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The Separation Of Compounds Of Different Polarity

Paper Type: Free Essay Subject: Biology
Wordcount: 2457 words Published: 8th May 2017

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Chromatography is a technique used to separate individual components in a mixture. The basic principle is, they all have a stationary phase (a solid, or a liquid supported on a solid) and a mobile phase (a liquid or a gas which carries the components of the mixture with it) The mobile phase flows through the stationary phase and carries Different components that travel at varied rates. Thin layer chromatography involves using a thin, uniform stratum of stationary phase coated onto a piece of metal or glass. The stationary phase contains a substance which fluoresces in UV light. The mobile phase is a suitable liquid solvent or mixture of solvents.

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PRINCIPLE AND MECHANISM INVOLVED

The stationary phase – silica gel

Silica gel is a supplement of silicon dioxide (silica). In a giant covalent structure the silicon atoms are joined via oxygen atoms. And, at the silica gel surface of the silicon atoms are attached to -OH groups. So, the surface contains Si-O-H bonds instead of Si-O-Si bonds. The surface part is depicted by the following structure

As the surface of silica gel has the -OH groups and is very polar, it can form two types of bonds like

Hydrogen bonds with suitable compounds around it

Van der Waals dispersion forces and dipole-dipole attractions.

Development of chromatogram and separation of compounds

The solvent that soaks up the plate first dissolves the compounds in the dots put on the base line. They are carried up the plate with the solvent moving upwards.

Carrying of compounds depends on two things:

The solubility of compound in mobile phase. (It is attraction between the molecules of the compound and those of the 5% diethyl ether in methyl benzene)

Sticking of compound to the stationary phase. (It is attraction between the molecules of the compound and the silica gel)

If the spot contains two compounds – one forms hydrogen bonds, and the other take part in weaker van der Waals interactions.

As the hydrogen bond having compound will stick to the surface of the silica gel more firmly than the other one, that one is more strongly adsorbed than the other.

Adsorption

The process by which molecules of a substance, such as a gas or a liquid, collect on the surface of another substance, such as a solid. The molecules are attracted to the surface but do not enter the solid’s minute spaces as in absorption.

There is a constant movement of a molecule between being adsorbed onto the silica gel surface and going back into solution in the solvent as adsorption isn’t permanent. Obviously the compound can only travel up the plate during the time that it is dissolved in the solvent. While it is adsorbed on the silica gel, it is temporarily stopped – the solvent is moving on without it. That means that the more strongly a compound is adsorbed, the less distance it can travel up the plate. It is very unlikely that both will hydrogen bond to exactly the same extent, and be soluble in the solvent to exactly the same extent. It isn’t just the attraction of the compound for the silica gel which matters. Attractions between the compound and the solvent are also important – they will affect how easily the compound is pulled back into solution away from the surface of the silica. The elution solvent passes the component during chromatography. If this reaches new adsorption places, it means that there will be equilibrium between the elution solvent and the adsorbent.

APPARATUS:

Chromatographic development tanks and paper

Plastic-backed silica get TLC plates (with and without fluorescent indicator )

Melting-point tubes

REAGENTS:

Phosphomolybdic acid, 12% in ethanol

Iodine

Chloroform solutions of squalene, Vitamin E (α-tocopherol), Cholesterol

Chloroform, methyl stearate, methyl oleate and one or more mixtures

Eluant solution (5% diethyl ether in methyl benzene)

PROCEDURE:

Pencil line is drawn at 1cm from the bottom of the TLC plates without fluorescent indicators and small drops of given known and unknown samples are spotted on them and are labelled. The plates are dried and then eluted with 5% diethyl ether in methyl benzene in ascending manner in a covered lined beaker.

After the solvent reached almost top of the plates, the chromatograms are again dried and then one plate Visualized by spraying with phosphomolybdic acid and heated and the other plate exposed to iodine in a covered tank.

It is repeated using a plate with fluorescent indicator and observed under a UV lamp. Also repeated with varying sample sizes.

Compared the effect of lining the tank with paper soaked in the eluting solvent with chromatograms developed in an unlined tank.

RESULTS:

The results obtained are

Plate 1

(Visualizing methods used UV & Phosphomolybdic acid with ethanol)

Distance travelled by the solvent front is 7.5cm

KNOWN SAMPLES

Sl. No.

SAMPLE NAME

Distance travelled

(in cm )

x/7.5

(in cm)

RF value

1

Methyl oleate

4.4

4.4/7.5

0.58

2

Squalene

6

6/7.5

0.8

3

Methyl stearate

4.5

4.5/7.5

0.6

4

Cholesterol

0.6

0.6/7.5

0.08

5

Vitamin -E

3.4

3.4/7.5

0.45

UNKNOWN SAMPLES

Sample A

Rf = 5.9/7.5=0.78

Rf= 0.6/7.5=0.08

As the Rf values are close to that of Squalene and Cholesterol, Sample A may contain Squalene and Cholesterol

Sample B

Rf=3.4/7.5=0.45

Rf=4.5/7.5=0.6

Rf=6/7.5=0.8

As the Rf values are close to that of Vitamin E, Squalene and Methyl stearate, Sample B may contain Vitamin E, Squalene and Methyl stearate

Sample C

Rf=0.6/7.5=0.08

Rf=4.5/7.5=0.6

As the Rf values are close to that of Cholesterol and Methyl stearate, Sample C may contain Cholesterol and Methyl stearate

Sample D

Rf=0.6/7.5=0.08

Rf=3.4/7.5=0.45

Rf=4.5/7.5=0.6

As the Rf values are close to that of Cholesterol, Vitamin E and Methyl stearate, Sample D may contain Cholesterol, Vitamin E and Methyl stearate

Plate 2

(Visualizing agent used is iodine)

Distance travelled by the solvent front is 7.5 cm

KNOWN SAMPLES

Sl. No.

SAMPLE NAME

Distance travelled

(in cm )

x/7.5

(in cm)

RF value

1

Methyl oleate

4.4

4.4/7.5

0.58

2

Squalene

6

6/7.5

0.8

3

Methyl stearate

4.5

4.5/7.5

0.6

4

Cholesterol

0.6

0.6/7.5

0.08

5

Vitamin -E

3.4

3.4/7.5

0.45

UNKNOWN SAMPLES

Sample A

Rf = 5.9/7.5=0.78

Rf= 0.6/7.5=0.08

As the Rf values are close to that of Squalene and Cholesterol, Sample A may contain Squalene and Cholesterol

Sample B

Rf=3.4/7.5=0.45

Rf=4.5/7.5=0.6

Rf=6/7.5=0.8

As the Rf values are close to that of Vitamin E, Squalene and Methyl stearate, Sample B may contain Vitamin E, Squalene and Methyl stearate

Sample C

Rf=0.6/7.5=0.08

Rf=4.5/7.5=0.6

As the Rf values are close to that of Cholesterol and Methyl stearate, Sample C may contain Cholesterol and Methyl stearate

Sample D

Rf=0.6/7.5=0.08

Rf=3.4/7.5=0.45

Rf=4.5/7.5=0.6

As the Rf values are close to that of Cholesterol, Vitamin E and Methyl stearate, Sample D may contain Cholesterol, Vitamin E and Methyl stearate Thus, the given unklnown samples contain:

Sample – A

Squalene, Cholesterol

Sample – B

Vitamin – E, Squalene, Methylstearate

Sample – C

Cholesterol, Methylstearate

Sample – D

Cholesterol, Vitamin – E, Methylstearate

DISCUSSION:

1) COMMENT ON THE RELATIVE POLARITIES OF THE FIVE COMPONENTS

The larger Rf of a compound, the larger the distance it travels on the TLC plate. The compound with the larger Rf is less polar because it interacts less strongly with the polar adsorbent on the TLC plate.

” As polarity increases, Rf decreases “

A molecules polarity depends primarily on

1. The extent to which it can hydrogen bond

2. The number of electronegative atoms

3. The polarizability of the bonds or atoms

4. The net dipole moment of the molecule.

Functional Group Polarity

Alkanes least polar.

Alkenes more polar than alkenes.

Conjugated Polyenes and Aromatic Compounds increasing polarity.

Ethers and alkyl halides are more polar than the above

Aldehydes, Ketones, and Esters large dipole moment due to the carbonyl.

Amines and Alcohols more polar

Carboxylic Acids are the most polar functional group

Functional groups in order of increasing polarity :

[ Hydrocarbons < ethers < tertiary amine < nitro < dialkyl amines < ketone < aldehyde < primary amine < alcohol < phenol < alkanoic acid < sulfonic acid ]

Relative polarity of 5 components

Methyl oleate ( Rf = 0.58) Methyl stearate ( Rf = 0.6 )

Here it can be seen that both the compounds differ in one alkene group ( double bond ), and methyl oleate with double bond (–CH2–) is more polar than the methyl stearate (–CH–).

Cholesterol ( Rf = 0.08) Squalene ( Rf = 0.8)

Here it can be seen that both the compounds differ in the ring structure and the number of double bonds. The cholesterol ( –OH ) is more polar and the squalene is least polar ( only – CH2– ).

Vitamin – E

It is a less polar compound ( — CH2– ).

Polarity sequence

[ CHOLESTEROL >> VITAMIN – E > METHYL OLEATE > METHYL STEARATE > SQUALENE ]

2) COMMENT ON THE METHODS OF VISUALIZATION USED,

EFFECT OF SAMPLE SIZE AND THE EFFECT OF USING AN UNLINED TANK

A ) Methods of visualization

The methods used showed good results in identifying the components, among them the UV- Method and the Iodine vapour Method are more reliable than Phosphomolybdic Acid Method.

Methods used are

Ultra violet method

Ultraviolet light of two different wavelengths is normally used. TLC plates can be supplied with the fluorescent compound zinc sulphide. The compounds present in the sample will show up as dark spots on the green background. The spots can be marked with a soft pencil.

Ultraviolet light of 356 nm is used to visualise aromatics and molecules with extended conjugated π-electrons. The compounds present in the sample will show up as purple spots.

Iodine vapour method

The staining of a TLC plate with iodine vapour is among the oldest methods for the visualization of organic compounds. It is based upon the observation that iodine has a high affinity for both unsaturated and aromatic compounds.

Preparation: To 100 mL wide mouth jar (with cap) is added a piece of filter paper and few crystals of iodine. Iodine has a high vapour pressure for a solid and the chamber will rapidly become saturated with iodine vapour.

Phosphomolybdic acid method

Phosphomolybdic acid stain is a good “universal” stain which is fairly sensitive to low concentrated solutions. It will stain most functional groups, however it does not distinguish between different functional groups based upon the coloration of the spots on the TLC plate. Most often, TLC’s treated with this stain will appear as a light green colour, while compounds of interest will appear as much darker green spots. Good for separation of lipids.

Preparation: Dissolve 10 g of phosphomolybdic acid in 100 mL of absolute ethanol.

B ) Effect of sample size :

Sample spots made using TLC capillaries should be no larger than 1-2 mm in diameter, because component spots in the developed plate will be no smaller than, and will usually be larger than, the size of the initial spot. If the initial spot is larger than 2 mm in diameter, then components with similar Rf values may not be resolved because their spots will be so large that they will overlap considerably and may appear to be one large spot. Small initial spots, on the other hand, maximize the potential of complete separation of components.

C ) Effect of using an unlined tank :

Unlined tank results in the uneven travelling of the component this results in the false identification of the compound. This uneven distribution can be termed as ” Edge effect “.

” Edge effect is a phenomenon in which the solvent moves more at the centre than the sides. It is over come by lining the tank “.

The beaker is often lined with some filter paper soaked in solvent. The reason for using lined tank is to make sure that the atmosphere in the beaker is saturated with solvent vapour. Saturating the atmosphere in the beaker with vapour stops the solvent from evaporating as it rises up the plate.

3) WOULD THE MEMBERS OF THE HOMOLOGOUS SERIES BE READILY SEPERATED BY THIS TECHNIQUE

YES, members of a homologous series can be readily separated with this method.

The homologous series found in this are

Squalene and Cholesterol

Methyl oleate and methyl stearate

Vitamin – E

A) Squalene and Cholesterol

Both forms a homologous series by small variations, the basic structure of both is Isoprene.

Although, they differ in some cases and with basic structure they can be easily separated by TLC Technique.

Both differ in their Retention Factor Values.

B) Methyl Stearate and Methyl Oleate

Both forms a homologous series by the basic structure with

—COOCH3, —CH3

Both differ in their Retention Factor Values.

C) Vitamin – E

Likewise, the Vitamin- E forms a series with them and can be easily separated.

Homologous series :

A homologous series is a series of organic compounds with a similar general formula, possessing similar chemical properties due to the presence of the same functional group, and shows a gradation in physical properties as a result of increase in molecular size and mass.

 

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