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Topic: Optical Rotation  (Read 18268 times)

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Offline pfnm

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Optical Rotation
« on: April 22, 2010, 01:49:26 AM »
Hi,

Working out the specific rotation for (S)-(-)-a-phenylethylamine sample resolved in the lab:

Measured 0.895g of the finalresolved product, dissolved in 10mL water and measured the sample in the polarimeter.

The value displayed was -0.148. Temperature is 22 degrees, pathlength 0.5dm.

So using specific rotation formula:

Specific rotation of a solution [α]D = α obs*100 / c l

Using c in g/100mL. I dissolved 0.895g in10mL,  so c = 8.95g per 100mL

α obs being -0.148degrees, and l is pathlength 0.5dm

So

100*-0.148/(8.95*0.5)=-3.307 specific rotation.

Then for optical purity the formula I have is
optical purity = specific rotation of sample / specific rotation for compound as given in literature

I find the literature value to be -40.3 for the (S)-(-) enantiomer.

So -3.307/-40.3x100%= 8.2%  opticalpurity.

Does mean an 8.2 percent excess of the (S)-(-) enantiomer?

Also, the value from the literature (which comes from orgsynth.org), isn't that for the pure compound,as in, the specific rotation for pure liquid rather than asolution?

Have I made any mistakes here? I've been atthisfor quite some time and any help would be appreciated.



Offline AC Prabakar

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Re: Optical Rotation
« Reply #1 on: April 27, 2010, 09:54:49 AM »
Your calculation part is correct.But you have told that after resolving the OR has been done.Still it is (SOR) -3.3 ONLY!!.
The calculated % of S and R-Isomer is 54.1% and 45.9% (ee8.2%) almost racemic!!!How is it possible for a resolved compound?
Is that compound is as pure single component?How do u resolved?

Offline pfnm

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Re: Optical Rotation
« Reply #2 on: May 01, 2010, 05:54:20 AM »
Hello,

Yes I'd written the numbers down incorrectly, the optical rotation is actually -0.548 not -0.148.

So with those values, I have a specific rotation of -12.25, and the value we are to use for the specific rotation of pure (S)-(-)-phenylethylamine is -30.5. (This is not the literature value but it is the value we are required to use from the lab book).

So -12.25/-30.5 = 40.15% optical purity, with 70.1% excess of (S) enantiomer and 29.9% of the (R) enantiomer.

There was a mistake at the end of the experiment, where a pasteur pipette was used to transfer the ethanol into the sample tube containing the 0.895g of solid product. The rubber bulb of the pasteur pipette had another substance in it, which leaked out into the ethanol and thus contaminated the sample. If this substance had some significant optical rotation perhaps it affected the observed rotation when the contaminated sample was placed in the polarimeter.

(that is, the actual optical purity of the substance, before contamination, may have been much higher or much lower).

Thank you


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