Silicon 1 Flashcards
What is a similarity and difference between silicon and carbon bonding?
- It can form 4 bonds
- But silicon bonds are longer and can impact the way silicon behaves
What type of reactions generally work well with silicon compounds?
- Nucleophilic Substitutions
- Nu attack at the tertiary centre of silicon
- Form an intermediate complex with 5 bonds (called an ‘ate’ complex)
- The Si-Cl bond then breaks, which is fast
How do the following to compounds react
(Pr = propyl = CH₂CH₂CH₃)
- E- from the C-Mg bond attack the Si
- Causing the Si-Cl bond to break
Why does Nucleophilic substitution happen as Sₙ², rather than Sₙ¹
(this is despite the Sₙ¹ Si intermediate being more stable too)
- Just too slow
- Si-Cl bond is quite a lot stronger than the equivalent C-Cl bond
- Therefore the silicon cation takes too long to form compared to the pentavalent intermediate)
- Hence cannot compete with the Sₙ² process
What generally causes the strength of Si-X bonds to increase
- Different in electronegativity, the larger this is, the stronger the bond is (ionic nature strengthing the bonds)
- Hence Si-Si bonds are weaker than C-C bonds
- But Si-OH and Si-F bonds are stronger than the carbon equivalents
How does the Cyanohydrin form on the RHS
- CN- attacks the carbonyl centre, causing the C=O bond to break
- E- from the broken C=O attack the K+
- Then adding acid, the O-K will form O-H instead
Why is this reaction reversible?
Because when the acid is added, the OH will reform, which means its no longer an anion, hence it is not donating electron density to stabilise intermediate
How does the following reaction occur
- LP on the carbonyl oxygen, attacks the Si
- E- from the Si-CN bond can now Nu attack the carbonyl carbon, causing the C=O to break
- (may need a lewis acid/base to do this reaction irl)
How does the following reaction occur - kinetic product
(LDA is a strong, bulky base)
- LDA (base) will deprotonate at the red H (not sterically hindered - kinetic product)
- E- are transferred to the adjacent C, forming C=C
- This causes the C=O to break, and E- from bond attacks the Li, forminh O-Li
- But O-Li bond is displaced with O-Si (strong O-Si)
How does the following reaction occur - thermodynamic product
(Et₃N - weak base)
- LP on carbonyl oxygen, attacks the Si, causing the Si-Cl bond to break
- Et₃N will deprotonate the blue H (thermodynamic), pushing E- onto the adjacent carbon forming C=C
- This causes the C=O to break
- (green reaction can occur but blue product is more stable and will accumulate over time)
We can use a carbanion (like Li-Me) to recover the enolate from silicon
What drives this reaction, when we have a really strong Si-O bond
The stability of the anion formed
We can use a fluoride (like Li-Me) to recover the enolate from silicon
What drives this reaction, when we have a really strong Si-O bond
It is the strength of the F-Si bond which is driving the reaction
How can electrophiles react with the enolate once it is formed?
- The electrons come down off the oxygen, reforming the C=O
- Causes the C=C bond to break, and the electrons go on to attack the electrophile
Silyl groups can be used as protecting groups for alcohols
What is a protecting group?
Protecting group provides a temporary block of otherwise reactive functionality
The strength of the Si-O bond is ideal for this application
Why are trimethylsilyl enols not typically used as protecting groups
Because trimethylsilyl enols are too unstable to water
If trimethylsilyl enols are not idea as a protecting group, which type of silyl group is?
More sterically bulky/hydrophobic silyl ethers are stable enough to survive several steps of synthesis
How will the following silyl group and the imidazole react
- The sp² Nitrogen on imidazole will attack the Si, causing the Si-Cl bond to break
How does the following Silyl-imidazole intermediate react with an alcohol?
- LP on the OH will attack the silicon, causing the Si-N bond to break
- Imidazole then deprotonated the alcohol
Trimethylsilyl can be a robust protecting group for which compound
Alkynes
Due to the fact that the C-Si bond is not easily hydrolysed by water, as carbanion is a poor leaving group
You would do this if you were trying to react somewhere else in the molecule
