December 2012

Heterocyclic Chemistry CONTENTS OVERVIEW Synthesis of heterocycles from 1,4-dicarbonyl compounds (furans, pyrroles, thiophenes and pyridazines) Synthesis of heterocycles from 1,3-dicarbonyl compounds (pyrroles, isoxazoles, pyrazoles and pyrimidines) Synthesis of heterocycles from 1,5-dicarbonyl compounds (pyridines and dihydropyridines)

Learning outcomes: At the end of the course you should be able to use reaction mechanisms to: 1.

2. 3.

4.

5.

Paal-Knorr synthesis Work out the furan, pyrrole or thiophene product arising from the reaction of 1,4diketones in the Paal-Knorr reaction. Knorr pyrrole synthesis Work out the product arising from a Knorr pyrrole synthesis. Pyrazoles, isoxazoles and pyrimidines Work out the product(s) arising from the reaction of 1,3-diketones with (substituted) hydrazine, hydroxylamine or amidines. Pyridines Work out the products arising from the reaction of 1,5-diketones with hydroxylamine. Hantzsch pyridine synthesis Work out the products arising from a Hantzsch pyridine synthesis.

Suggested reading Organic Chemistry, 1st Edition, J. Clayden, N. Greeves, S. Warren and P. Wothers, Oxford University Press. Mainly Chapter 44 (Chapters 29 and 30 in 2nd ed.). Aromatic Heterocyclic Chemistry, D. T. Davies, (Oxford Primer No. 2): (QD 400 D2) useful in parts More advanced texts: Heterocyclic Chemistry, 2nd Ed., T. L. Gilchrist and Heterocyclic Chemistry, 3rd Ed., J. A. Joules, K. Mills and G. F. Smith (QD 400 J8) See http://www.hull.ac.uk/php/chsanb/teaching.html for past exam paper questions with answers. Any comments on or feedback on the usefulness of these notes are gratefully received. Dr AN Boa, [email protected], 01482 465022.

December 2012

Introduction This course is closely linked to the “Bifunctional Chemistry” lectures. You will find out how 1,n-dicarbonyl species (n = 3, 4 and 5), made using bifunctional chemistry, can be converted into a range of heterocycles using simple condensation reactions. There is some key knowledge prerequisites needed for success with this course. It is therefore worthwhile reviewing your notes for Dr Eames’ “Chemistry of the carbonyl group” before starting these notes. Specifically this course relies heavily on the following: 1. Condensation of an amine nucleophile with a ketone or aldehyde using acid catalysis to yield an imine, oxime or hydrazone. 2. Acid-catalysed elimination of an alcohol. Also needed is familiarity of the following topics: 3. Tautomerism (keto-enol and imine-enamine) 4. Aldol-type condensation reactions (e.g. the Knoevenagel condensation) 5. The conjugate addition reaction Topics 3, 4 and 5 are covered in detail in the bifunctional chemistry course. It is therefore wise to tackle at least the first half of that course before tackling this material. In all cases full mechanisms are shown for the specific examples covered. The notes are split into small sections but, as the course overview above indicates, these can be further grouped into the chemistry of 1,3-, 1,4- and 1,5-dicarbonyl compounds. Parts 2 and 3 cover reactions of 1,4- dicarbonyl compounds, parts 5 and 6 cover reactions of 1,3- dicarbonyl compounds with part 7 covering 1,5-dicarbonyl compounds. Parts 4 and 8 cover syntheses which are modified versions of reactions seen earlier. These are more versatile syntheses and allow for the preparation of a wider range of unsymmetrically substituted heterocycles. Slide 1a Numbering begins with the heteroatom of highest priority and proceeds either clockwise or anticlockwise depending on other heteroatoms or substituents. Deal with the heteroatoms in the ring first and then the substituents such that the numbers for the substituents are kept as low as possible. 4 5

3 O 1

N2

ISOXAZOLE

4 NOT

5 N

3 O 2

1

ISOXAZOLE

December 2012 Cl 6

4 5 6

N 3

2 N 1 PYRIMIDINE

5

4 N 1

2 N 3 PYRIMIDINE 4

BUT

5

N 3

6

2 N 1 4-CHLOROPYRIMIDINE and not 6-CHLOROPYRIMIDINE

Slide 1b Re-read your notes on the chemistry of the carbonyl group! Slide 2a Some material is linked to the bifunctional chemistry course. More information on these reactions can be found in those course notes. Reference to that material is indicated in these notes like this: Bif3a = bifunctional chemistry notes slide 3a. Slide 2b The production of the furan (or thiophene, pyrrole) using acid catalysis is the thermodynamic process; the stable aromatic ring is made (this is an important feature in heteroaromatic syntheses). The base catalysed route is the kinetic process (faster reaction). The latter process, an aldol reaction, proceeds via the less substituted enolate which is produced faster than the more substitted enolate (Bif6b). Slide 3a Step 1 is an acid catalysed enolisation (Bif6ab). In the cyclisation step, protonation of the C=O makes the carbonyl carbon more electrophilic and thus susceptible to attack by the enol OH. To finish off the reaction there is a proton tansfer step which assists in neutralising charge (remember the cyclisation step is acid catalysed) and it also helps to eliminate water. In such cyclisations the ring size is important. Generally 5 membered rings are formed fastest (even if 6 membered rings are more stable). Note that the final reaction product has a molecular formula which is short of H2O compared to the starting diketone; the reaction could be described as a cyclodehydration. Slide 3b The Paal Knorr method for making thiophenes and pyrroles is essentially the same as that for making furans. However instead of an enol as the key cyclisation precursor, there is a thioenol or enamine (the structure in blue on the left). These can be easily made using the same starting material. Slide 4a This slide shows some examples of the Paal Knorr method. To spot these reactions look first for a 1,4-diketone (or 1,4-ketoaldehyde etc). Then note whether H2S (thiophene) or ammonia/ammonium ion/primary amine (pyrrole) is a reactant. Use these examples to practice drawing the reaction mechanism using the information in slides 3a or 3b as a template.

December 2012 Next try out the following problems: A

PROBLEMS – Paal Knorr synthesis Work out the products for the following heterocyclic syntheses, using your lecture notes if needed, and identifying any key intermediates in the sequence.

1

conc. HCl O

O

2

O

EtNH2 H+ cat.

O

3

O

NH3 H+ cat.

O

4

H2S H+ cat.

O O

5

O

CH3NH2 Ph O

6

H3C

H+ cat.

O conc. HCl

O

H2S

7

H+ cat. O O

December 2012 Slide 4b The start of this slide shows the use of an approach called retrosynthetic analysis to examine how the pyridazine may be made (cf. Bif29a,b). This is a “paper chemistry” method used to analyse possible ways to make a molecule and does not represent actual transformations (note the use of special retrosynthesis arrows). What this reveals is that a pyridazine might be formed by oxidation of a dihydropyridazine. This dihydropyridazine contains two linked enamine-like groups, each of which is the tautomeric form of a C=N compound (a bis-hydrazone). We have already seen how imines can be made by condensation of a nitrogen nucleophile (slide 3b) and using hydrazine (NH2NH2) with a 1,4-diketone will afford a hydrazone. Slide 5a Once a mono-hydrazone has formed then the free NH2 can cyclise onto the second carbonyl group to form another C=N bond (with loss of anther molecule of water). The bis-hydrazone (or tautomer) can be easily oxidised to the aromatic pyridazine. Slide 5b Look back at the key cyclisation step in slide 5a: a 1,6 ring closure is seen. So what of the fast 1,5- ring closure mentioned above (slide 3a)? It turns out that the monohydrazone can ring close in a 1,5- fashion (a Paal-Knorr reaction) and thus 1aminopyrroles are often seen as by-products of the reaction. Slide 6a The Paal-Knorr synthesis is entirely dependent on the availability of 1,4-dicarbonyl species. In cases of more sustituted variants their preparation is in fact not trivial. The Knorr pyrrole synthesis is a related reaction which can cicumvent this problem. It takes a [3+2] approach of making the heterocyclic ring instead of the Paal-Knorr’s [4+1] approach.

C

C

C

C

C N H Paal-Knorr [4+1]

N H

C

C C NH

N H

Knorr pyrrole [3+2]

The retrosynthetic analysis shows that instead of dealing with two imine condensations (Paal-Knorr), the enamine C=C could have come from elimination of an alcohol. This alcohol is β- to an ester group (two in this example) and this is reminiscent of an aldol product. EtO2C H

EtO2C H

CH3 H

H3C

CO2Et O

NH

Paal-Knorr intermediate

CH3

H3C

add water

CO2Et O

NH2

EtO2C H H3C O

CH3 OH H CO2Et NH2

-hydroxycarbonyl

December 2012 To summarise, the Knorr pyrrole synthesis involves an imine condensation step, an aldol step and an elimination of water. As seen on the next slide the order is not necessarily as implied from the retrosynthetic analysis, and the aldol-like reaction involves an enamine instead of the more commonly encountered enol. Slide 6b The mechanism of the Knorr pyrrole synthesis is shown in full. The condensation of the amine with the ketone (imine/enamine formation) is in fact the first step as addition to C=O bond is normally a rapid reaction. Next is an enamine aldol-type reaction followed by loss of water give the pyrrole. In this example the classical (“symmetric”) Knorr synthesis is shown. Note that the R1 and R2 groups are present on both fragments. The amine component can in fact be made from the ketone component by a process of oxime formation followed by reduction (Bif10b, and 7a below). Thus two moles of ketone are reacted with one mole of sodium nitrite/acid to form a 50:50 mixture of starting material and oxime. Selective reduction of the latter gives the aminoketone in the presence of the unreacted ketone starting material and then the Knorr cyclisation can proceed from this point. Slide 7a The Knorr pyrrole synthesis requires an aminoketone as one reactant, and a ketone (or aldehyde) with an active methylene group (acidic α-CH2) as the other. In the examples above the aminoketone has the same carbon skeleton as the active methylene component; indeed the aminoketone can be made from the active methylene component in situ. However this is not a requirement and the aminoketone may be made separately. This approach is useful as it allows for the preparation of unsymmetrically substituted pyrroles. This slide shows some methods for making aminoketones, both of which rely on enol chemistry (Bif11a,Bif13a) to introduce the α-functionality. Slide 7b This slide shows some examples of the Knorr pyrrole synthesis. Use these examples to practice drawing the reaction mechanism by using the information in slide 6b as a template. Remember that one component must have an α-aminoketone unit, and the other component a carbonyl group with an active methylene (acidic CH 2) group ‘next door’. Most often these reactants are 1,3-diketones or β-keto esters. Next try out the following problems: B

PROBLEMS – Knorr pyrrole synthesis Work out the products for the following heterocyclic syntheses, using your lecture notes if needed, and identifying any key intermediates in the sequence. CO2Me

1 O NH2

+ O

December 2012

2

O +

O H2N

CO2H

O

3 O

O +

H2N

O2N

4 How could you make the following pyrroles using the Knorr method? In each case identify the appropriate amino ketone and 1,3-dicarbonyl precursors. H 3C

CO2CH3

H3C

5

N H

CH3

O CH3

N H

CH3

6 CO2CH3 CH3

N H

7

O O N H

Slide 8a Isoxazoles and pyrazoles are 5-membered heterocycles with two neighbouring heteroatoms (N and O for isoxazoles, and two Ns for pyrazoles). They can be made from 1,3-dicarbonyl reactants using hydroxylamine (NH2OH) or hydrazine (NH2NH2). respectively. The pyrazoles can be made by double imine formation, whereas isoxzoles are formed by one imine condesation and an addition to the second C=O followed by loss of water.

December 2012 Slide 8b The difference in the mechanisms for the synthesis of isoxazoles and pyrazoles, as perhaps implied above, are in fact negligible. This is shown in the common mechanism illustrated in this slide. Note how the reaction is like the Paal-Knorr reaction in mechanism, but like the Knorr reaction in that it is a 3+2 process (instead of 4+1). Slide 9a With unsymmetrical 1,3-diketones isomeric isoxazoles may be formed as either ketone may be attacked initially by the NH2 of hydroxylamine. This is not a problem for pyrazoles as hydrazine itself is symmetrical. The two products which may be formulated are in any case tautomeric structures. Thus it is the thermodynamic stabilty of each tautomer whiich will determine which is isolated (or if there is a mixture) rather than which C=O group is attacked first. However, if an N-subsituted hydrazine is used (RNH-NH2), then the situation is like hydroxylamine. Isomeric (not tautomeric) products may be obtained and these cannot interconvert. Slide 9b In the case of isoxazole and pyrazole formation, the problem of isomer formation described above will not be an issue if one of the carbonyl groups is sufficiently more electrophilic than the other. The more electrophilic C=O will be attacked first by the NH 2 of hydroxylamine or the less hindered NH2 of RNH-NH2. Slide 10a Pyrimidines are six-membered aromatic rings with two nitrogen atoms in a 1,3 relationship. They can be made from 1,3-dicarbonyl compounds by a mechanism which is identical to the synthesis of pyrazoles (2 × imine formation), but in place of hydrazine (NH2NH2) an amidine is used RC(=NH)NH2. This is a 3+3 type cyclisation process. The related pyrimidin-2-ones can be made using urea H2NC(=O)NH2 as the nitrogen containing component. The reactions can be conducted using either acid or base catalysis. Slide 10b This slide shows the mechanism of the pyrimidine synthesis reaction Slide 11a Some examples are shown on this slide. It is important to note that in making pyrimidines (from amidines), pyrimidin-2-ones (from urea), or 2-aminopyrimidines (from guanidine) the final structure from each may be in a different tautomeric form than perhaps expected from another example. The ability or not to tautomerise may change the mechanism slightly, although the essential condensation steps (between N and C=O) will be the same. The actual tautomeric structure obtained will in most cases be dictated by which is the thermodynamically stable and this is not obvious to the novice (i.e. you wont need to know). As with the case of N-alkylhydrazines used in pyrazole syntheses, use of N-alkylureas will mean that tautomeric structures are not accessible at all compared to if an unsubstitued urea had been used.

December 2012

R

R H2N

O

N

+

N

NH

NH

H2 N

O

R NH2

NH

R

R

R

R

R

R

H2N

O

N O

H2 N

O

R

R

R

preferred tautomer

O

O

+

R

no tautomer!

N

HN R

R

R

N

N

H2N

O

OH

NH

R

O

N

O

+

N preferred tautomer

R

Next try out the following problems: C

PROBLEMS – Heterocycles with two heteroatoms from 1,3-dicarbonyls Work out the products for the following heterocyclic syntheses, using your lecture notes if needed, and identifying any key intermediates in the sequence.

1

O

O

NH2NH2 H+ cat.

O

2

O NH2OH NO2

MeO

NH O

3

O H2N

H3C

CH3 H+ cat.

CH3

4

H O

CH3

O

CH3NHNH2 H+ cat.

H+ cat.

December 2012

Ph

O

NH2OH

5

H+ cat. O

O

O MeNH-NH2 H+ catalyst

6 N Ph

NH

7

O

O

H2N

H+ cat.

N Ph

Slide 11b Building on the Paal-Knorr mechanism, this slide shows how a 1,5-diketone can be used to make a 1,4-dihydropyridine from an amine (typically ammonia). H 4 3

N pyridine

2

1N

H 1,4-dihydropyridine

Synthesis of the aromatic pyridine in one go by this method would require a 1,5diketone with a double bond, and this would need to be of the cis configuration in order to get the ring to form. So it is often easier to make the 1,4-dihydropyridine and oxidise this to the aromatic pyridine in a separate step. Slide 12a A slightly modified approach to pyridines uses hydroxylamine as the nitrogen nucleophile. The N-hydroxy-1,4-dihydropyridine is initially produced but this can then eliminate water under the acidic condition used for the reaction. I.e. the extra double bond is introduced in the cyclised product and so the separate oxidation step mentioned in the previous slide is not needed.

December 2012 Slide 12b The Hantzsch synthesis of dihydropyridines is a multi-component approach which attempts to bypass the problem of making complicated substituted 1,5-diketones. Symmetrical 1,4-dihydropyridines can be made in a one flask process (slide 13a) using a β-keto ester, aldehyde and amine. Certain 1,4-dihydropyridines like this are important cardiovascular drugs, e.g. Nifedipine and the Hantzsch method is very useful for their preparation. Unsymmetrical 1,4-dihydropyridines can be made by separating the reaction into distinct steps (slide 13b) for each part of the overall transformation. Thus an aldol-type process produces an α,β-unsaturated carbonyl compound which can then be subjected to a conjugate addition reaction using an enamine made from a different β-keto ester. Slide 13a This slide shows the mechanism of the Hantzsch pyridine synthesis. Note how the long mechanism is in fact made up of a series of much simpler steps that should all familiar from earlier parts of this course (an indeed previous years). This typifies heterocyclic chemistry mechanisms; if you know your basic reactions and processes (imine formation, tautomerisation, aldol, conjugate addition etc) then the overall transformation will be much more meaningful and easier to follow. Slide 13b For unsymmetrical 1,4-dihydropyridines note how the enone is made separately from the enamine component and thus can have a different substitution pattern. In this case the second β-keto ester has been pre-condensed with ammonia introducing the nitrogen atom early on. When these two are brought together the enamine reacts as a nucleophile in the same conjugate addition step as the enolate seen above (13a). You will remember from the bifunctional chemistry course how enamine conjugate addition reactions are similar to enol/enolate reactions (Bif36b). Slide 14a This slide shows some examples of the Hantzsch synthesis. Use these examples to practice drawing the reaction mechanism by using the information in slide 13a/b as a template. Also be sure to work out how and fron what the two components, i.e. the enone and enamine, are made from simpler precursors. Next try out the following problems: D

PROBLEMS – Pyridines and the Hantzsch synthesis Work out the products for the following heterocyclic syntheses, using your lecture notes if needed, and identifying any key intermediates in the sequence.

1

O

O

NH2OH H+

December 2012 2 NH2OH O

H+ cat.

O

3 NH2OH

O

H+ cat. O

O

4

pH 8.5

O H3C

+ H

2 eq.

5

O

NH3

Ph

O

H3C

1 eq.

CO2Me

pH 8.5

O +

NH3

H

Ph

N 2 eq.

1 eq.

CH3

6

O

EtO2C

pH 8.5 +

Ph

H2N

O

O

7

pH 8.5

EtO2C Ph

Ph

+ O

H2N

CH3

Slides 14b, 15a and 15b These slides depict a succinct summary of the reactions covered in the course organised on the basis of 1,n-dicarbonyl precursor.

December 2012 Past paper questions In the final exam two thirds (15 marks) of the heterocyclic / heteroaromatic question will come from Prof Boyle’s lectures, and one third from this course (10 marks). Q1 For TWO of the following parts work out the structure of the unknown product. Mechanisms are not required but you should draw the key intermediates to explain your answer. O

conc. HCl B

(i)

C12H20O

O

NH3

C

(ii) H+ cat.

O O

O

CO2Et Ph

(iii)

C10H15N

PhCH2NH2 D C26H31NO2

+

H cat.

O

(2 × 5 marks) Q2

Consider the scheme below: O

NH2NH2

O

HN

N +

cat. H+ D

C

O

MeNHNH2

O

cat. H+ F

E C6H10N2

G C11H11N3O2

NO2

(i)

Draw the structure of heterocycle E which is produced with D in the reaction of 1,3-diketone C with hydrazine.

(i)

What is the relationship between D and E?

(ii)

Work out the structure of G from reaction of 1,3-diketone F with methylhydrazine, and explain why G is the only heterocycle formed in this reaction. (3, 1, 6 marks)