Synthesis of Heteroaromatics Flashcards
Identify the product formed in the reaction shown below
Answer = C
What are the 3 factors which allows for the synthesis of Heteroaromatic systems?
- Formation on strong C-O, C-N and C-S bond (enthalpy)
- Formation of a more stable (aromatic) 5- and 6-membered ring
- Intramolecular reactions are favoured over intermolecular reactions (don’t have to wait for the molecules to diffuse together - entropy)
the synthesis of pyrroles starts off with which compound?
1,4-dicarbonyl and amines
What are the two main steps within carbonyl synthesis
- Step 1: would be to form an enamine from a carbonyl compound and amine
- Step 2: nitrogen forms a enamine on each side of the ring to form pyrrole
Paal Knorr Synthesis, is the easiest way to make 5-membered heterocycles (pyrrole)
What are the first 4 steps?
- NH₃ (Nu) attack the carbonyl carbon, breaking the C=O dbl bond
- PT between the NH₃ and the O-
- intramolecular nucleophilic attack from the newly added NH₂ and the other carbonyl carbon, breaking the C=O double bond
- PT between NH₂ and O-
- Then uses acid to eliminate water as a leaving group
Paal Knorr Synthesis, is the easiest way to make 5-membered heterocycles (pyrrole)
What happens after the first 4 steps
- Subsequent deprotonation on the 3 and 4 positions
- And using acid to loose water as a leaving group on the 2 and 5 positions
What would you need for the synthesis of furan?
- 1,4-dicarbonyl and an acid (H⁺)
What would you need for the synthesis for thiophene
1,4-dicarbonyl and P₄S₁₀ (or Lawesson’s reagent)
(driving force is the strong P-O bond)
If 1,4-dicarbonyls are not readily available for heteroaromatic synthesis, what instead can we use?
We can combine 2-carbon units
(benefit because with breaking it up into pieces we can have more variety through different substituents)
What is the need for a halogen on one of the reagents?
- An enamine is made from one of the reagents, then is used to displace the halogen on the other reagent (as a leaving group)
- Overall forming a 1,4-dicarbonyl from two carbon fragments
Describe the first 3 steps of pyrrole synthesis from two carbon units
- Enamine formation from one of the carbonyl fragements and NH₃
- Amine group on the enamine forms a C=N dbl bond causing e- from the alkene bond to attack the electrophilic carbon on the other carbon unit
- Causes chlorine to leave
- Imine units internally attacks the carbonyl carbon and breaks C=O dbl bond
Describe what happens after the first 3 steps in the synthesis of pyrrole from 2 carbon units
- Subsequent loss of hydroxyl groups and deprotonation
The following molecule below is a Azole
Suggest 2 different carbon fragments required to form it
- One fragment with a carboxyl group and a halide
- Another fragment with an amine and the desired heteroatom
6-membered heteroaromatics can also be synthesised from dicarbonyls
How would you make the 1,5-dicarbonyl required required for the synthesis
1,5-dicarbonyl must be made by conjugate addition and/or Aldol reaction
How would these two carbon units react to form a 1,5-dicarbonyl
- The O-H bond will break, allowing the C=O double bond to reform
- Result in e- from the C=C being pushed off, and will attack the terminal alkene carbon
- e- from broke C=C will move down the carbon chain, resulting in the C=O bond to break
What will 1,5-dicarbonyl react with to form pyridine?
Hydroxylamine
(with the elimination of one proton and 2 hydroxyl groups)
You can also use ammonia to synthesis pyridine with 1,5-dicarbonyl
What is the difference in this reaction?
1,4-dihydropyridine is formed first
Which then readily oxidises in the air to the aromatic pyridine
Identify the product formed in the reaction shown below
Answer = D
Intramolecular electrophilic substituion can be used to allow the formation of another ring
Using the example below, how does this occur
- E⁻ from the aromatic ring is used to attack the carbonyl carbon, resulting in chlorine leaving
If the side chain contains a heteroatom, the new ring formed is heterocyclic
Explain this using the example below
- The LP on the heteroatom will attack the electrohilic carbon, resulting in the leaving group being booted off
- This will then allow intramolecular electrophilic attack
Describe the mechanism for the second part of the reaction for the synthesis of a heterocylic
- E⁻ from the heteroatom will form a double bond with the adjacent C, resulting in the aromatic E⁻ attacking the electrophilic carbon
- Causing the C=O dbl bond to break
- Then followed by the elimination of the hydroxy group
Fischer Indole Synthesis allows the formation of indoles
(once the imine is formed the synthesis is slightly different)
What is the main difference in the last step
Ammonia is the leaving group, rather than the usual water
