A compound Z with molecular formula $\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ reacts with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$ to give a solid, insoluble in alkali. Identify Z.
$\mathrm{Z}\left(\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}\right)$ is an aliphatic amine. On reaction with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$ (Hinsberg's reagent), it gives a product insoluble in alkali. It means that the product does not have a replaceable H -atom attached to the N - atom. So, compound Z is a secondary amine (ethyl methyl amine).
A primary amine, $\mathrm{RNH}_2$ can be reacted with $\mathrm{CH}_3-\mathrm{X}$ to get secondary amine, $\mathrm{R}-\mathrm{NHCH}_3$ but the only disadvantage is that $3^{\circ}$ amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where $\mathrm{RNH}_2$ forms only $2^{\circ}$ amine?
$$\mathrm{\mathop {RN{H_2}}\limits_{\Pr imary\,a\min e} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Carbyla\min e\,reaction}^{KOH/CHC{l_3}}} \mathop {RNC}\limits_{Alkyl\,Iso\,cyanide} \buildrel {{H_2}/Pd} \over \longrightarrow \mathop {RNHC{H_3}}\limits_{Secondary\,a\min e}}$$
Primary amines show carbylamine reaction in which two H -atoms attached to N -atoms of $\mathrm{NH}_2$ are replaced by one C-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.
Complete the following reaction.
The reaction exhibits azo-coupling reaction of phenols. Benzene diazonium chloride reacts with phenol in such a manner that the para position of phenol is coupled with diazonium salt to form p-hydroxy azobenzene.
Why is aniline soluble in aqueous HCl ?
Aniline is soluble in aqueous HCl due to formation of ionic anilinium chloride.
Suggest a route by which the following conversion can be accomplished.
Complete conversion can be performed as
