Diazonium salt containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilisation between the benzene nucleus and $$N$$ - atom. Diazonium ion act as a electrophile.
42.
Aniline when diazotised in cold and then treated with $$N, N$$ -dimethylaniline gives a coloured product. The structure of this product is
TIPS/Formulae : In the formation of cyanohydrin number of carbon atoms in parent chain increases by one.
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl+alc.\,KCN\xrightarrow{{{100}^{\circ }}C}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CN+KCl\]
44.
Which of the following is more basic than aniline?
Benzylamine, $${C_6}{H_5}C{H_2} - \ddot N{H_2}$$ is more basic than aniline because benzyl group $$\left( {{C_6}{H_5}C{H_2} - } \right)$$ is electron donating group due to $$ + I$$ - effect So, it is able to increase the electron density of $$N$$ of $$ - N{H_2}$$ group. Thus, due to higher electron density, rate of donation of free pair of electron is increased, i.e. basic character is higher. Phenyl and nitro group are electron withdrawing groups, so they decreases the electron density on $$N$$ of \[-\ddot{N}{{H}_{2}}\] group. Hence, they are less basic than aniline.
45.
Oxidation of aniline with manganese dioxide and sulphuric acid produces
Only pure aliphatic $${1^ \circ }$$ amines can be prepared by Gabriel synthesis. Aromatic $${1^ \circ }$$ amines such as aniline, toluidines, etc., cannot be prepared by this method because aryl halides $$\left( {{C_6}{H_5}Cl\,\,{\text{or}}\,\,{C_6}{H_5}Br} \right)$$ do not undergo nucleophilic substitution ( i.e., cleavage of $$C-X$$ bond in haloarene is quite difficult ) with potassium phthalimide under ordinary conditions to give $$N$$ - phenyl phthalimide.
49.
A compound $$(X)$$ with molecular formula $${C_3}{H_9}N$$ reacts with $${C_6}{H_5}S{O_2}Cl$$ to give a solid which is insoluble in alkali. $$(X)$$ is
A
$$C{H_3}C{H_2}C{H_2}N{H_2}$$
B
\[\underset{\begin{smallmatrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,| \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|
\end{smallmatrix}}{\mathop{C{{H}_{3}}-N}}}\,\]
C
$$C{H_3} - NH - C{H_2}C{H_3}$$
D
\[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
C{{H}_{3}}\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,N{{H}_{2}}\]
Since the compound reacts with benzenesulphonyl chloride to give a product which is insoluble in alkali, it shows there is no $$H$$ attached to $$N$$ in the product. Hence, the compound $$X$$ is a secondary amine. \[C{{H}_{3}}-NH-{{C}_{2}}{{H}_{5}}+{{C}_{6}}{{H}_{5}}S{{O}_{2}}Cl\] \[\to \underset{N\text{-Ethyl-}N\text{-methylbenzene sulphonamide}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix}
| \\
\,\,\,\,{{C}_{2}}{{H}_{5}}
\end{smallmatrix}}{\mathop{-N-}}\,S{{O}_{2}}{{C}_{6}}{{H}_{5}}}}\,\]
50.
Choose the correct statement from the following.