Question

Specific rotation of L-alanine in water (at 25 °C) is +2.8. Suppose you have 3.00 g...

Specific rotation of L-alanine in water (at 25 °C) is +2.8. Suppose you have 3.00 g of mixture of L and D-alanine. You dissolved it in 10 mL of water, and measured rotation in 10 cm cell. The observed rotation was +0.12. Calculate the ee% for the mixture. How many grams of D- and L-alanine do you have in the sample?

Optical purity (ee%) = observed specific rotation/ specific rotation of the pure enantiomer x 100= (+0.12 / +2.80 ) x 100 = 4.29 %

I got the optical purity - I hope that 4.8% is correct - but, I do not know how to get how many grams of D- and L- alanine are in the sample. How do you do this?

Homework Answers

Answer #1

Yes. Optical purity = observed specific rotation x 100/sp. rotation of pure sample

= +0.12 x100/+2.80

= 4.28%

That is 4.28 is pure + isomer and the remaining 95.72 is racemised.

Hence of the 95.72percent racemic mixture , half is the + isomer = 95.72/2=47.86

and the rmaining half is - isomer.

Thus total (+) isomer = 4.28 + 47.86= 52.14%

and the (-) isomer = 47.86%

Thus the amount of (+) isomer = 3.00x 52.14 /100= 1.5642 g[ L- alanine]

hence the amount of (-) isomer = 3.00x 47.86/100 = 1.4358 g

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