DIY History | Transcribe | Scholarship at Iowa | Phenylbromethylbenzenesulfonamid and phenylbromethylamin by Carl Leopold von Ende, 1893 | Phenylbromethylbenzenesulfonamide and Phenylbromethylamin by Carl Leopold von Ende, 1893, Page 4

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Phenylbromethylbenzenesulfonamid and phenylbromethylamin by Carl Leopold von Ende, 1893

Phenylbromethylbenzenesulfonamide and Phenylbromethylamin by Carl Leopold von Ende, 1893, Page 4

[page]3.[/page]
their sulphates with aluminium sulphate to form alums.
     For distinguishing the primary, secondary and tertiary amins, nitrous acid is used, it being applied indirectly. The salt of the amin is treated with KNO[subscript]2[/subscript]. In the case of a primary amin, as CH[subscript]3[/subscript] * NH[subscript]2[/subscript] , this reaction takes place:

CH[subscript]3[/subscript] * NH[subscript]2[/subscript] + HNO[subscript]2[/subscript] = CH[subscript]3[/subscript]OH + N[subscript]2[/subscript] + H[subscript]2[/subscript]O

giving the alcohol, free nitrogen and water.
     In the case of secondary amins

(CH[subscript]3[/subscript])[subscript]2[/subscript]NH + HNO[subscript]2[/subscript] = (CH[subscript]3[/subscript])[subscript]2[/subscript]N[branch1]N[/branch1] = O + H[subscript]2[/subscript]O

giving us a nitrosamin, which is an oily body insoluble in water.
     The tertiary amins give no reaction whatever with HNO[subscript]2[/subscript]. They readily combine with alkyl iodids, giving an iodid of a compound ammonium, as 

N(CH[subscript]3[/subscript])[subscript]4[/subscript]I

Tetramethylammoniumiodid.

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[page]3.[/page] their sulphates with aluminium sulphate to form alums. For distinguishing the primary, secondary and tertiary amins, nitrous acid is used, it being applied indirectly. The salt of the amin is treated with KNO[subscript]2[/subscript]. In the case of a primary amin, as CH[subscript]3[/subscript] * NH[subscript]2[/subscript] , this reaction takes place: CH[subscript]3[/subscript] * NH[subscript]2[/subscript] + HNO[subscript]2[/subscript] = CH[subscript]3[/subscript]OH + N[subscript]2[/subscript] + H[subscript]2[/subscript]O giving the alcohol, free nitrogen and water. In the case of secondary amins (CH[subscript]3[/subscript])[subscript]2[/subscript]NH + HNO[subscript]2[/subscript] = (CH[subscript]3[/subscript])[subscript]2[/subscript]N[branch1]N[/branch1] = O + H[subscript]2[/subscript]O giving us a nitrosamin, which is an oily body insoluble in water. The tertiary amins give no reaction whatever with HNO[subscript]2[/subscript]. They readily combine with alkyl iodids, giving an iodid of a compound ammonium, as N(CH[subscript]3[/subscript])[subscript]4[/subscript]I Tetramethylammoniumiodid.