Azide Chemistry Flashcards
What is an azide
linear, polyatomic anion with the formula N³⁻
What are the two resonance forms of an Azide?
How to this ylide react with water to form the following stable products
- LP on nitrogen attack a proton on water
- E- from broken O-H attack the phosphorus cation
- Nitrogen then attack proton of the OH, causing new P=O to form
- And weak N-P bond to break
In a Staudinger reduction of azides, it is reacted with triphenylphosphate to form….
- Attack N with no charge by LP on phosphorus
- LP on other N attacks the now positive phosphorus
- 2+2 cyclisation
Azide will act as a……
undergoing which type of reaction?
Nucleophile
So sₙ² reactions mean that chirality can be transferred via inversion
What other way can we reduce an azide to an amide which is not chemoselective
Using LiAlH₄ OR Pd + H₂
But if there is an ester/ketone (LiAlH₄) or an alkene (Pd + H₂) within the molecule these will also be reduced as well
If you wanna form an imine from an aldehyde, you might undertake the Aza-Wittig reaction
What is the mechanism for this?
- Attack using LP on N to δ⁺C
- E- from C=O attack positive phosphorus
- In intermediate 2+2 cyclisation
- Producing an always trans alkene
Why during an Aza-Witting reaction is a trans alkene always formed?
- N LP can invert
- So R group always trans to alkyl
This is another example of a Aza-Wittig reaction
What intermediate is formed from the following reagents and how does it form the final product?
(notes the imine final product cannot be made through condensation)
In the first step of this reaction an Azaphosphonium ylide is formed which is nucleophilic
Describe the mechanism for this reaction?
- Nu attack of N, followed by elimination of OEt
- Elimination of PPh₃ using water, forming amine group
- OEt knocks of OH, to form triphenylphosphate oxide
The following reaction is called a Straudinger Ligation
- Phosphate attacks the azide, forming new P-N bond
- LP on N attacks δ⁺C of carbonyl, which subsequently breaks the C-OMe bond
- N deprotonates water, and a new P-O bond is formed from -OH ion
- P-OH is deprotonated using -OMe
This scheme below shows a way to form an aziridine for an epoxide
Describe the mechanism
Note: HC-CN = solvent
