A student measured the optical activity of an unknown sugar at two concentration

Question

A student measured the optical activity of an unknown sugar at two concentrations. The results of his measurements are shown below. Given that the sample cell had a path length of 10.0 cm, calculate the specific rotation for the unknown sugar. A) -10.5 degree B) +25.6 degree C) +79.5 degree D) -105 degree E) +256 degree Diff: 3 Section: 5.4 Compounds that rotate the plane of polarized light clockwise are called _______. Diff: 1 Section: 5.4 Would a 50:50 mixture of (2R, 3R)-2, 3-dibromobutane and (2R, 3S)-2, 3-dibromobutane be optically active? Explain. Diff: 2 Section: 5.4 Can one predict whether a compound with a single as carbon is dextro- or levorotatory based on the R/S assignment at this asymmetric carbon? Explain briefly. Diff: 2 Section: 5.4 A newly natural product was shown to be optically active. If a solution of 2.0 g in 10 mL of ethanol in a 50 cm tube gives a rotation of +2.57 degree, what is the specific rotation of this natural product? Diff: 3 Section: 5.4

Explanation / Answer

The specific optical rotation is calculated with formula

[]D25 = obs /c*l

where

[]D25 is specific optical rotation measurement conducted at 25 oC using the D-line of the sodium lamp (=589.3 nm)

obs is observed optical rotation
c is the concentration of the solution in grams per milliliter
l is the length of the tube in decimeters (1 dm=10 cm)

62)

given that,

for sample one

solution of 2.0g in 10ml

c = 2/10 = 0.2 g/ml

l = 10 cm = 1 dm

obs = +159.10

since ghost reading is 1800 form true reading

obs = +159.10 - 1800 = -20.90

hence

[]D25 = obs /c*l

[]D25 = -20.9 / 0.2*1

[]D25 = -104.50    ...........(1)

for sample two

solution of 5.0g in 10ml

c = 5/10 = 0.5 g/ml

l = 10 cm = 1 dm

obs = +127.80

since ghost reading is 1800 form true reading

obs = +127.80 - 1800 = -52.20

hence

[]D25 = obs /c*l

[]D25 = -52.2 / 0.5*1

[]D25 = -104.40    ...........(2)

taking mean of (1) and (2),

[]D25 = (-104.5-104.4) / 2

[]D25 = -104.450

66)

given,

solution of 2.0g in 10ml of ethanol

c = 2/10 = 0.2 g/ml

l = 50 cm = 5 dm

obs = +2.570

hence

[]D25 = obs /c*l

[]D25 = +2.57 / 0.2*5

[]D25 = +2.570

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