Question

Chem. 251 ** ICE-3 ** Page 2 3. What is the specific rotation of each of the following solutions? Show all of your work. See McMurry, sec. 5-3 and chap. 5 PPT slides 10-13. a) 1.00 g of sample is diluted to 5.00 mL. A 3.00 mL aliquot is placed in a tube that is 1.0 cm long. The observed rotation is +0.45°. b) A 0.20 g sample is diluted to 2.0 mL and placed in a 10 cm tube. The observed rotation is -3.2º. 4. Identify all the chiral centers with an asterisk (*) in each molecule and determine the absolute fimatinn ar Ror See McMurry ser 5-5 and chap. 5 PPT slides 16-22

Answer 1

$[\alpha]^{{T}} _{\lambda }$ = specific rotation = $\frac{observed rotation}{concentration\times path length in dm}$

1) given concentration = 1.00g/5mL = 0.2 g/mL

Observed rotation = 0.45°

Path length = 1cm = 0.1 dm

Specific rotation = 0.45 / (0.2×0.1) = + 22.5°  

b) concentration = 0.2g/2mL = 0.1g/mL

Path length = 10cm = 1dm

Observed rotation = -3.2°

Specific rotation = -3.2 / (0.1×1) = -32°