Nucleophilic Aromatic Substitution of Heteroaromatics Flashcards
(27 cards)
Identify the product formed in the reaction shown below
answer = A
It would be D, but it is not possible to delocalise the charge onto the nitro group
What is an imine
a compound with a C=N double bond
Where the carbon of the C=N is electrophilic
If the following imine is subject to nucleophilic attack, what occurs
The halide is readily displaced
* N=C double bond will break when Nu attack
* But then reform which will boot of the Cl
If the following imine is added to a base, what will occur?
The α-protons are acid
* Base will react with α-proton, causing the electron to be push back onto carbon and then a C=C bond to form
* Causes the imine C=N double bond to break
Why does nucleophilic attack occur relatively easily in pyridines
Because of their imine-like properties
(same for other 6-membered heteroaromatics)
The aromatic pi-system is electron-deficient because of the electronegative N (electrophilic)
Why does N-Alkylation or N-acylation dramatically increase the ease of attack
Increasing the number of N atoms, makes the ring even more electron deficient, and more susceptible to nucleophilic attack
Pyridine is most electrophilic at which positions
the two and four positions
(nitrogen should have a negative charge)
The Chichibabin reaction in a nucleophilic aromatic substitution on pyridine to form 2-aminopyridine
(other powerful nucleophiles react by a similar mechanism)
What is the mechanism for this reaction
- ⁻NH₂ will attack at the carbon at the 2 position (remember imine) breaking the C=N double bond, and pushing the e- back onto N
- N=C reforms, which boots of the H on the tertiary carbon
- formed in high yield upon quenching sodium salt with water
What affect does a halide substituent in the 2 position on pyridine have on the rate of nucelophilic aromatic substitution
They are more reative towards nucleophiles
How does 2-chloropyridine react when added to the methoxide nucleophile?
- The methyoxy will attack the imine carbon, causing the C=N double bond to break and e- will transfer onto N
- C=N double bond will reform and boot off the chlorine
How does phenyllithium act as a nucleophile
C-Li bond break, resulting in a negative charge on the carbon
How do resonance effect enable nucleophilic aromatic substitution of 4-chloropyridine
Negative charge will delocalise down onto nitrogen
Then C=N double bond will reform and boot of Cl
Why is 3-chloropyridine much less reactive for nucleophilic aromatic substitution?
Because the negative charge cannot delocalise onto nitrogen
(however still considerably more reactive than chlorobenzene)
Pyridine-N-oxide is more susceptible to electrophilic attack than pyridine
It is also more susceptible to…
Nucleophilic attack
How does the reaction occur when pyridine-N-oxide is added to POCl₃
- LP on O will attack the phosphate forming a strong O-P bond (nucleophilic attack) and a Cl- being lost from the POCl₃
- The Cl- will then attack the pyridine at the 2 position causing the C=N double bond to break
- A proton on leaves off carbon 2, which allows the C=N double bond to reform and the N-O bond is broken
Pyridines readily oxidise to corresponding N-oxide
What about pyrroles?
They are very unstable and usually just degrade
Why do 5-membered heterocycles readily undergo electrophilic attack?
5-membered heterocycles are “pi-excessive” (higher pi-electron density) meaning they are more nucelophilic than benzene
Why is nucleophilic attack rare for simple pyrroles
How is it overcome?
Difficult on electron rich rings like pyrrole
More common when electron-withdrawing groups are present
What is an Azole
Azoles are a class of five-membered heterocyclic compounds containing a nitrogen atom and at least one other non-carbon atom
This compound was froma previous example pf nucleophilic substitution
How can we alter the substituents on the thiophene to make nucleophilic attack easier
- By forming an azole with a C=N double bond, nucleophic attack is easier
- Better LGs
How does Nucleophilic attack on the following compound occur?
- The nucleophile attacks the electrophilic carbon, which causes the C=N double bond to break and the electron to be pushed back onto N
- The C=N bond reforms which boots off the bromide
Identify the product(s) formed in the follwoing reaction
Answer = A + C
For the following electrophilic additions, the top reaction will direct in the 3 position because the aromaticity of benzene is not disturbed, and the bottom reaction will meta direct because the carbonyl is electron withdrawing
How can we extend these reactivity modes available
lithiation
n-Butyllithium can deprotonate aromatic compounds and form lithiated rings (nucleophiles) with can then react with electrophiles
5-membered herteroaromatic with ortho-lithiate where?
lithiation occurs next to the heteroatom
* A preliminary bond is formed between the heteroatom and the lithium
* the butyl part of the chain will deprotonate on the adjacent carbon and the LP now on that carbon is the used to form a bond with lithium
