Amines

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Vishal Baghel

Contributor-Level 10

Pkb value is the negative logarithm of the basicity constant (Kb) .i.e., pKb = -logKb

Evidently, smaller the value of pKb , stronger is the base (strong tendency to donate electrons). Aliphatic amines(R-NH2) are more basic(tendency to donate electrons) than aromatic amines(C6H5NH2) because of the following reasons:

=>In aliphatic amines, alkyl groups are present. Alkyl groups are electron releasing groups, hence they increase the elecron density of N-atom and thus is easily available to donate electrons. This poperty makes aliphatic amines more basic.

As a result of resonance, the lone pair of electrons on the nitrogen atom gets delocaliz

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Vishal Baghel

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The best test for distinguishing methyl amine and dimethylamine is the Carbylamines test.

Carbylamine Test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines.

In this case, Methylamine (which is an aliphatic primary amine) gives a positive carbylamine test while dimethylamine wont.

Hinsberg's reagent (benzenesulphonyl chloride, C6H5SO2Cl). can be used to distinguish secondary and tertiary amines.

Hinsberg Test: Secondary amines react with Hinsberg's reagent to form a product that is insoluble in an alkali. For example, N, N? diethylamine reacts

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Vishal Baghel

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1 - Methylethanamine

The root name is based on the longest chain with the -NH2 attached. The chain is numbered so as to give the amine unit the lowest possible number. The longest chain is ethane chain which is further suffixed with 'amine'.

2 - Propan-1-amine

The longest chain here is propane. The naming is such that amine unit should get a the lowest possible number. Propane-1-amine can also be written as 1-propylamine.

3 - N−Methyl-2-methylethanamine

The chain is numbered so as to give the amine unit the lowest possible number. The other alkyl group is treated as a substituent, with N as the locant. The N locant is listed before numeri

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Vishal Baghel

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1- When 3-methylaniline treated with NaNO2 + HCl it gets converted into chlorine complex.

When that complex reacted with HBF4 It gets converted into Barium Fluoride complex. This complex reacts with NaNO2 in presence of copper to give 3-Nitrotoluene.

When aniline reacts with Br2 water it gets converted into 2,4,6 tribromobenzamine. When this further reacted with NaNO2/HCl it forms Chloride complex. This complex forms 1,3,5 tribromobenzene after treating with H3PO2 in presence of water.

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Vishal Baghel

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The different isomers of the molecular formula: C3H9N are given in the table. However only 1° amines will liberate nitrogen gas on the treatment with h=the nitrous acid are given as follows:

Note: Only primary amines liberate nitrogen gas on treatment with nitrous acid.

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Vishal Baghel

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When aniline is treated with benzoyl chloride in the presence of base it gets converted into N-Phenylbenzamide.

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Vishal Baghel

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On excessive alkylation with methyl iodide aniline gets converted into N, N-Trimethylanilinium iodide. After reacting it with sodium carbonate it get converted into N, N-Trimethylanilinium carbonate.

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