C

next to FG


.Retrosynthesis

Which bond would you disconnect first?(please remember what I have just written)

O Oab c

de


OHCO2H

ab c

FGI

.Retrosynthesis

Which bond would you disconnect next?(think about selectivity and how easy it is form the nucleophile)


.Retrosynthesis

Once the alkyne is installed the rest of the retrosynthesis is ok.-synthesis-

O O OHCO2H

C–O

lactonisation

FGI

reductionOH

CO2H

1,1-C–C

CO OH

OH

≡

CO O

OH

1,1-C–C

O


.Retrosynthesis

The synthesis is quite simple. Just note we need two equivalents of butyllithium for the second deprotonation due to the relatively acidic alcohol.-next pattern-

HH

i. NaNH2ii. PrCHO

OH

i. 2 x BuLiii. CO2iii. HO

OH

OHO

i. H2, Pd/Cii. HO

O O


.Retrosynthesis

not the correct virus...©groovelock@Flickr

Cl

OMe

OO

Et

OEt

( )5

arildone(anti-polio & herpes

simplex virus)


.Retrosynthesis

C–X

malonate(1,3-dicarbonyl)

Identify FG & patterns connecting them (guidelines 1 & 2)The new pattern is 1,2-C–C and corresponds to enolate chemistry -pattern-

ether

next to FG(1,2-C–C)

Cl

OMe

OO

Et

OEt

( )5


.1,2-C–C

A carbonyl group should always be one of the first places you look to simplify a molecule, either by 1,1-C–C disconnections and oxidation or 1,2-C–C disconnection.-retrosynthesis-

R R2

O 1,2-C–C

R R2

O

≡ ≡

RR2

OBr


.Retrosynthesis

Which bond would you disconnect first?

Cl

OMe

OO

Et

OEt

a b


.Retrosynthesis

©looseends@Flickr

Oindustrial precursor

to β-caroteneterminology guidelines aromatics aliphatics two group patterns C–C bonds

.Retrosynthesis

allylic(activated

electrophile)

allylic

Identify FG & patterns connecting them (guidelines 1 & 2)The new pattern is 1,2-C–C and corresponds to enolate chemistry -retrosynthesis-

alkene

next to FG(1,2-C–C)

O

ketone


.Retrosynthesis

Looks fairly straight forward.Are there any problems with this in the forward sense? -yes- -no-

O

1,2-C–C

O

≡ ≡

O

Br


So where would you start your retrosynthesis?Disconnection or functional group interconversion?

.Retrosynthesis

Oa

b FGI


.Retrosynthesis

The synthesis is quick and efficient.-next pattern-

Br

OEt

O ONaOEt

OCO2Et

NaOH

OCO2H

H+heat

O


...the aldol reaction

...1,3-diO

As soon as you see an alcohol 1 carbon from a carbonyl group you should think about the aldol reaction.-example-

.1,3-diO (aldol)

R R3

O OH

R2

1,3-diO1 2 3 R

O

R2R3

OH

≡ ≡

R

O

R2R3

O


.Retrosynthesis

©itspaulkelly@Flickr

OMe

OH

OOH

gingerol(hot flavour of ginger)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)The new pattern is 1,3-diO and corresponds to aldol chemistry -retrosynthesis-

alcohol

1,3-diO(aldol)

ketone

OMe

OH

OOH


OMe

OH

OOH

a b

.Retrosynthesis

So, where would you start your retrosynthesis?


.Retrosynthesis

Readily prepared by the Mukiayama aldol reaction.-another example-

OMe

OH

O i. (TMS)2NLiii. TMSCl OMe

OH

OTMS

hexanalTiCl4

OMe

OH

OOH


.Retrosynthesis

MeO

EtO2C

HO

thromboxane antagonist intermediate


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)Ester is key but remember the problem of self-condensation -retrosynthesis-

ester

1,3-diO(aldol)

allyl

MeO

EtO2C

HOalcohol 1,2-C–C


.Retrosynthesis

Where would you start your retrosynthesis?

MeO

EtO2C

HO

FGIa

b

c


MeO

EtO2C

HO

CO2Et

ab

.Retrosynthesis

What would be the next step of the retrosynthesis?


.Retrosynthesis

-next example-

CO2EtEtO2C

Br

NaOEtCO2EtEtO2C

NaOEt

MeO

O

MeO

EtO2C

HO

CO2Eti. NaOHii. H+, heat

MeO

EtO2C

HO


...the aldol condensation

...α,β

Enones can readily be formed form the dehydration of 1,3-hydroxyketones (and related molecules)...Or we can perform the disconnection in one step...-example-

.α,β (the aldol condensation)

this is the same as

R

O

R" R

O

R"

OH

R

O O

R"

FGI 1,3-diO

R

O

R" R

O O

R"

α,β

aldol condensation


.Retrosynthesis

©Thijs van Exel@Flickr

H

O

oxanamide intermediate(tranquiliser)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)Enone is key to simplifying this problem -retrosynthesis-

alkene

α,β

aldehyde

enone

H

O


{

.Retrosynthesis

FGI allows aldol reaction (or 1,3-diO)...Alternatively...-one step-

H

O FGI

dehydrationH

OOH

1,3-diO

H

OOH≡2 x H

O


.Retrosynthesis

α,β disconnection gives us the two aldehydes in one go.It is the same thing but misses out some of the thought processes (so for advanced students only?)-synthesis-

H

Oα,β

aldol condensation

H

OO


.Retrosynthesis

Simple really!-example-

H

O

ONaOEt

H

O


.Retrosynthesis

FHN

Ocinflumide

(muscle relaxant)terminology guidelines aromatics aliphatics two group patterns C–C bonds

{.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)Remove reactive functionality and then look at unsaturated system -retrosynthesis-

α,β-unsaturated

α,β

amide

FHN

O C–Namide


.Retrosynthesis

FHN

O

a b



.Retrosynthesis

The synthesis requires a malonate to prevent self-condensation.Otherwise, it is fairly straightforward.-another example-

FO

CO2HHO2Cheat

FOH

OSOCl2

FCl

O

H2N

FHN

O



N O

O

Ndoxpicomine(analgesic)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)These are the obvious patterns but there is another we should consider. -hidden pattern-

amine

C–N

acetal

2 x C–Oacetal

N O

O

N


.Retrosynthesis

The 1,3-diX relationship between heteroatoms suggests that we should consider conjugate addition and hence formation of an α,β-system. -hidden pattern-

amine acetal

N O

O

N1

23


N O

O

N O

.Retrosynthesis

The 1,3-diX relationship between heteroatoms suggests that we should consider conjugate addition and hence formation of an α,β-system. -retrosynthesis-


.Retrosynthesis


N O

O

N

a

bc d

e


N O

O

N

EtO

OEtb

c

.Retrosynthesis

Which should be the next disconnection?


.Retrosynthesis

And the complete synthesis.There are other ways of making amines as we shall see...-new pattern-

N

O

CO2EtEtO2Cbase

N O

OEtO

OEt

HN

N O

OEtO

OEtN

LiAlH4

N OH

OH

N

CH2=OBF3N O

O

N


...nitrile chemistry

...1,3-aminoalcohols

Unsubstituted methylene amines can be readily prepared from nitriles-example-

.1,3-aminoalcohols (nitrile chemistry)

R

OH

R"

NH2 1,3-aminoalcohol

R

OH

R"

NH2

≡ ≡

R

O

R"

Nkey: no substituent



MeO

NHO

Venlafaxine(anti-depressant)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)It contains 1,3-aminoalcohol pattern so we should know what to do... -retrosynthesis-

amine

1alcohol

1,3-aminoalcohol

MeO

NHO 2

3 {


.Retrosynthesis


MeO

NHO

a

b

cd

e

f


...the Mannich reaction

...1,3-aminoketones

Three-component coupling reaction to form 1,3-aminoketones-example-

.1,3-aminoketones (Mannich reaction)

R'

O

R"

NR R 1,3-aminoketone

Mannich reaction R'

O

R"

NR R

≡ ≡

R'

O

R"

NR R

≡R'

O

R"

ONH

R R



Ph N

O

MeO

nisoxetine analogue(anti-depressant)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)Disconnections should lead to 1,3-aminoketone pattern-retrosynthesis-

amine1

ether

1,3-aminoketone

2 3 {Ph N

O

MeO


Ph N

O

MeOab

c

d f

e

.Retrosynthesis



.Retrosynthesis

Synthesis is quick and easy.-next pattern-

Ph

O

HNMe2H2C=OH+

Ph N

O

NaBH4

Ph N

OH

Ph N

ClSOCl2

HO

MeO

base

Ph N

O

MeO


...the Claisen reaction

...1,3-diCO

Formation of 1,3-diketones-example-

.1,3-diCO (Claisen reaction)

R' R"

O O 1,3-diCO

Claisen reaction R' R"

O O

≡ ≡

R' R"

O O

RO


.Retrosynthesis

Ph

N

Tazadolene(anti-depressant)



O O

Ph

intermediate


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)New disconnection is our 1,3-diCO-retrosynthesis-

1

ketone

1,3-diCO

23

O O

Ph

ketone

{terminology guidelines aromatics aliphatics two group patterns C–C bonds

O O

Phab

.Retrosynthesis

So, which bond is it going to be?


.Retrosynthesis

-next pattern-

ONH

O

N

O

O

PhCl

O O

Ph

BnNH2H2(g)catalyst

NH2 OH

Phdehydration

NH2

Ph(BrCH2CH2)2

N

Ph


...conjugate addition

...1,5-diO

Formation of 1,5-diketones-example-

.1,5-diO (conjugate addition)

R' R"

O O 1,5-diO1

23

45 R'

O

R"

O

≡ ≡

R'

O

R"

O



NH

N O

O

Et

rogletimide(sedative)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)New disconnection is 1,5-diO so all other disconnections uncover this.-retrosynthesis-

1 carbonyl

1,5-diO

23

imide/amideNH

N O

O

Et

45


.Retrosynthesis


NH

N O

O

Et

a

b

c

c


.Retrosynthesis

The synthesis is relatively straight forward. Use of acrylamide instead of ethyl acrylate leads to a more practicable synthesis.-new pattern-

CO2EtN

baseEtBr

CO2EtN

Et

NH2

O

base

NEt

O

NH2OEt

baseNH

N O

O

Et


...alkene synthesis

...C=C

Already seen this one.-more C=C-

.C=C (α,β)

this is the same as

R

O

R" R

O

R"

OH

R

O O

R"

FGI 1,3-diO

R

O

R" R

O O

R"

α,β

aldol condensation


Wittig is a reliable method for forming C=C (but remember stereochemistry)There are many methods for the formation of alkenes: ring closing metathesis or cross metathesis, Julia olefination, Peterson reaction, Tebbe’s reagent etc.-example-

.C=C (Wittig reaction)

R

R' R'"

R" C=C

Wittig

R

R' R'"

R"

≡ ≡

R

R' R'"

R"PPh3 O


.Retrosynthesis

©Tadeeej@Flickr

Cl

OHOH

phenaglycol(tranquiliser)


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)Dihydroxylation will get us back to an alkene.-retrosynthesis-

dihydroxylation

diolCl

OHOH

{


.Retrosynthesis

Dihydroxylation will get us back to an alkene.At this point we have a lot of choices...-retrosynthesis-

Cl

OHOH

FGI

dihydroxylation

Cl


.Retrosynthesis

Here are two ways of using the Wittig reaction. Both get back to easily prepared starting materials.We don’t even have to use the Wittig reaction...-retrosynthesis-

Cl C=C

C=C

Ar ≡

≡

Ar ≡

≡

Ar

Ar

PPh3

O

O

Ph3P


.Retrosynthesis

The alkene could also be formed by dehydration giving us this possible route or...-the following-

ClFGI

dehydration

Ar

OH

1,1-C–C

1,1-C–C

1,1-C–C

Ar OBrMg

Ar

O

BrMg

O BrMg Ar


.Retrosynthesis

Even with these seven possible routes, the industrial synthesis is different...-industrial retrosynthesis-

ClFGI

dehydration

ArOH

1,1-C–C

1,1-C–C

Ar MgBr

O

ArOEt

OBrMg


.Retrosynthesis

...and it doesn’t even involve the formation of an alkene!-practice-

Cl

OHOH

1,1-C–CCl

OEtOH

O

FGIhydrolysis

Cl

OH

N1,1-C–C

Cl

O

NaCN



Ph Ph

OH

Start by identifying functional groups and patterns...-patterns-


.Retrosynthesis

Identify FG & patterns connecting them (guidelines 1 & 2)-retrosynthesis-

1,1-C–C

alcohol

Ph Ph

OH

1,1-C–C

1,2-C–C

FGI


.Retrosynthesis


Ph Ph

OHFGI1 2

3


.Retrosynthesis

it is the only way to improve.Just pick any molecule and see how many different ways you can think

of making it (with known reactions)

-last slide-

practice!terminology guidelines aromatics aliphatics two group patterns C–C bonds

.RetrosynthesisRead:

Stuart Warren - The Disconnection ApproachThe first edition was the book to read for retrosynthesis. I assume that

the second edition is as good if not better...

A good place to practice is this wonderful site from Arizona State University:

http://www.asu.edu/courses/chm332/retrosynthesis.html

This pdf file was written whilst listening to a lot of music (it took a while to put this together) and I’ll just list a few:

...and you will know us by the trail of dead ‘tao of the dead’black dog productions ‘bytes’

broadcast ‘ha ha sound’swans ‘swans are dead’ (the swans are, of course, the greatest rock band of all time (narrowly pipping the bad seeds))

mark kozelek ‘what’s next to the moon’

©Gareth Rowlands (except the pictures, which are accredited to their rightful owners)


http://www.amazon.co.uk/Organic-Synthesis-Disconnection-Stuart-Warren/dp/0470712368/ref=sr_1_1?ie=UTF8&qid=1301264192&sr=8-1

http://www.amazon.co.uk/Organic-Synthesis-Disconnection-Stuart-Warren/dp/0470712368/ref=sr_1_1?ie=UTF8&qid=1301264192&sr=8-1




Breaking the left-hand bond could give you a nucleophilic benzene fragment. I would write more but this is only the demonstration.

OH



Disconnecting the right-hand bond gives the possibility of a nucleophilic methyl fragment. Definitely possible.

OH

CH3


Retrosynthetic analysis: example

We try to simplify the molecule with each step backwards. For this to work, we need to have an understanding of lots of simple chemical reactions. The more we known, the more options we have and the simpler synthesis is.

OH2N FGI

OO2N

C–N

OC–C

Cl

O

AlCl3


synthon: example

Disconnection of a simple C–C bond gives two pairs of synthons.

From your knowledge of carbonyl reactivity and simple aromatic chemistry you should be able to identify which pair of synthons relates to a known reaction. Choose one and see if you are correct...

OC–CC–C

O O


synthon: example

Wrong choice!

Aromatic rings are normally electron rich. They need a powerful electron withdrawing group as a substituent before you can attempt nucleophilic aromatic substitution (actually, you can perform nucleophilic aromatic substitution if you had a very good leaving groups, a diazonium ion but this leads us to the next problem).

The carbon of the carbonyl group is normally electrophilic not nucleophilic. Of course, we can reverse the inherent polarity of the carbonyl group and start discussing the concept of umpolung chemistry.

O


synthon: example

Good going!

A benzene ring can be considered electron rich and hence nucleophilic (like an anion).

The carbon of the carbonyl group is inherently electrophilic (like a cation, hence the synthon) due to the polarity of the C=O bond or the electronegativity of the oxygen.

Now the question is, what would be the actual reagents or synthetic equivalents.

O


functional group interconversion FGI: example

A simple problem, how would we make aniline?-choose- the bond to disconnect first.

NH2



Hopefully, the C–N bond looks the most promising, even without covering the material on retrosynthetic analysis!

This disconnection gives a nucleophilic benzene synthon and an electrophilic ammonia synthon. Benzene is electron rich so can be considered to be the synthetic equivalent for the first synthon. There is no obvious synthetic equivalent for electrophilic ammonia.

We must find a new approach -NEXT-

NH2NH2C–N

NH2


NH2

NH2

FGI

NO2


Functional group interconversion FGI of the amine to a nitro group does not simplify our molecule (exactly the same number of atoms, just different ones).

But, it does set the scene for a very simple undergraduate reaction, that of nitration, and thus makes the rest of the retrosynthetic analysis simpler. -NEXT-

NH2

X


NH2

NH2

FGI

NO2C–N

O N O


We can now imagine the C–N disconnection. We get two synthons, the nucleophilic benzene ring & the nitronium cation. If you remember simple aromatic chemistry, this is the postulated intermediate in nitration & thus has a reliable forward reaction. We can now carry out the synthesis.

NH2

X


NH2C–N

NH2

FGI

NO2C–N

O N O≡

HNO3, H2SO4

SnHCl


Start retrosynthesis

NH2C–N

NH2

FGI

NO2C–N

O N O≡

NH2

X



bromobenzene would be the synthetic equivalent

NH2

Br

C–N

Br

NH2

No synthetic equivalentthis is a bad

disconnection


NH2

Br

C–BrBr

NH2


aniline is highly activated towards electrophilic aromatic substitution

Br2/Feis the synthetic

equivalentnow

-identify problems-



NH2Br2/Fe

NH2Br Br

BrAmine of aniline is very effective at activating aromatic ring so we get multiple additions.

Will a FGI aid any disconnections?



Bromination of a aniline amide (acetanilide in this case) normally only occurs once. The amide readily undergoes hydrolysis to regenerate the desired amine. Thus FGI (amine to amide) overcomes the problematic multiple bromination. Therefore, the retrosynthesis is...

NHAcBr2/Fe NaOH

NHAc

Br

NH2

Br


NH2

Br

FGI

hydrolysis

NHAc

Br

C–BrNHAc

bromination

FGIacetylation

NH2FGI

reduction

NO2C–N

nitration


Remember: writing the process below the retrosynthesis makes it clear that the steps you propose are possible. The synthesis would be...


NH2

Br

FGI

hydrolysis

NHAc

Br

C–BrNHAc

bromination

FGIacetylation

NH2FGI

reduction

NO2C–N

nitration

HNO3/H2SO4 Fe/HCl

Ac2O

Br2, FeNaOH




Br2/Fe is the synthetic equivalent

Nitro group is meta-directing (& strongly deactivating)

Poor choice

NO2

Br

C–Br

bromination

NO2

Br



HNO3/H2SO4 is the synthetic equivalent

Bromide is activating and ortho/para-directing. So bromobenzene

is the synthetic equivalent

NO2

Br

C–N

nitrationNO2

Br

-full retrosynthesis-


NH2

Br

FGI

reduction

NO2

BrC–N

nitration

Br

C–Br

bromination


and the -synthesis-


.Simple retrosynthesisNH2

Br

FGI

reduction

NO2

BrC–N

nitration

Br

C–Br

bromination

Br2, Fe

HNO3/H2SO4

Sn/HCl

next-example-



identify potential -problems-

Br

OH OH

BrC–Br

bromination

ortho-directing with activating alkyl

group is promising

bromination with Br2/Fe is known reaction



try other disconnection

Br

OH OH

BrC–Br

bromination

NO regioselectivityactivation would most likely lead to

ortho & para-bromination



consider FGI as aid retrosynthesis

Br

OHOHC–C

nucleophilic addition

Br

Good electrophilic synthon. Synthetic equivalent is the

carbonyl group

selective formation of nucleophile (Grignard reagent) in presence on

bromide would be hard


Br

O O

BrC–Br

bromination


try other disconnection

ketone is meta-directing and deactivates the ring

poor choice

bromination with Br2/Fe is known reaction



look at the retrosynthesis in full

synthetic equivalent for electrophilic carbonyl is a carboxylic acid derivative

bromide is ortho,para-directing & slightly

activating

Br

OOC–C

Friedel-Crafts

acylation

Br


X.Disconnection a or d

Poor choices. Neither disconnection simplifies the problem. In both cases we still have the majority of the molecule to prepare. Either disconnection b or c would be better as they split the molecule in half & lead to a convergent synthesis.

Cl

S

Clb c


.Convergent vs linear synthesis

A linear synthesis build a molecule up stepwise. Unfortunately, the maths is against you, even if you get 80% yield per step, in just three steps you are down to a 33% yield. With a convergent synthesis...

33%A B

AB

CD

EF

AB

CD

E

AB

CD

AB

C80% 80%

80%

80%

80%

A + B


51%.Convergent vs linear synthesis

A convergent synthesis builds the molecule in units. Therefore the longest linear sequence is far smaller & the mathematics allows a higher yield; at 80% per step, this convergent synthesis gives 51% yield compared to 33% by the linear sequence.

A B80%

A + B

C D80%

C + D

E F80%

E + F

80% AB

CD

80% AB

CD

EF


.Disconnection b

no synthetic equivalents

Cl

S

Clb

Cl S

Cl

Cl S

Cl

Cl HS

ClBr

≡

no reliable reaction*


.alright, I lied...

XHS R

Pd(0), phosphine,

baseS

R

Currently, the study of palladium(0)-catalysed reactions for the formation of C–C, C–N, C–O and C–S bonds is an area of intense study (Nobel Prize this year (2010) went to Heck, Suzuki & Negishi for their work with Pd). Formation of thiols has not seen as much research as the other areas but there are still some good papers out there...

But, we have not taught you this chemistry so cannot expect you to know it, thus I’ll ignore it for the time being.


.Disconnection c

no good synthetic equivalents

Cl

S

Cl

Cl

Cl

Cl

Cl

Cl Br

ClSH

≡

c

S S

-synthesis-terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection c

Cl

S

Cl

Cl

Cl

Cl

Cl

Cl Br

ClSH

≡

c

S S

sulfur is a good nucleophile (lone pairs) & a thiophenol is readily

deprotonated

benzylic position is an activated electrophile

-synthesis-terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection a

Hopefully, you are happy that the phenol portion should be the nucleophilic portion & the benzylic moiety should be the electrophile.

There are two problems with this disconnection. The first is a minor quibble, it does not really simplify the problem.

Can you see what the second, and more important problem is?

O

NHBn O PhaC–O

O

NHBn OPh


O

NHBn OPh

.Disconnection a

Chemoselectivity - there are two nucleophiles in this molecule & we cannot be certain where the electrophile would add (the oxygen as desired or the nitrogen).

You need to try another disconnection...


.Disconnection b

Hopefully, you are happy that the phenol portion should be the nucleophilic portion, leaving a carbon-based electrophile.

This disconnection is good because it cuts molecule in half, simplifying the synthesis considerably, but...

...there is a serious problem. Can you see what it is?

O

OBnb

BnHN C–O

O

OBnBnHN


.Disconnection b

Chemoselectivity - the electrophile has the chance to react with two nucleophiles and thus we have issues of selectivity.

In fact, cyclisation (intramolecular attack of the amine on to the electrophilic carbon) would give an aziridine, a potentially nasty functionality found/formed in mustard gases. On the other hand, aziridines, like epoxides, can be very

useful in synthesis...

Lets try again...

O

OBnBnHN


.Disconnection d

Very poor choice!Firstly, this disconnection does not simplify the problem.Secondly, any subsequent disconnections will involve an alkylation in the presence of an amine. This is bad.

Go to the next slide to see why...

OBn

O

HN

Ph

dC–N

Ph

OBn

O

HN


.Disconnection d

chemoselectivityTwo nucleophiles

O

O

H2N Ph

H2N OBn

O

O

O

NH2Ph

chemoselectivityTwo nucleophiles


.Always plan ahead...

...it will avoid problems further down the sequence....

El laberinto del fauno ©PictureHouse


.Disconnection c

Good choice!This disconnection may not split the molecule in half but it is chemically possible and it greatly simplifies the synthesis.

This leads to another good guideline...

OBn

O

HN

BnC–N

BnHN

c OBn

O≡ ≡

BnNH2

OBn

O

Br


.Disconnection e

Poor choice!Chemoselectivity is the issue. There are two electrophiles in two different molecules. Hard to control which will react.

Lets try the other disconnection...

O

O

Br

Ph

eC–O O

O

Br

Ph

≡ ≡

OH

O

Br

Ph

Br


.Disconnection f

Good choice!Splits molecule in half. Whilst there are two electrophiles they are on the same molecule, so if we use this reagent in excess statistics should give us the product of a single addition.

The full retrosynthesis is...

OBn

O

Brf

C–O OBn

O

Br

≡ ≡

OBn

HO

Br

Br


.Retrosynthesis

The starting material is a di-nucleophile but the functionality is in the same molecule so once again we can employ the trick of using an excess to ensure mono-benzylation.

The full synthesis is...

O

OBnNHBn C–N

O

OBnBr

C–O

BnNH2

HO

O Br

BrPh

C–O

HO

OH

Ph

Cl


.Disconnection a

Poor choice!No reliable reaction (that we have taught you).

Try again...

N

N

Ph

OOMe

F

aC–N

N

N

Ph

OOMe

F


.Disconnection b

Not the best choice!The forward reaction may be possible but potential problem of chemoselectivity (N- vs. O-alkylation of amide) and possibility of amide acting as either a base or a nucleophile. There is a better disconnection.

Try again...

N

N

Ph

OOMe

Fb

C–N N

Ph

N

OOMe

F


.Disconnection c

Good choice!The forward reaction is a simple amide formation. It simplifies the problem by removing amide and ether functionality.

What will the next disconnection be?

N

N

Ph

OOMe

F

cC–N

N

N

Ph

F

OOMe


.Retrosynthesis


N

HN

Ph

F

ad

FGI


.Disconnection a

Still a Poor choice!No reliable reaction (that we have taught you).

Try again...

N

HN

Ph

F

a

C–NN

HN

Ph

F


.Disconnection d

Hello! What is this section about?The alkylation of amines is problematic; over alkylation can occur. This reaction might work, steric hindrance might prevent a second alkylation but humour me and have another go...

Try again...

N

HN

Ph

F

d

C–N N

Ph

HN

F


.FGI

GoodFGI prevents multiple alkylations and simplifies the C–N bond forming step (condensations are easy reactions).

Le ts look a t the who le retrosynthesis...

N

HN

Ph

F

FGI N

N

Ph

Fimine reduction

C–Namine

condensation

N

O

Ph

H2N

F


OPh

OHa C–O O

Ph

OH≡

≡

HOPh

OH

Br

.Disconnection a

Possibly...But, can you see the potential problem with this reaction?

If you can’t, the problem is shown here.

Try again?


.Problem a

Two nucleophilic alcohols; so chemoselectivity is an issue.Alkylation could occur on either.

Try again?

HOPh

OH


.Disconnection b

Good...Splits molecule in half. But still have two alcohols that could cause trouble.

Furthermore, the bromide is not stable...

What could we use instead?

OPh

OHb C–O

O

Ph

OH

≡

≡

OH

Ph

OHBr


OOH

NH

a

OOH

NH

≡

≡

NHO

OHC–O

.Disconnection a

Possibly...Phenoxide is a good nucleophile. But there is a possible problem...

If you can’t see the problem it is here.

Try again?


NHO

.Problem a

Two nucleophiles; so chemoselectivity is an issue.Alkylation could occur on either.

Try again?

OH


.Disconnection b

Good...This disconnection removes the reactive functionality (the amine) first.

Still haven’t got a simple starting material.

More retrosynthesis is required.

OOH

NH

b

C–NO

OH

NH

≡

≡O

H2N

O


.Problem

Two electrophiles

But the reaction of both gives an almost identical product...

OCl Nuc


.Problem

But isn’t this the same compound?

Well, if you don’t care about stereochemistry it is...but for the rest of us...

OClNuc O

Nuc

Cl NucO O

Nuc


.Problem

If you start with a single enantiomer of the epoxide, the two different mechanisms actually give the two different enantiomers of the final product.

OH

OCl

OObase

OH

OCl

OObase


.Answer

Normally, the epoxide reacts first.

In fact, depending on the strength of the base you can isolate the alkoxide intermediate.

-Next example-

OCl Nuc


.Disconnection a

Why would you choose this? Have I taught you nothing?How many reactions do you know for making N–N bonds?How does this simplify the problem?

-Try again- (or get a drink and come back to this later as you really aren’t thinking about it)

ON

O

H2N

N O

a

ON

O

N O

H2NO

ON

O

N O

H2NO


.Disconnection d

PossiblyEasy bond to make. Removes an amine. You might go this route but...

It does not make use of 1,2-diX disconnection which would simplify the problem quicker.

-Look for a route that will give you the 1,2-diX disconnection-

ON

O

H2N

N Od

ON

O

H2NN O


.Disconnections b & c

Remove carbamate.Two simple C–X bonds that can be readily made in one step (& there are safer reagents than phosgene, this is just to keep things simple)

-Now we have the 1,2-diX precursor but one question still-

ON

O

H2N

N O

b c

ONOH2N

N O

O

ClCl

OHNHH2N

N O

≡≡1

1

2


.Disconnections f

Disconnection f is fine but...

Can you see what is wrong with the subsequent disconnection?

-problem-

OHNHH2N

N Of

1,2-diX OHNHH2N

N O

≡≡H2NHN HN

OOeNH2NH2

ClO ?

-try again-


.Problem

Multiple additions could easily occur.

Furthermore, if we add hydrazine early we have to carry the additional reactive functionality through the entire synthesis

-try again-

NH2

NH2Cl

O

two electrophilic

centres

two nucleophilic

centres


.Disconnections e

Disconnection e is good

Removes reactive hydrazine functionality early (or adds late in synthesis).Minimises possibility of multiple additions

-full retrosynthesis & synthesis-

OHNHH2N

N Oe 1,2-diX OH

N ONHH2N

≡≡N ONH2

H2NO

fC–X

ClO

HNO


.Disconnection b

Disconnection b is a poor choice.

You remove one carbon, this hardly simplifies the problem.How would you make the necessary Grignard reagent (or its equivalent)?

Whilst possible it would be better if -you tried again-

OH C–C OH

≡ ≡

MgBr O


.Disconnection a

Disconnection a could be a good choice.

Split molecule in half and the epoxide is a good electrophile.But, stereoselective formation of the anion hard. Deprotonation is not selective. Formation of alkenyl halide or equivalent complex.

Whilst possible it would be better if -you tried again-

OH C–COH

≡ ≡

O?


OH FGI OH C–C

OH

≡≡

H

O

C–CBr

H

H

reduction

-Synthesis-

.FGIUse of an FGI allows simple alkyne chemistry to be employed.In the forward sense the reactions is a stereospecific reduction.Rest of retrosynthesis follows standard chemistry.


.Disconnection a

Disconnection a is problematic.

Deprotonation of an alkene is hard and stereoselectivity would be an issue.

Whilst possible, it would be better if -you tried again-

O

Oa

C–C

O

O

( )7


.FGI

A good choice but to get the trans alkene we need to perform a dissolving metal reduction and it is possible that the ester functionality will not survive.

-try again-

O

O

FGIFGI

reduction

O

O


O

O C–OOH( )7

FGIreduction

OH( )7C–C

OH( )7C–CHH

Br OH( )7

.Disconnection b

Disconnection b is good.

Remove the reactive ester first. Then we can do our FGI to give us an alkyne that allows C–C bond formation, twice!

The actual synthesis is slightly different and shown -here-


.Disconnection a

Disconnection a is a good choice.

Readily available starting materials & a reliable reaction.

Now lets try a real example -here-

PhOH 1,1-C–C

PhOH

Me

≡ ≡

PhO

BrMg Me


.Disconnection b

Disconnection b is a very good choice.

Readily available starting materials, splitting the molecule in half & a reliable reaction.


PhOH 1,1-C–C

PhOH

≡ ≡

Ph MgBrO


.Disconnection c

Disconnection c is a very good choice.

Readily available starting materials & a reliable reaction.


PhOH 1,1-C–C

PhOH Me

Me

≡ ≡

PhO

OMe

BrMg MeBrMg Me


.Disconnection b

Disconnection b is a poor choice.

Whilst possible, it is a single group disconnection and would leave us having to prepare the halide. Two group disconnections cause greater simplification.

How could we perform a two group disconnection on this molecule?

NPh

OH

Ph

C–X

Ph

OH

Ph

N


.Disconnection a

Disconnection a is a good choice.The 1,2-C–C disconnection sets up the 1,3-diX disconnection.The full synthesis is -here-

NPh

OH

Ph

2 x1,1-C–C

N

OHPhPh

N

O

≡ ≡

MeOPh MgBr

Ph MgBr

1,3-diX

O

MeO

N≡O

MeO

HN


.Disconnection a

Disconnection a is not a good choice.It will make the product but only as a minor component.-What is the major product?--Try again-

OClC–C

OCl

≡ ≡

OCl

Cl


.Problem: regioselectivity

Problem: para substitution is favoured over ortho substitution.-the solution-

OCl

ClFeCl3

ClCl

PhO

O

Ph

88 : 11


.Disconnection b

Disconnection b is a good choice.There is no issue of regioselectivity during addition and the ketone will deactivate product so no over addition.-synthesis-

OClC–C

Cl

≡ ≡O

Cl O

Cl


.Disconnection a

Disconnection a is a poor choice.Why? Why would you choose this bond? Whilst it is actually possible to go down this route, it is not obvious, it is not anything we have taught you & it certainly does not follow on from the previous slides!-try again-

O O

1,1-C–C

O O


.Disconnection b

Disconnection b is a poor choice.You cannot be serious! I admire your creativity but really, this isn’t helping (or following my hints). What are you going to use as an electrophile? Reverse the synthon and it is still looking hard.-try again-

O O

1,1-C–C

O O


.Disconnection c

Disconnection c is not the best choice.Whilst possible, this won’t be the simplest solution, there is one much easier bond to form. One bond which we teach you in the first year...-try again-

O O

1,1-C–C

O O


.Disconnection e

Disconnection e is a poor choice.Whilst there are a number of interesting ways this might be achieved, it certainly isn’t straight forward. It ignores my hints and a very simple reaction.-try again-

O O O O

C–X


.Disconnection d

Disconnection d is a good choice.Esterification is a reliable reaction (& good at forming medium rings).The disconnection leaves an alcohol and an acid to use for further simplification.-next disconnection-

O O

C–X

O O≡

OHCO2H


.Disconnection a

Disconnection a is a poor choice.It would be hard to get selectivity in the addition of the Grignard reagent to the aldehyde in the presence of the acid.-try again-

OHCO2H

1,1-C–C

OHCO2H

≡ ≡

MgBrOCO2H


.Disconnection b

Disconnection b is a poor choice.Formation of the Grignard reagent in the same molecule as an acid could cause selectivity issues.-try again-

OHCO2H

1,1-C–C

OH CO2H≡ ≡

CO2HO

BrMg


.Disconnection c

Disconnection c is not the best choice.Whilst it would be possible to form the Grignard reagent, there are better ways to achieve this. The use of carbon dioxide to form the acid is good.-try again-

OHCO2H

1,1-C–C

OH

O

OH

≡ ≡

OHMgBr C OO


OHCO2H

FGI

reductionOH

CO2H

.Functional group interconversion

Functional group interconversion FGI is the best route. By incorporating the alkyne we can now readily simplify the rest of the molecule-retrosynthesis-


.Disconnection a

Disconnection a is ok but not the best choice...The problem is the alkylation of the diketone; we have a compound with 2 x electrophilic carbons and 2 x acidic protons. Could get cyclisation.-try again-

Cl

OMe

OO

Et

Et O

C–X

ether formation

Cl

OMe

OH

BrO

Et

Et O

a

1,2-C–C

BrBr

Et Et

O O

H H


.Disconnection b

Disconnection a is goodOnly potential problem arises with the dibromide (adding to phenols) but this can be minimised by controlling the stoichiometry-next example-

Cl

OMe

OO

Et

Et O

C–X

ether formation

1,2-C–C

BrBr

b

Et Et

O O

H H

Cl

OMe

OBr

Cl

OMe

OH


.Problem: self-condensation

Why did you say no?Surely, by now you have realised that I won’t have asked the question if the answer had been no...-the problem-

WRONG!terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Problem: self-condensation

Acetone is prone to self-condensation (it adds to itself readily)-the solution-

O O BASE

O OH

Oand/or


.Solution: functional group interconversion

β-Keto ester behaves like acetone then we can remove the unwanted ester by decarboxylation-retrosynthesis-

O ≡O

OEt

O


.Disconnection a

Why the f*$* have you chosen this disconnection?Did you read any of the previous slides?-try again-

O

1,2-C–C

O

The same as before!Really, that’s your choice...I give up...



First, we need the ester to prevent self-condensation.Then retrosynthesis is straightforward.-synthesis-

O

FGI

decarboxylationO

CO2Et

1,2-C–C

OCO2Et

≡Br OEt

O O


.Disconnection a

Disconnection a leads to a quick simplification of the problem-synthesis-

OMe

OH

OOH

1,3-diO

OMe

OH

OOH

≡ ≡

OMe

OH

OO


.Disconnection b

Why would you choose this bond in a section about the aldol reaction?-try again-

OMe

OH

OOH

C–C

OHOMe

OH

O

≡ ≡

O ?

OHOMe

OH

O

≡ ≡

? OMe

OH

O

X


MeO

EtO2C

HO

FGI

decarboxylation

MeO

EtO2C

HO

CO2Et


Functional group interconversion makes enolate formation easier. This makes subsequent disconnections easier.Prevents the self-condensation of ethyl acetate .-retrosynthesis-


.Disconnection a

Disconnection a is problematicPotentially, we could alkylate the alcohol with the allyl bromide.-try again-

MeO

EtO2C

HO

CO2Et

1,2-C–C

MeO

EtO2C

HO

CO2Et

≡ ≡

MeO

EtO2C

HO

BrCO2Et

H


.Disconnection b

Disconnection b is a good choiceReverse of the aldol reaction removes reactive alcohol & splits molecule in two. Both halves readily prepared.-synthesis-

MeO

EtO2C

HO

CO2Et

1,3-diOEtO2C

CO2Et

MeO

HO

≡ ≡

CO2EtEtO2C

MeO

O1,2-C–CCO2EtEtO2C

Br


.Disconnection a

Disconnection a is badUnlikely to be able to control chemoselectivity of allylation & you will probably observe allylation of the alcohol.Even if this step worked the next disconnection would fail due to self condensation!-try again-

MeO

EtO2C

HO

1,2-C–C

MeO

EtO2C

HO

≡ ≡

MeO

EtO2C

HO

H Br


.Disconnection b

Disconnection b seems okSplits molecule in two.Need FGI to add ester & prevent self-condensation.Shows that there is more than one right answer!-my choice of synthesis-

MeO

EtO2C

HO1,3-diO EtO2C

MeO

HO

≡ ≡

EtO2CMeO

OFGI

decarboxylation

EtO2C

CO2Et

1,2-C–C

BrCO2EtEtO2C


.Disconnection c

Disconnection c is a bad choiceUnlikely to be able to control chemoselectivity of Grignard reaction; addition to either carbonyl or deprotonation of acidic malonate-like position.-try again-

MeO

EtO2C

HO1,1-C–C

MeO

EtO2C

HO

≡ ≡

EtO2C

O MeO

BrMg


.Disconnection a

Disconnection a might work but...Potentially the amide functionality could disrupt the condensation.Cyclopropane rings are often reactive (ring strain). -try again-

FHN

O

α,β

aldol condensation F

O

HN

O


.Disconnection b

FHN

O

C–N

amide FCl

O

H2N

FGI

FOH

O

α,β

Knoevenagel condensation

FO

OH

O

Disconnection b betterRemove reactive functionality first then we can disconnection α,β-system. -synthesis-


.Disconnection a

Disconnection a is problematicWhilst the addition of the Grignard is attractive, the iminium species will tautomerise to an enamine, which is not electrophilic. -try again-

N O

O

N

1,1-C–C

NO

O

N

≡ ≡

NMgBr

O

O

N


.Disconnection b

Disconnection b is problematicI’m not sure if we would observe the desired substitution. It is likely that the amine would act as a base and cause elimination. If it didn’t how would we make the chloride (and this is quite useful as it will lead us to a faster route)-further disconnections-

N O

O

N

C–NN O

O

N

≡ ≡

HNN O

O

Cl


.Disconnection b: next steps

Halide from alcoholAlcohol allows simple 1,1-C–C disconnection, which takes us back to an aldehyde (good) and a strange Grignard reagent. How would we make this? -further disconnections-

N O

O

Cl

FGIN O

O

OH

1,1-C–C

NO

O

OH

≡N

OO

O

BrMg


.Disconnection b: next steps

Making the Grignard would be hard but simple FGI allows us to use a malonate. Problem: competing Knoevenagel condensation but this is actually good and shows us how we should proceed... -Try again (using this knowledge)-

N O

O

OH

C–O

acetal formation

N OH

OH

OH

NCO2Et

CO2Et

OH

1,2-C–CN

O

CO2EtEtO2C

FGI


.Disconnection c

Disconnection c is ok-ishAddition to the iminium ion should work. The iminium is dervied from the aldehyde. Making the Grignard reagent is going to be tougher-further disconnections-

N O

O

N

1,1-C–CN

O

O

N

≡ ≡

N

NO

O

BrMg


.Disconnection c: the next steps

FGI will allow us to use a malonate instead of the difficult Grignard.Our biggest problem would be competing Knoevenagel condensation but we could use to this to our advantage...-try again-

O

O

BrMg≡

O

O C–O

acetal formation

OH

OH

CO2Et

CO2Et

FGI

≡O

O


.Disconnection d

Disconnection d is rubbishThere is no reason to choose this disconnection. It does not simplify the problem. It does not have useful synthetic equivalent. If you honestly chose this there is a good chance you are going to fail...-try again-

N O

O

N

C–ON O

O

N


.Disconnection e

Disconnection e is a good choiceIn the introduction to this molecule, I said that the 1,3-diX disconnection would be useful. So the first thing we need to do is add the carbonyl group so that we set up the correct pattern-next step-

N O

O

N

2 x C–O

acetal formation

N OH

OH

N

FGIreduction

N O

O

N

EtO

OEt


.Disconnection c

Disconnection c is a possibilityThis might work. Should be able to get addition to iminium ion and hopefully, elimination of the amine will not be a problem but a more reliable route exists-try again-

N O

O

N

EtO

OEt

1,2-C–CN O

O

N

EtO

OEt

≡ ≡

N

NO

OEtO

OEt


.Disconnection b

Disconnection b is the best choiceAll reliable reactions that will lead to the desired product.-synthesis-

N O

O

N

EtO

OEt

1,3-diX N O

O

N

EtO

OEt

≡

N O

OHN

EtO

OEt

α,βN

OOO

EtO OEt


.Disconnection a

Why would you choose this disconnection? It does not simplify the problem and there is no reliable method of making this bond. You are going to fail... -try again-

why?terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection b

Disconnection b is ok-ishCuts molecule in half, which is good. Might have trouble opening the epoxide but certainly worth considering...Except this is a section on 1,3-aminoalcohols so use that disconnection!-try again-

MeO

NHO

MeO NHO

≡ ≡

MeO

MgBr

NO

(for practice, work out how to make the epoxide...)


.Disconnection c

Disconnection c is a bad choiceI can see where you are coming from but NO! How are you going to make the Grignard reagent (or anion)?-try again-

MeO

NHO

MeO

N

HO

≡ ≡

OMeO

NBrMg


.Disconnection d



.Disconnection e

Disconnection e is a bad choiceDoesn’t simplify the problem and doesn’t even set-up the disconnection we have been discussing!-try again-

MeO

NHO

MeO

NHO

≡ ≡

MeO

HNHO

Cl


.Disconnection f

Disconnection e is a good choiceThe initial disconnection may only remove methyl groups from the end of the molecule but it sets-up the 1,3-aminoalcohol disconnection.How would you make the nitrile? -new pattern-

MeO

NHO

2 x C–N

MeO

NH2HO

FGIreduction

MeO

HO

N1,3-aminoalcohol

MeON

O


.Disconnection a



.Disconnection b

Why would you choose this disconnection? It does not simplify the problem and there is no reliable method (that we have taught you) of making this bond. You are going to fail... -try again-


.Disconnection c

Disconnection c is a good choiceRemoved large part of molecule and one step closer 1,3-aminoketone pattern. How would we make this pattern from here?-next step-

Ph N

O

MeO

C–OPh N

O

MeO

≡ ≡

Ph NHO

MeOCl


.Disconnection c

The chloride can easily be taken back to the required ketone and thus we can use the Mannich disconnection.-synthesis-

Ph N

Cl FGIPh N

OH

FGIreduction

Ph N

O1,3-aminoketone

Mannich reaction

Ph

O

H

O

HHN


.Disconnection d

Disconnection d...really?Do any of these look like plausible synthons?-try again-

Ph N

O

MeOPh N

O

MeO

Ph N

O

MeO


.Disconnection e

Disconnection e not the best choiceYes, the disconnection works but it doesn’t simplify the problem or aid in simplifying the problem. It is a waste of time.Unless you were looking at an asymmetric synthesis and required an allylic alcohol (look at Sharpless’s synthesis of the related molecule, Prozac)-try again-

Ph N

O

MeO

Ph N

O

MeO


.Disconnection f

Disconnection f not the best choiceWe can complete the synthesis as shown...but it is long dull synthesis.No advantages to this route, it just shows there is more than one answer.-try again-

Ph N

O

MeO

Ph NH2

O

MeO

Ph

O

MeO

N

2 x C–O FGI

reduction

HO

MeO

Ph

ClN

FGI

Ph

OHN

1,1-C–CPh

O

N

C–O


.Disconnection a

Disconnection a is not good...Chemoselectivity is going to be an issue; which side will form the enolate & which site is the better electrophile-try again-

O O

Ph

O O

Ph

≡

O O

PhOEt

1,3-diCO


.Disconnection b

Disconnection b worksNo issue of chemoselectivity. I hope you can see why I only choose one set of synthons...-synthesis-

O O

Ph1,3-diCO

O O

Ph

≡ ≡

OO

PhCl


.Disconnection a

Disconnection a is not good...We have not taught you any reaction to make the requisite C–aryl bond (this is potentially a way of doing it but that is for a more advanced class) -try again-

NH

N O

O

EtNH

NO

O

Et


.Disconnection b

Disconnection b is ok...Removing the ethyl group is possible. It does not simplify our problem but on the plus side, the late stage addition of this substituent would allow the synthesis of analogues. I will not go through this route but you will hopefully see its similarity to the completed route (just swap two steps).-try again-

NH

N O

O

Et

1,2-C–CNH

N O

O

Et

≡ NH

N O

O

Et I


.Disconnection c

Disconnection c is goodRemoving the nitrogen sets up the 1,5-diO disconnection.But which bond, x or y, is it?

NH

N O

O

Et

2 x C–NCO2Et

CO2Et

NEt

NH3

x

y


.Disconnection y

Disconnection y is not your best choiceWhilst it is possible, it can’t proceed by a 1,5-diO disconnection and thus I don’t think you have been paying much attention...-try again-

CO2Et

CO2Et

NEt

1,5-diOCO2Et

CO2Et

NEt

≡ ≡

CO2Et?


.Disconnection x

Disconnection x is goodMolecule can simply be prepared by enolate chemistry.-synthesis-

CO2Et

CO2Et

NEt

1,5-diOCO2EtCO2Et

NEt

≡ ≡

CO2EtCO2EtN

Et1,2-C–C

CO2EtN

Et Br


.Disconnection 2

Disconnection 2Which is the correct set of synthons?

Ph Ph

OH Ph

OHPh

Ph

OHPh

1,1-C–C


.Disconnection 2

Disconnection 2-try again-

Ph Ph

OH Ph

OHPh

Ph

OHPh

1,1-C–CNo!You want an electrophilic carbon on alcohol so that you can use an aldehyde...its the most standard reaction, please learn it (or you will fail)


Ph Ph

OH Ph

OHPh

Ph

OHPh

1,1-C–C

≡ ≡

Ph

OPh

BrMg

.Disconnection 2

Disconnection 2Simple one step synthesis.-try a different disconnection- -finish-


.Disconnection 1

Disconnection 1This is the only set of synthons for this disconnection (unless you do an FGI first) and leads to a simple one step synthesis.-try a different disconnection- -finish-

Ph Ph

OH 1,1-C–CPh

Ph

OH

≡ ≡

Ph MgBrPh

O


.Disconnection 3

Disconnection 3Which is the correct set of synthons?-try a different disconnection- -finish-

Ph

OH

Ph

Ph Ph

OH1,2-C–C

Ph

OH

Ph


.Disconnection 3

Disconnection 3Possible but needs some work (protecting alcohol for instance)-try a different disconnection- -finish-

Ph

OH

Ph

okterminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection 3

Disconnection 3The epoxide gives yet another good route. So many ‘correct’ answers...-try a different disconnection- -finish-

Ph

OH

Ph

Ph Ph

OH1,2-C–C

≡ ≡

Ph

OH

Ph

PhPh

OBrMg


.Functional Group Interconversion

Functional Group Interconversion Opens up another useful collection of disconnections.Which one would you choose?

Ph Ph

OH FGI

reduction Ph Ph

Oa b c


.Disconnection c

Disconnection cWhich set of synthons is best?-try a different disconnection- -finish-

Ph Ph

OPh

1,3-C–CPh

O

PhPh

O


PhPh

O.Disconnection c

-try again-

No!How would you make the nucleophile (hint: you could not start with a halide, it would be unstable).


.Disconnection c

Disconnection cConjugate addition is a good method for the formation of this molecule. But, from a practical point of view you would probably need some copper present to encourage 1,4 over 1,2-addition.-try a different disconnection- -finish-

Ph Ph

OPh

1,3-C–C

≡ ≡

BrMgPh

Ph

O

PhPh

O

Ph

O


.Disconnection b

Disconnection bWhich set of synthons is best?-try a different disconnection- -finish-

Ph Ph

O Ph

1,2-C–CPh

O

PhPh

O


.Disconnection b

-try again-

How would you get chemoselectivity in the nucleophilic addition to a compound containing a carbonyl group?

PhPh

O

No!terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection b

Disconnection bSimple enolate chemistry offers the best route for this disconnection.-try a different disconnection- -finish-

Ph Ph

O Ph

1,2-C–C

≡ ≡

Ph

Ph

O

PhPh

O

Ph

OBr


Ph Ph

O1,1-C–C

Ph

OPh

Ph

OPh

.Disconnection a

Disconnection aWhich set of synthons is best?-try a different disconnection- -finish-


Ph

OPh

.Disconnection a

-try again-

Really, you that was the best choice? So what is your nucleophile and electrophile going to be? Do you remember any aromatic chemistry?

No!terminology guidelines aromatics aliphatics two group patterns C–C bonds

.Disconnection a


Disconnection aTo be honest, I don’t think that Friedel Crafts will work here. Cyclisation (or intramolecular Friedel Crafts) is going to compete...-try a different disconnection- -finish-

Ph Ph

O1,1-C–C

≡ ≡

PhH

Ph

OPh

Ph

OPh

Ph

O

Cl

retrosynthesis: 123.312

Education

group patterns cc bonds

terminology target

set of guidelines

various bonds

retrosynthetic analysisohi

terminology xya synthon

general idea

analysisgareth j rowlands123