Compounds Containing Nitrogen

$ \text { The reaction involved is } $

Conversion of benzene diazonium chloride to chlorobenzene is called as Sandmeyer reaction.

Image The correct sequence of reaction involved is the given coversion is

So, the processes involved in given sequence are bromination, reduction and carbylamine reaction.

Initially lone pair electron of –NH₂ attack on electrophilic carbon, after then lone pair electron of oxygen attacks leading to formation of cyclic compound.

The +M effect of NH₂ is stabilizing the carbocation.

Propanamide $\mathrm{\buildrel {B{r_2}/KOH} \over \longrightarrow}$ Ethylamine

Aromatic amines are less basic than aliphatic amines. Among given aliphatic amines, $2^{\circ}$ amine is most basic, followed by $3^{\circ}$ amine and $1^{\circ}$ amine. Therefore the correct basic strength $\left(\mathrm{K}_{\mathrm{b}}\right)$ order of the given amines is

$ \underset{(C)}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCH}_2 \mathrm{CH}_3}>\underset{(D)}{\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}}>\underset{(B)}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2}>\underset{(A)}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2} $

The $\mathrm{pK}_{\mathrm{b}}$ order of the given amines will be just the opposite of their basic strength order. The correct matching is

$A-I I I, B-I V, C-I I, D-I$

A. Propanamine      N–Methylethanamine

B. Hexan–2–one      Hexan–3–one

C. Ethanamide     Hydroxyethanimine

D. o–Nitrophenol    p–nitrophenol

$\therefore$ Statement $\mathrm{I}$ is true but statement $\mathrm{II}$ is false.

Firstly, benzoic acid reacts with the mixture of conc. $\mathrm{H}_2 \mathrm{SO}_4$ and conc. $\mathrm{HNO}_3$ to give $m$-nitrobenzoic acid $(X)$.

In second step, $\mathrm{NO}_2$ group is reduced to $\mathrm{NH}_2$ group in presence of $\mathrm{Sn} / \mathrm{HCl}$.

In last step, $Y$ is first converted to diazonium salt in presence of $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ which is followed by reaction with $\mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{HCl}$ to give $m$-chlorobenzoic acid $(Z)$.

The $s p$-hybrid N -atom is most electron withdrawing due to maximum $s$-character, more electronegative element has less tendency to donate electrons and

therefore is least basic. Therefore, $R \mathrm{C} \equiv \mathrm{N}$ is least basic.

Out of $R-\mathrm{NH}_2$ and $\mathrm{RN}=\mathrm{CH} R, R-\mathrm{NH}_2$ has $s p^{3-}$-hybrid N -atom and less electronegative in nature and therefore is more basic.

Therefore, correct order of decreasing basicity.

$ R-\mathrm{NH}_2>R-\mathrm{N}=\mathrm{CH} R>R-\mathrm{C} \equiv \mathrm{~N} $

i.e. $\mathrm{III}>\mathrm{I}>\mathrm{II}$

Benzylamine reacts with chloroform and alcoholic KOH to give an foul-smelling substance i.e. benzyl isocyanide as product.

$X$ is a secondary amine, on reaction with $p$-toluene sulphonyl chloride gives sulphonamide which is insoluble in alkali.

Alkyl nitriles on acidic hydrolysis gives alkyl carboxylic acid.

Alkyl nitriles can be converted to $1^{\circ}$ amine by the reaction of $\mathrm{LiAlH}_4$ followed by hydrolysis.

$ \text { From Eq. (i) and Eq. (ii), } X=\text { Nitrile. } $

Gabriel's phthalimide reaction and Hoffmann-bromamide reaction is used in the preparation of aliphatic primary amines.

(i) Gabriel phthalimide reaction It is a chemical process that converts primary alkyl halides into primary amine in presence of strong base.

(ii) Hoffmann-bromamide reaction It is an organic reaction used to convert a primary amide to primary amine using a halogen, base, water and heat.

Firstly, aniline reacts with acetic anhydride to give acetanilide $(A)$. Then, acetanilide is treated with bromine in acetic acid to give $p$-bromoacetanilide $(B)$ as major product.

Acetic acid is heated with $\mathrm{P}_2 \mathrm{O}_5$ to give ethanoic anhydride $(M) . M$ reacts with aniline to give acetanilide $N$. At last, $(N)$ undergoes nitration to give $p$-nitroacetanilide which on acidic hydrolysis give $p$-nitroaniline $(R)$ as major product.

Acetamide reacts with NaOH and $\mathrm{Br}_2$ to give methyl amine $(X)$. In next step, methyl amine undergoes diazotisation reaction in presence of $\mathrm{NaNO}_2$ and HCl , which on further hydrolysis give methanol $(Y)$ as major product.