Experiment 2: Polarimetry



  1. Part A
  2. Aim
  3. Method
  4. Results
  5. Questions
  6. Conclusion
  7. Part B
  8. Aim
  9. Method
  10. Results
  11. Questions
  12. Conclusion
  • Bibliography

Practical Date:

Submission Date:

Group 2

Demonstrator: Lori





An Enzyme catalysed resolution of racemic amino acids using papain


To resolve a racemic mixture of D, L- Benzoylalanine and then use polarimetry to quantitate and confirm identity of the product, Benzoyl –L-alanine anilide.



Refer to Laboratory Exercises in Applied Biochemistry: Semester 1, 2012.

Experiment 2, pp 23 – 32.




Benzoyl Chloride D/L- Alanine D/L- Benzoyl alanine

M Wt = 135.5 M Wt =89 M Wt = 193


10.8g/ 135.5= 0.07moles 6.2g/89 = 0.07moles Expect 0.07 x193 = 13.51g



Under ideal conditions, we expect 13.51g of D/L –benzoylalanine in this reaction.

Since only ONE HALF of this product is the L-isomer (benzoylalanine), the expected theoretical value is 13.51/2 = 6.755g.


To calculate :

% yield in this reaction = Amount of dried and weighed D/L Benzoyl Alanine x 100

Theoretical amount calculated from starting reagents





D/L Benzoyl alanine + Aniline L- Anilide (precipitate) + D- Anilide (soluble)

M Wt = 193 M Wt = 93 M Wt = 268


0.02moles 0.05 moles x 93 = 4.65g 0.02moles


Amount of Benzoyl alanine required = 0.02 moles x 193 = 3.86 g

Expectedyield = 0.02 moles x 268 = 5.36g

Therefore, from the reaction we expect 5.36g of D/L – Benzoyl alanine anilide

From this product, only half of it is L- isomer (Benzoyl –L-alanine anilide).


The expected yield is 5.36g/2 = 2.68g.


% yield in this reaction = Amount of dried and weighed Benzoyl –L-alanine anilide x 100

Theoretical amount calculated from starting reagents



% yield in this reaction = 2.3g/2.68gx 100 = 85.82 %


The Benzoyl –L-alanine anilide was driedand 0.67g was weighed and dissolved in chloroform in a 25ml volumetric flask. The slightly cloudy solution was directly filtered through a filter paper that was fluted in a glass filter funnel. The filtrate was placed in a 1 decimetre polarimetry tube and sealed with rubber washer and glass mirror.

The concentration of the Benzoyl –L-alanine anilide is 0.67g/25ml x 4 = 2.68g/100ml.


TheSpecific optical rotation of the equivalent 2.68g/100 mL chloroform solution of the Papain–resolved Benzoyl –L-alanine anilide was now completed.


In a 1 decimetre tube, a pure chloroform solution was first placedand 5 individual readings were made to establish that there is no optical rotation of plane-polarised sodium light.

Three readings on the left and right vernier of the Papain–resolved Benzoyl –L-alanine anilidewere taken and the results tabulated as follows



Blank chloroform Test L-benzoyl –L-alanine anilide
Vernier (L) Vernier ( R)Vernier (L) Vernier (R)
0.05°0.10°178.10 (-1.90°) 178.10 (-1.90°)
0.00° 0.05°178.10 (-1.90°) 178.10 (-1.90°)
0.05°0.10°178.10 (-1.90°) 178.10 (-1.90°)
Average L & R vernier = 0.075°rotationAverage L & R vernier = – 1.90° rotation



The specific rotation for the purified Benzoyl-L-alanine anilide is determined as follows:


20 100 x a(observed vernier rotation corrected)

[a] = Length of cell (dm) x Concentration in g/100ml


Here the observed vernier angular rotation ais –1.90° – 0.075°= -1.975°

Concentration of Benzoyl-L-alanine anilide = 2.68g/100 mL of chloroform

Length of polarimeter tube is 1 decimetre =1dm


20 100 x -1.975°

[a] = 1.00dm x 2.68g/100 mL


Specific rotation of the L –benzoyl-L-alanine anilide = -73.69°

This is similar to the literature value since the literature value is -74° +/- 1.5°




The solvent is methanol –chloroform (3:1) for the development of the chromatogram on a silica gel plate with l254nm


Solvent front


The developing solvent is

MeOH-chloroform (3:1)

One spot is visible under

UV light at l254nm









Baseline application




Calculate Rf value of the compound = Distance moved to the centre of the Spot (mm)

Distance moved by the solvent Front (mm)

Rf value of this compound = 68mm = 0.9067



  1. (a) Catalytic activity of papain

Papain catalysed hydrolysis obeys Michaelis-Menten kinetics.

Papain catalyses the process of hydrolysis through a three-step reaction pathway. Below is Michaelis complex illustrating the above:



E + S ES ES’ E + P2 + P1




ES is Michaelis-Menten Complex (Enzyme Substrate).

ES’ is acyl enzyme. ES’ is formed through thiol group of cysteine-25.

P2 refers to the acid.

P1 is the amine or alcohol moiety of the substance that has undergone hydrolysis.

Menten parameter, Km and Kcatobtained from steady steady-state reaction rates are in line with those illustrated in the equation below:

V0 = Kcat. [E][So]/(Km + [So])

Kcat and Km are related this form

Kcat. = K+2(K+3)/(K+2 + K+3)


Km = K+3(K-1 + K+2)/K+2 (K+2+K+3)

In case the rate limiting step is deacylation (K+3) then;

Kcat = K+3

If K-1 > K+2.

This means that:

Km = K-1K+3/K+1K+2 = K+1K+3/K+2

In case the rate limiting step is acylation (K+2) then;

Kcat. = K+2 and,

But if K-1> K+2, then

Km = K+1/K+1 = K3


Cataylsis of papain is promoted by the active site thiol and imidazole groups as they interact with the substrate. The tetrahedral groups are assumed to play a vital role in this process comparable to an analogy of simple amides and esters hydrolysis. The hydrolytic step of deacylation is generally base catalyzed implying that the breakdown step is catalyzed by an acid. This principle applies to acylation where its formation is base-catalyzed and breakdown of the tetrahedral intermediate acid-catalyzed.

(b) PapainSubtrateSpecificity

Papain specificity is broad for acyl-L-amino acid derivatives. The active site of papain is thought to be a cleft in the enzyme. This site consists of seven subsites which can accommodate one amino acid each. Specificity of papain is seen in the acylation step, K+2. According to the Model building, is essential for binding of peptide bonds that are adjacent to the bond that will be cleaved. Papain has a low preference for S3 site with hydrophobic and bulky amino acids.

Two critical binding sites for papain are very vital for its activity. The structure of papain contains hydrophobic pockets in its cleft. The surface of these clefts is made up of side chains consisting of valine-133, tyrosine-67, proline-68, tryptophan-69, phenylalanine-207, alanine-160, and valine-157. The two bonds are C=O group and the side chain. This is the major reason for specificity of the enzyme for residues of L-amino acid.L-alanine, stereospecificity for this site only happens when there is actual contact with the enzyme. This is because L-alanine cannot bind to the hydrophobic bond. However, even though the side chain of L-Alanine cannot bind the hydrophobic pocket, stereospecificity is assured.

(c) Binding Site

The binding site of lysozyme consists of six subsites. These sites can bind six sugar residues. The binding site of lysozymes is similar to tapain in that it has acleft between its domain. It also acts by causing the substrate to adopt a conformation that is strained. The lysozyme binds the six-unit sugar distorting the fourth sugar breaking the glycosidic bond.

(d) Essence of thiol group

In alkylation, the thiol group of cysteine-25 residue acts as the nucleophile while in the activation mechanism it is the leaving group. There is a lot of evidence implying that thiol group is essential in the activity of the enzyme papain. It is readily deactivated by oxygen, hydrogen peroxide, mercuric salts and oxygen. Reactivation of the enzyme with cyanide ions, thiols, and hydrogen sulphide in EDTA presence. N-ethylmaleimide and iodoacetate irreversibly inhibit papain while the hydrolysate form of papain is irreversibly inhibited by iodoacetate found in S-carbomethyl cysteine residue. From this outlined properties, papain belongs to the group of enzymes that belong to the category of cysteine enzymes that depend on the thiol group for enzymic activity.











Experimental yield of L – anilide is 2.3g. The percentage yield is 85.82% and the specific optical rotation is -73.69°


We resolved a racemic mixture of D,L – Benzoylalanine by using papain to catalyse the synthesis of Benzoyl- L- anilide. The product was relatively pure (0.9067) on TLC and had a specific optical rotation of -73.69°. We achieved a yield of 85.82%.


Determination of the Sucrose Content of condensed milk by Polarimetry Determination


To use polarimetry to determine the sucrose content of Nestle’s condensed milk.



Refer to Laboratory Exercises in Applied Biochemistry: Semester 1, 2012.

Experiment 2, pp 23 – 32.




Blank distilled water Direct Sucrose solution reading (D)
Vernier (L) Vernier (R)Vernier (L) Vernier (R)
0.10°0.10°+11.60° +11.60°
Average L & R vernier = 0.095° rotationAverage L & R vernier = +11.68° rotation (D)




Blank distilled water Invert Sugar solution reading (I)
Vernier (L) Vernier ( R)Vernier (L) Vernier (R)
0.00° 0.20°178.00 (-2.00°) 178.20 (-1.80°)
0.15° 0.20°178.20 (-1.80°) 178.30 (-1.70°)
0.00°0.05°178.40 (-1.60°) 178.40 (-1.60°)
0.05°0.10°178.40 (-1.60°) 178.50 (-1.50°)
0.10° 0.10°178.40 (-1.60°) 178.50 (-1.50°)
Average L & R vernier = 0.095° rotationAverage L & R vernier = -1.67° rotation (I)



Calculation based on the observations above can be made as follows to calculate the content of sucrose in the condensed milk tested

w = weight of sample taken (31.80g)

F = percentage of fat in the sample (manufacturer claims 9%)

P = percentage of protein (n x 6.38) in the sample (manufacturer claims 8.5%)

V = volume to which the sample is diluted before filtration (here 250 mL)

v = correction in mLof precipitatevolume produced during clarification

D = observed direct (D) polarimeter reading (+11.68° rotation – 0.095° rotation =+11.585°)

I = observed invert (I) polarimeter reading (-1.67° rotation – 0.095° rotation =-1.765°)

L = length of dm of polarimeter tube (used above for measurements a 2dm tube)

Q = inversion divisor factor (0.8825)


v =correction in ml for the volume of precipitate produced during clarification process


v = w ([(F x 1.08) + ( P x 1.55) ]


v = 31.80 ( [ ( 9.00 x 1.08) + (8.5 x 1.55) ]


v = 7.28061 mL


Now the % Sucrose in the sample tested can be calculated as:


% Sucrose = D- (5/4 x I) x (V-v) x V

Q V l x w



= +11.585° – (5/4 x-1.765°) x (250 – 7.28061) x 250

0.8825 250 2.00dm x 31.80

= 15.6275 x 0.9709 x 3.9308

= 59.64 % sucrose

The manufacturer claims on the label that the range is 42-45% of sucrose in the product

This product does not comply with the manufacturer’s claims.


  1. Calculation

One gram of an organic substance was dissolved in 50.00ml of water. The solution in a 2.0 dm tube read +2.676° in a polarimeter, while distilled water in the same tube read + 0.016°. Calculate the specific rotation of the substance.


Specific rotation is determined by the following equation:


20 100 x a(observed rotation corrected)

[a] = Length of cell (dm) x Concentration in g/100ml


Here the observed angular rotation a +2.676° – (+ 0.016°) = +2.66°

Concentration of the organic substance = 1g/50ml x 2 = 2g/100ml

Length of polarimeter tube is 2 decimetre = 2dm




20100 x +2.66°

[a] = 2.00 dm x 2.0g/100 mL




Specific optical rotation = +66.5°

  1. Effect of Temperature on Optical Rotation

Temperature changes affect the rotation in this way. First, the temperature at which the solvent was prepared profoundly affects the rotation of a compound. Preparation of the solvent at a certain temperature and then measured at a different one, an error will occur due to the thermal expansion of this solvent. An excellent example is water which for every 5 degrees Celsius, increases in volume by 0.1%.

Secondly, change in the temperature of the sample will affect the rotation of the compound under investigation. This comes about because of the change of the sample cell length under use. This change in length arises due to the thermal expansion of the cell material i.e. stainless steel or glass. For good results to be obtained, this has to be taken into consideration.

  1. Circularly Polarized Light and Circular Dichroism

Circularly polarized light: – light has both electrical and magnetic components. Both of these properties possess vector quantities, perpendicular to each other. Thus, plane polarized light has two vectors of equal length rotating circularly but in opposite directions. These make up the circularly polarized light, which consists of the LCP Light, and RCP Light. The LCP Light and RCP Light can be absorbed to a different extent by the same substance.

Circular dichroism is the differential absorption, which exhibited by LCP Light and RCP Light.Optically active molecules, exhibit it in the form of absorption bands. Circular dichroism principle finds use in the study of biological molecules. Proteins, being biological molecules that possess both levorotatory and dextrorotatory components, can be studied. The secondary structure of these proteins, will impact a distinct Circular dichroism to its various respective molecules. Hence, the double helix, which makes up the nucleic acids and the alpha helix that makes up the proteins, have distinct Circular dichroism spectral signatures that represent their respective structures. Thus, any little change can be detected since that the protein has distinct Circular dichroism spectral signature.


The content of sucrose in Nestle’s condensed milk was 59.64%


Martin, A. N., Sinko, P. J., &Singh, Y., 2011. Martin’s physical pharmacy and pharmaceutical sciences: physical chemical and biopharmaceutical principles in the pharmaceutical sciences. 6th ed., 50th anniversary ed. Baltimore, MD: Lippincott Williams & Wilkins.


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