Resolving The Pure Enantiomers Of Phenylethylamine Environmental Sciences Essay

The intent of this research lab was to decide the pure enantiomorphs of ( ± ) -?-phenylethylamine ( racemic ) mixture, by dividing their diasteriomeric derived functions utilizing ( + ) -tartaric acid. The differing enantiomorphs form different salts with acids. Two molecules that are enantiomorphs have about indistinguishable physical and chemical belongingss although this may be true, the salts that are formed after the reaction with acid have distinguishable belongingss. Some salts are less soluble [ ( + ) ( – ) ] than others, and hence crystallize from the mixture in a about pure stereoisomeric signifier. When utilizing NaOH as a strong base to handle the salt, it allows for the isolation of the enantiomorph ( Lab Manual, 2007 ) . Polarimetry is a common method used to separate between enantiomorphs, based on their ability to revolve the plane of polarized visible radiation in opposite waies ( + and – ) . This allows the perceiver to find the enantiomeric pureness, and hence the composing of the mixture ( Wade, 2007

Chemical Chemical reaction:

( – ) -amine ( + ) -amine less soluble salt [ ( – ) ( + ) ] : crystallizes more soluble salt [ ( + ) ( + ) ] remains in solution

2NaOH

+ 2H2O

( – ) -?-phenylethylamine ( Lab Manual, 2007 )

Procedure:

Alternatively of utilizing a 50 milliliter beaker to boil the amine solution in, we used a 50 milliliter Erlenmyer flask

For the remainder of the proceduce refer to pg. 18, 22-24 ( Lab Manual, 2007 )

Observations:

The crystals were given a 4 hebdomad crystallisation period and afterward, the ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate salt was observed to be a white crystalline solid, and the methyl alcohol was a crystalline liquid. Two really distinguishable beds were seeable following the reaction with the NaOH ( strong base ) and add-on of the methylene chloride ( CH2Cl2 ) . The top bed was translucent in some topographic points and opaque in others, really cloudy, white liquid, while the bottom bed was crystalline and besides liquid. The attendant mixture following the three separate extractions was close to transparent

Consequences:

Table 1: Experimental Datas: Multitudes and Optical Rotations

Mass

Filter Paper

0.58 g

Filter Paper + Initial Crystal Sample

8.25 g

Recovered Crystal Sample

7.67 g

50 milliliters Erlenmeyer Flask with 2 boiling rocks

39.75 g

50 milliliters Erlenmeyer Flask with Amine merchandise and 2 boiling rocks

42.63 g

Amine merchandise

2.88 g

Optical Rotation

Specific Rotation of ( – ) -?-phenylethylamine

-31.8o

Table 2: Experimental Raw Given Data

Volume of ( ± ) -?-phenylethylamine

10.0 milliliter

Density of ( ± ) -?-phenylethylamine

0.9395 g/mL

Molecular Weight of ( – ) -?-phenylethylamine

121.8 g/mol

Molecular Weight of ( + ) -tartaric acid

150.09 g/mol

[ ? ] D ( – ) -?-phenylethylamine

-40.4o ± 0.2o

Table 3: Multitudes, Moles, Optical Purity, and % Output

Mass ( ± ) -?-phenylethylamine

9.40 g

Gram molecules ( ± ) -?-phenylethylamine

0.0776 mol

Gram molecules ( – ) -?-phenylethylamine

0.0388 mol

Gram molecules of tartaric acid:

0.0388 mol

Percentage Output of ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate

73.1 %

Percentage Output of ( – ) -?-phenylethylamine

61.3 %

Optical Purity

83.7 %

Calculations:

% Output of ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate:

Mass ( ± ) -?-phenylethylamine

Gram molecules ( ± ) -?-phenylethylamine

m ( ± ) -?-phenylethylamine = denseness ten volume

= 0.9395 g/mL X 10 milliliter

= 9.40 g

N ( ± ) -?-phenylethylamine = mass/molecular weight

= 9.40 g/ 121.18 g/mol

= 0.0776 mol

Gram molecules ( – ) -?-phenylethylamine and tartaric acid:

N ( – ) -?-phenylethylamine = 0.0776 mol/ 2

= 0.0388 mol

*Racemic mixture so divided by 2*

( half of entire moles )

N ( + ) -tartaric acerb = N ( – ) -?-phenylethylamine

= 0.0388 mol

Theoretical Output of

( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate:

Actual Output of

( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate:

m ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate

= n x M

= 0.0388 mol X ( 121.18 g/mol + 150.09 g/mol )

= 10.5 g

m ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate

= Mass filter paper + initial crystal sample – Mass filter paper

= 8.25 g – 0.58 g

= 7.67 g

Percentage Output of ( – ) -?-phenylethylamine- ( + ) -hydrogen tartrate:

% Output = ( Actual Yield / Theoretical Yield ) X 100 % i? Actual ( what was obtained after experiment )

= ( 7.67 g / 10.5 g ) X 100 % i? Theoretical ( the mass that should hold been

= 73.1 % obtained if all aminoalkane was extracted )

% Output of ( – ) -?-phenylethylamine:

Theoretical Output of ( – ) -?-phenylethylamine

Actual Output of ( – ) -?-phenylethylamine

Since the initial mixture was racemic:

m ( – ) -?-phenylethylamine = m ( ± ) -?-phenylethylamine / 2

= 9.40 g / 2

= 4.70 g

m ( – ) -?-phenylethylamine = mflask w/ amine+ rocks -mflask w/ rocks

= 39.75 g – 42. 63 g

= 2.88 g

Percentage Output of ( – ) -?-phenylethylamine

% Output = ( Actual Yield / Theoretical Yield ) X 100 % i? Actual ( what was obtained after experiment )

= ( 2.88 g / 4.70 g ) X 100 % i? Theoretical ( the mass that should hold been

= 61.3 % obtained if all aminoalkane was extracted

Optical Purity of Sample:

Theoretical Optical Purity:

Actual Optical Purity:

Optical Purity

= -40.4o ± 0.2o

Specific Rotation ( [ ? ] D ) :

=Optical Rotation [ ? ( observed ) ] / c * 1

= -31.8o / ( 1.0 diabetes mellitus x 0.94 g/mL )

= -33.8o

Optical Purity:

= ( Actual optical pureness obtained/ theoretical optical pureness ) X 100 %

= -33.8o / -40.4o x 100 %

= 83.7 %

Discussion:

When the ( + ) -tartaric acid was added to the racemic mixture, ( ± ) -?-phenylethylamine, ( – ) -amine- ( + ) -hydrogen tartrate, and ( + ) -amine- ( + ) -hydrogen tartrate salts were formed. The ( – ) -amine- ( + ) -hydrogen tartrate was much less soluble in methyl alcohol, and hence crystallized out of the solution ( Lab Manual, 2007 ) . This method of separation was proven to be rather successful, as the per centum output of this crystallisation was 73.1 % , which is comparatively high. The presence of drosss, every bit good as the inability to wholly crystallise the salt from methyl alcohol most probably attributed to any disagreements. It is besides possible that although the ( – ) ( + ) salt is less soluble than the other salts, it still has some kind of solubility, and hence crystallizes instead easy ( hence the compulsory 2 hebdomad waiting period, in our instance it was 4 hebdomads ) . As good, the other salts, despite their high solubility in methyl alcohol, may hold still crystallized really somewhat over the long waiting period, adding to drosss

Addition of NaOH resulted in the formation of two distinguishable beds: a white, cloudy aqueous bed ( top ) , and a clear aminoalkane bed ( underside ) , and allowed for the isolation of ( – ) -?-phenylethylamine ( Lab Manual, 2007 ) . The add-on of 5 milliliter of H2O to the flask confirm that the top bed was the aqueous bed, since it increased comparative to the bottom bed and the H2O was absorbed here ( Lab Manual, 2007 ) . The aqueous bed consisted of the ( – ) -amine, along with Na tartrate, and H2O, while the aminoalkane bed included any drosss. The Na tartrate readily dissolved in H2O, while methylene chloride ( CH2Cl2 ) was added to fade out ( – ) -?-phenylethylamine ( boiling point ~ 186oC ) , since it had a lower boiling point ( 40oC ) , and could easy be removed through warming ( Synthesis and declaration of alpha-phenyethylamine.

After a filtration procedure, including a series of extractions, there was per centum output of 61.3 % for the ( – ) -?-phenylethylamine, which is a lower output than the original 73.1 % , bespeaking that there was a loss of aminoalkane during the 2nd portion of the experimental process. The chief cause of this mistake was the inadvertent disposal of much of the aminoalkane bed, in which a little sum of ( – ) -?-phenylethylamine was still present. The presence of some drosss may hold besides affected consequences, nevertheless, they would hold alternatively increased the output and lead to deceptive consequences. Another possible cause of mistake is the little escape out of the glass stopper on the separatory funnel when the solution was shaken. There was a spot of solution that leaked out the underside or squirted out the top when let go ofing the force per unit area in the funnel. Subsequently, the mistake that well lowered the output of the merchandise greatly increases the optical pureness of the mixture. The ascertained rotary motion of the concluding sample was -31. 8o ( levorotatory, left manus rotary motion ) and the specific rotary motion was -33.8o compared with the empirical specific rotary motion of -40.4o ± 0.2o ( Lab Manual, 2007 ) . The attendant optical pureness was 83.7 % , which is well high. Aside from the antecedently mentioned disposal of the organic bed, legion other mistakes, such as the presence of drosss may hold contributed to divergences in the optical pureness. The negative ( antagonistic clockwise ) rotary motion basically confirmed that the enantiomorph being isolated was the ( – ) -?-phenylethylamine, and the high optical pureness demonstrated that the extraction was accomplished with much success and considerable truth, since the concluding merchandise was chiefly ( – ) -amine, despite the comparatively low output.