Ncert Solutions Chemistry Class 12th

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Vishal Baghel

Contributor-Level 10

1- C6H5NH2 + CHCl3 + KOH

It is a carbylamine reaction in which a isocyanide compound is formed along with side products of potassium chloride.Basically the name of reaction is given is due to formation of a foul smelling compound called as isocyanide.

2- C6H5N2Cl + H3PO2 + H2O

Benzenediazonium chloride is a very reactive compound which oxidises hypophosphorous acid to hypophosphoric acid and the reactant is reduced to benzene.

3- C6H5NH2 + H2SO4 (conc.)

Aniline undergoes sulphonation to anillium hydrogensulphate.

4- C6H5N2Cl + C2H5OH

Aniline is very activating group which undergoes reaction to give ortho and para product. But in acidic medium

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alok kumar singh

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7.20

SO2 acts as a reducing agent when passed through an aqueous solution containing Fe (III) salt. It reduces Fe (III) to Fe (II) i.e., ferric ions to ferrous ions.

2Fe3+ + SO2 + 2H2O? 2Fe2+ + SO4 2- + 4H+

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Vishal Baghel

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It is given that compound 'C' having the molecular formula, C6H7N is formed by heating compound 'B' with Br2and KOH. This is a Hoffmann bromamide degradation reaction (in which isocyanide compound is formed). Therefore, compound 'B' is an amide and compound 'C' is an amine. The only amine having the molecular formula, C6H7N is aniline. Therefore, compound 'B' (from which 'C' is formed) must be benzamide.

Further, benzamide is formed by heating compound 'A' with aqueous ammonia. Therefore, compound 'A' must be benzoic acid. The given reactions can be explained with the help of the following equations:

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alok kumar singh

Contributor-Level 10

7.19

One of the quantitative method for estimating O3 gas is when ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This iodine can be titrated against a standard solution of sodium thiosulphate.

2I (aq) + H2O (l) + O3 (g)? 2OH (aq) + I2 (s) + O2 (g)

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Vishal Baghel

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(i) Ethyl iodide reacts with NaCN gives a substitution reactions to give propanitrile upon partial hydrolysis gives B, upon reaction with sodium hydroxide gives C.

 

(ii) Benzenediazoniumchloride gives nucleophilic substitution reactions gives A, upon hydrolysis the CN ion is replaced by OH ion which is less better leaving group gives B upon heating with ammonia gives C.

 

(iii) Ethylbromide gives nucleophilic substitution reactions gives B, upon reduction gives B followed by reacting with nitrous acid, i.e oxidation gives propanol.

 

(iv) Nitrobenzene upon reduction with iron/acid gives A, reacting with sodium nitrite giv

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alok kumar singh

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7.18

Since ozone is not very stable at room temperature; it decomposes to give oxygen molecule and nascent oxygen on heating. The nascent oxygen, being a free radical is very reactive. Hence, it acts as a powerful oxidizing agent. The equation for the above reaction is given by,

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Vishal Baghel

Contributor-Level 10

(i) Carbylamine reaction

Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

For example,

 

(ii) Diazotisation

Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diaz

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Vishal Baghel

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Primary amine: A primary (1°) amine is an amine that has the following general structural formula.

R= alkyl or aryl group

Secondary amine: A secondary (2°) amine is an amine that has the following general structural formula.

R1 and R2= alkyl or aryl group

Tertiary amine: A tertiary amine is an amine that has the following structure

R1, R2 and R3 are alkyl or aryl groups

Identification of Primary, Secondary and Tertiary amines

Primary, secondary and tertiary amines can be identified by the following test:

Hinsberg's test: This is an excellent test for the identification of primary, secondary and tertiary amines. In this test, the amine is shaken with benzenesulphonyl chloride ( Hinsberg's reagent) in the presence of an excess of aqueous KOH solution when

(i) A primary amine gives a clear solution which on acidification gives an N-alkylbenzene sulphonamide which is soluble in

Due to the presence of strong electron withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as a proton. So it is acidic and dissolves in alkali.

(ii) A secondary amine reacts with Hinsberg's reagent to give a sulphonamide which is soluble in There is no H-atom attached to the N-atom in the sulphonamide Therefore it is not acidic and soluble in alkali.

(iii) A Tertiary amine does not react with Hinsberg's reagent at all

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alok kumar singh

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7.17

(i) C2H4 + O2→ 2CO2 + 2H2O

Ethane in reaction with oxygen gives carbon dioxide and water.

(ii) 4Al + 3 O2→ 2Al2O3

Aluminium on reaction with oxygen gives alumina.

 

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