Question

1. A sample of 2-butanol has a specific rotation of +3.25'. Determine the % ee and the molecular composition of this sample. The specific rotation of pure (+)-2-butanol is +13.0°. (0.5 point) sofie 2. A sample of 2-butanol (see question 1) has a specific rotation of -9.75°. Determine the % ee and the molecular composition of this sample. (0.5 point)

Answer 1

1.

Given, Observe specific rotation of 2-butanol = +3.25 degree

Pure specific rotation of 2-butanol = +13 degree

Now,

• % of enantiomeric excess(ee) = Observe specific rotation / Pure specific rotation * 100

= +3.25 / +13 * 100

=(+) 0.25 * 100

= (+) 25%

Hence, ee = 25% and it is excess of (+) isomer

• ee 25% means (+) and (-) isomers = (100 - 25)% = 75% ( it is racemic mixture of (+) and (-) isomers)
• (-) isomer = 75/2 % = 37.5%
• (+) isomer = (25 + 75/2)% = 52.5%

Because ee is added to (+) isomer as it is excess of (+) isomer.

2.

Given,

Observe specific rotation of 2-butanol = -9.75 degree

Pure specific rotation of 2-butanol = + 13 degree

Now,

• % enantiomeric excess(ee) = - 9.75/+13 * 100

= (-) 0.75 *100

= (-) 75%

Hence, ee = 75% and it is excess of (-) isomer

Now,

• ee = 75% means (+) and (-) isomers = (100 - 75)% = 25% (it is racemic mixture of (+) and (-) isomers)
• (+) isomer = 25/2 % =12.5%
• (-) isomer = (75 + 25/2)% = 87.5%

Here ee = 75% is added to (-) isomer as it is excess of (-) isomer.

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