Question

A 0.120 M solution of an enantiomerically pure chiral compound D has an observed rotation of 0.24° in a 1-dm sample container. The molar mass of the compound is 140.0 g/mol.

a) what is the specific rotation of D?

b) what is the observed rotation if this solution is mixed with an equal volume that is 0.15 M in L, L is the enantiomer of D?

c) what is the observed rotation if the solution of D is diluted with an equal volume of solvent?

d) what is the specific rotation of D after the diltuion descibed in part (C)?

e) what is the specific rotation of L, the enantiomer of D, after the dilution described in part C?

f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of D and 0.005 mole of L? Assume a 1 dm path length

Answer 1

(a) specific rotation (degrees) = observed rotation (degrees)/concentration (g/ml) x parth length (dm)

Here, concentration = 0.12 mol/L = 0.12 * 140 /1000 = 0.0168 g/ml

observed rotation = +0.24 degree for enantiomer D

path length = 1 dm

Calculate = S.Rotation = 0.24/ 0.0168 *1 = 14.29°

(b)

If mixed with equal amount 0.15M D concentration = 0.15 * 2* 140/(2*1000)= 0.021

rotation must be 0

(c) If diluted with equal amount of solvent

V = V1+´V2 = 2000

Concentration = 0.12* 140/2000 = 0.0084 g/ml

So the concentration of solution has halved

Oberved rotation would become = +14.29x 0.0084= 0.120036 degrees

(d) specific rotation after the dilution in part (c) specific rotation

14.29 / 0.120036 =11. 9

(e) The specific rotation of L enatiomer after dilution =

Simply the inverse --> = 14.29° -> = -14.29°

(f) Excess of D = 0.005 mol

concentration = 0.005 x 140/100 = 0.007 g/ml

So, observed rotation = 14.29x 0.007 = 0.10003 degrees