Tests for Nitrogen Containing Compounds Questions in English

(A) The yellow stain on the skin caused by $HNO_3$ is due to the nitration of proteins present in the skin. This reaction produces a yellow-colored compound known as $xanthoprotein$. This is a characteristic test for proteins,often referred to as the $xanthoproteic$ test.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,$K_2[HgI_4]$.
In the presence of ammonia or ammonium salts,the active species $HgI_4^{2-}$ reacts to form a brown precipitate known as the iodide of Millon's base.
The chemical reaction is:
$2[HgI_4]^{2-} + NH_4^+ + 4OH^- \rightarrow NH_2Hg_2OI + 7I^- + 3H_2O$
Thus,the active species involved in the detection is $[HgI_4]^{2-}$.

(B) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$.
It is prepared by adding potassium iodide $(KI)$ to mercuric iodide $(HgI_2)$ to form potassium tetraiodomercurate$(II)$ $(K_2HgI_4)$,which is then made alkaline by adding potassium hydroxide $(KOH)$.
The chemical composition is $K_2HgI_4 + KOH$.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which is prepared by adding potassium iodide to a solution of mercury$(II)$ iodide.
The chemical reaction is: $HgI_2 + 2KI \to K_2[HgI_4] \rightleftharpoons 2K^+ + [HgI_4]^{2-}$.
Thus,the active species in Nessler's reagent is the tetraiodomercurate$(II)$ ion,$[HgI_4]^{2-}$.

(B) The reaction described is the carbylamine reaction,which is a characteristic test for primary amines $(R-NH_2)$.
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic $KOH$ or $NaOH$ to form an isocyanide (carbylamine),which has a foul,unpleasant odour.
$C_6H_5NH_2$ (Aniline) is a primary amine.
The reaction is: $C_6H_5NH_2 + CHCl_3 + 3NaOH \to C_6H_5NC + 3NaCl + 3H_2O$.
Therefore,$X$ is $C_6H_5NH_2$.

(B) The Carbylamine reaction is a diagnostic test for primary amines.
In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH(alc.) \to R-NC + 3KCl + 3H_2O$.
Since chloroform is a trihalogen methane,option $B$ is the correct description.

(D) Biuret is a compound formed by heating urea at $180^{\circ}C$,which results in the condensation of $2$ molecules of urea.
The peptide bonds in Biuret give a positive result for the test; hence,the reagent is named so.
It is considered a general test for compounds (proteins and peptides) having two or more peptide $(CO-NH)$ bonds.

(B) Benzene sulphonyl chloride $(C_6H_5SO_2Cl)$ is known as Hinsberg’s reagent.
It is used to distinguish between primary,secondary,and tertiary amines.

(D) The Hinsberg's reagent is $C_6H_5SO_2Cl$ (benzenesulfonyl chloride).
It reacts with primary,secondary,and tertiary amines to form different products:
$1$. Primary amines form $N$-alkylbenzenesulfonamide,which is soluble in alkali.
$2$. Secondary amines form $N,N$-dialkylbenzenesulfonamide,which is insoluble in alkali.
$3$. Tertiary amines do not react with the reagent.
Because these products have different solubilities in alkali,the Hinsberg's method is primarily used for the separation of a mixture of primary,secondary,and tertiary amines.

(A) The reaction described is the Carbylamine reaction,which is a characteristic test for primary amines ($1^{\circ}$ amines).
Primary amines react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides $(RNC)$,which have a foul smell.
The chemical equation is: $RNH_2 + CHCl_3 + 3KOH \to RNC + 3KCl + 3H_2O$.

(A) The carbylamine reaction is a characteristic test for primary $(1^o)$ amines.
In this reaction,an aliphatic or aromatic primary amine is heated with chloroform $(CHCl_3)$ and an alcoholic solution of potassium hydroxide $(KOH)$ to form an isocyanide or carbylamine,which has a foul smell.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.

(A) Primary amines undergo the carbylamine reaction when treated with $CHCl_3$ and alcoholic $KOH$ to produce isocyanides (carbylamines),which have a characteristic foul smell.
$R-NH_2 + CHCl_3 + 3KOH \text{ (alc.)} \rightarrow R-NC + 3KCl + 3H_2O$.
Secondary and tertiary amines do not give this reaction,allowing them to be distinguished.

(C) The mustard oil reaction is a characteristic test for primary amines.
In this reaction,a primary amine is heated with carbon disulfide $(CS_{2})$,which forms a dithiocarbamic acid intermediate.
Upon subsequent heating with mercuric chloride $(HgCl_{2})$,this intermediate decomposes to yield an alkyl isothiocyanate,which has a characteristic smell similar to mustard oil.
Therefore,the reagent used is $CS_{2}$.

(B) Secondary nitro compounds $(R_2CHNO_2)$ react with nitrous acid $(HNO_2)$ to form pseudonitroles $(R_2C(NO)NO_2)$.
These pseudonitroles are crystalline solids.
When treated with an alkali like $NaOH$,they dissolve to form a blue-coloured solution.
$(R)_2CHNO_2$ $\xrightarrow{HNO_2} (R)_2C(NO)NO_2$ $\xrightarrow{NaOH} \text{Blue solution}$

(C) Primary amines $(1^o)$ react with nitrous acid $(HNO_2)$ to form aliphatic diazonium salts,which are unstable and decompose to evolve nitrogen gas $(N_2)$ and form alcohols: $R-NH_2 + HNO_2 \to R-OH + N_2 + H_2O$.
Secondary amines $(2^o)$ react with nitrous acid $(HNO_2)$ to form $N$-nitrosoamines,which are yellow oily compounds: $R_2NH + HNO_2 \to R_2N-N=O + H_2O$.

(B) The correct answer is $B$.
Aniline undergoes diazotisation with $HNO_2$ at $0-5 \ ^{\circ}C$ to form benzene diazonium chloride,which then couples with phenol to form an azo dye (a yellow-orange precipitate).
Methyl amine $(CH_3NH_2)$ does not form a stable diazonium salt under these conditions and does not undergo the coupling reaction with phenol.

(B) $C_6H_5SO_2Cl$ is known as Hinsberg's reagent.
Primary amines $(R-NH_2)$ react with it to form $N$-alkylbenzenesulphonamide,which is soluble in alkali due to the acidic hydrogen on the nitrogen atom.
Secondary amines $(R_2NH)$ react with it to form $N,N$-dialkylbenzenesulphonamide,which does not contain any acidic hydrogen and is therefore insoluble in alkali.
Tertiary amines do not react with Hinsberg's reagent.
Thus,the correct answer is $B$.

(D) . Anilinium hydrogen chloride $(C_6H_5NH_3^+Cl^-)$ is an ionic salt that dissociates in water to produce chloride ions $(Cl^-)$.
When $AgNO_3$ is added to the solution,the $Cl^-$ ions react with $Ag^+$ ions to form a white precipitate of silver chloride $(AgCl)$:
$Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) \text{ (white precipitate)}$.
$p$-chloroaniline does not contain ionic chloride,so it does not form a white precipitate with $AgNO_3$.

(D) Tollen's reagent is a mild oxidizing agent used to detect the presence of aldehydes and certain other reducing groups.
$CH_3CHO$ (acetaldehyde) is an aldehyde and reduces Tollen's reagent to silver mirror.
$HCOOH$ (formic acid) contains an aldehydic group and acts as a reducing agent,thus reducing Tollen's reagent.
$C_6H_5NHOH$ (phenylhydroxylamine) is a strong reducing agent and reduces Tollen's reagent.
$C_6H_5NO_2$ (nitrobenzene) does not contain a reducing group capable of reducing Tollen's reagent.
Therefore,the correct answer is $D$.

(B) Liebermann's nitroso reaction is a characteristic test for $2^{\circ}$ (secondary) amines. In this reaction,the secondary amine reacts with nitrous acid $(HNO_2)$ to form a yellow-colored $N$-nitrosoamine. This nitrosoamine,when warmed with phenol and concentrated $H_2SO_4$,produces a green or blue color,which turns red upon dilution with water and then green or blue again upon adding an alkali like $NaOH$.

(B) The carbylamine test is a characteristic reaction for primary $(1^\circ)$ amines (both aliphatic and aromatic).
$2, 4$-dimethylaniline is a primary aromatic amine $(Ar-NH_2)$,so it reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
$N, N$-dimethylaniline is a tertiary $(3^\circ)$ amine.
$N$-methyl-$o$-methylaniline is a secondary $(2^\circ)$ amine.
$p$-methylbenzylamine is a primary amine,but the question specifically refers to the structure provided in the image,which is $2, 4$-dimethylaniline (also known as $2, 4$-xylidine).

(D) The Carbylamine test is a chemical test used for the detection of primary $(1^o)$ amines.
When aliphatic or aromatic primary amines are heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,they form isocyanides (carbylamines),which have a characteristic foul smell.
Secondary $(2^o)$ and tertiary $(3^o)$ amines do not undergo this reaction and therefore do not give the Carbylamine test.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$C_6H_5CN$ (Benzonitrile) does not contain these groups.
$RNH_2$ (Primary amines) do not contain these groups.
$CH_3OH$ (Methanol) does not contain these groups.
Therefore,all of the given substances do not give the iodoform test.

(D) . $(1)$ Anilinium hydrochloride is an acid salt and liberates $CO_2$ gas from $NaHCO_3$ solution. However,$p-$chloroaniline is basic and does not liberate $CO_2$.
$(2)$ Anilinium hydrochloride contains ionic chloride $(Cl^-)$ which reacts with $AgNO_3$ to form a white precipitate of $AgCl$. $p-$chloroaniline contains covalently bonded chlorine,so it does not give a white precipitate with $AgNO_3$.

(D) The reaction described is the Liebermann's nitroso reaction.
In this reaction,a secondary amine reacts with nitrous acid $(HNO_2)$ to form a nitroso amine $(R_2N-N=O)$.
When this nitroso amine is heated with concentrated $H_2SO_4$,it undergoes hydrolysis to regenerate the secondary amine.

(D) The Hofmann mustard oil reaction is a characteristic test for primary amines $(R-NH_2)$.
In this reaction,a primary amine reacts with carbon disulfide $(CS_2)$ followed by heating with mercuric chloride $(HgCl_2)$ to form an alkyl isothiocyanate,which has a characteristic mustard-like smell.
Secondary amines $(R_2NH)$ and tertiary amines $(R_3N)$ do not contain the necessary hydrogen atoms on the nitrogen to form the isothiocyanate intermediate.
Among the given options,$N-$Methyl$-1-$propanamine is a secondary amine $(CH_3CH_2CH_2NHCH_3)$,therefore it does not give the Hofmann mustard oil reaction.

(A) The reaction of amines with benzenesulfonyl chloride $(C_6H_5SO_2Cl)$ is known as the Hinsberg test.
$1^{\circ}$ amines $(R-NH_2)$ react with benzenesulfonyl chloride to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen atom attached to the nitrogen. This makes it soluble in alkali $(NaOH)$.
$2^{\circ}$ amines $(R_2NH)$ react to form $N,N$-dialkylbenzenesulfonamide,which does not contain any acidic hydrogen atom attached to the nitrogen,and thus it is insoluble in alkali.
$3^{\circ}$ amines do not react with benzenesulfonyl chloride.
In the given options,option $A$ is an $N$-substituted sulfonamide $(R-SO_2NH-R')$,which has an acidic hydrogen on the nitrogen atom,making it soluble in $NaOH$.

(C) $NaNO_2 + HCl$ generates $HNO_2$ (nitrous acid) in situ.
$C_6H_5NH_2$ (aniline) forms a diazonium salt.
$C_6H_5OH$ (phenol) undergoes electrophilic substitution (nitrosation).
$(CH_3)_2CHNO_2$ (a secondary nitroalkane) reacts with $HNO_2$ to form a pseudonitrole.
$(CH_3)_3CNO_2$ (a tertiary nitroalkane) does not have an $\alpha$-hydrogen atom,so it cannot react with $HNO_2$.

(D) $1$. Acidic hydrolysis: $CH_3CN + 2H_2O + H^+ \rightarrow CH_3COOH + NH_4^+$. This is a correct property.
$2$. Alkaline hydrolysis: $CH_3CN + OH^- + H_2O \rightarrow CH_3COO^- + NH_3$. This is a correct property.
$3$. Tautomerization: $CH_3-C \equiv N \rightleftharpoons CH_2=C=NH$. It does not form methyl isocyanide $(CH_3NC)$ via tautomerization.
$4$. Carbylamine reaction: The carbylamine reaction is characteristic of primary amines $(R-NH_2)$,not nitriles $(R-CN)$. Therefore,$CH_3CN + CHCl_3 + KOH$ does not produce carbylamine.
Both options $C$ and $D$ are incorrect properties for ethanenitrile,but in the context of standard chemistry questions,the carbylamine reaction is specifically a test for primary amines,making $D$ the most distinct incorrect property.

(B) The reaction of an aromatic nitro compound $(Ar-NO_2)$ with $Zn/NH_4Cl$ (a mild reducing agent) produces an aromatic hydroxylamine $(Ar-NHOH)$.
$Ar-NO_2 + Zn/NH_4Cl \rightarrow Ar-NHOH + ZnO$
$Ar-NHOH$ is a strong reducing agent that reduces Tollens' reagent (ammoniacal silver nitrate) to metallic silver $(Ag)$,which appears as a black precipitate.
Therefore,the compound $A$ must contain a $-NO_2$ group.

(B) Aniline undergoes diazotization to form a diazonium salt,which then reacts with phenol to form an azo dye. Methylamine does not undergo diazotization to form a stable diazonium salt,so it does not form an azo dye. Thus,this test distinguishes between the two.

(B) The carbylamine test (or isocyanide 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 potassium hydroxide $(KOH)$ solution to form an isocyanide (carbylamine),which has a foul,offensive smell. The reaction is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.

(C) The reaction described is the dye test,which involves the formation of a diazonium salt followed by an azo coupling reaction with $\beta$-naphthol to form an azo dye.
Primary aromatic amines $(ArNH_2)$ react with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form stable diazonium salts,which then undergo coupling with $\beta$-naphthol to produce a bright orange or red dye.
Option $A$ is a tertiary amine,which forms a nitroso compound,not a diazonium salt.
Option $B$ is a secondary amine,which forms an $N$-nitrosoamine.
Option $C$ is $p$-toluidine,a primary aromatic amine. It forms a diazonium salt with $NaNO_2/HCl$,which then couples with $\beta$-naphthol to form a bright dye.
Option $D$ is benzylamine,an aliphatic primary amine. Its diazonium salt is highly unstable and decomposes to form alcohol and nitrogen gas,thus it does not form a stable dye.

(C) An azo compound contains the functional group $R-N=N-R'$,where $R$ and $R'$ are aryl groups.
$A$. Methyl orange is an azo dye containing the $-N=N-$ linkage.
$B$. Benzenediazonium chloride contains the $-N^+=N \ Cl^-$ group,which is a diazonium salt,not an azo compound.
$C$. Phenolphthalein is a phthalein dye,not an azo compound.
$D$. $p$-Hydroxyazobenzene contains the $-N=N-$ linkage.
Since both $B$ and $C$ are not azo compounds,and the question asks for the most appropriate answer in the context of common chemistry curriculum,$C$ is a classic example of a non-azo indicator dye,while $B$ is a diazonium salt. However,strictly speaking,neither $B$ nor $C$ are azo compounds. Given the options,$C$ is the most distinct non-azo compound.

(B) The presence of a nitro group $( -NO_2 )$ in aromatic compounds like nitrobenzene can be detected using the $Mulliken$ and $Barker$ test. In this test,the nitro compound is reduced to a hydroxylamine $( -NHOH )$ using zinc dust and ammonium chloride. The resulting hydroxylamine then reduces Tollens' reagent to metallic silver,which confirms the presence of the nitro group.

(D) The Hinsberg reagent is $C_6H_5SO_2Cl$ (benzenesulfonyl chloride).
It is used to distinguish and separate primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines.
Primary amines react to form a sulfonamide that is soluble in alkali.
Secondary amines react to form a sulfonamide that is insoluble in alkali.
Tertiary amines do not react with the Hinsberg reagent.
Therefore,it is used for the separation of amines.

(C) $3^o$ amines do not react with Hinsberg reagent $(C_6H_5SO_2Cl)$ because they lack an acidic hydrogen atom attached to the nitrogen atom.

(A) 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.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$
Secondary and tertiary amines do not give this test.

(B) The carbylamine test is given by primary amines ($R-NH_2$ or $Ar-NH_2$).
Among the given options,$2, 4-$dimethylaniline is a primary aromatic amine.
Therefore,it reacts with chloroform and alcoholic $KOH$ to produce an isocyanide (carbylamine) with an offensive smell.

(C) The $Carbylamine$ reaction (also known as the $Isocyanide$ test) is a specific test used to identify primary amines $(R-NH_2)$.
When a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it produces an isocyanide (carbylamine),which has a characteristic foul,offensive smell.
Secondary and tertiary amines do not give this test.

(C) $NaNO_2 + HCl \to HNO_2 + NaCl$
$H_2NCONH_2 + 2HNO_2 \to 2N_2 + CO_2 + 3H_2O$
Urea reacts with nitrous acid $(HNO_2)$ generated in situ to evolve $N_2$ and $CO_2$ gases,which results in brisk effervescence.

(A) The dye test is a characteristic reaction of primary aromatic amines.
When primary aromatic amines like $Aniline$ $(C_6H_5NH_2)$ are treated with nitrous acid $(HNO_2)$ at $0-5 \ ^{\circ}C$,they form a diazonium salt.
This diazonium salt then undergoes a coupling reaction with phenols or other aromatic amines to form azo dyes,which are colored compounds.
Since $Aniline$ is a primary aromatic amine,it gives the dye test,whereas aliphatic amines like $Methylamine$ and $Ethylamine$ do not form stable diazonium salts at these temperatures,and $Diphenylamine$ is a secondary amine.

(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 alcoholic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which has a foul,unpleasant smell.
Among the given options,$Aniline$ $(C_6H_5NH_2)$ is a primary aromatic amine.
The reaction is: $C_6H_5NH_2 + CHCl_3 + 3KOH_{(alc)} \rightarrow C_6H_5NC + 3KCl + 3H_2O$.

(C) The reaction of nitro compounds with nitrous acid $(HNO_2)$ depends on the presence of $\alpha$-hydrogen atoms.
$1^o$ nitro compounds (containing two $\alpha$-hydrogens) react with $HNO_2$ to form nitrolic acids,which dissolve in $NaOH$ to give a red solution.
$2^o$ nitro compounds (containing one $\alpha$-hydrogen) react with $HNO_2$ to form pseudonitroles,which dissolve in $NaOH$ to give a blue solution.
$3^o$ nitro compounds do not contain any $\alpha$-hydrogen atoms,and therefore,they do not react with nitrous acid.
In the given options:
$(A)$ $CH_3CH_2CH_2NO_2$ is a $1^o$ nitro compound with $\alpha$-hydrogens.
$(B)$ $(CH_3)_2CHCH_2NO_2$ is a $1^o$ nitro compound with $\alpha$-hydrogens.
$(C)$ $(CH_3)_3CNO_2$ is a $3^o$ nitro compound with no $\alpha$-hydrogen atoms.
$(D)$ $CH_3C(=O)CH(CH_3)NO_2$ is a $2^o$ nitro compound with an $\alpha$-hydrogen.
Thus,$(CH_3)_3CNO_2$ does not react with nitrous acid.

(C) The isocyanide test (carbylamine reaction) is given only by primary $(1^{\circ})$ amines,both aliphatic and aromatic.
Secondary $(2^{\circ})$ and tertiary $(3^{\circ})$ amines do not give this test.
In the given options,$CH_3-NH-CH_2-CH_3$ is a secondary amine,therefore it does not give the isocyanide test.

(B) The carbylamine test is a characteristic reaction given only by primary $(1^{\circ})$ amines,both aliphatic and aromatic.
$(a)$ $N,N$-Dimethylaniline is a tertiary $(3^{\circ})$ amine.
$(b)$ $2,4$-Dimethylaniline is a primary $(1^{\circ})$ aromatic amine.
$(c)$ $N$-methyl-$o$-methylaniline is a secondary $(2^{\circ})$ amine.
$(d)$ $p$-Methylbenzylamine is a primary $(1^{\circ})$ aliphatic amine.
Since both $(b)$ and $(d)$ are primary amines,they will give a positive carbylamine test.

(B) The carbylamine reaction is a characteristic test for primary aliphatic and aromatic amines.
$CH_3NH_2$ (methylamine),$PhNH_2$ (aniline),and $EtNH_2$ (ethylamine) are all primary amines and will give the carbylamine reaction.
$CH_3CONH_2$ (acetamide) is an amide,not a primary amine,and therefore it does not give this reaction.

(C) Aniline is a primary aromatic amine,while benzylamine is a primary aliphatic amine.
$1$. Dye-azo test: Aniline reacts with nitrous acid $(HNO_2)$ at $0-5 \ ^\circ C$ to form a benzene diazonium salt,which then couples with $\beta$-naphthol to give an orange-red dye.
$2$. Benzylamine reacts with nitrous acid to form a diazonium salt,but it is highly unstable and decomposes immediately to form benzyl alcohol and nitrogen gas.
$3$. Therefore,the dye-azo test is the standard chemical method to distinguish between primary aromatic and aliphatic amines.

(C) The carbylamine reaction is a characteristic test for primary amines ($1^{\circ}$ amines).
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and an alcoholic base (like $KOH$) to form an isocyanide (carbylamine),which has a foul,offensive smell.
Among the given options:
$A$. Urea $(NH_2CONH_2)$ is an amide,not a primary amine.
$B$. $CH_3CONH_2$ is an amide,not a primary amine.
$C$. $C_2H_5NH_2$ is an aliphatic primary amine ($1^{\circ}$ amine).
Therefore,only $C_2H_5NH_2$ will give the carbylamine reaction.