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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.

(a) Crystallisation: In this process the impure solid is dissolved in the minimum volume of a suitable solvent. The soluble impurities pass into the solution while the insoluble ones left behind. The hot solution is then filtered and allowed to cool undisturbed till crystallisation is complete. The crystals are then separated from the mother liquor by filtration and dried.
Example: crystallisation of sugar.

(b) Distillation: The operation of distillation is employed for the purification of liquids from non-volatile impurities. The impure liquid is boiled in a flask and the vapours so formed are collected and condensed to give back pure liquid in another vessel. Simple organic liquids such as benzene toluene, xylene etc. can be purified.

(c) Chromatography: Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid the basis is adsorption and when it is a liquid the basis is partition. Chromatography is generally used for the separation of coloured substances such as plant pigments or dyestuffs.

Inductive effect: The inductive effect refers to the polarity produced in a molecule as a result of higher electronegativity of one atom compared to another. Atoms or groups which lose electron towards a carbon atom are said to have +I Effect. Examples of +I effect are (Electron releasing)
(CH₃)₂C—, (CH₃)₂CH—, CH₃CH₂— CH₃— etc.
Those atoms or groups which draw electron away from a carbon atom are said to have -I Effect. Examples of -I effect are:
NO₂, F, Cl, Br, I, OH etc.

Electromeric effect: The electromeric effect refers to the polarity produced in a multiple bonded compound as it is approached by a reagent.

The atom A has lost its share of electron pair and B has gained this share.
As a result A acquires a positive charge and B a negative charge. It is a temporary effect and takes place only in the presence of a reagent.
(a) –I effect as shown below:
As the number of halogen atoms decreases, the overall -I effect decreases and the acid strength decreases accordingly.

(b) +I effect as shown below:
As the number of alkyl groups increases, the +I effect increases and the acid strength decreases accordingly.

11.18

The hydroboration-oxidation reaction is a two-step reaction that converts an alkene into a neutral alcohol by the net addition of water across the double bond. The hydrogen and hydroxyl group are added in a syn addition leading to the cis configuration. Hydroboration- oxidation is an anti-Markovnikov reaction, with the hydroxyl group attaching to the less substituted carbon. In first step Addition of Hydroborate group is done and in next step, it is oxidized by hydrogen peroxide.

For example: - When propene undergoes hydroboration-oxidation reaction, then it produces propan-1-ol as product. In this reaction diborane i.e., (BH₃)₂ reacts with propene, which in result generates trialkyl borane as an addition product. Then trialkyl borane is oxidised, by using hydrogen peroxide in the presence of aqueous sodium hydroxide to form alcohol, as final product.