Resolution of the Enantiomers of Ibuprofen lab questions a)The optical rotation

Question

Resolution of the Enantiomers of Ibuprofen lab questions

a)The optical rotation of a sample of 2-butanol is measured to be obs = -0.35º. The specific rotation for pure (+)-2-butanol is []D = 13.52° ml/g dm. If the cell path length was 0.6 dm and the concentration of 2-butanol in the sample was 0.15 g/ml, calculate the specific rotation and the percentage of the (+) and (-) isomers in your sample.

b)You have a sample that you know is composed of 21% R isomer and 79% S isomer. What is the % optical purity of the sample? If you know that the specific rotation for the pure S isomer is [] = +75°, what would you expect the observed specific rotation for your sample to be?

Explanation / Answer

a)

observed rotation is = -0.35

and specific rotation of pure sample = +13.52

specific rotation calculated for the sample = observed rotation / concentration(g/mL) x length of cell (dm)

= -0.35 /( 0.15 g/mL x 0.6 dm)

= -3.89

optical purity = observed specific rotation x 100 / specific rotation of pure enentiomer

= 3.89 x100/13.52

= 28.77 %

Since the rotation is -ve , the sample has 28.77% (-) isomer and the remaining(71.23) is racemised.

That is the amount of ( - ) isomer in the sample is = 28.77 + (71.23/2)

= 64.385 %

and the amount of ( + ) isomer in the sample = 35. 615 %

b) The sample has 21% R- isomer and 79% S - isomer.

so optical purity = |% of one isomer - % of other isomer| x100/ total

= [79-21] x100/100

= 58

Which shows the mixture has 58% of S- isomer and the remaining is racemised.(21 R and 21 s)

specific rotation of pure S - isomer = +75

then using the relation

optical purity = observed specific rotation x 100/ specific rotation of pure sample we calculate the oberved rotaion.

Thus observed sp. rotation = optical purity x sp. rotaion of pure sample /100

= 58 x (+75) /100

= +43.5

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