Organic Compounds Containing Nitrogen MCQ Questions & Answers in Organic Chemistry | Chemistry

Electrophilic substitution takes place at electron rich positions. The ring attached with $$- NH$$  will have rich electron density due to resonance. As $$ortho$$  - position is blocked, the electrophile attacks the $$para$$  - position.

\[\underset{\begin{smallmatrix} \text{Benzene} \\ \text{diazoniumchloride} \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}{{N}_{2}}Cl}}\,\xrightarrow[\text{or}\,\,Cu+HCl]{CuCl/HCl}\underset{\text{Chloro benzene}}{\mathop{{{C}_{6}}{{H}_{5}}Cl}}\,+{{N}_{2}}\]
The above reaction is known as Sandmayer’s reaction.

Given reactions indicate that $$B$$  has \[{{1}^{\circ }}N{{H}_{2}}\]   group, and thus \[A,{{C}_{2}}{{H}_{3}}N,\]   should be \[C{{H}_{3}}C\equiv N.\]   Hence $$C$$  should be \[C{{H}_{3}}C{{H}_{2}}NC\]
\[\underset{A}{\mathop{C{{H}_{3}}C\equiv N}}\,\xrightarrow{\text{reduction}}\underset{B}{\mathop{C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}}}\,\xrightarrow[KOH]{CHC{{l}_{3}}}\underset{C}{\mathop{C{{H}_{3}}C{{H}_{2}}N\overset{\to }{\mathop{=}}\,C}}\,\]

There is no free hydrogen in tertiary amines hence they do not form salts and are not soluble in acids.

Benzylamine is most basic. In others the basic character is suppressed due to Resonance ( see applications of resonance ).

\[C{{H}_{3}}-N\equiv C+4\left[ H \right]\xrightarrow{LiAl{{H}_{4}}}\underset{\text{Dimethylamine}}{\mathop{C{{H}_{3}}NHC{{H}_{3}}}}\,\]
On catalytic reduction or with lithium aluminium hydride $$\left( {LiAl{H_4}} \right)$$   or with nascent hydrogen, alkyl isocyanide yield $${2^ \circ }$$ amine whereas cyanide gives $${1^ \circ }$$ amine on reduction.

Key Idea (i) Presence of electron withdrawing substituent decreases the basicity while the presence of electron releasing substituent like, $$ - C{H_3}, - {C_2}{H_5},$$   etc, increases the acidity.
(ii) $$HN{O_2}$$  converts $$ - N{H_2}$$  group of aliphatic amine into $$ - OH$$  while that of aromatic amines into $$ - N = NCl.$$
Since, phenyl group is a electron withdrawing group, it decreases the basicity. Alkyl group, on the other hand, being electron releasing, increases the basicity. Thus, alkyl amines are more basic as compared to aryl amines as well as ammonia.
\[R-\underset{\text{Alkyl amine}}{\mathop{N{{H}_{2}}}}\,\xrightarrow{HN{{O}_{2}}}R-OH\]
Thus, $$HN{O_2}$$  (nitrous acid) converts alkyl amines to alcohols.
But aryl amines react with nitrous acid to form diazonium salt.
\[\underset{\text{Aryl amine}}{\mathop{{{C}_{6}}{{H}_{5}}N{{H}_{2}}}}\,\underset{\left( 273-278\,K \right)}{\mathop{\xrightarrow[0-{{5}^{\circ }}C]{HN{{O}_{2}}}}}\,{{C}_{6}}{{H}_{5}}-N_{2}^{+}Cl\]
$${\text{at}}\,\,{0^ \circ }{\text{ - }}{5^ \circ }C\,\,{\text{temperature}}$$
$$NaN{O_2} + HCl \to HN{O_2} + NaCl$$
Thus, $$HN{O_2}$$  does not convert aryl amines into phenol.

TIPS/Formulae : $$- NH$$  is an activating group whereas as group is a deactivating group.
Hence electrophilic substitution will be governed by
the ring having group.