Conjugate additions 2 Flashcards
What is met by Reterosynthesis
Essentially doing a reaction in reverse
What a synthons
Imaginary idea reagents, that don’t often exist in reality
What are the synthetic equivalents of these synthons
- Enone on the left, as the C=C allows the carbon to be electron deficient
- Dicarbonyl remains the same
If we want a 1,4-addition to occur over a 1,2-addition, what are the reaction conditions required?
reflux
A 1,4-conjugate addition initally gives enolate (or enol if in acid)
If we introduce a suitable electrophile to the reaction…
Then we can functionalise the enone twice
Electrophile adds where C=C would protonate (can allow synethsis to be more efficient)
What is the final product of this reaction
- BuLi will knock of the iodine to form a C-Li bond (will react like Grignards - Nucleophilic)
- Then adding CuI turns the alcohol lithium into an alcohol copper
How will following two molecules react
- 1,4-addition
- Electrons from Cu attack the δ⁺ C=C
- causes the electrons to transfer to the adjacent carbon and the C=O bond to break
How do the following reagent react
- LP comes down off oxygen, causing the e- from the double bond to attack the δ⁺C-I
- (The sterochemistry of these reactions can be really easily controlled due to the substituents only adding in certain places)
How would you break down the following into its synthetic equivalents?
- Break first bond to form a 1,5-dicarbonyl
- Break second bond to form an enone and a cyclic dicarbonyl
How will the follow reaction occur
- EtO acts as a base, deprotonating on the ring
- E- from the C-H bond move the adjacent C-C forming C=C and break C=O
- Forms an enolate
How to the following reagents interact
- 1,4-addition
- LP comes down from oxygen, reforming C=O, and breaks adjacent C=C
- E- from C=C attack the δ⁺ C of the C=C
- Causes electrons to move onto adjacent carbon and break C=O bond
How do the following reagent interact
- LP comes down off oxygen, reforming C=O and breaks the C=C bond
- Electrons from the C=C attack the H of the alcohol
How do the following reagent react
- The LP comes down off oxygen, causing the C=C bond to break
- E- from C=C bond attack the δ⁺C of the other C=O, causing it to break
- LP now oxygen deprotonates the alcohol
How do you reform the enoate from this point?
- Deprotonation using -OEt, leads to formation of C=C and breaking of C=O
- LP from oxygen reforms C=O, and E- from C=C transfer to the adjacent carbon
- This causes the C-OH bond to break
- (the C-OH bond cannot break simply through the deprotonation due to it being a poor leaving group)
In the previous reaction forming the steriod rings, the EtOH will swap position of the enolates
Why?
The reagent on the right would form a 4-member ring, if it attacked the carbonyl
Due to the high strain of the ring, the reverse reaction of this is thermodynamically fast
