Question

Question 2: An aqueous solution containing 4.23 g of an optically pure compound, (+)-Porgsareawesome (MW=213.4 g/mol) was diluted to 700.0 mL with water and placed in a polarimeter tube 15.0 cm long. The measured rotation was +90' at 25.0'C. a) On the answer sheet provided, set up the equation to give the compound's specific rotation - you do not need to do any actual math! Hint: Make sure to use proper units. Use the equation at left as your guide: Separately, pure (+)-One RingToRule Them has a known specific rotation of +120'. What would be the observed specific rotation of the following mixtures? b) 25% (+)-OneRingToRule Them and 75% (-)-OneRingToRule Them c) 50% (+)-OneRingToRule Them and 50% (-)-One RingToRule Them d) 66.7% (+)-OneRingToRule Them and 33.3% (-)-One RingToRule Them

Answer 1

a)

Specific rotation of a compound is calculated as follows:

a =

Where

at temperature of T=25 C.

at = specific rotation

X = wavelength of light source used

g= measured rotation

1 = path length of polarimeter in dm

C = concentration in g/dm3

Now, amount of the compound = 4.23 g

Volume of solution = 700.0 mL = 0.700 L =

0.700 dm

Hence, Concentration C can be calculated as

4.23 9 6 .043g/dmº 0.700 dm3

The path length l is given as

1 = 15.0 cm = 1.50 dm

The measured rotation is

a = +90°

Hence, we can calculate the specific rotation as follows:

+90% 6.043 g/dm3 X 1.50 dm 9.93° 9-1. dm?

Note that the specific rotation can be calculated in other units too. Generally the unit
g-+ dmor g-1 cm
is dropped and only the degrees are mentioned.

b) (+)-OneRingToRuleThem has a specific rotation of

+120°

Observed specific rotation is related to the enantiomeric excess and the specific rotation of pure enantiomer as follows:

enantiomeric excess = observed rotation specific rotation of pure enantiomer - X 100

Our mixture is 25% (+) isomer and 75% (-) isomer.

Hence, the enantiomeric excess = 75%-25% = 50 % excess of (-) isomer.

Hence, we take specific rotation of for the pure compound.

—120°

Hence, we can calculate the observed rotation as

enantiomeric excess = observed rotation – x 100 specific rotation of pure enantiomer observed rotation 50 = - x 100 120° 50 X - 120° → observed rotation = - = -60° 100

c)

Here, we have 50% (+) isomer and 50% (-) isomer. If (+) isomer rotates the light clockwise, the (-) isomer will rotate the light counterclockwise by the same amount. Hence, they will cancel the effect of each other.

Hence, net observed rotation will be ${\color{Red} 0^\circ }$ .

Note: doing the calculation with enantiomeric excess of 0 will give the same answer.

d)

Here, we have 66.7% of (+) isomer and 33.3 % of (-) isomer.

Hence, enantiomeric excess is

66.7-33.3 = 33.4 % of (+) isomer.

Hence, the observed rotation can be calculated as

observed rotation enantiomeric excess = – x 100 specific rotation of pure enantiomer observed rotation 33.4 = - +120° 33.4 x 120° → observed rotation = - x 100 - +40° 100