Experiment #6: Polarimetry Resolution of (∓)-Racemate: (∓)-ɑ-Phenylethylamine
OBJECTIVE
All chiral molecules have a mirror image of themselves. While they have the same physical properties, they interact with other chiral molecules completely different. Since there are many chiral molecules in organics, this can lead to the need to make optically pure products. If the chemist does not, it can lead to many problems including death. In this experiment, the separation of two enantiomers was attempted. First was the creation of the racemic mixture of phenylethylamine-(+)tartaric acid was made last week.
(-)-ɑ-Phenylethylamine (+)-tartarate
(+)-ɑ-Phenylethylamine (+)-tartarate
This allowed the (+) enantiomer to be soluble in methanol while the (-) crystallized out in methanol. The methanol and solute where separated from the crystals. The crystals were then dissolved in 75 ml of 8% NaOH to separate the amine from the tartaric acid. Ethyl acetate was used to separate the amine from the …show more content…
Weight of (-)amine used to make the salt, (g)
1.88
b.
The weight of the crystals isolated in step 5, (g)
2.84
c.
Theoretical weight of the crystals
4.16
d.
%Recovery of the crystals
68.3%
e.
The weight of the free amine determined in Step 12
4.51
f.
The weight of the free amine divided by the number of students in your group
1.13
g.
%Recovery of free amine
60%
h.
Observed rotation, ɑ, by each student Average ɑ
-20, -17, -20, -19 -19
i.
Measured Path length, l [decimeters, dm]
0.7
j.
CALCULATED SPECIFIC ROTATION [alpha]
-28
k.
Calculated % Optical Purity, ee
68.5%
l.
Actual compostition of the Resolved (-) Amine in %
84.75%
a) (4ml*.94 g/ml)/2= 1.88 g
c) 1.88g+(4.55g/2)= 4.16 g
d) 2.84g/4.16g * 100%= 68.3%
f) 4.51g/4 students= 1.13 g
g) 1.13g/ 1.88g * 100%= 60%
h) (-20-17-20-19)/4 = -19
i) -19/(.97 g/ml * .7 dm) = -28
j) -28/-40.3= 69.5%
j) 69.5+(½ * (100-69.5))= 84.75
OBJECTIVE
All chiral molecules have a mirror image of themselves. While they have the same physical properties, they interact with other chiral molecules completely different. Since there are many chiral molecules in organics, this can lead to the need to make optically pure products. If the chemist does not, it can lead to many problems including death. In this experiment, the separation of two enantiomers was attempted. First was the creation of the racemic mixture of phenylethylamine-(+)tartaric acid was made last week.
(-)-ɑ-Phenylethylamine (+)-tartarate
(+)-ɑ-Phenylethylamine (+)-tartarate
This allowed the (+) enantiomer to be soluble in methanol while the (-) crystallized out in methanol. The methanol and solute where separated from the crystals. The crystals were then dissolved in 75 ml of 8% NaOH to separate the amine from the tartaric acid. Ethyl acetate was used to separate the amine from the …show more content…
Weight of (-)amine used to make the salt, (g)
1.88
b.
The weight of the crystals isolated in step 5, (g)
2.84
c.
Theoretical weight of the crystals
4.16
d.
%Recovery of the crystals
68.3%
e.
The weight of the free amine determined in Step 12
4.51
f.
The weight of the free amine divided by the number of students in your group
1.13
g.
%Recovery of free amine
60%
h.
Observed rotation, ɑ, by each student Average ɑ
-20, -17, -20, -19 -19
i.
Measured Path length, l [decimeters, dm]
0.7
j.
CALCULATED SPECIFIC ROTATION [alpha]
-28
k.
Calculated % Optical Purity, ee
68.5%
l.
Actual compostition of the Resolved (-) Amine in %
84.75%
a) (4ml*.94 g/ml)/2= 1.88 g
c) 1.88g+(4.55g/2)= 4.16 g
d) 2.84g/4.16g * 100%= 68.3%
f) 4.51g/4 students= 1.13 g
g) 1.13g/ 1.88g * 100%= 60%
h) (-20-17-20-19)/4 = -19
i) -19/(.97 g/ml * .7 dm) = -28
j) -28/-40.3= 69.5%
j) 69.5+(½ * (100-69.5))= 84.75