Chemistry

11.22

There are many ways to this conversion. Two of them are given below:-

(a) In the above conversion, the chlorobenzene is treated with a base such as NaOH, KOH etc. (strong base). The base abstracts the hydrogen from the C-2 position (it can also abstract the hydrogen from the C-6 position, as both are equally acidic) leaving the negative charge at that position.

In the next step Cl^- leaves, leaving behind the positive charge at that carbon. Both the negative charge and positive charge forms a bond resulting Benzyne as the intermediate.

After the formation of the Benzyne intermediate OH^- of the base attacks at the C-1 position and further, the H⁺ attacks to stabilize the negative charge thus resulting in the phenol.

The reaction can start from benzene also. Chlorination of benzene gives Chlorobenzene. When it is further treated with NaOH and H₂O at 350°C it results in the formation of sodium phenoxide which on further treatment H⁺ gives Phenol as the final product.

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:
Weight of organic compound = W gm
weight of silver halide = x g.

% of halogen =

Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaSO₄ so obtained the percentage of sulphur can be calculated.

% of sulphur =  x

Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phosphoric acid is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄.12MoO₃, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.

% of phosphorus =,

Where, molar mass of (NH₄)₃PO₄.12MoO₃ = 1877 g.

If phosphorus is estimated as Mg₂P₂O₇

% of P =   

Lassaigne's test: In organic compounds, nitrogen, sulphur and halogens are covalently bonded. Their detection in 'Lassaigne's test' is possible if they are in the ionic form. This can be achieved by fusing the organic compound with sodium metal.

Chemistry for test for nitrogen: 

Sodium fusion extract is boiled with ferrous sulphate and acidified with sulphuric acid. Sodium cyanide reacts with ferrous sulphate and forms sodium hexacyanoferrate (II).  On heating with sulphuric acid, some ferrous is oxidized to ferric hexacyanoferate (II) Fe₄ [Fe (CN)₆]₃ which is prussian blue in colour.

Chemistry of the test for sulphur:

Acetic acid is added to sodium fusion extract. Complete precipitation of sulphur in the form of lead sulphate occurs which is black in colour. Also, the sodium fusion extract is treated with sodium nitroprusside to obtain violet color. However if N and S are present, then instead of NaCN, NaSCN is obtained. This gives blood red colour on reaction with Fe³⁺ ions.

Chemistry of test for halogens:

Sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. However, if nitrogen and sulphur both are present, then Lassaigne's extract is boiled to expel nitrogen and sulphur which would otherwise interfere in the test for halogens.