Electrophilic Aromatic Substitution of Heteroaromatics Flashcards
Define how pyridine will react
- Pyridine is nucleophilic on the nitrogen, due to the LP being delocalised
- It reacts with a variety of electrophiles, Lewis Acids and metals
Pyridine does not readily undergo electrophilic aromatic substituition
Why?
- Kinetics: the kinetic product is made through attack at N, which forms a cationic intermediate that is innert to further electrophilic attack
- Thermodynamics: the nitrogen lowers the HOMO compared to benzene = ring is electron deficient and less nucleophilic
Which site of pyridine is favoured for electrophilic attack
- The electronegative N effectively withdraws electron density away from the carbon
- Hence looking at the Whetland intermediates, the 3-position is favoured as the site of electrophilic attack
What are the issues with electrophilic aromatic substitutions of pyridine
How can we overcome this
- Yields are often poor
(e.g. Friedel-Craft reactions usually fail) - Useful electrophilic aromatic substitution occurs on activated pyridines (e.g. electron donating groups)
If you have an Electron withdrawing and an Electron donating group on the same system, who wins?
The effects of the electron donating group will always win
(Hence adding an EDG to pyridine electrophilic substitution can occur)
Useful electrophilic aromatic substitution occurs on activated pyridines
An example is N-oxides
in this stable dipolar species, oxygen acts as…
- a protecting group (blocks reaction on nitrogen)
- activating group (O electrons are delocalised around the ring: raises HOMO, and increasing nucleophilicity)
Pyridine N-oxide can attack electrophiles in which positions
on para- and ortho-positions
How do you oxidise nitrogen if you want electrophilic aromatic substitution to occur?
using either mCPBA or H₂O₂
How would the nitration of Pyridine N-oxide occur
- Oxygen electrons will delocalise within the aromatic ring resulting in…
- Electron from a double bond will attack nitronium ion (at para-position)
- C-H bond breaks reforming C=C double bond, restoring aromaticity and electron density returns to oxygen
How can you remove the oxygen from the pyridine N-oxide, once the electrophilic aromatic substitution has occured
- Reduction with PPh₃, PCl₃, P(OMe)₃ using phosphorus to knock the oxygen off
- The driving force is the strong O=P bond
The following compounds are Quinoline and Isoquinoline
They undergo electrophilic aromatic substitution
Where on the rings does this occur
- On the more electron-rich benzene ring due to benzene being more nucelophilic than pyridine
- Usually at positions 5 and 8
Pyrrole, furan and thiophene are ‘electron-rich’ or ‘electron-poor’ compared to benzene
Electron rich compared to benezene
They readily undergo electrophilic aromatic substitution
Why are 5-membered Heteroaromatic so electron rich
Because the LP on the heteroatom can delocalise into the atomatic ring
Resulting in negative charge which can delocalise across carbons on the ring
Hence this is why electrophilic aromatic substitution can readily occur
Which position is generally preferred in the electrophilic aromatic substituion of 5-membered Heteroaromatic and why?
- 2-substitution is generally preferred
due to stabilisation of the carbocation - Attacl om the 2-position results in 3 resonance forms = more stable
Out of the main 3 5-membered heteraromatics we have looked at, which one reacts the fastest during electrophilic aromatic substitution
Pyrrole reacts the fastest (N)
Followed by Furan (O) and Thiophene (S)
Difference in reactivity due to:
* Electronegative (O>N»S)
* And high resonance stabilisation of Thiophene
Pyrrole is so reactive that it can be difficult to selectionaly functionalise
How can we over come this
You have to reduce reactivity or change reaction condition if you want pyrrole to only react in one position
The Vilsmeier-Haack reaction is a (Lewis) acid-free alternative to selectively mono-formylate (add and aldehyde group to) electron-rich aromatics
How is the electrophile formed?
- React dimethylformamide with phosphorus oxychloride
- The nitrogen electron delocalise forming a C=N bond and the electrons from the C=O double bond attack the δ⁺P
- This forms a bond between O-P and knocks a chloride off
- The chloride then attacks the imine carbon (C=N), causing the C-O bond to break
The Vilsmeier-Haack reaction is a (Lewis) acid-free alternative to selectively mono-formylate (add and aldehyde group to) electron-rich aromatics
How does pyrrole attack the nucleophile?
- Electrons from the nitroge delocalise into the ring
- Causes electrons from the C=C bond to attack the imine carbon (C=N)
- Results in chloride as a leaving group
Substituion can also occur of the NH of the pyrrole
Best yields are obtained using?
How will this compound react with the pyrrole?
- Best yeilds are obtained with a potassium salt
- It will act as a base, deprotonating the NH
Why can standard Friedel-Crafts conditions be used for Furan?
Because Furan is a lot less reactive towards electrophiles than pyrrole
Why can Furan also undergo electrophilic additions as well as electrophilic aromatic substitutions
- Furan is ‘weakly aromatic’
- Meaning if strong bonds (e.g. C-O) can be formed, addition reactions that breaks aromaticity may be preferred over substitution
How might the electrophilic addition of furan occur then?
- The electrons in furan will delocalise within the ring, resulting in electrons from a C=C bond to attack the δ⁺Br
- C-Br bond is formed at the 2-position
- The MeOH will attack the carbonyl carbon, breaking a C=O, and LP return to oxygen
- oxygen uses LP to reform C=O, resulting in the breaking of the C-Br
- Another MeOH will attack the newly formed carbonyl carbon, breaking the C=O
Furan can also act as a diene in which type of reaction?
Diels-Alder reactions
An indole is shown below
Where does electrophilic aromatic substitution occur and why?
- Electrophilic aromatic substituion occurs on the pyrrole ring (reactive pyrrole ring, relatively unreactive benzene ring)
- Favours the 3-position (aromaticity of benzene ring is not disturbed)
