How is aniline prepared commercially? Describe carbylamines test for primary amines.


Aniline is prepared commercially by the reduction of nitrobenzene using iron and hydrochloric acid.



Because of the presence of hydrochloric acid in the reaction mixture actually aniline hydrochloride (C
6H5NH3+Cl-) is obtained. This on treatment with aqueous sodium carbonate gives aniline.



Carbylamine test for primary amines:
Primary amines, both aliphatic and aromatic, give carbylamines having a pungent smell, when heated with chloroform and alcoholic potash.


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Explain: In aqueous solution order is Et2NH,>Et3N > EtNH2.


In aqueous solution, the Kb order is Et2NH > Et3N > EtNH2
This may be explained by the following factors:
(a)    Steric factor: The size of the alkyl group is more than that of hydrogen and, therefore, it hinders the attack of acid on the amine and therefore, basic strength decreases. Now crossing of alkyl group increases from primary to tertiary amines. As a result, their basic strength decreases. This is called steric hindrance. According to this effect, alkyl amines should be least basic.

(b)    Solvation of ions: When amines are dissolved in water, they undergo hydration through hydrogen bonding. The protonated amines form hydrogen bonds with water molecules and release energy called hydration energy. Now, greater the extent of hydrogen bonding in protonated amine, more will be its stabilization and consequently, greater will be the tendency of amine to change into a cation and greater will be the strength of the corresponding amine. As can be seen below, the hydration of protonated amine due to hydrogen bonding decreases as:
Et NH3+ > Et2 NH2+ > Et3NH+

The tertiary ammonium ion is less hydrated than secondary ammonium ion, which is less hydrated than primary ammonium ion. Thus, tertiary amines have less tendency to form ammonium ion and consequently are less basic.
As a consequence of combining effects of inductive effect and solvation, the secondary amines are the strongest bases among amines and basic strength varies as:
Et2NH > Et3 N > Et NH2

 
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Identify the substances A and B in each of the following sequences of reactions:



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How can you separate a mixture of primary, secondary and tertiary amines? Write chemical reactions involved in the process.


Hinsberg test is employed to separate primary, secondary and tertiary amines from a mixture. In this test the mixture of amines is treated with benzene sulphonyl chloride C6H5SO2Cl (Hinsberg’s reagent) followed by treatment with aqueous KOH (5%) solution and then shaken with ether in a separatory funnel.
(a) The primary amine reacts forming mono alkyl sulphonamide which is soluble in alkali forming a potassium salt which forms the lower aqueous layer.



(b) The secondary amine reacts forming dialkylsulphonamide which is insoluble in alkali.



The tertiary amines does not react at all.
Both dialkyl sulphonamide and tertiary amine go into the upper ether layer. While the monoalkyl sulphonamide remains in the lower aqueous layer. The two layers are separated, the aqueous layer containing the potassium salt of mono-alkyl sulphonamide is hydrolysed with concentrated hydrochloric acid when primary amine is regenerated as its hydrochloride.

Primary amine is recovered from the hydrochloride by distillation with alkali.

RNH2.HCl+ KOH → RNH2 + KCl + H2O

The ether layer distilled when tertiary amine distills over. The residue of dialkyl sulphonamide is hydrolysed with concentrated hydrochloric acid when secondary amine is regenerated as its hydrochloride.

C6H5SO2NR2 +H2O + HCI → C6H5SO2OH + R2NH.HCl

The hydrochloride on distillation with alkali gives the free secondary amine
R2NH.HCl + KOH → R2NH + KCl + H2O

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Write the structure of the isomeric aromatic amines with formula C7H9N and give their IUPAC names.


Five Isomeric aromatic amines are possible. These are given below.


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