Organic Compounds Containing Nitrogen
The correct order of reactivity towards the electrophilic substitution of the compounds aniline $$(I)$$, benzene $$(II)$$ and nitrobenzene $$(III)$$ is:
Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline.
(i) Methylamine and dimethylamine :
Carbylamine test.
Methyl amine, on heating with alc. KOH solution and chloroform forms foul smelling methyl isocyanide. This test is not given by dimethylamine.
(ii) Secondary and tertiary amines :
Secondaty amines give libermann nitrosoamine test.
On heating with nitrous acid (prepared in situ), they give yellow coloured oily N-nitrosoamine.
Tertiary amines do not give such test.
(iii) Ethylamine and aniline :
Azo dye test.
Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphtol (in alkaline solution) forms brilliant orange or red dye. Ethylamine will not form dye. It will give brisk effervescence (due to liberation of nitrogen gas) but solution remains clear.
(iv) Aniline and benzylamine :
Azo dye test.
Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphtol (in alkaline solution) forms brilliant orange or red dye. Benzylamine will not give such test.
(v) Aniline and N-methylaniline :
Carbylamine test.
Aniline, on heating with alc. KOH solution and chloroform forms foul smelling methyl isocyanide. This test is not given by N-methylaniline.
The correct order of reactivity towards the electrophilic substitution of the compounds aniline $$(I)$$, benzene $$(II)$$ and nitrobenzene $$(III)$$ is:
When semicarbazide reacts with a ketone(or aldehyde) to form semicarbazone. Only one nitrogen atom of semicarbazide acts as a nucleophile and attack the carbonyl carbon of the ketone. The product of the reaction consequently is $${ R }_{ 2 }C=N-NH-CO{ NH }_{ 2\\ }$$ rather than $${ R }_{ 2 }C=NCONH-{ NH }_{ 2 }$$. What factor account for the fact that two nitrogen atoms of semicarbazide are relatively non nucleophilic?
The correct order of reactivity towards the electrophilic substitution of the compounds aniline (I) benzene (II) and nitrobenzene (III) is
The carbylamine reaction is:
Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Primary and secondary amines react with acid chloride or acid anhydride to form:
Following reaction gives two products. $$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ NH }_{ 2 }\xrightarrow { ({ CH }_{ 3 }CO{ ) }_{ 2 }O,\ heat }A+B$$ Write the structures of the products formed.
(A) is
Give the structures of $$A,B$$ and $$C$$ in the following reaction: $${ CH }_{ 3 }COOH\xrightarrow [ \Delta ]{ { NH }_{ 3 } } A\xrightarrow [ ]{ { Br }_{ 2 }+KOH } B\xrightarrow [ ]{ CH{ Cl }_{ 3 }+NaOH } C$$
Ethylamine reacts with acetyl chloride to form: