Horza2002

If only that was true Reach....

No problemo. Im not 100% sure of the mechanism, however thioester bonds are normally formed in this manor (i.e. in polyketide synthase enzymes)

The ubiquitin has a carboxylic acid which is activated by ATP with a Mg2+ ion that helps activate the molecule. The activated phosphoester is attacked by the thiol of the E1 enzyme. As far as I can see, thats what you've asked for.

http://en.wikipedia.org/wiki/Ubiquitin

There is alot of information on this webpage about the mechanisms if you want.

ubiq.pdf

I don't think you will ever have As subsituted for P in a biological system. Im pretty sure that As is so toxic because it takes the place of P in compounds like ATP where it then rapidly hydrolyses degrading the molecule making it useless for the organism. So for an organism to have As bonds then it'l never survive.

I would have guessed that the sterics were the major issue controlling where the protection occured. Espcially using TBDPS groups which are sterically large as well! 70% over three steps is pretty good as well! I've also found that to given my limited experience (only done 4 months of my PhD so far) that slight difference in reactiity can be useful for selectivity..although so far I don't get great selectivity in some of my reactions.

Mississippichem, when you remove the silyl ether protecting groups, do you use acidic conditions to do it? If you have some pH sensitive metals, then you can also use flouride to remove the silyl group. Adding NaF is often enough to remove them, but there are some better flouride releasing groups as well (e.g. TBAF). The flouirde adds to the silicon (forming an extremely strong Si-F bond) followed by elimination of the alcohol. That might be useful for you if using some pH sensitive metals...not sure how adding a fluoride donor would affect metals though...

You said you had some experince with this sort of chemistry? Is there a major difference betweent he reactivity of the primary compared to the secondary alcohol. I always thought that they would be too similar to get them to react differently; I tend to avoid them in my reaction routes. Is it a case of carefully adding 1 equivalent and temperature control to selectively get the primary portection.

Yer, I doubt ekay will read this, but its good to have a synthetic discussion...there tend not to be too many around here!

From what I've read and using my chemistry knowledge, what they mean is that they have found bacteria that can survive very high levels of As (higher than would be fatal for humans). However, given that any As bond formed in a biolical system would hydrolyse very rapidly (they are rather weak), its unlikely that the bacteria would have any of them in te first place...but isolating them would be nearly impossible as well.

I agree, methylate firt would be so you don't have to worry about accidentally eliminating any of the other hydroyl groups.

I've just had an idea how to make it from Maleic anyhdride (easily and cheaply avaliable).

The first step involves the di-hydroxylation of the alkene with osmium tetroxide (giving a racemic mixture)

Melthylation of these two hydroxyl groups should be easy enough by deprotonating (NaH) them and then treating them with methyl iodide.

Conversion of one of the carbonyls to an alkene could then be done using one equivalent of Tebbe olefination (uses the metal carbene generated from titanocene dichloride and trimethyl aluminium).

Reduction of this alkene with hydrogen gas on palladium should be easy enough.

Reduction of the carbonyl (I can't remember the conditions that I've used to do this before) and finally methylating again like for the diol.

Again though, as I don't know what stereochemistry you want, this route might not work.

idea.doc

Sterically, the 6 position would be favoured for the tosylation because its a primary alcohol where as all the others are secondary. However in practise, you'd probably end up with a mixture of tosylated products. The problem here is that primary and secondary hydroxyl groups reactivity is still very similar so getting seletvity would be a problem.

Seeing as no stereochemistry is shown, you could oxidise all of the hydroyl group to the ketones (for the secondary) and carboxylic acid (for the primary). Then convert the acid to a methyl ester and then reduce all the ketones back to hydroxyl groups with sodium borohydride. This is not exactly a great way to do it as you would lose all stereocontrol of the reaction not to mention the potential for transesterification reactions leading to polymers.

You might be able to selectively protect the secondary alcohol here because there are essentially two sets of 1,2-diols. With this, you could potentially make two acetonides with them. This would then allow you to use a different protecting group for the primary alcohol (say benzyl). You could then remove the acetonide and you'd have selectively protected the primary. However, again this route is not exactly efficient at making sure you doubley protect the two 1,2-diols.

Arr ok then, I always assumed it was due to sodium present that make it that colour!

The coloured flame you get from using copper is closer to green in my opinion than blue. The normal colour quoted for GCSE students is that it gives a greeny/blue colour. As CaptainPanic has said, impurities in the staff itself will give there own colours off with; bright orange will be the most prominant due to the trace anounts of sodium in the material.

Alot of the "glass" that is bought from lab suppliers is borosilicate glass. Depending on what you want it for, NMR tubes are good quality borosilicate glass (unless your going to be doing boron NMR in which case you'll need a much higher quality grade).

As Mississippichem has said, its to do with how you'r determinging the stereochemistry. Unforuntantly, D/L, +/- and R/S labells are all unrelted to ach other

As far as I'm aware, there is no other way to assign R/S to structures becasue the R/S notation comes from giving the substituents there ordering priorities. I don't build models for organic stereocentres, its just practise of being able to rotate the object in my head in 3D...just practise. A meso compound is one that contains stereogenic centres, but also contains an internal plain of symmetry....so if you can spot a plane of symmetry, then you have a meso compound.

If one enantiomer is R, then by definition, the other has to be S. IN this case, the iodine has highest priority, followed by bromine, chlorine and finally hydrogen. Rotate the molecule so that the hydrogen is pointing away from you; the priorities are then going clockiwise so the top left is indeed R.

Well the basisity of a solution depends on the concentration of H3O+ ions (i.e. the same as what makes a solution acidic). This means that the ions that make something acidic/basic will only stay in the aqueous phase. If you extracted with DCM, then the majoirty of the ions would have stayed in the aqueous layer and so you wouldn;t change the pH.

Also, what do you mean by having an oil? Is your product and oil or does it just have some solvent trapped in it preventing it becoming a solid. If its the first, the it wont dissolve very well in water anyway so when you added it to the NaOH not much will happen. If you want the oil to mix better with water, then try dissolving it in THF (it easily mixes with water and then easy enough to vac of at the ened so you can reextract your oil). But in term of making the oil basic in the organic phase, I don't think your going to get that to happen unless you use an organic base (say triethyl amine, DMAP, DBU, etc) because NaOH just wont dissolve in the DCM.

Don't really understand what your asking, but ninhydrin will react with an N-H bonds in your molecule (whether they are primary/secondary amines or amides).

Tbh, making it from glucose would be pretty difficult and also probably very ineffcient. If I where you, I'd try to come up with another way to make it.

Ive attached the mechanism for borane reduction. It works best for electron rich substrates because of the electron-deficient boron atom in the reagent. Amides are easily reduced with this method...esters are slower...carboxlylic acids are a little tricky but it works ok. I've managed to get around 55% yields doing reductions like this; not great but good. Depending on your molecule, its often easier to make the methyl ester and then use LiAlH4 to reduce the ester instead...but like i said it depends on the other groups present in your molecule.

borane.pdf

Borane (BH3) is one reagent that will reduce a carboxylic acid to the corrosponding alcohol. I've never had much luck doing it, but according to the literature, it can be done

Why do you think borohydride will stop at the aldehyde? Borohydride is usually used to reduce a aldeyde or ketone in the presence of other groups (e.g. esters). If you manage to reduce the carboxylic acid with borohydride, then you most certainly will reduce the aldehyde down to the alcohol (aldehydes are extremely reactive).

O and yes, if you going to use aluminium hydride, then make sure there is NO water ANYWHERE!

Mississippichem has explained it well. I just want to point out that hydrogen does have higher energy levels than 1s. However, they are so high in energy that they are hardly ever filled. So in principle, you can have H(-3) but you'd be extremely hard pushed to make it

What do you mean by a powerful solvent? That doesn't mean anything

Its been a while since ive studied photosynthesis, but as I remember, once the electrons are excited out of the magnesium atom, they are captured by proteins which are then used to drive reactions to make sugars. Eventually, new electrons are used to replace those removed from the magnesium so that the process can carry on again.

http://en.wikipedia.org/wiki/Photosynthesis

Colour would be the easiest way to determin the identity of many of the metal ions. And a flame test should help you with the likes of sodium, calcium, potassium and barium that don;t have a colour.