JEE Main Chemistry Important Questions Purification and Characterisation Of Organic Compounds

JEE Chemistry Purification And Characterisation Of Organic Compounds Questions cover techniques and methods used to isolate, refine, and analyse chemical substances to understand their properties and composition. It is one of the important topics for JEE Main exams. Check out the table below for more information.

To reach your goals, prepare for your JEE Main 2025 exams using Vedantu’s Purification and Characterisation Important Questions PDF and the JEE Main Chemistry Syllabus 2025.

Focus Areas from Purification and Characterisation for JEE Main 2025

The Purification and Characterisation in JEE Main, certain topics can boost your JEE Main 2025 scores. Knowing these topics will make your preparation and focus for the exam easier. You can find the list of significant subjects below. But before you get the Purification and Characterisation Important Questions for JEE Main, you can review this list of significant topics.

Crystallisation

This process is based on the differing solubilities of the compound and impurities in an appropriate solvent. At low temperatures, impurities are only slightly soluble or insoluble, while at high temperatures, both the compound and impurities dissolve. When the solution is then cooled, the pure compound crystallises, leaving the impurities dissolved in the solution, which is removed. Repeating this process several times achieves a high level of purification.

Distillation

This technique helps separate:

• Volatile liquids from non-volatile impurities.

• Liquids with significantly different boiling points.

Liquids with different boiling points evaporate at different temperatures. For example, chloroform (boiling point 334 K) and aniline (boiling point 457 K) can be separated by distillation. The liquid with the lower boiling point boils first, is collected, and condensed into a pure form. Then, the liquid with a higher boiling point is collected by further heating.

If the boiling points of two liquids are close, simple distillation doesn’t work well.

There are two main types of distillation:

(i) Fractional Distillation

When the boiling points of liquids in a mixture are similar, simple distillation can't effectively separate them, as both liquids form vapours in the same temperature range. In fractional distillation, vapours pass through a fractionating column before condensing, allowing the liquid with the higher boiling point to condense first. This method is widely used in the petroleum industry to separate crude oil into different components.

(ii) Steam Distillation

This technique separates compounds that are volatile in steam and don’t mix with water. Here, the combined vapour pressures of the organic liquid and water equal atmospheric pressure, allowing the organic liquid to evaporate at a temperature lower than its boiling point. This technique is effective for mixtures like aniline and water, allowing separation at a temperature close to or below 373 K.

Chromatography

This technique is used to separate mixtures into their components, purify substances, and test their purity.

In this process, the mixture is applied to a stationary phase (solid or liquid), and a pure solvent, solvent mixture, or gas moves over it, acting as the mobile phase.

Chromatography is divided into different types based on its principles:

Adsorption Chromatography

Adsorption chromatography separates components based on how much they stick to an adsorbent. Common adsorbents include silica gel and alumina. As the mobile phase moves over the stationary phase, the mixture’s components move by different distances. Two main types are:

• Column Chromatography: Here, a glass column is packed with adsorbent, and the mixture is placed on top. A liquid or liquid mixture flows down the column, separating components based on their adsorption levels. Strongly adsorbed components stay near the top.

• Thin Layer Chromatography (TLC): A thin layer of adsorbent (silica gel or alumina) is spread on a glass plate, dried, and used to separate mixtures. The mixture is applied as a small spot near the base, and when the plate is placed in a closed jar with eluant, the components move up with the eluant based on their adsorption levels. Separation is measured by the retardation factor (Rf):

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

Partition Chromatography

Partition chromatography separates components based on how they divide between two liquid phases.

• Paper Chromatography: This is a partition method where special chromatography paper holds water as the stationary phase. The mixture is spotted at the bottom, and a solvent (mobile phase) rises, carrying components to different heights based on partitioning. The resulting paper, called a chromatogram, shows separated components as spots at various distances.

Sublimation

This process involves heating a solid so that it changes directly into vapour without becoming a liquid first.

It’s used to purify compounds that can turn into vapour easily, separating them from impurities that don’t sublimate.

Differential Extraction

This technique is used to purify an organic compound that dissolves better in an organic solvent than in water.

In this method, the organic and aqueous solvents do not mix with each other. It’s simpler to isolate the compound from the organic solvent through distillation or evaporation than from the water.

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 the 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 ferric ferrocyanide) which is Prussian blue. 

(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 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)

(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 the 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 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.

Let the mass of the organic compound be M g, and the mass of water and carbon dioxide produced by 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 the 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

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

Let the mass of organic compound = M g 

The 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 does not apply 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 a 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 a known amount is heated with fuming nitric acid, and 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}  $ %

Learn with Purification and Characterisation Important Questions and Score High in JEE Main 2025

Purification and Characterisation in JEE Main is the field of isolating and refining substances while analysing their properties to gain insights into their composition and behaviour. The weightage of Purification and Characterisation in JEE Main is around 2-3%. This means that there are typically 1-2 questions asked from this chapter. Students can also expect direct questions from this section, which makes it one of the scoring topics in Purification and Characterisation Important Questions for JEE Main. Here are the few points that make the chapter high weightage.

• Foundation for Advanced Concepts: Understanding purification and characterisation techniques is fundamental for comprehending advanced topics in chemistry and other scientific disciplines.

• Application in Real World: The chapter's topics, such as chromatography and spectroscopy, have widespread applications in research, industry, and environmental analysis, making the content relevant and practical.

• Analytical Skills: Mastering techniques like qualitative and quantitative analysis hone students' analytical skills, which are valuable in problem-solving across various domains.

• Basic Laboratory Skills: The chapter covers basic laboratory techniques that are not only important for exams but also lay the groundwork for practical work in higher education and research.

• Weightage in Competitive Exams: Due to its application-oriented nature, questions related to purification and characterisation often appear in competitive exams like JEE Main, making it a high-yield topic.

• Conceptual Understanding: Students who grasp the underlying principles of purification and characterisation can apply their knowledge to solve a wide range of problems in chemistry.

• Interdisciplinary Relevance: The chapter's concepts extend to biochemistry, materials science, and pharmaceuticals, demonstrating its interdisciplinary significance.

• Focus on Purity: With an increasing emphasis on chemical purity in industries and research, the ability to effectively purify and characterise substances is of paramount importance.

• Critical Analysis: Techniques like spectroscopy require critical interpretation of data, enhancing students' ability to draw meaningful conclusions from experimental results.

• Strong Conceptual Base: Students who excel in this chapter often possess a strong conceptual base, making it easier for them to tackle other intricate topics in chemistry.

• Concepts Over Memorisation: This chapter requires an understanding of concepts rather than memorisation, aligning with the shift towards conceptual learning in modern education.

Purification and Characterisation Formulas to Learn with Important Questions

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