Synthesis of Biomolecules Flashcards
What are important factors in the synthesis of peptides?
The reagents must be pure and no racemisation can occur during the reactions to preserve the stereochemistry of the reagents. The reaction should also only be between the 2 wanted reagents, not been the same reagent.
Give the outline synthesis of a dipeptide in solution.
Protect the N-terminus of one amino acid and protect the C-terminus on the other. Then activate the carboxyl group and couple the group. Once the amide bond is formed the groups are deprotected.
Outline the two different ways to protect the N-terminus, giving reagents and protection and deprotection synthesis.
Outline the three different ways to protect the C-terminus, giving reagents and protection and deprotection synthesis.
When coupling the protected amino acids, why can’t an acid chloride be used? Draw a diagram.
A side reaction can occur where the protection carbonyl in the enol form can act as an internal nucleophile and form a five-membered ring. This loses the stereospecifity.
Draw a synthesis of amino acid coupling giving reagents and mechanisms. What are the next steps that can be taken?
For the dipeptide, just deprotection is required. To extend to a tripeptide, one side is deprotected and coupled with another half protected amino acid, then deprotected.
Describe the original resin used to synthesis peptides on a solid phase and describe its functionalitiy. How is the first amino acid introduced?
It is a crosslinked polystyrene (ethene with Ph group). The crosslinks, which make the polymer insoluble is a styrene monomer with two alkene groups para to each other. Some of the Ph groups have chloromethyl groups which acts as the proctecting group for the carboxylic group.
A soluble, organic base such as Cs2CO3 is also required in the attaching of the amino acid.
What goes into the choice of solvent for polymerisation on the solid phase? What is required for the coupling reaction? Why is this chosen?
Using a solvent that would usually dissolve the linear polymer will result in pores in the bead that can be filled by the monomer. The coupling reaction requires a coupling agent that doesn’t form a solid as it needs to be washed away. DIC, like DCC except two isopropyl groups. DIPEA, diisopropylethylamine is also used a non-nucleophilic base.
Draw the mechanism of the cleaving of the peptide from the polymer bead, giving the reagents.
How can the solid resins be manipulated to be safer? What considerations must be made?
Linkers are added which replace the Cl group with a O-Ph-MeOH group. This requires an esterification to add but can be removed with TFA (therefore Boc cannot be used as the N-protection for new amino acids). TFA is still bad so if a meta-OMe group is added, 0.5% TFA can be used.
What are the benefits and limits of solid-phase peptide synthesis?
Pros: Can be automated very easily to save months of manual work.
No work-ups required as reagents can be washed away.
Cons: No purification during synthesis, only after cleaving from resin.
1 mmol of peptide requires 1 gram of resin and the resin will become large in solvent.
Define the different types of saccharide, a carbohydrate and a glycoside.
Saccharides can have the prefix of mono, di and oligo meaning 1, 2 and more than 2 respectively.
A carbohydrate is one or several sugar units.
A glycoside is any compound with a carbohydrate group.
Draw the cyclisation of glucose from the long chain aldehyde and draw the structure of the cyclised sugar.
Describe the anomeric effect, how it can be used and the reason it occurs.
The anomeric carbon will typically be in the alpha form due to the equitorial position being the most stable. However when the OH is transformed to an OMe the alpha, more sterically strained version is favoured. This is also true when the OHs are all transformed to acetyls.
One theory is that hyperconjugation allows the non-bonding electron pair to share its electron density to the LUMO of the OMe. This stabilises the oxygen by forming a simple MO. Evidence for this is the shortening of the bond compared to the equitorial position. The other theory is that the dipoles are minimised.
How do sugars link? How are they labelled?
Via an ether bond from the glycosidic bond from one sugar and any of the OH groups from the other sugar. There is two types, alpha and beta which refer to the orientation of the anomeric carbon. The disaccharides are labelled 1-x where x is the position of the OH on the second sugar.
Give the non-selective protecting groups for hydroxyl groups on carbohydrates and their addition and removal reagents.
Give a protecting group that selectively protects primary alcohol groups and its addition and removal reagents.
Give the synthesis and activation of bromine as a coupling agent, describing the stereochemistry outcomes.
Give the synthesis and activation of thioethers as a coupling agents, describing the stereochemistry outcomes.
Describe and draw a diagram to explain how solvent choice can affect stereochemical outcomes when acetates or benzoates can’t be used for NGP.
Describe in detail how to form a disaccharide with 1,2-cis groups. Why is the complex method required?
For the alpha product, non-NGP protecting groups can be used but for the beta product, neither the anomeric effect or the NGP can be used. Hence intramolecular tethering is used.
