Compounds Containing Nitrogen

The reaction sequence involved is as follows.

$ \text { The reaction involved is as follows } $

$ \text { The major product } Z \text { is benzene. } $

Hinsberg's reagent is used to distinguish between primary, secondary and tertiary amines from a mixture.

$ \text { The complete reaction sequence involved is as follows } $

Aliphatic and aromatic primary amines give positive carbylamine reaction. Thus, among the given options,

will give positive test with a mixture of chloroform and alc. KOH solution.

The complete reaction is as follows.

Thus, $ X $ and $ Y $ are $ \left(\mathrm{NaOH}+\mathrm{Br}_{2}\right) $ and [(i) $ \mathrm{LiAlH}_{4} $

(ii) $ \mathrm{H}_{2} \mathrm{O} $ ]

The Wurtz-Fittig reaction is a chemical process that produces substituted aromatic compound from aryl halides, alkyl halide in presence of dry ether and sodium metal. The complete reaction is as follows.

The complete reaction is as follows.

Product $(Z)$ is an aldehyde.

Ethyl aldehyde is only aldehyde which gives iodoform test.

Hence, statements I and III are correct.

The lesser the $\mathrm{p} K_b$ value, more stronger will be the base.

Hence, the correct match is A-II, B-III, C-IV, D-I.

The reaction involved is as follows,

Thus, reagent/chemical $X$ and $Y$ are

DIBAL - $\mathrm{H}, \mathrm{H}_2 \mathrm{O}$ and

Statement given in B is correct which $A$ and $C$ are incorrect. The correct form of A and C are:

(A) Aniline can form stable diazonium salt at 273-283 K.

(C) Formula of Hinsberg reagent is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$.

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.

(A) When an amide is treated with bromine in aq. solution of NaOH , degradation of amide takes place leading to formation of primary amine, i.e. Hoffmann's bromamide reaction.

(B) Nitriles can be converted to $1^{\circ}$ amines by reaction with $\mathrm{LiAlH}_4$.

(C) Hinsberg reagent is an alternative name for benzene sulphonyl chloride.

Used for distinction of $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ amines.

(D) Carbylamine reaction is effective only for primary amines.

Benzene sulphonyl chloride react with dimethyl amine to give $\mathrm{N}, \mathrm{N}$-dimethyl benzene sulphonyl amide.

(A) Hofmann bromamide reaction Primary amines are formed.

Nitrile formed on reduction with $\mathrm{H}_2 / \mathrm{Pt}$ gives $1^{\circ}$-amine.

(C) $1^{\circ}$ amines reacts with carbonyl compounds to form enamine which on reduction with $\mathrm{H}_2$ / Pt gives secondary amines.

(D) Gabriel phthalamide reaction It gives only primary amines.

Cyanides on reaction with $\mathrm{Na}-\mathrm{Hg}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$, undergo reduction reaction to produce primary amine.

The Stephen's reaction is used to prepare aldehydes from nitriles or cyanides. The reduction of nitriles with stannous chloride and hydrogen chloride in ethyl acetate solvent produces imine intermediate. The hydrolysis of this intermediate with water gives the corresponding aldehyde.

$\mathrm{HI} /$ Red P is a strong reducing agent. It reduces alcohols, aldehydes, ketones to alkanes.

$ \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HI}+\text { Red } \mathrm{P} \longrightarrow\mathrm{CH}_3-\mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}+2 \mathrm{I}_2 $

So, finally the CN gets converted to methyl ( $-\mathrm{CH}_3$ ) group. CN contains triple bond between carbon and nitrogen so it is $s p$-hybridised, whereas ' C ' in $-\mathrm{CH}_3$ is $s p^3$ hybridised.

In Hofmann bromamide degradation reaction, an amide reacts with bromine and an aqueous solution of sodium hydroxide and result in the production of primary amine. The number of ' $C$ ' in product (amine) is always one less than present in parent amide (reactant).

$ \underset{\text { (General amide) }}{R-\mathrm{CO}-\mathrm{NH}_2}+\underset{\text { (Bromine) }}{\mathrm{Br}_2}+\underset{\text { (Sodium hydroxide) }}{4 \mathrm{NaOH}} \longrightarrow R-\mathrm{NH}_2+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NaBr}+2 \mathrm{H}_2 \mathrm{O}$

$ \underset{\text { Hexanamide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CONH}_2}+\mathrm{Br}_2+\mathrm{NaOH} \longrightarrow \underset{\text { Pentanamine }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2}+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NaBr}+2 \mathrm{H}_2 \mathrm{O}$

$ \text { In the given reaction, the final product is as follows : } $

In step (i), we get diazonium product, which go for coupling reaction to give the final product as shown in the reaction. Hence, option (d) is the correct answer.

$ \text { The major product in the following reaction is } $

(i) $\mathrm{Sn} / \mathrm{HCl}$ show reduction.

(ii) $\mathrm{Br}_2$ show addition

(iii) Show formation of diazonium salt.

(iv) Show, the final product.

Hence, option (c) is the correct answer.

Step I Amide react with bromine and aqueous KOH to form amine.

Step II Amine react with $\mathrm{NaNO}_2, \mathrm{HCl}$ followed by hydrolysis form alcohol.