Tests for Nitrogen Containing Compounds Questions in English

(D) In the dye test,the phenolic $-OH$ group is converted to the phenoxide ion $-O^{\ominus}$ in the presence of a base.
This phenoxide ion is a strong activating group that increases the electron density on the aromatic ring,thereby facilitating Electrophilic Aromatic Substitution $(EAS)$ with the diazonium salt.
This activation is only possible in an alkaline solution.
Therefore,the correct option is $(D)$.

(C) Hinsberg's reagent is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.
It reacts with $1^{\circ}$ and $2^{\circ}$ amines because they possess at least one hydrogen atom attached to the nitrogen atom.
$(A)$ $C_6H_5NH_2$ is a $1^{\circ}$ amine (reacts).
$(2)$ $C_6H_5N(CH_3)_2$ is a $3^{\circ}$ amine (does not react).
$(C)$ $CH_3NH_2$ is a $1^{\circ}$ amine (reacts).
$(4)$ $N(CH_3)_3$ is a $3^{\circ}$ amine (does not react).
$(E)$ $(C_6H_5)_2NH$ is a $2^{\circ}$ amine (reacts).
Therefore,$A, C$ and $E$ react with Hinsberg's reagent.

(C) The carbylamine test is a characteristic reaction for primary $(1^{\circ})$ amines.
Aliphatic or aromatic primary amines react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines),which have a foul,pungent smell.
Secondary $(2^{\circ})$ and tertiary $(3^{\circ})$ amines do not give this test.
In the given options:
$A. CH_3CH_2NH_2$ is a primary amine.
$B. (CH_3)_2NH$ is a secondary amine.
$C. CH_3NH_2$ is a primary amine.
$D. (CH_3)_3N$ is a tertiary amine.
$E. C_6H_5NH_2$ is a primary amine (aniline).
Therefore,$A, C,$ and $E$ are primary amines and will give a positive carbylamine test.

(D) The isocyanide test (also known as the carbylamine reaction) is a characteristic test for primary $(1^{\circ})$ amines.
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and an alcoholic base $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
Secondary $(2^{\circ})$ and tertiary $(3^{\circ})$ amines do not undergo this reaction because they lack the necessary hydrogen atoms on the nitrogen to form the isocyanide structure.
Among the given options:
$1$. Cyclohexylamine $(C_6H_{11}NH_2)$ is a primary amine.
$2$. Amino benzene (Aniline,$C_6H_5NH_2$) is a primary amine.
$3$. Ethyl amine $(C_2H_5NH_2)$ is a primary amine.
$4$. $N,N$-Dimethyl aniline $(C_6H_5N(CH_3)_2)$ is a tertiary amine.
Therefore,$N,N$-Dimethyl aniline does not give the isocyanide test.

(D) The correct matches are as follows:
$1$. $A.$ Carbylamine test is used to detect primary amines like $IV.$ Aniline.
$2$. $B.$ Bayer's test (alkaline $KMnO_4$) is used to detect unsaturation,such as in $III.$ Ethylene.
$3$. $C.$ Iodoform test is given by compounds containing $CH_3CO-$ group,such as $II.$ Acetone.
$4$. $D.$ Phthalein dye test is used to detect $I.$ Phenol.
Therefore,the correct sequence is $A-IV, B-III, C-II, D-I$.

(D) The carbylamine test is a characteristic reaction given only by primary $(1^{\circ})$ amines,whether aliphatic or aromatic.
In this reaction,the primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.
Among the given options:
$A$. Ethylamine $(CH_3CH_2NH_2)$ is a primary amine.
$B$. Sec. butylamine $(CH_3CH_2CH(NH_2)CH_3)$ is a primary amine.
$C$. Isopropylamine $((CH_3)_2CHNH_2)$ is a primary amine.
$D$. Dimethylamine $((CH_3)_2NH)$ is a secondary $(2^{\circ})$ amine.
Therefore,Dimethylamine does not give the carbylamine test.

(D) The reaction with benzene sulphonyl chloride (Hinsberg's reagent) is used to distinguish between primary,secondary,and tertiary amines.
Primary amines $(R-NH_{2})$ react with benzene sulphonyl chloride to form $N$-alkylbenzene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen.
This acidic hydrogen makes the product soluble in excess of potassium hydroxide $(KOH)$,resulting in a clear solution.
Secondary amines $(R_{2}NH)$ form $N,N$-dialkylbenzene sulphonamides,which lack an acidic hydrogen and are insoluble in $KOH$.
Tertiary amines $(R_{3}N)$ do not react with benzene sulphonyl chloride.
Among the given options,$CH_{3}NH_{2}$ is a primary amine,so it forms a clear solution.

(A) Hinsberg's reagent is benzene sulphonyl chloride.
Its molecular formula is $C_6H_5SO_2Cl$.
This reagent is used to distinguish primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines.

(A) The reaction with $p$-toluene sulphonyl chloride (Hinsberg reagent) is used to distinguish between $1^\circ$,$2^\circ$,and $3^\circ$ amines.
$1^\circ$ amines $(R-NH_2)$ react to form an $N$-alkyl $p$-toluene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen. This makes the product soluble in alkali (forming a clear solution).
Upon acidification of this clear solution,the $N$-alkyl $p$-toluene sulphonamide precipitates out as an insoluble compound.
$2^\circ$ amines form a product that is insoluble in alkali,and $3^\circ$ amines do not react with the reagent.
Therefore,the primary amine $C_2H_5NH_2$ is the correct answer.

(C) Hinsberg's reagent is benzenesulfonyl chloride.
Its chemical formula is $C_6H_5SO_2Cl$.

(A) The reagent used to distinguish between primary,secondary,and tertiary amines is Hinsberg reagent,which is Benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.
$1$. Primary $(1^{\circ})$ amines react with Benzenesulfonyl chloride to form $N$-alkylbenzenesulfonamide,which is soluble in alkali due to the presence of an acidic hydrogen atom attached to the nitrogen.
$2$. Secondary $(2^{\circ})$ amines react to form $N$,$N$-dialkylbenzenesulfonamide,which is insoluble in alkali because it lacks an acidic hydrogen atom.
$3$. Tertiary $(3^{\circ})$ amines do not react with Benzenesulfonyl chloride.

(A) The Hinsberg reagent is $C_6H_5SO_2Cl$,which is known as benzene sulphonyl chloride.
It is used to distinguish between primary,secondary,and tertiary amines.

(C) In the first step,acetaldehyde $(CH_3-CHO)$ reacts with chlorine $(Cl_2)$ and calcium hydroxide $(Ca(OH)_2)$ to undergo the haloform reaction,producing chloroform $(CHCl_3)$.
$2CH_3-CHO + 3Cl_2 + Ca(OH)_2 \rightarrow CHCl_3 + (HCOO)_2Ca + 2H_2O$ (Note: The reaction produces $CHCl_3$ as the haloform).
In the second step,chloroform $(CHCl_3)$ reacts with aniline $(C_6H_5-NH_2)$ and alcoholic $KOH$ (Carbylamine reaction) to form phenyl isocyanide $(C_6H_5-NC)$.
$C_6H_5-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5-NC + 3KCl + 3H_2O$

(A) Hinsberg's reagent is $C_6H_5SO_2Cl$,which is known as benzenesulphonyl chloride.
It is used to distinguish between primary,secondary,and tertiary amines.

(D) $Aniline$ is a primary amine $(C_6H_5NH_2)$,while $N$-methylaniline is a secondary amine $(C_6H_5NHCH_3)$.
Primary amines react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines),which have a foul smell. This is known as the Carbylamine test.
Secondary amines do not give this test.
Therefore,chloroform and alcoholic $KOH$ are suitable reagents to differentiate between them.

(B) The carbylamine test is a chemical test for the detection of primary amines. In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and an alcoholic solution of potassium hydroxide $(KOH)$. The reaction produces an isocyanide (also known as a carbylamine),which has a characteristic foul smell. The general reaction is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$. Among the given options,the general product is an isocyanide $(R-NC)$.

(B) Hinsberg reagent is $C_{6}H_{5}SO_{2}Cl$ (benzene sulphonyl chloride).
It is used for the detection and differentiation of primary,secondary,and tertiary amines.

(B) The reaction of an aromatic nitro compound $(Ar-NO_2)$ with $Zn / NH_4Cl$ is a selective reduction that produces an aromatic hydroxylamine $(Ar-NHOH)$.
$Ar-NO_2 + Zn / NH_4Cl \rightarrow Ar-NHOH + ZnO$
Aromatic hydroxylamines are strong reducing agents and can reduce Tollen's reagent (ammoniacal silver nitrate) to metallic silver $(Ag)$,which appears as a black precipitate.
$Ar-NHOH + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow Ar-NO + 2Ag(s) + 4NH_3 + 2H_2O$
Therefore,the functional group present in compound "$A$" is the nitro group $(-NO_2)$.

(C) The reagent used to distinguish between primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$,commonly known as the Hinsberg reagent.
$1$. Primary $(1^{\circ})$ amines react with benzenesulfonyl chloride to form $N$-alkylbenzenesulfonamide,which is soluble in alkali due to the presence of an acidic hydrogen atom on the nitrogen.
$2$. Secondary $(2^{\circ})$ amines react to form $N$,$N$-dialkylbenzenesulfonamide,which is insoluble in alkali because it lacks an acidic hydrogen atom on the nitrogen.
$3$. Tertiary $(3^{\circ})$ amines do not react with benzenesulfonyl chloride in the absence of an acidic hydrogen atom on the nitrogen,and thus remain insoluble in the reaction mixture.

(D) The test described is the carbylamine test,which is a characteristic reaction of primary amines ($R-NH_2$ or $Ar-NH_2$).
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has an extremely unpleasant odor.
Option $A$ is a secondary amine $(C_6H_5NHCH_3)$.
Option $B$ is a tertiary amine $(C_6H_5N(CH_3)_2)$.
Option $C$ is a quaternary ammonium salt $([C_6H_5N(CH_3)_3]^+X^-)$.
Option $D$ is a primary amine $(C_6H_5CH_2NH_2)$,which will react with $CHCl_3$ and $KOH$ to give a positive carbylamine test.

(D) $Hinsberg's$ reagent,which is $benzenesulfonyl$ $chloride$ $(C_6H_5SO_2Cl)$,is used to distinguish between primary,secondary,and tertiary amines.
$1$. Primary amines react with $Hinsberg's$ reagent to form $N-alkylbenzenesulfonamide$,which is soluble in alkali.
$2$. Secondary amines react to form $N,N-dialkylbenzenesulfonamide$,which is insoluble in alkali.
$3$. Tertiary amines do not react with $Hinsberg's$ reagent.

(D) Aniline is an aromatic primary amine,while cyclohexylamine is an aliphatic primary amine.
The azo dye test is specific to aromatic primary amines.
When aniline reacts with nitrous acid $(HNO_2)$ at $0-5 \ ^{\circ}C$ followed by coupling with $\beta$-naphthol,it forms a bright orange or red azo dye.
Cyclohexylamine,being aliphatic,does not form a stable diazonium salt at low temperatures and consequently does not give the azo dye test.
Therefore,the azo dye test is used to distinguish between them.

(D) The carbylamine reaction is also known as the $Hoffman$ isocyanide synthesis. It is the reaction of a primary amine,chloroform,and a base to synthesize isocyanides or carbylamines.
The dichlorocarbene intermediate is very important for this conversion.
The carbylamine reaction is not given by secondary and tertiary amines,so primary amines can be distinguished from secondary and tertiary amines.
The chemical equation is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$

(C) The reaction described is the $Carbylamine$ reaction,which is a characteristic test for primary amines ($R-NH_2$ or $Ar-NH_2$).
When a primary amine is heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which is characterized by a highly offensive or nauseating smell.
Among the given options,$PhNH_2$ (aniline) is a primary amine.
The reaction is: $PhNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} PhNC + 3KCl + 3H_2O$.
Thus,$PhNH_2$ produces the nauseating smell.

(A) The reduction of nitrate ions $(NO_3^-)$ to ammonia $(NH_3)$ by zinc $(Zn)$ dust in an alkaline medium is a standard laboratory test for nitrates.
The chemical reaction is: $NaNO_3 + 4 Zn + 7 NaOH \rightarrow 4 Na_2ZnO_2 + NH_3 + 2 H_2O$.
Here,'$A$' represents the alkaline medium,which is $NaOH$ (caustic soda).

(A) Hinsberg's reagent is benzenesulphonyl chloride $(C_6H_5SO_2Cl)$.
$1^{\circ}$ amines react to form $N$-alkylbenzenesulphonamides,which are soluble in alkali due to the acidic hydrogen on the nitrogen atom.
$2^{\circ}$ amines react to form $N,N$-dialkylbenzenesulphonamides,which lack acidic hydrogen and are therefore insoluble in alkali.
$3^{\circ}$ amines do not react with Hinsberg's reagent.
Let us analyze the given amines:
$1$. Aniline $(1^{\circ})$: Soluble in alkali.
$2$. $N$-Methylaniline $(2^{\circ})$: Insoluble in alkali.
$3$. Methanamine $(1^{\circ})$: Soluble in alkali.
$4$. $N,N$-Dimethylmethanamine $(3^{\circ})$: No reaction.
$5$. $N$-Methylmethanamine $(2^{\circ})$: Insoluble in alkali.
$6$. Phenylmethanamine $(1^{\circ})$: Soluble in alkali.
$7$. $N$-Propylaniline $(2^{\circ})$: Insoluble in alkali.
$8$. $N$-Phenylaniline $(2^{\circ})$: Insoluble in alkali.
$9$. $N,N$-Dimethylaniline $(3^{\circ})$: No reaction.
$10$. Allylamine $(1^{\circ})$: Soluble in alkali.
$11$. Isopropylamine $(1^{\circ})$: Soluble in alkali.
The $2^{\circ}$ amines are: $N$-Methylaniline,$N$-Methylmethanamine,$N$-Propylaniline,and $N$-Phenylaniline.
Total number of alkali-insoluble sulphonamides = $4$.

(A) . Cyclohexanol: It is a secondary alcohol,which is tested by the Lucas test $(ZnCl_2 + conc. HCl)$. Thus,$A-III$.
$B$. Cyclohexylamine: It is a primary amine,which is tested by Hinsberg's reagent $(Benzenesulfonyl chloride)$. Thus,$B-I$.
$C$. Cyclohexanecarbaldehyde: It is an aldehyde,which is tested by Tollen's test $(Ammoniacal silver nitrate)$. Thus,$C-IV$.
$D$. Phenol: It is tested by the Phthalein dye test (reaction with phthalic anhydride in the presence of conc. $H_2SO_4$). Thus,$D-II$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.