Qualitative & Quantitative Analysis of Organic Compounds | Chemistry for JEE Main & Advanced PDF Download

What is Qualitative Analysis?

Quantitative analysis is an analysis method used to determine the number of elements or molecules produced during a chemical reaction. Organic compounds are comprised of carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur and halogens. The various methods used for the measurement of percentage composition of elements in an organic compound are explained here.
Qualitative analysis is the analysis of the species present in a given compound. For example, if a compound is taken, the qualitative analysis would be more focused on finding the elements and the ions present in the compound rather than study as to how much they are present.

Detection of C and H

C and H are detected by heating the compound with CuO in a dry test tube. They are oxidised to CO₂ and H₂O respectively. If the CO₂ turns lime water milky, and H₂O turns anhydrous CuSO₄ blue, then the presence of C and H is confirmed.
Detection of Carbon and Hydrogen

Test for Phosphorous

The organic compound is heated with an oxidising agent to oxidise phosphorous to phosphate. The solution is then boiled with concentrated HNO₃ and treated with ammonium molybdate. Yellow precipitate confirms the presence of phosphorous.
The reaction is given below,
Na₃PO₄ + 3HNO₃ → H₃PO₄ + 3NaNO₃
H₃PO₄ + 12(NH₄)₂MoO₄ + 21HNO₃ → (NH₄)₃PO₄.12MoO₃ + 21NH₄NO₃ + 12H₂O
Quantitative analysis is more towards finding out how much of the elements are present, that is, their amounts.

Estimation of C and H

Liebig’s Combustion Method

A known mass of the compound is heated with CuO. The carbon present is oxidised to CO₂ and hydrogen to H₂O. The CO₂ is absorbed in KOH solution, while H₂O is absorbed by anhydrous CaCl₂ and they are weighed.
Percentage of C = 12/44 x ((Mass of CO₂)/(Mass of compound)) x 100
Percentage of H = 2/18 x ((Mass of H₂O)/(Mass of compound)) x 100
Liebig’s combustion method

Estimation of Halogens

Carius Method

A known mass of the compound is heated with Conc. HNO₃ in the presence of AgNO₃ in a hard glass tube called Carius tube. C and H are oxidised to CO₂ and H₂O. The halogen forms the corresponding AgX. It is filtered, dried and weighed.
Estimation of Halogens by Carius method

Percentage of X = ((Atomic mass of X)/(Molecular mass of AgX))x((Mass of AgX)/(Mass of the compound))x100

Calculations:

Let the mass of the given organic compound be m g.
Suppose the mass of AgX formed = m₁ g.
We know that 1 mol of AgX consists of 1 mol of X.
So, in m₁ g of AgX , mass of halogen =

Percentage of halogen =

Estimation of Sulphur

A known mass of the compound is heated with conc. HNO₃ in the presence of BaCl₂ solution in Carius tube. Sulphur is oxidised to H₂SO₄ and precipitated as BaSO₄. It is then dried and weighed.
Percentage of S= ((Atomic mass of S)/(Molecular mass of BaSO₄)) x (( Mass of BaSO₄)/(Mass of the compound)) x 100

Calculations:

Suppose the mass of organic compound = mg
Let the mass of barium sulphate formed = m₁ g
We know that 32 g sulphur is present in 1 mol of BaSO₄
Therefore, 233 g BaSO₄ contains 32 g sulphur:
⇒ M₁ g of BaSO₄ contains  of sulphur
Percentage of sulphur =

Estimation of Phosphorus

A known mass of the compound is heated with HNO₃ in a Carius tube, which oxidises phosphorous to phosphoric acid. It is then precipitated as ammonium phosphomolybdate ((NH₄)3PO₄.12MoO₃) by adding NH₃ and ammonium molybdate ((NH₄)₂MoO₄). It is filtered, dried and weighed.
Percentage of P =
((Atomic mass of P)/(Molecular mass of (NH₄)3PO₄.12MoO₃)) x ((Mass of (NH₄)3PO₄.12MoO₃)/(Mass of compound)) x 100

Estimation of Nitrogen

Estimation of Nitrogen by Dumas Method

A known mass of the compound is heated with CuO in an atmosphere of CO₂, which yields free nitrogen along with CO₂ and H₂O.
C_xH_yN_z + (2x+ 0.5y) CuO → xCO₂ + 0.5y H₂O + 0.5z (N₂) + (2x+ 0.5y)Cu
The gases are passed over a hot copper gauze to convert trace amounts of nitrogen oxides to N₂. The gaseous mixture is collected over a solution of KOH which absorbs CO₂, and nitrogen is collected in the upper part of the graduated tube.
Estimation of Nitrogen by Dumas Method

Let the volume of N₂ collected be V₁ mL.
Then, volume of N₂ at STP = (P₁V₁ x 273)/ (760 x T₁) = V mL
Where P₁ and V₁ are the pressure and volume of N₂.
P₁ = Atmospheric pressure – aqueous tension
22.4 L of N₂ weighs 28 g,
Therefore, V ml of N₂ weighs
= (28 x V)/22400 grams
Percentage of N would be,
= (28/22400) x (V/ Mass of compound) x 100

Estimation of Nitrogen by Kjeldahl Method

A known mass of an organic compound is (0.5 g) is mixed with K₂SO₄ (10 g) and CuSO₄ (1.0 g) and conc.H₂SO₄ (25 mL), and heated in a Kjeldahl’s flask.
CuSO₄ acts as a catalyst, while K₂SO₄ raises the boiling point of sulphuric acid. The nitrogen in the compound is quantitatively converted to (NH₄)₂SO₄. The resulting mixture is reacted with an excess of NaOH solution and the NH₃ so evolved, is passed into a known but the excess volume of standard acid.
The acid left unreacted is estimated by titration with some standard alkali. Thus the percentage of nitrogen can be calculated.
Estimation of Nitrogen by Kjeldahl Method

The reactions are given below:

• C + H + S → CO₂ + H₂O + SO₂
• N → (NH₄)₂SO₄
• (NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2NH₃ + 2H₂O
• 2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Calculation of the percentage of N
Let the mass of the organic compound be mg.
Volume of H₂SO₄ (Molarity M) = V mL
The volume of NaOH of molarity M used for titration excess of H₂SO₄= V₁ mL
mEq of excess H₂SO₄ = mEq of NaOH
= MV₁ mEq
Total mEq of H₂SO₄ taken = 2MV
mEq of H₂SO₄ used for neutralisation of NH₃ = (2MV – MV₁)
Therefore,
mEq of NH₃ = (2MV-MV₁)
1000 mEq or 1000 mL of NH₃ solution contains = 17 g of NH₃ (or) 14 g of N
Therefore,
(2MV-MV₁) mEq of NH₃ contains
=(14 x (2MV-MV₁)) / 1000 g of N
Percentage of N = ((14x(2MV - MV₁)) x (100/(1000 x m)))

Estimation of Oxygen – Aluise’s method

A known mass of the compound is decomposed by heating it in the presence of N₂ gas. The mixture of gases so produced is passed over red hot coke. This is done so that all the O₂ is converted to CO. This mixture is heated with I₂O₅ in which CO is oxidised to CO₂ liberating I₂.
The reactions are given below,
Organic compound → Other gaseous products + O₂
2C + O₂ → 2CO
I₂O₅ + 5CO → 5CO₂ + I₂
Percentage of O = ((Molecular mass of O₂/Molecular mass of CO₂) x (Mass of CO₂/Mass of the compound) x 100

Types of Organic Reactions

Organic reactions are the chemical reactions that are undergone by organic compounds (the chemical compounds containing carbon). A few important types of organic reactions are illustrated below.

There are mainly five types of organic reactions:

• Substitution reaction
• Elimination reaction
• Addition reaction
• Radical reactions
• Oxidation-Reduction Reactions.

Substitution Reactions

In a substitution reaction, one atom or a group of atoms is substituted by another atom or a group of atoms to form a new substance. Let us take an example of a C-Cl bond, in which the carbon atom has a partial positive charge due to the presence of highly electronegative chlorine atom. In a nucleophilic substitution reaction, the nucleophile must have a pair of electrons and also should have a high affinity for the electropositive species as compared to the substituent which was originally present.

Elimination reaction

There are some reactions which involve the elimination or removal of the adjacent atoms. After these multiple bonds are formed and there is a release of small molecules as products. One of the examples of elimination reaction is the conversion of ethyl chloride to ethylene.
CH₃CH₂Cl →   CH₂=CH₂ + HCl
In the above reaction, the eliminated molecule is HCl, which is formed by the combination of  H⁺ from the carbon atom which is on the left side and Cl^– from the carbon atom which is on the right side.

Addition reactions

Addition reaction is nothing but just the opposite of elimination reaction. In an addition reaction, the components A and B are added to the carbon-carbon multiple bonds and this is called addition reaction. In the reaction given below when HCl is added to ethylene, it will give us ethylene chloride.
HCl + CH₂=CH₂ → CH₃CH2Cl

Radical reactions

Many of the organic reactions involve radicals. Addition of a halogen to a   saturated hydrocarbon involves free radical mechanism. There are three stages involved in a radical reaction i.e. initiation, propagation, and termination. Initially when the weak bond is broken initiation of the reaction takes place with the formation of free radicals. After that when the halogen is added to the hydrocarbon a radical is produced and finally, it gives alkyl halide.