[f1987]

I have an exam and a recurring question of this nature keeps coming up.
Two compounds an amine and a phosphine, one produces one spot the other
produces two spots when analyed by TLC. Explain.
OR
Two alcohols are given, isomers of eachother and asked to expain why one produces one spot, the other two.
Any help much appreciated.

[adam]

Hi,  I don't understand well the question?
Do you analyze the products by chiral TLC or normal TLC??
If it's a chiral TLC, 2 spots could be mean you have a chiral molecule, one spot for each enantiomer. 1 spot could be mean that you have an achiral molecule.
If it's an achiral TLC, 2 spots means two diferrent products...maybe the phosphine and the phosphine oxide (the phosphine could be oxidized when you are running the TLC).
The second question:
The same if you are using a chiral TLC.
And if you are using an achiral TLC, two spots could be a mixture of products (maybe a diastereoisomers).
Hope this can help you!

[phil81]

If it's chiral TLC, the first question could refer to the conformational flip of amines vs. phosphines. The flip is fast for tertiary amines at room temperature, which explains their achirality. Phosphines (as well as sulfoxides) flip slowly; therefore PR(1)R(2)R(3) is chiral.

Hope this helps. Maybe it would be easier if you posted the exact structures.

[adam]

I'm agree with you phil81!

[Dan]

Yeah the amine/phosphine question is almost certainly about flipping.

Chiral TLC may not be required. If the amine/phosphine has an additional stereogenic centre (with defined configuration) then inverting the centre at N/P will give diastereoisomers which could be distinguished by standard TLC.

eg. (2R)-N-methyl-2-methyl-propylamine and corresponding phosphine. If the stereochemistry at C2 is defined then you have two possible diastreomers depending on the relative configuration at the heteroatom.

I suspect the alcohols question is a case of enantiomers vs diastereoisomers.

Perhaps you could post specific questions, it would clear up any ambiguity.

[f1987]

Wow thank you all so much for you help. I have attached 3 questions of this form which I was referring to.

[Dan]

Ok, so the first one is ring flipping of the amine leading to rapid interconversion of the two enantiomers of that amine. So chral TLC shows one spot for the amine, because the molecules are interconverting rapidly compared to the timescale of your TLC, so you see one spot which is effectively an average of the two enantiomers.

The phosphine doesn't racemise rapidly, so you can resolve them as this interconversion is slow relative to the timescale of TLC - 2 spots.

The reaction with MeI produces the ammonium/phosphonium salts which do not interconvert readily.

The next one is about flipping again. The amine will have scrambled stereochemistry at the N centre, and so the cyclisation gives two possible diastereoisomers which can be separated on standard TLC plates.
The stereochemistry at the P centre in the phosphine will not scramble quickly, so only one product is formed.

Last question. Identify all the possible structures of the products. I count 2 for the first, 4 for the second. Remember that you will not resolve enantiomers on silica, only (in the context of this question)diastereoisomers.
For relative intensities think about the Felkin-Ahn model.
This one's alot of work for 4 marks...

[f1987]

Cheers for that Dan.