[jena]

Hi,

I'm having problems figuring out optical rotations. Does anyone know of a site that deals with optical rotations for a question like this one for exampel:

What is optical rotation value of a 1:4 mixture of enantiomers given that the rotation value of pure enantiomer would be 100 degrees?


Could someone please help with this question I posted a similiar question like this before and no one responded to it. I hope I'm not being rude by asking if someone could at least respond to this I really need help

Please help and Thank you  

[dexangeles]

well, isn't it that enantiomers' rotations are equal in magnitude, but opposite in sign?

[movies]

Yeah, pretty much.  So in the case of a 1:4 mixture you essentially have 40% racemic material and 60% of one enantiomer in excess.

The equation you need is alpha = [alpha]*c*l

alpha is the observed rotation, [alpha] is the specific rotation, c is the concentration of the sample, and l is the pathlength of the instrument.

[alpha] won't change because you have the same compound and all that.  You assume that l is the same too.  So what has changed?  The effective c is different because only 60% of the material will be productively rotating the plane polarized light (the racemic material doesn't do anything).  So if the effective c is decreased from 1 (arbitrarily) to 0.6, then the observed rotation should decrease from 100 to 60.  So in essence the value of c is artificially high because you have a equal amount of material (10 mg in each sample) but in one sample all of the material is rotating light whereas in the other only 60% is rotating light.

I think that's right, but I'm not really that experienced in dealing with optical rotations.