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Determination of Components of a Given Organic Compound

The first step in the investigation of an organic compound after it has been obtained in a pure state is to carry out its qualitative analysis, i.e., to detect the various elements present in it. The elements which commonly occur in organic compounds are carbon (present in all compounds), hydrogen (present in most of the compounds), oxygen, nitrogen, halogens, sulphur and in a few cases phosphorus and metals.

 

DETECTION OF CARBON AND HYDROGEN

Carbon must be present in a compound under examination if it is definitely known to be an organic compound (Fig. 38.1). Similarly, hydrogen too, is generally present. Both these elements can be tested together. A small quantity of the dry and powdered substance is mixed with freshly ignited cupric oxide. The mixture is heated in a well-dried hard glass test-tube provided with a delivery tube having a bulb at one end. The other) end of the tube is dipping under lime water taken in another test tube. Carbon present in the compounds is oxidised to carbon dioxide which turns lime water milky.

The hydrogen present in the compound is oxidised to water which condenses on the cooler parts of the test-tube or in the bulb of delivery tube. The drops turn white anhydrous copper sulphate placed in the bulb into blue.

If the compound to be analysed is a volatile liquid or a gas, then the vapour of the liquid or the gas itself is passed through heated copper oxide and the liberated gases are tested for carbon dioxide and water vapours.

 

DETECTION OF NITROGEN

Nitrogen in an organic compound is detected by the following tests:

(i) Sodalime Time

A pinch of an organic substance is heated strongly with sodalime (NaOH + CaO) in a test tube. The evolution of ammonia gives the indication of nitrogen.

limitation. A large number of organic compounds such a nitro and diazo compounds do not liberate ammonia under these conditions.

(ii) Lassaigne’s Test

This test is a confirmatory test for nitrogen in organic compounds. This test involves two steps:

(a) Preparation of Lassaigne’s extract. A small piece of a sodium metal is heated gently in a fusion tube till it melts to a shining globule. At this stage, a small amount of substance is added and the tube is heated strongly. The red hot tube is plunged into distilled water contained in a china dish. The contents are boiled for some time, cooled and then filtered. The filterate is known as sodium extract or Lassaigne ‘s extract.

(b) Test for nitrogen. The Lassaigne’s-extract is usually alkaline. If not, it may be made alkaline by the addition of a few drops of a dilute solution of sodium hydroxide. To a part of sodium extract a small amount of a freshly prepared ferrous sulphate solution is added and the contents are warmed. A few drops of ferric chloride solution are then added to the contents and the resulting solution is acidified with dilute hydrochloric acid. The appearance of a prussian blue colouration confirms the presence of nitrogen in the organic compound. The chemical reactions involved are as follows:

If the organic compound contains both nitrogen and sulphur, sodium sulphocyanide may be formed during fusion. This reacts with ferric ions to form ferric sulphocyanide which gives a blood red colouration.

DETECTION OF SULPHUR

The presence of sulphur in the organic compound can be detected by the following tests:

 

(i) Oxidation Test

The organic compound is fused with a mixture of potassium nitrate and sodium carbonate. Sulphur, if present, gets oxidised to sodium sulphate.

The fused mass is extracted with water, the contents are boiled and filtered. The filtrate contains sodium sulphate. The filtrate is acidified with dilute hydrochloric acid and a solution of barium chloride is added. Formation of a white precipitate indicates the presence of sulphur.

(ii) Lassaigne’s Test

Lassaigne’s extract is prepared as described in the case of nitrogen. The extract contains sodium sulphide formed by the reaction between sulphur (present in the compound) and sodium.

2Na + S à Na2S

The Lassaigne’s extract is divided into two parts and following tests are performed:

(a) Lead acetate test. One part of the extract is acidified with acetic acid and then lead acetate solution is added. Formation of a black precipitate confirms the presence of sulphur in the organic compound.

(b) Sodium nitroprusside test. A few drops of sodium nitroprusside solution are added to another part of the Lassaigne’s extract. The appearance of purple colouration confirms the presence of sulphur:

DETECTION OF HALOGENS

The presence of halogens in the organic compound is detected by the following tests:

(i) Beilstein’s Test

 A copper wire flattened at one end is heated in the oxidising flame of Bunsen burner. The heating is continued till it does not impart blue colour to flame. The hot end of copper wire is now touched with the organic substance and is once again kept in flame. The reappearance of green or blue colour indicates the presence of halogens in the organic compound.

Limitations.                (a) Substances such as urea, thiourea, etc., do not contain halogens, but give this test.

(b) It does not tell as to which halogen is present in the organic compound.

(ii) Lassaigne’s Test

A small amount of the compound is fused with metallic sodium as described earlier. The halogen present in the compound combines with sodium to form sodium halide which passes into the sodium extract.

A part of the Lassaigne’s extract is boiled with dilute nitric acid to expel all the gases if evolved. The silver nitrate solution is added to the resulting solution.

If any of the halogen is present, a precipitate is formed.

(a) A white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound.

(b) A dull yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.

(c) A bright yellow precipitate, completely insoluble in ammonium hydroxide solution indicates the presence of iodine in the organic compound.

Function of nitric acid. In case, nitrogen and sulphur are present along with the halogens in the organic compound, the Lassaigne’s extract contains sodium sulphide (Na2S) and sodium cyanide (NaCN) along with sodium halide. Nitric acid decomposes sodium cyanide and sodium sulphide which otherwise form precipitate with silver nitrate and thus, interfere with the test.

(iii) Special Test For Bromine and Iodine

Acidify the Lassaigne’s extract with nitric acid and add a few drops of freshly prepared chlorine water. Free Br2 and I2 will be evolved. Shake this solution with carbon disulphide or carbon tetrachloride. The liberated halogens (Br2 or I2) will dissolve in it giving as orange or brown colour for bromine and violet colour for iodine

DETECTION OF PHOSPHORUS

A very few organic compounds contain phosphorus. Its presence is detected by fusing the compound with an oxidizing mixture of sodium carbonate and potassium nitrate or with sodium peroxide alone when an alkali phosphate is formed. The fused mass is extracted with water and filtered. The filtrate containing sodium phosphate is heated with cone. HNO3 and an excess of ammonium molybdate solution is added. A yellow ppt. or colouration if formed if phosphorus is present. The yellow precipitate is ammonium phosphomolybdate, (NH4)3[PMo12O40] or (NH4) 3PO4.12 MoO3.
 

DETERMINATION OF EMPIRICAL AND MOLECULAR FORMULA ON THE BASIS OF GIVEN DATA

The empirical formula of a compound is the simplest formula which expresses the simple whole number ratio of the atoms of constituent elements present in the molecule. For example, CH2O is the empirical formula of acetic acid, it expresses that the simplest whole number ratio between carbon, hydrogen and oxygen atoms present in one molecule of acetic acid is 1 : 2 : 1. The empirical formula does not tell us the exact number of various atoms of different elements present in one molecule of the substance. However, the exact number of atoms of different elements present in one molecule is always a simple multiple of ratio of atoms in the empirical formula.

 

The formula which gives the actual number of atoms of various elements present in the molecule of the substance is termed the molecular formula.

= (CH3COOH)            or         (CH2O) x 2      [here n = 2]

The value of ‘n’ can be obtained by dividing the molecular mass by empirical formula mass.

n = Molecular mass /  Empirical formula mass

The percentage composition of the compound can be calculated from its empirical formula. For example,

Empirical formula of acetic acid= CH2O

Empirical formula mass               = (12 + 2 + 16)

= 30

Percentage of C = 12 / 30 x 100 = 40.00

Percentage of H = 2 / 30 x 100 = 6.66

Percentage of O = 16 / 30 x 100 = 53.34

Thus, the simplest formula of a substance capable of expressing its percentage composition can be called its empirical formula.

Calculation of empirical formula. Empirical formula is calculated from the percentage composition. The steps involved in the calculation are as follows:

Step 1. The percentage of each element is divided by its atomic mass. This gives the relative number of different atoms present in the molecule.

Step 2. The relative numbers of different atoms obtained in Step 1 are divided by the lowest one amongst them as to get simple ratio of atoms present in the molecule.

 Step 3. The values obtained in Step 2 may or may not be whole numbers. In case one or more values are fractional, these are multiplied by a suitable integer to get simplest ratio in whole numbers. Minor fractions are neglected.
 

Step 4. The symbols of each element present are written side by side in a line with the number of atoms as determined in Step 2 or Step 3 as subscripts to the lower comer of each.

This gives the empirical or simplest formula.

Knowing the empirical formula, the molecular formula can be ascertained if the molecular mass of the substance is known. It may be the same as the empirical formula of the substance or an exact multiple of it.

Molecular formula = n x (Empirical formula)

The value of ‘n’ can be determined if molecular mass of the substance is known.

n = Molecular mass / Empirical formula mass