heterocyclic chemistry guidance note and problems
DESCRIPTION
jTRANSCRIPT
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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. Paal-Knorr synthesis
Work out the furan, pyrrole or thiophene product arising from the reaction of 1,4-diketones in the Paal-Knorr reaction.
2. Knorr pyrrole synthesis Work out the product arising from a Knorr pyrrole synthesis. 3. Pyrazoles, isoxazoles and pyrimidines Work out the product(s) arising from the reaction of 1,3-diketones with (substituted)
hydrazine, hydroxylamine or amidines. 4. Pyridines
Work out the products arising from the reaction of 1,5-diketones with hydroxylamine.
5. 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.
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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.
ISOXAZOLE
1
2
34
5 NO
ISOXAZOLE
1
2
3
4 5
NO
NOT
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December 2012
1
2
3
4
5
N
N
6
PYRIMIDINE
1
2
3
4
5
N
N
6
PYRIMIDINE
1
2
3
4
5
N
N
6
4-CHLOROPYRIMIDINEand not
6-CHLOROPYRIMIDINE
Cl
BUT
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.
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December 2012
Next try out the following problems: A
1 2 3 4 5 6 7
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.
O O
conc. HCl
EtNH2
H+ cat.O
O
O
O
NH3
H+ cat.
H2S
H+ cat.O O
Ph
O
O
CH3NH2
H+ cat.
O
OH3C
conc. HCl
OO
H2S
H+ cat.
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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 1-
aminopyrroles 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-Knorrs [4+1] approach.
NH
NH
C
CC
CNH
C
CC
CNH
Paal-Knorr[4+1]
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 - to an ester group (two in this example) and this is reminiscent of an aldol product.
O
EtO2C
H3C
CH3
CO2Et
H
NH2 O
EtO2C
H3C
CH3
CO2Et
H
NH2
OHadd water
-hydroxycarbonyl
H
O
EtO2C
H3C
CH3
CO2Et
H
NH
H
Paal-Knorr intermediate
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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 -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
-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 -aminoketone unit, and the other component a carbonyl group with an active methylene (acidic CH2) group next door. Most often these reactants are 1,3-diketones or -keto esters.
Next try out the following problems: B 1
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
NH2
OO
+
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December 2012
2 3 4 5 6 7
H2N
O
O
+CO2H
O
H2N
OO
+
O2N How could you make the following pyrroles using the Knorr method? In each case identify the appropriate amino ketone and 1,3-dicarbonyl precursors.
NH
CO2CH3H3C
H3C CH3
NH
CH3
CH3
O
NH
CO2CH3
CH3
NH
O
O
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.
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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 NH2
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.
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December 2012
O
O
R
R
H2N
NH
H2N
+
NH
N
NH
R
R
O
O
R
R
H2N
O
H2N
+
NH
N
O
R
R
O
O
+
H2N
O
HN
R
N
N
O
R
N
N
NH2
R
R
R
R
R
R
N
N
OH
R
R
preferred tautomer
preferred tautomer
no tautomer!
Next try out the following problems:
C
1 2 3 4
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.
NH2NH2O O
H+ cat.
NH2OH
OO
NO2MeOH+ cat.
H3C CH3
O O
CH3
H2N CH3
NH
H+ cat.
CH3NHNH2H
OOH+ cat.
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December 2012
5 6 7
NH2OH
O
OPh
H+ cat.
N
O O
Ph
MeNH-NH2H+ catalyst
H2N
NH
H+ cat.N
O O
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).
NN
H
Hpyridine 1,4-dihydropyridine
1
2
3
4
Synthesis of the aromatic pyridine in one go by this method would require a 1,5-diketone 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.
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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 -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 ,-unsaturated carbonyl compound which can then be subjected to a conjugate addition reaction using an enamine made from a different -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 -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
1
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.
NH2OH
H+
O O
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2 3 4 5 6 7
NH2OH
OO H+ cat.
O
O
NH2OH
H+ cat.
OH3C
pH 8.5O
Ph
O
H3C
H NH3
+
2 eq. 1 eq.
CO2MepH 8.5OO
Ph H NH3
+
2 eq. 1 eq.N
O
EtO2C
Ph H2N
pH 8.5
+
CH3 O
O
EtO2C
Ph H2N CH3
pH 8.5+
O
Ph
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.
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December 2012
Past paper questions In the final exam two thirds (15 marks) of the heterocyclic / heteroaromatic question will come from Prof Boyles 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.
Bconc. HCl
O
O C12H20O
CNH3
H+ cat. C10H15N
DPh
O
O
CO2Et
C26H31NO2
PhCH2NH2
H+ cat.
O O
(i)
(ii)
(iii)
(2 5 marks) Q2 Consider the scheme below:
D
NH2NH2OE
C6H10N2
O
O O
NO2
MeNHNH2G
C11H11N3O2
C
F
cat. H+
cat. H+
HN N
+
(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)