Purification and Characterisation of Organic Compounds Chapter - Chemistry JEE Main

This chapter helps in understanding the principles of purification techniques such as chromatography, sublimation, distillation, etc. on the basis of the chemical nature of the organic compounds. After the extraction of organic compounds, they need to be purified based on the nature of organic compounds and impurities.

In this chapter, we will study the techniques and methods for the purification of organic compounds which are important for the JEE competitive exams and this article will help you understand various techniques for purification according to the nature of organic compounds which are important for the JEE exam.

JEE Main Chemistry Chapters 2024

Important Topics of Purification and Characterisation of Organic Compounds Chapter

• Classification of organic compounds

• Chromatography

• Chromatography

• Sublimation

• Crystallisation

Important Definitions of Purification and Characterisation of Organic Compounds

Methods of Purification of Organic Compounds

The organic compounds synthesised in the laboratory or  extracted from a natural source are purified. Purification methods of organic compounds are based on the nature of the compound and the impurity present in it.  The various techniques used for purification are as follows:      

1. Sublimation 

2. Crystallisation 

3. Distillation 

4. Differential extraction 

5. Chromatography 

The purity of purified compounds is ascertained by determining their melting or boiling point as the pure compounds have sharp melting points and boiling points. These days the purity  of an organic compound is determined on the basis of different types of chromatographic and spectroscopic techniques. 

• Sublimation

• It is the process in which the solid is directly converted into its vapour state by heating without passing it through the liquid state.

• This process is used for the purification of sublimable compounds from their non-sublimable impurities.

• Crystallisation

• This process is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. 

• Impurities are little soluble or insoluble in the solvent at low temperature but highly soluble at high temperature. At high temperatures, the compound and impurities dissolve, when the solution is cooled to a very low temperature, the pure compound crystallises and the impurities remain dissolved in the solution, which is discarded. This process is done multiple times to purify the compound to the highest level. 

• Distillation:

• This technique is used to separate:

• Volatile liquids from non-volatile impurities and 

• The liquids which have sufficiently high differences in their boiling points. Liquids which have different boiling points vaporise at different temperatures. 

• Chloroform (b.p. 334 K) and aniline (b.p. 457 K ) can be easily separated by the distillation process.

• The compound of lower boiling point boils first and is collected in the receiver and condensed to get the pure compounds from the mixture and the same process is done for the compound of higher boiling point by heating the mixture further.

• If the difference in boiling points of two liquids is not high, then a simple distillation method cannot be used to separate them. 

There are two types of distillation which are:

(i)  Fractional Distillation

• The mixture in which the difference in boiling points of liquids is not much, simple distillation cannot be used to separate them as the vapours of such liquids formed have approximately the same temperature range and are condensed simultaneously. 

• In this technique, vapours of a liquid mixture before condensation are passed through a fractionating column.

• The vapours of the liquid which has a higher boiling point condense before the vapours of the liquid which has a lower boiling point.

• In the petroleum industry, the fractional distillation technique is used to separate different fractions of crude oil.

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(ii) Steam Distillation 

• This technique is used to separate compounds which are steam volatile and are immiscible with water. 

• In steam distillation, the sum of vapour pressures due to the organic liquid (p1) and that due to water (p2) becomes equal to the atmospheric pressure (p), i.e., p = p1 + p2 when the liquid boils. As p1 is lower than p, the organic liquid evaporates at a lower temperature than its boiling point. 

• Thus, if one of the components in the mixture is water and the other is a water insoluble component, then the mixture will boil close to or below 373 K. 

• Aniline is separated from the aniline-water mixture by steam distillation technique.

• Differential Extraction

• This is the technique which we use to purify the organic compound which is more soluble in the organic solvent than in the aqueous solvent. 

• In this process, the two solvents  aqueous and organic are immiscible with each other. 

• It is easier to obtain a compound from an organic solvent than from an aqueous solvent by distillation or by evaporation to get the compound. 

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• Chromatography 

• This  technique is used to separate mixtures into their components, purify compounds, and also to test the purity of compounds. 

• In this technique, the mixture of components is applied to a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas, acts as a moving phase and is allowed to move slowly over the stationary phase. T

On the basis of the  principle involved, chromatography is classified into different categories: 

1.  Adsorption Chromatography 

• Adsorption chromatography is based on the degree of adsorption of  different components on an adsorbent. 

• In adsorption chromatography, the commonly used adsorbents are silica gel and alumina. 

• When a mobile phase is allowed to move over a stationary phase (or adsorbent), the components present in the mixture move by varying distances over the stationary phase. 

• The two common types of chromatographic techniques, based on the principle of differential adsorption, are as follows:

• Column chromatography

• Thin layer chromatography 

Column Chromatography: In column chromatography, a column of adsorbent (stationary phase) is packed in a glass tube which is used to separate the components from the mixture. The mixture is then adsorbed on the adsorbent and is placed on the top of the adsorbent column packed in a glass tube. A suitable eluant which can be a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree of adsorption of the component on the adsorbent, complete separation takes place. The most easily adsorbed substances remain near the top and others come down to various distances in the column.

Thin Layer Chromatography: In thin-layer chromatography (TLC), the separation of substances of a mixture takes place over a thin layer of an adsorbent which is  coated on a glass plate. A thin layer of an adsorbent such as alumina or silica gel is spread over a glass plate and allowed to dry. This dried plate is known as thin-layer chromatography plate or chromaplate. The solution of the mixture which is to be separated is applied as a small spot approximately 2 cm above one end of the TLC plate. 

When the TLC plate is placed in a closed jar containing the eluant at the bottom, the eluant starts rising up the plate and the different components of the mixture move along with the eluant to different distances depending on the degree of adsorption of the components and so separation takes place. 

The relative adsorption of components in the mixture is expressed by its retardation factor, which is given by:

Rf = $\dfrac{\text{Distance moved by the substance from base line (x) }}{\text{Distance moved by the solvent from base line (y) }}$

2.  Partition Chromatography

• The principle of partition chromatography is based on the difference in the partitioning of components of a mixture between stationary and mobile phases in the liquid phase.

• Paper chromatography is one of the methods of partition type chromatography.

• A special quality paper is used in paper chromatography, which is known as a chromatography paper. The chromatography paper contains water trapped in it. Water acts as the stationary phase. 

• The solution of the mixture is spotted at the base of the strip of chromatography paper, the strip is then suspended in a suitable solvent or a mixture of solvents. This solvent acts as a mobile phase and rises up the paper.

• The paper selectively retains different components according to their difference in a partition in the two phases. The developed  paper strip is known as a chromatogram. The coloured components of the mixture can be seen as spots at different heights from the initial spot position on the chromatogram. 

Qualitative Analysis of Organic Compounds

Organic compounds majorly contain carbon and hydrogen elements, in addition to oxygen, nitrogen, sulphur, halogens, and phosphorus. These elements in the organic compound can be detected by the following methods:

• Detection of Carbon and Hydrogen 

• Carbon and hydrogen elements in an organic compound are detected by oxidising the elements in the compound with copper (II) oxide. 

• Carbon present in the compound is oxidised to carbon dioxide on heating with copper (II) oxide, which is tested with lime water, and develops turbidity due to the formation of CaCO3.

C    +     2 CuO    $\longrightarrow$     CO2   +  Cu

CO2    +    Ca(OH)2   $\longrightarrow$   CaCO3     $\downarrow$   +  H2O

• Hydrogen present in the compound is oxidised to water on heating copper (II) oxide, which is tested with white anhydrous copper sulphate, and turns blue due to formation of  CuSO4. 5 H2O. 

H2           +   CuO   $\longrightarrow$         Cu    +     2 H2O  

5 H2O     +    CuSO4   $\longrightarrow$     CuSO4 . 5 H2O  

• Detection of Other Elements

• Elements other than carbon and hydrogen such as nitrogen, sulphur, halogens, and phosphorus present in an organic compound are detected by "Lassaigne's test". 

• Lassaigne's extract: The organic compound is fused with sodium metal to convert the covalent form into an ionic form.

Na   +   C   +   N  $\longrightarrow$         NaCN 

2 Na        +       S  $\longrightarrow$      Na2S 

Na           +         X $\longrightarrow$        NaX

The cyanide, sulphide, and halide ions formed on the fusion of organic compounds with sodium can be extracted from the fused mass by boiling the extract with distilled water, this extract is known as sodium fusion extract. 

(a) Test for Nitrogen: The sodium fusion extract is heated with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of the Prussian blue colour confirms the presence of nitrogen in the organic compound.

6CN￣  +      Fe2+ $\longrightarrow$    [Fe(CN)6]4- 

3[Fe(CN)6]4-    +     4  Fe3+ $\longrightarrow$    Fe4[Fe(CN)6]3.xH2O 

• Sodium cyanide first reacts with iron(II) sulphate to form sodium hexacyanoferrate(II).

• On heating with conc. sulphuric acid, some iron(II) ions are oxidised to iron(III) ions, which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (or ferriferrocyanide) which is Prussian blue in colour. 

(b) Test for Sulphur 

• For sulphur detection, the sodium fusion extract is reacted with acidified with acetic acid and then lead acetate is added to it. The  black precipitate formation of lead sulphide indicates the presence of sulphur. 

S2-    +     Pb2      $\longrightarrow$         2 PbS  (Black ppt.)

• Sodium fusion extract on treating with sodium nitroprusside (Na2[Fe(CN)5NO]), the appearance of a violet colour indicates the presence of sulphur. 

S2-  +   [Fe(CN)5NO]2-   $\longrightarrow$    [Fe(CN)5NOS]4- (Violet complex)

(c) Test for Nitrogen and Sulphur

• When both nitrogen and sulphur are present in the organic compound, the sodium extract will form sodium thiocyanate, which gives blood red colour on boiling with iron(II) sulphate and concentrated sulphuric acid.

Na   +  C   +  N   + S    $\longrightarrow$    NaSCN 

Fe3+     +     SCN￣   $\longrightarrow$     $\left[Fe(SCN)]^{2+} \right]$(Blood red)

(d) Test for Halogen

The sodium fusion extract, for halogen test, is acidified with nitric acid and then treated with silver nitrate. 

• A white precipitate, which is soluble in ammonium hydroxide, confirms the presence of chlorine. 

• A yellowish precipitate, sparingly soluble in ammonium hydroxide, confirms the presence of bromine. 

• A yellow precipitate, insoluble in ammonium hydroxide, confirms the presence of iodine.

If nitrogen or sulphur is also present in the organic compound, then the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne's test as these ions would otherwise interfere with the silver nitrate test for halogens. 

(e) Test for Phosphorus

• The compound is heated with an oxidising agent  such as sodium peroxide so that the phosphorus in the organic compound is oxidised to phosphate. 

• The solution is then boiled with nitric acid and treated with ammonium molybdate. A yellow precipitate or yellow colouration indicates the presence of phosphorus. 

Na3PO4 + 3 HNO3 $\longrightarrow$   H3PO4  +  3NaNO3 

H3PO4   +    12(NH4)2MoO4   +   21 HNO3   $\longrightarrow$   21 NH4NO3    +    12H2O      +    (NH4)3PO4.12MoO3  (Ammonium phosphomolybdate - yellow ppt.)

You can read more about Qualitative analysis from Vedantu’s page on Principles of Qualitative Analysis.

Quantitative Analysis

In quantitative analysis, the percentage composition of elements present in an organic compound is determined by the methods based on the following principles:

• Carbon and Hydrogen 

• Both carbon and hydrogen in organic compounds are estimated by heating the organic compound in the presence of excess of oxygen and copper(II) oxide such that the carbon and hydrogen elements  in the organic compound are oxidised to carbon dioxide and water, respectively. 

CxHy   +   (x + y/4) O2   $\longrightarrow$   xCO2   +    (y/2)H2O 

• The mixture is passed through a U-tube containing anhydrous calcium chloride and the mass of water produced is determined by passing the mixture through a U-tube. 

• Carbon dioxide gas is absorbed in another U-tube containing a conc. solution of potassium hydroxide. The increase in the mass of calcium chloride and potassium hydroxide gives the mass of CO2 and water from which the mass of carbon and hydrogen can be calculated.

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Let the mass of organic compound be M g, the mass of water and carbon dioxide produced be Mw and Mc g, respectively;

Percentage of carbon = $\dfrac{12\times Mc\times 100}{44\times  M} $  

Percentage of hydrogen = $\dfrac{2\times Mw \times 100}{18 \times  M}$

• Nitrogen

There are two methods for estimation of nitrogen: 

1. Dumas Method: The nitrogen-containing organic compounds are heated with copper oxide to give carbon dioxide, water, and free nitrogen.

CxHyNz    +     (2x + y/2) CuO   $\longrightarrow$   xCO2   +    (y/2)H2O  +  (z/2) N2 + (2x + y/2) Cu

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The nitrogen gases produced are collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected at the top in the graduated tube.

Let the mass of organic compound = M g 

Volume of nitrogen collected = V1 ml 

Room temperature = T1K 

Pressure of nitrogen = P1

Volume of nitrogen = V1

Volume of nitrogen at STP, V  =   $\dfrac{P_{1} \times V_{1} \times 273}{760 \times T_{1}}$ 

We know  that at STP 22400 mL N2 weighs 28 g. 

V mL N2 at STP weighs  =    $\dfrac{28 \times V }{22400}$ g

Percentage of nitrogen  =    $\dfrac{28   \times   V  \times 100}{22400  \times  M}$ 

2. Kjeldahl's Method: The nitrogen-containing organic compound is heated with concentrated sulphuric acid due to which the nitrogen present  in the compound gets converted to ammonium sulphate. The resulting mixture is then heated with excess sodium hydroxide. The evolved ammonia gas is absorbed by an excess of the standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction which can be done by estimating unreacted sulphuric acid left after the absorption of ammonia.

Organic compound + H2SO4   $\longrightarrow$   (NH4)2SO4   +   2NaOH —> Na2SO4   +   2NH3   +   2H2O

2NH3   +   H2SO4  $\longrightarrow$     (NH4)2SO4 

Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and the nitrogen present in the ring (such as pyridine) as nitrogen of these compounds does not change to ammonium sulphate.

• Halogens

Carius method

• It is the method for the quantitative estimation of halogen in which a known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate preset in a hard glass tube known as carius tube, in a furnace. 

• Carbon and hydrogen elements present in the organic compound are oxidised to carbon dioxide and water, respectively.

• The halogen present in the compound forms the corresponding silver halide (AgX), which is filtered, washed, dried, and weighed.

Let the mass of organic compound = M g 

Mass of AgX formed = Mx g 

1 mol of AgX contains 1 mol of X. 

Mass of halogen in Mx g of AgX =   $\dfrac{\text{atomic mass of ‘X’ $\times$ M$_x$  }}{\text{molecular mass of AgX}}$

Percentage of halogen  =  $\dfrac{\text{atomic mass of ‘X’ $\times$ M$_x$  $\times$ 100}}{\text{molecular mass of AgX $\times$ M}}$

• Sulphur 

An organic compound of  known mass is heated in a Carius tube with an oxidising agent such as sodium peroxide or fuming nitric acid, the sulphur present in the compound is oxidised to sulphuric acid. The sulphuric acid is precipitated as barium sulphate by adding excess barium chloride solution in water. The white precipitate of barium sulphate is filtered, washed, dried, and weighed. Then the  percentage of sulphur can be calculated from the mass of barium sulphate by the using following formula:

Let the mass of organic compound = M g 

Mass of barium sulphate formed = m g 

We know that 1 mol of BaSO4 =  233 g BaSO4   =   32 g sulphur 

m g BaSO4 contains  $\dfrac{32 \times m }{233}$  g sulphur

Percentage of sulphur   =     $\dfrac{32  \times m  \times 100}{233 \times M}$ g

• Phosphorus

The organic compound containing phosphorus of known amount is heated with fuming nitric acid, the phosphorus present in the compound is oxidised to phosphoric acid. On adding ammonia and ammonium molybdate, the formed phosphoric acid is precipitated as ammonium phosphomolybdate, (NH4)3PO4.12MoO3. Phosphoric acid can also be precipitated as MgNH4PO4 by adding magnesia mixture which on heating yields Mg2P2O7. 

Let the mass of organic compound  =   M g 

Mass of ammonium phosphomolybdate  =   m g

Molar mass of (NH4)3PO4.12MoO3   =    1877 g 

Percentage of phosphorus   =   $\dfrac{31 \times m \times 100}{1877 \times M}$ %

If phosphorus is estimated as Mg2P2O7, 

Mass of Mg2P2O7  =   m g

Molar mass of (NH4)3PO4.12MoO3   =    222 g 

Percentage of phosphorus   =    $\dfrac{62 \times m \times 100}{222 \times  M}  $ %

Solved Examples from the Chapter

Example 1: Which method is used for the purification of aniline ? 

(a) Steam distillation 

(b) Simple distillation 

(c) Distillation under reduced pressure 

(d) Solvent extraction 

Solution: (a) Aniline is an organic compound which is steam volatile and immiscible in water, hence steam distillation method is used for the purification of aniline.

Key points: The organic compound which is steam volatile and also immiscible in water, can be purified by steam distillation.

Example 2: Why is the fusion of organic compounds carried out with sodium in the Lassaigne test? 

(a) To increase ionisation of compound 

(b) To increase the volume of compound 

(c) To increase the reactivity of compound 

(d) To convert covalent compound into ionic compound 

Solution:  (d) In Lassaigne’s test, the organic compound is fused with sodium metal to convert the covalent compound into the ionic compound. In ionic compounds, the elements can be detected easily.

Key points: Ionic compounds easily dissociate into ions and hence can give the ions test for element detection more easily than the covalent compounds. Organic compounds are covalent compounds and to convert  it into ionic compounds, they are fused with sodium metal.

Calculations of Empirical Formulas and Molecular Formulas

Calculating empirical and molecular formulas is a fundamental skill in chemistry and is particularly important for JEE Main students. Empirical formulas represent the simplest whole number ratio of atoms in a compound, while molecular formulas provide the actual number of each type of atom in a molecule. In this discussion, we will explore how to calculate empirical and molecular formulas and work through numerical problems in organic quantitative analysis.

Empirical Formula:

An empirical formula represents the simplest whole number ratio of atoms in a compound. To calculate the empirical formula, follow these steps:

Obtain Data: Begin with the given data, which typically includes the mass or percentage composition of each element in the compound.

Convert to Moles: Convert the mass or percentage composition of each element to moles. You can do this by dividing the mass by the molar mass of the element.

Determine the Ratio: Find the mole ratio by dividing the number of moles of each element by the smallest number of moles calculated. This ensures that you have the simplest whole number ratio.

Write the Empirical Formula: Use the mole ratios to write the empirical formula using subscripts.

Example: Calculating the Empirical Formula

Suppose a compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. To determine the empirical formula, follow these steps:

Convert the percentages to moles:

Moles of carbon (C) = 40.0 g / 12.01 g/mol = 3.33 moles

Moles of hydrogen (H) = 6.7 g / 1.01 g/mol = 6.64 moles

Moles of oxygen (O) = 53.3 g / 16.00 g/mol = 3.33 moles

Determine the mole ratio:

C:H:O = 3.33 moles : 6.64 moles : 3.33 moles

Simplify the ratio to 1:2:1

Write the empirical formula:

$CH_2O$

So, the empirical formula of the compound is $CH_2O$.

Molecular Formula:

The molecular formula provides the actual number of each type of atom in a molecule. To calculate the molecular formula, you need additional information—the molar mass of the compound. Follow these steps:

Calculate the molar mass of the empirical formula.

• Determine the empirical formula mass by adding the molar masses of the atoms in the empirical formula.

• Divide the given molar mass of the compound by the empirical formula mass to find the ratio.

• Multiply the subscripts in the empirical formula by the ratio to find the molecular formula.

Example: Calculating the Molecular Formula

Suppose you have a compound with an empirical formula of $CH_2O$ and a molar mass of 90 g/mol. To determine the molecular formula, follow these steps:

Calculate the empirical formula mass:

Empirical formula mass = (1 × 12.01 g/mol) + (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 30.03 g/mol

Determine the ratio:

90 g/mol (given molar mass) / 30.03 g/mol (empirical formula mass) ≈ 3

Multiply the subscripts in the empirical formula by the ratio:

$(C_1H_2O_1)_3 = C_3H_6O_3$

So, the molecular formula of the compound is $C_3H_6O_3$.

Numerical Problems in Organic Quantitative Analysis:

Organic quantitative analysis involves determining the composition, purity, and quantity of organic compounds. Here are some numerical problems commonly encountered by JEE Main students:

1. Determining Mass Percentage:

Given the mass of each element in a compound, calculate the mass percentage of each element and the compound's empirical formula.

2. Molecular Formula Calculations:

Given the empirical formula and the molar mass of a compound, calculate the molecular formula.

3. Limiting Reactants:

Calculate the limiting reactant and the amount of a product formed in a chemical reaction.

4. Stoichiometry:

Determine the number of moles of reactants and products in a chemical reaction.

5. Concentration Calculations:

Calculate the concentration of a solute in a solution in terms of molarity, molality, or mass percent.

Example Problem:

Suppose you have 2.0 g of acetic acid ($CH_3COOH$) and react it with excess ethanol ($C_2H_5OH$) to produce ethyl acetate ($CH_3COOC_2H_5$) in a chemical reaction. Calculate the limiting reactant, the maximum amount of ethyl acetate produced, and the mass of the excess reactant remaining.

By working through problems like this, JEE Main students can develop a strong understanding of stoichiometry, chemical reactions, and quantitative analysis in organic chemistry.

JEE Main Purification and Characterisation of Organic Compounds Solved Questions

Question 1: The formation of violet colour when sodium nitroprusside is added to sodium extract indicates the presence of 

1. Nitrogen 

2. Sulphur 

3. Oxygen 

4. Halogens

Solution: (b) Sodium fusion extract of sulphur containing organic compounds, on treatment with sodium nitroprusside (Na2[Fe(CN)5NO]), the  appearance of a violet colour indicates the presence of sulphur. 

S2-  +   [Fe(CN)5NO]2-   $\longrightarrow$    [Fe(CN)5NOS]4- (Violet complex)

Question 2: Which of the following elements in an organic compound cannot be detected by Lassaigne's test? 

1. N 

2. S

3. Cl

4. H. 

Solution: (d) H and C cannot be detected by Lassaigne's test. Carbon and hydrogen in an organic compound are detected by heating the compound with copper (II) oxide. 

Question 3: The principle of column chromatography is 

1. Gravitational force 

2. Capillary action

3. Differential adsorption of the substances in the solid phase.

4. Differential absorption of the substances in the solid phase. 

Solution: The principle of column chromatography is differential adsorption of the substance on a solid phase adsorbent. Depending upon the degree of adsorption of the component on the adsorbent, complete separation of components takes place.

Practice Questions

Question 1: Refining of petroleum involves the process of 

1. Simple distillation 

2. Fractional distillation 

3. Steam distillation 

4. Distillation under reduced pressure. 

Answer: (b) Fractional distillation 

Question 2: Separation of two substances by fractional crystallisation depends upon their difference in 

1. Densities

2. Solubilities 

3. Melting point 

4. Boiling points 

Answer: (b) Solubilities 

JEE Main Chemistry Purification and Characterisation of Organic Compounds Study Materials

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Conclusion

Organic compounds synthesised in laboratories or extracted need to be purified. In this article, we described the various methods of purification of organic compounds. The various methods of qualitative and quantitative analysis of the elements in the organic compounds are also discussed in this article.

This branch of organic chemistry is useful for the practical analysis of organic compounds and this chapter is one of the easiest chapters of organic chemistry for the JEE examination.