Compounds Containing Nitrogen

Then P, Q and R will be

Given

Now from S to major products its given.

The basicity of an amine depends on the ability of its nitrogen atom to donate lone pair of electrons in an acid-base reaction. Resonance, inductive effect and steric hindrance etc. plays a very important role in deciding the basicity of compound.

Among the given compounds, (III) is most basic due to the presence of electron donating methyl group which increases the electron density on nitrogen. Compound (IV) is less basic than (III). Hybridisation of nitrogen atom in aniline (I) and pyrrole (II) is sp³ and sp² , respectively, therefore, (I) is more basic than (II). Thus, the correct order of basicity is

(d) : Greater the electron density on nitrogen, more basic is the compound. Thus, order of basicity is :

IV > I > II > III

The conjugate acid of IV is stabilized by resonance with lone pairs on both $\ddot N{H_2}$ groups.

The conjugate acid of I is stabilized by resonance with lone pair on $-$$\ddot N{H_2}$ group and by hyperconjugation of $-$CH₃ group.

The conjugate acid of II is stabilized by resonance with lone pair on

group.

In compound III, the lone pair of nitrogen is involved in aromaticity. So, it is least basic.

The reaction is

The reactions involved are as follows:

The reactant, phthalimide, undergoes an acid-base reaction with KOH. In this reaction, the proton from the nitrogen reacts with the OH^- from the base to form water.

The resulting salt then undergoes a substitution reaction with p-bromobenzyl chloride to form the major product. This reaction proceeds via an S_N1 mechanism due to the stability of the p-bromobenzyl cation.

Azo dye test is given by aromatic primary amines.

The reaction taking place is Hoffmann bromamide degradation followed by benzoylation.

(i) It is an example of electrophilic substitution reaction which results in the formation of a coupled product. The nucleophilic nitrogen attacks at electron rich carbon of phenol at para position as follows:

$ A \rightarrow R \text { and } S $

(ii) The reaction represents Pinacole-pinacolone rearrangement. In this reaction, the intermediate is carbocation.

The reaction is represented as follows:

$ B \rightarrow T $

(iii) It is an example of addition reaction to carbonyl group, where lithium aluminium hydride $\left(\mathrm{LiAlH}_4\right)$ adds hydrogen across carbon and oxygen of carbonyl group thus reducing it to an alcohol. Both R and S enantiomers will be formed. Hence, racemic mixture will be obtained as carbonyl carbon becomes chiral.

(C) $ \to $ (P, Q)

(iv) It is an example of nucleophilic substitution, where electron rich sulphur (due to lone pairs) displaces chlorine on the ring forming a bicyclic compound.

$ D \rightarrow R $

(A)

(B) Reaction with DIBAL-H and alkaline hydrolysis:

(C) Reduction with Pd-C/H$_2$:

$C{H_3} - CH = CH - C{H_2}OH\buildrel {Pd - C/C{H_2}} \over \longrightarrow C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$

(D) Foul smell on treatment with $CHC{l_3} + KOH$:

$RN{H_2}\buildrel {CHC{l_3} + KOH} \over \longrightarrow \mathop {RNC}\limits_{Foul\,smell} + KCl + {H_2}O$

Let us discuss the given statements one by one.

Statement 1 : Aniline reacts with NaNO$_2$/HCl at 273 K to form colourless diazonium salt.

Benzene diazonium chloride is then coupled with $\beta$-naphthol to yield a dark red or orange coloured dye as shown in the reaction below:

Thus, the statement 1 is false as the colour of the precipitate formed is orange red not blue. As we have already seen in the above mentioned reactions that the colour compound (dye) formed in the reaction of aniline with NaNO$_2$/HCl at 0$^\circ$C was due to the successive coupling of diazonium salt with $\beta$-naphthol. Hence, statement 2 is true.

As a result, Statement 1 is False and Statement 2 is true.

A, B, C, D will give NaCN in their sodium fusion extract. Adding FeSO$_4$ will give Fe$_4$[Fe(CN)$_6$]$_3$ which is Prussian blue in colour.

B and C have Phenol group which gave positive result to FeCl$_3$ test.

A, B, C, D both give precipitate with AgNO$_3$ but only A and C will give pure white precipitate.

A and D can only give hydrazone derivative in reaction with aldehyde.

When a primary amine is heated with chloroform and KOH , a nitrogen containing compound carbylamine or isocyanide is formed.

The complete chemical reaction is shown below:

The products obtained during reaction are methyl isocyanide, potassium chloride and water.

When compound (i), 4-chlorobenzamide is treated with KOH and $\mathrm{Br}_2$, one hydrogen atom from $-\mathrm{NH}_2$ group is replaced by bromine atom to produce compound (ii), which is N-bromo derivative.