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

$$p$$ - Chloroaniline and anilinium hydrogen chloride can be distinguished by \[AgN{{O}_{3}}.\]   Anilinium hydrogen chloride will give white $$ppt$$  of $$AgCl$$  on reaction with \[AgN{{O}_{3}},\]  but $$p$$ - chloronoaniline will not react with it because $$Cl$$  is directly attached to benzene nucleus.

$${{\text{3}}^ \circ }$$ amines do not react with Hinsberg's reagent $$\left( {{C_6}{H_5}S{O_2}Cl} \right)$$    as there are no freely available hydrogens on the amine group which can donate electrons to the sulphur of the Hinsberg reagent.

II is not an acceptable canonical structure because nitrogen has 10 valence electrons in the structure. Anilinium ion exists in two canonical structures only, which are I and III.

Nitrobenzene and aniline hydrochloride have electron- withdrawing ( \[-N{{O}_{2}}\]  and \[-\overset{+}{\mathop{N}}\,{{H}_{3}}\]   ) groups, hence these will undergo electrophilic substitution with difficulty. Aniline and \[N-\]  acetylaniline (acetanilide) have electron – releasing groups, however \[-NHCOC{{H}_{3}}\]    is less electron- releasing than \[-N{{H}_{2}}\]  due to delocalisation of lone pair of electron on $$N$$  toward carbonyl group. Hence aniline ( having \[-N{{H}_{2}}\]  ) will undergo electrophilic substitution most easily.

In amines, $$N$$  is \[s{{p}^{3}}\]  hybridised and thus has pyramidal shape. In the given structure, since the three alkyl groups are different, and the fourth corner of the pyramid is occupied by lone pair of electrons, the molecule is chiral. However, the two enantiomers of the amine are not resolvable because of their rapid interconversion through a transition state having planar structure ( \[s{{p}^{2}}\]  hybridised nitrogen )

As $$(A)$$  gives alcohol on treatment with nitrous acid thus, it should be primary amine. $${C_3}{H_9}N$$  has two possible structures with $$ - N{H_2}$$  group.
$$C{H_3} - C{H_2} - C{H_2} - N{H_2}$$       \[\text{or}\,\,\,C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\, \\ C{{H}_{3}}\,\, \end{smallmatrix}}{\mathop{-CH-}}\,N{{H}_{2}}\]
As it gives isopropylmethyl amine thus it should be isopropyl amine not $$n$$ - propyl amine.

(a) When chlorination is done earlier than nitration, chlorobenzene formed at first step would introduce \[-N{{O}_{2}}\]  group in ortho-position, not in $$m$$ - position.
(b) Again if \[-N{{O}_{2}}\]  group is reduced earlier than the chlorination step, \[-N{{H}_{2}}\]  group formed on reduction will again introduce \[-Cl\]  in $$o$$ - position.

Higher the stability of the conjugate acid, more is the basic character of the parent amine.

The conjugate acid is stabilized by resonance with two different $$ - N{H_2}$$  groups.

The conjugate acid is stabilized by resonance with one $$ - N{H_2}$$  group. Hence as compared to $$IV$$  it is less basic.

Lone pair is not involvd in aromaticity. Hence more available

Lone pair is involved in aromatic sextet. Hence not available.
Hence the correct order of basic strength is IV > I > II > III