Question

optical activity problems.. please show all steps i dont understand

You have a sample (Sample X) which is a mixture of +/- Carvone. The solution was made by dissolving 4.50 g of the sample in enough methanol to bring the volume of solution to 10,0 ml. Some of the solution is placed in a 100 cm polarimeter cell and its optical rotation is measured at 25°C using light of the sodium D line wavelength (589.6 nm). The observed rotation is +22.2. What is the specific rotation of this sample? 5) An enantiomerically pure sample of the Senantiomer of Carvone has a specific rotation of + 15.5'. What is the % enantiomeric excess of Sample X from problem 4? 6) What percentage of Sample X (problem 5 above) is the Senantiomer? What percentage of the sample is the R enantiomer?

Answer 1

4)

Specific rotation = observed rotation / (density of the solution in g/mL x length of the polarimeter cell in dm)

Now,

Mass of the sample (m) = 4.50 g

Volume of the solution (V) = 10.0 mL

Density of the solution = m/V = 4.50 g/10.0 mL = 0.45 g/mL

Length of the cell = 100 cm = 10 dm    ( $\because$ 1 cm = 0.1 dm)

Thus, specific rotation = +22.2^o/(0.45 x 10) = +4.93^o

Hence, the specific rotation of the sample = +4.93^o

5)

% enantiomeric excess = (specific rotation of sample x 100)/specific rotation of pure enantiomer

                                      = (+ 4.93^o x 100)/15.5^o

                                      = 31.8 %

This implies that the solution contains 31.8 % of S isomer in excess.

6)

% racemic mixture = 100 % - % enantiomeric excess

                               = 100 % - 31.8 %

                               = 68.2 %

Now, a racemic mixture contains R and S isomer in 1 : 1 ratio.

Hence, % of R in racemic mixture = % of S in racemic mixture = 68.2 %/2 = 34.1 %

Hence, the total % of S isomer in the sample = (31.8 + 34.1) % = 65.9 %

The total % of R isomer in the sample = 34.1 %