Tests for Nitrogen Containing Compounds Questions in English

(B) The reaction of primary amines with $CHCl_3$ and alcoholic $KOH$ is known as the carbylamine reaction. This reaction produces isocyanides (carbylamines),which have a characteristic offensive smell.
Both aliphatic and aromatic primary amines undergo this reaction.
However,in the case of $p$-aminobenzoic acid,the presence of the acidic $-COOH$ group and the basic $-NH_2$ group in the same molecule leads to the formation of a zwitterion.
Due to the formation of this zwitterion,the $-NH_2$ group is not available to react with $CHCl_3 / KOH$ under normal conditions,and thus it does not produce the offensive smell of a carbylamine.

(A) $1$. The reaction of benzene diazonium chloride with phosphinic acid $(H_3PO_2)$ yields benzene $(A)$.
$2$. Benzene reacts with $SO_3$ (sulfonation) to form benzene sulfonic acid ($B$,$C_6H_5SO_3H$).
$3$. Benzene sulfonic acid reacts with $PCl_5$ to form benzene sulfonyl chloride ($C$,$C_6H_5SO_2Cl$).
$4$. Benzene sulfonyl chloride reacts with ethylamine $(EtNH_2)$ to form $N$-ethylbenzene sulfonamide ($D$,$C_6H_5SO_2NHC_2H_5$).
$5$. Evaluating the statements:
- $A$ (benzene) is $NOT$ formed when phenol is heated with $Zn$ dust; rather,benzene is formed,but the statement implies a specific synthesis pathway. Actually,phenol + $Zn$ dust gives benzene,so statement $B$ is correct.
- $B$ (benzene sulfonic acid) when fused with $NaOH$ followed by acidification gives phenol. This is a standard reaction. Statement $C$ is correct.
- $C$ $(C_6H_5SO_2Cl)$ contains chlorine. Statement $D$ is correct.
- $D$ $(C_6H_5SO_2NHC_2H_5)$ contains an acidic hydrogen on the nitrogen atom,making it soluble in alkali. Therefore,the statement '$D$ is insoluble in alkali' is incorrect.

(A) The carbylamine reaction (also known as the isocyanide test) is a diagnostic test used to identify primary 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 characteristic foul smell. Among the given options,only $CH_3CH_2CH_2NH_2$ (n-propylamine) is a primary amine. The others are secondary or tertiary amines,which do not undergo this reaction.

(D) Primary,secondary,and tertiary amines can be distinguished using the Hinsberg reagent (benzenesulfonyl chloride,$C_6H_5SO_2Cl$).
- Primary amines form a sulfonamide that is soluble in alkali.
- Secondary amines form a sulfonamide that is insoluble in alkali.
- Tertiary amines do not react with the Hinsberg reagent.
Additionally,$HNO_2$ (nitrous acid) also distinguishes them:
- Primary aliphatic amines react with $HNO_2$ to evolve $N_2$ gas.
- Secondary amines form yellow oily $N$-nitrosoamines.
- Tertiary amines form soluble nitrite salts.
Therefore,both $A$ and $B$ are correct.

(D) The isocyanide test (carbylamine reaction) is given only by primary $(1^{\circ})$ amines,both aliphatic and aromatic.
$A$: Cyclohexylamine is a primary amine $(R-NH_2)$.
$B$: Isopropylamine $(CH_3)_2CHNH_2$ is a primary amine $(R-NH_2)$.
$C$: Aniline $(C_6H_5NH_2)$ is a primary aromatic amine $(Ar-NH_2)$.
$D$: Dimethylamine $(CH_3NHCH_3)$ is a secondary $(2^{\circ})$ amine $(R_2NH)$.
Since secondary amines do not undergo the carbylamine reaction,$D$ is the correct answer.

(A) The isocyanide test (carbylamine reaction) is given only by primary amines ($R-NH_2$ or $Ar-NH_2$).
In reaction $(A)$,benzamide $(C_6H_5CONH_2)$ is reduced by $LiAlH_4$ to benzylamine $(C_6H_5CH_2NH_2)$,which is a primary amine and gives a positive isocyanide test.
In reaction $(B)$,$NaBH_4$ does not reduce amides.
In reaction $(C)$,$N$-acetylaniline $(C_6H_5NHCOCH_3)$ is reduced to $N$-ethylaniline $(C_6H_5NHCH_2CH_3)$,which is a secondary amine and does not give the isocyanide test.
In reaction $(D)$,$NaBH_4$ does not reduce amides.

(C) The carbylamine test is a characteristic reaction given only by primary $(1^{\circ})$ amines,both aliphatic and aromatic.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.
Dimethylamine is a secondary $(2^{\circ})$ amine,and $N$-methylethanamine is also a secondary $(2^{\circ})$ amine.
Aniline $(C_6H_5NH_2)$ is a primary $(1^{\circ})$ aromatic amine,therefore it gives the carbylamine test.

(A) The isocyanide test is a characteristic reaction for primary $(1^{\circ})$ amines. Aliphatic or aromatic primary amines,when heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,form isocyanides (carbylamines),which have a foul smell.
In option $(A)$,benzamide $(C_6H_5CONH_2)$ is reduced by $LiAlH_4$ to form benzylamine $(C_6H_5CH_2NH_2)$,which is a primary amine. Thus,it gives a positive isocyanide test.
$C_6H_5CONH_2$ $\xrightarrow{LiAlH_4} C_6H_5CH_2NH_2$ $\xrightarrow{CHCl_3 + KOH} C_6H_5CH_2NC$ (Isocyanide).
$NaBH_4$ is a weaker reducing agent and does not reduce amides to amines. Option $(C)$ involves the reduction of an $N$-substituted amide,which would yield a secondary amine,not a primary amine.

(D) Primary,secondary,and tertiary amines are distinguished by the Hinsberg reagent test.
The Hinsberg reagent is benzene sulphonyl chloride,represented as $C_6H_5SO_2Cl$.

(C) The solubility profile of the compound '$X$' is as follows:
$1$. Insoluble in water: Indicates it is likely a non-polar or high molecular weight organic compound.
$2$. Insoluble in $5\% \,HCl$: Indicates it is not a basic compound (like an amine).
$3$. Insoluble in $10\% \,NaOH$: Indicates it is not an acidic compound (like a carboxylic acid or a phenol).
$4$. Insoluble in $10\% \,NaHCO_3$: Confirms it is not a strong acid (like a carboxylic acid).
Evaluating the options:
- Oleic acid is a carboxylic acid,so it is soluble in $NaOH$ and $NaHCO_3$.
- $o-$Toluidine is an amine,so it is soluble in $HCl$.
- $m-$Cresol is a phenol,so it is soluble in $NaOH$.
- Benzamide $(C_6H_5CONH_2)$ is a neutral compound. It is insoluble in water,$HCl$,$NaOH$,and $NaHCO_3$. Therefore,it matches the given solubility profile.

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

(D) The reaction with $HNO_2$ (nitrous acid) depends on the functional group present:
$1$. $CH_3CH(CH_3)NO_2$ is a secondary nitroalkane. It reacts with $HNO_2$ to form a pseudonitrole.
$2$. $CH_3CH(CH_3)NHCH_3$ is a secondary amine. It reacts with $HNO_2$ to form an $N$-nitrosoamine.
$3$. $C_6H_5OH$ (phenol) reacts with $HNO_2$ to form p-nitrosophenol.
$4$. $(CH_3)_3CNO_2$ is a tertiary nitroalkane. Tertiary nitroalkanes do not have an $\alpha$-hydrogen atom,which is required for the reaction with $HNO_2$ to form a pseudonitrole. Therefore,it does not react with $HNO_2$.

(A) The Hinsberg reagent is benzene sulfonyl chloride $(C_6H_5SO_2Cl)$.
Primary amines $(R-NH_2)$ react with the Hinsberg reagent to form $N$-alkylbenzene sulfonamide $(C_6H_5SO_2NHR)$.
This product contains an acidic hydrogen atom attached to the nitrogen atom,which makes it soluble in alkali (e.g.,$NaOH$).
Secondary amines $(R_2NH)$ form $N,N$-dialkylbenzene sulfonamide,which does not contain an acidic hydrogen and is insoluble in alkali.
Tertiary amines $(R_3N)$ do not react with the Hinsberg reagent.
Therefore,$CH_3-CH_2-NH_2$ (a primary amine) is the correct answer.

(A) The isocyanide test is a characteristic reaction for primary $(1^{\circ})$ amines. When a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which has a characteristic foul smell.
$1$. In option $A$,benzamide $(C_6H_5CONH_2)$ reacts with $LiAlH_4$ (a strong reducing agent) to form benzylamine $(C_6H_5CH_2NH_2)$,which is a primary amine and thus gives a positive isocyanide test.
$2$. $NaBH_4$ is a weaker reducing agent and does not reduce amides to amines.
$3$. $N$-phenylacetamide $(C_6H_5NHCOCH_3)$ is a secondary amide; its reduction would yield a secondary amine,which does not give the isocyanide test.
Therefore,the correct reaction is benzamide with $LiAlH_4$.

(C) Let's analyze each option:
$1$. $C_6H_5NO_2$ (Nitrobenzene) is commonly known as oil of mirbane. This is correct.
$2$. $CH_3-N=C=S$ (Methyl isothiocyanate) has a mustard oil-like smell. The formula provided in the question was incorrect ($CH_3-S^{-}C \equiv N$ is not the standard representation for mustard oil). However,$C_6H_5SO_2Cl$ is the standard Hinsberg reagent.
$3$. $C_6H_5SO_2Cl$ (Benzenesulfonyl chloride) is the Hinsberg reagent. This is correct.
$4$. $R-N_2^{\oplus}Cl^{-}$ (Diazonium salt) undergoes diazo coupling reactions. This is correct.
Given the context of multiple correct statements,the most standard textbook identification for the Hinsberg reagent is $C_6H_5SO_2Cl$ and Nitrobenzene as oil of mirbane. In many competitive contexts,if only one must be chosen,$C_6H_5SO_2Cl$ is the most definitive chemical reagent name.

(D) Ammonia reacts with Nessler's reagent to form a brown precipitate known as the iodide of Millon's base.
The chemical reaction is:
$2K_2HgI_4 + NH_3 + 3KOH \to H_2N-HgO-HgI + 7KI + 2H_2O$
Here,$K_2HgI_4$ is Nessler's reagent,and the product $H_2N-HgO-HgI$ forms a brown precipitate.

(C) The reaction of nitroalkanes with nitrous acid $(HNO_2)$ is used to distinguish between primary,secondary,and tertiary nitroalkanes.
Primary nitroalkanes $(R-CH_2-NO_2)$ react with $HNO_2$ to form nitrolic acids,which dissolve in $NaOH$ to give a red color.
Secondary nitroalkanes $(R_2CH-NO_2)$ react with $HNO_2$ to form pseudonitroles,which are blue in color.
Tertiary nitroalkanes do not react with $HNO_2$.
The compound $CH_3-C(NO)(NO_2)-CH_3$ is a pseudonitrole,which exhibits a blue color.

(C) The carbylamine reaction is a characteristic test for primary $(1^o)$ amines,both aliphatic and aromatic.
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 $(2^o)$ and tertiary $(3^o)$ amines do not contain the $-NH_2$ group required for this reaction.
Analyzing the options:
$(A)$ $H_2C=CH-CH_2-NH_2$ is a $1^o$ amine.
$(B)$ $C_6H_5-NH_2$ (Aniline) is a $1^o$ amine.
$(C)$ $C_6H_5-NH-R$ is a $2^o$ amine.
$(D)$ Cyclohexylamine is a $1^o$ amine.
Therefore,option $(C)$ will not give the carbylamine reaction.

(C) Primary and secondary amines can be distinguished using the $Hinsberg$ reagent ($benzenesulfonyl$ $chloride$,$C_6H_5SO_2Cl$).
$1.$ Primary amines react with $Hinsberg$ reagent to form $N$-alkylbenzenesulfonamide,which is soluble in alkali due to the presence of an acidic hydrogen atom on the nitrogen.
$2.$ Secondary amines react with $Hinsberg$ reagent to form $N,N$-dialkylbenzenesulfonamide,which is insoluble in alkali because it lacks an acidic hydrogen atom on the nitrogen.
$3.$ Tertiary amines do not react with $Hinsberg$ reagent.
Therefore,the $Hinsberg$ reagent is the correct method for distinguishing between primary and secondary amines.

(D) $1.$ $CHCl_3 / Alc. KOH$ (Carbylamine test): Only $1^o$ amines give a foul-smelling isocyanide,while $2^o$ amines do not.
$2.$ $NaNO_2 / HCl$: $1^o$ aliphatic amines form unstable diazonium salts that evolve $N_2$ gas,whereas $2^o$ amines form $N$-nitrosoamines (yellow oily liquids).
$3.$ $PhSO_2Cl$ (Hinsberg reagent) then $NaOH$: $1^o$ amines form a sulfonamide soluble in $NaOH$,while $2^o$ amines form a sulfonamide insoluble in $NaOH$.
Since all three reagents can distinguish between $1^o$ and $2^o$ amines,the correct answer is $D$.

(A) Ethyl amine $(CH_3CH_2NH_2)$ is an aliphatic amine,while aniline $(C_6H_5NH_2)$ is an aromatic amine.
$1$. $Br_2/H_2O$: Aniline reacts with bromine water to form a white precipitate of $2,4,6$-tribromoaniline,whereas ethyl amine does not show this reaction.
$2$. $CHCl_3/KOH$: Both undergo the carbylamine test,but the test is generally used to distinguish primary amines from secondary/tertiary amines,not specifically aliphatic from aromatic.
$3$. $C_6H_5SO_2Cl$ (Hinsberg's reagent): Both react,but the solubility of the resulting sulfonamide in alkali differs due to the nature of the substituents.
However,$Br_2/H_2O$ is the most characteristic test to distinguish between aliphatic and aromatic primary amines.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,$K_2[HgI_4]$.
When ammonia $(NH_3)$ is added to Nessler's reagent,it forms a brown precipitate known as the iodide of Millon's base.
The chemical reaction is: $2K_2[HgI_4] + NH_3 + 3KOH \rightarrow HgO \cdot Hg(NH_2)I + 7KI + 2H_2O$.
The brown precipitate is $HgO \cdot Hg(NH_2)I$.

(B) Diethyl oxalate is used for distinguishing $1^o, 2^o$ and $3^o$ amines because they react differently.
$1^o$ amines form crystalline substituted oxamides.
$2^o$ amines form liquid diethyl oxamic esters.
$3^o$ amines do not react with diethyl oxalate because they lack a replaceable hydrogen atom.
$RNH_2 + (COOC_2H_5)_2 \to (CONHR)_2 + 2 C_2H_5OH$ ($1^o$ amine,crystalline oxamide)
$R_2NH + (COOC_2H_5)_2 \to (CONR_2)(COOC_2H_5) + C_2H_5OH$ ($2^o$ amine,liquid oxamic ester)
$R_3N + (COOC_2H_5)_2 \to \text{No reaction}$ ($3^o$ amine)

(B) Benzamide $(C_6H_5CONH_2)$ is an amide and is neutral in nature,so it does not react with cold dilute $NaOH$ or $HCl$.
Benzylamine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine and is basic in nature.
It reacts with cold dilute $HCl$ to form a salt,$C_6H_5CH_2NH_3^+Cl^-$,whereas benzamide does not react.
Therefore,cold dilute $HCl$ can be used to distinguish between them.

(A) The Lassaigne's test (sodium fusion test) is used to detect nitrogen in organic compounds.
Benzene diazonium chloride $(C_6H_5N_2Cl)$ is unstable and decomposes upon heating to release $N_2$ gas before the fusion process is complete.
Consequently,it does not give the characteristic Prussian blue color test for nitrogen.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(A) Hinsberg's reagent (benzene sulfonyl chloride,$C_6H_5SO_2Cl$) is used to distinguish between primary,secondary,and tertiary amines.
$1$. Primary $(1^\circ)$ amines react to form $N$-alkylbenzene sulfonamides,which contain an acidic hydrogen on the nitrogen atom and are therefore soluble in alkali.
$2$. Secondary $(2^\circ)$ amines react to form $N,N$-dialkylbenzene sulfonamides,which lack an acidic hydrogen on the nitrogen atom and are insoluble in alkali.
$3$. Tertiary $(3^\circ)$ amines do not react with Hinsberg's reagent.
In the given options,$(CH_3)_2CH-NH-CH(CH_3)_2$ is a secondary amine. It reacts with Hinsberg's reagent to form an $N,N$-disubstituted sulfonamide that has no acidic hydrogen,making it insoluble in alkali.

(N/A) $(i)$ Methylamine and dimethylamine can be distinguished by the carbylamine test. Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test,while dimethylamine does not.
$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC + 3KCl + 3H_2O$
$(CH_3)_2NH + CHCl_3 + 3KOH \xrightarrow{\Delta} \text{No reaction}$
$(ii)$ Secondary and tertiary amines can be distinguished by Hinsberg's test using benzenesulphonyl chloride $(C_6H_5SO_2Cl)$. Secondary amines react to form a product insoluble in alkali,whereas tertiary amines do not react.
$(iii)$ Ethylamine and aniline can be distinguished using the azo-dye test. Aniline (aromatic primary amine) reacts with $HNO_2$ at $0-5^{\circ}C$ to form a diazonium salt,which couples with $2$-naphthol to form an orange dye. Ethylamine (aliphatic primary amine) does not form a stable diazonium salt and evolves $N_2$ gas.
$(iv)$ Aniline and benzylamine can be distinguished by their reaction with nitrous acid $(HNO_2)$. Benzylamine reacts with $HNO_2$ to form an unstable diazonium salt,which decomposes to give benzyl alcohol and $N_2$ gas. Aniline forms a stable diazonium salt at $0-5^{\circ}C$.
$(v)$ Aniline and $N$-methylaniline can be distinguished using the carbylamine test. Aniline (primary amine) gives a positive carbylamine test,while $N$-methylaniline (secondary amine) does not.

(N/A) Primary,secondary and tertiary amines can be identified and distinguished by Hinsberg's test. In this test,the amines are allowed to react with Hinsberg's reagent,benzenesulphonyl chloride $(C_6H_5SO_2Cl)$. The three types of amines react differently with Hinsberg's reagent.
$1$. Primary amines $(R-NH_2)$ react with benzenesulphonyl chloride to form $N$-alkylbenzenesulphonamide,which is soluble in alkali due to the acidic hydrogen atom attached to the nitrogen atom.
Reaction: $C_6H_5SO_2Cl + R-NH_2 \rightarrow C_6H_5SO_2NHR + HCl$
$2$. Secondary amines $(R_2NH)$ react with Hinsberg's reagent to give $N,N$-dialkylbenzenesulphonamide,which is insoluble in alkali because there is no hydrogen atom attached to the nitrogen atom.
Reaction: $C_6H_5SO_2Cl + R_2NH \rightarrow C_6H_5SO_2NR_2 + HCl$
$3$. Tertiary amines $(R_3N)$ do not react with Hinsberg's reagent at all because they lack a hydrogen atom on the nitrogen atom to form a sulphonamide.

Carbylamine reaction: Aliphatic and aromatic primary amines,when heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,form isocyanides or carbylamines,which are foul-smelling substances.
Secondary and tertiary amines do not undergo this reaction.
This reaction is known as the carbylamine reaction or isocyanide test and is used as a diagnostic test for primary amines.
General reaction:
$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$
Example with methanamine:
$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3-NC + 3KCl + 3H_2O$
Example with aniline:
$C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5-NC + 3KCl + 3H_2O$

Benzenesulphonyl chloride $(C_{6}H_{5}SO_{2}Cl)$ is known as Hinsberg's reagent.
It is used to distinguish between primary,secondary,and tertiary amines.
Primary amines react with Hinsberg's reagent to form $N$-alkylbenzenesulphonamide,which is soluble in alkali.
Secondary amines react with Hinsberg's reagent to form $N,N$-dialkylbenzenesulphonamide,which is insoluble in alkali.
Tertiary amines do not react with Hinsberg's reagent.
The reaction of diethylamine (a secondary amine) with benzenesulphonyl chloride is shown below:
$C_{6}H_{5}SO_{2}Cl + HN(C_{2}H_{5})_{2} \rightarrow C_{6}H_{5}SO_{2}N(C_{2}H_{5})_{2} + HCl$

(C) The reaction of an amine with $C_6H_5SO_2Cl$ (Hinsberg reagent) is used to distinguish between primary,secondary,and tertiary amines.
Primary amines $(RNH_2)$ react to form a sulfonamide that is soluble in alkali due to the acidic hydrogen on the nitrogen atom.
Secondary amines $(R_2NH)$ react to form a sulfonamide that is insoluble in alkali because there is no acidic hydrogen on the nitrogen atom.
Tertiary amines $(R_3N)$ do not react with $C_6H_5SO_2Cl$.
Since the product is a solid insoluble in alkali,$Z$ must be a secondary amine.
Given the molecular formula $C_3H_9N$,the secondary amine is ethylmethylamine $(CH_3NHC_2H_5)$.
The reaction is: $CH_3NHC_2H_5 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2N(CH_3)C_2H_5 + HCl$.

(A) The reaction of $CHCl_{3} +$ alc. $KOH$ is the Carbylamine test,which is characteristic of primary amines ($-NH_{2}$ groups).
Adenine contains a primary amine group ($-NH_{2}$ attached to the ring).
Lysine is an amino acid with two amino groups,one of which is a primary amine ($-NH_{2}$ at the end of the side chain).
Therefore,both Adenine and Lysine will give a positive Carbylamine test.

(A) The correct matches are as follows:

$Test/Method$	$Reagent$
$(i)$ Lucas test	$(d)$ Conc. $HCl + ZnCl_2$
$(ii)$ Dumas method	$(c)$ $CuO/CO_2$
$(iii)$ Kjeldahl's method	$(e)$ $H_2SO_4$
$(iv)$ Hinsberg Test	$(a)$ $C_6H_5SO_2Cl + aq. KOH$

Therefore,the correct sequence is $(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)$.

(B) The carbylamine test is a characteristic reaction for primary amines (both aliphatic and aromatic).
When a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which has a foul,offensive smell.
Secondary and tertiary amines do not contain the necessary $-NH_2$ group to form an isocyanide and therefore do not give this test.
Among the given options,aniline $(C_6H_5NH_2)$ is a primary amine and will give the carbylamine test.

(B) The reaction sequence is a characteristic test for primary amines using Hinsberg's reagent $(PhSO_2Cl)$.
$1$. $A$ is a primary amine $(RNH_2)$.
$2$. $A$ reacts with $PhSO_2Cl$ to form $N$-alkylbenzenesulfonamide ($B$,$PhSO_2NHR$).
$3$. $B$ has an acidic hydrogen on the nitrogen atom,so it reacts with $KOH$ to form a water-soluble potassium salt ($C$,$PhSO_2N^-R K^+$).
$4$. The salt $C$ undergoes alkylation with ethyl iodide $(C_2H_5I)$ to form $N$-ethyl-$N$-alkylbenzenesulfonamide ($D$,$PhSO_2N(C_2H_5)R$).
Therefore,$A$ is $RNH_2$ and $D$ is $PhSO_2N(C_2H_5)R$.

(C) Hydrazoic acid $(N_{3}H)$ reacts with $NaOH$ to form sodium azide $(NaN_{3})$ and water. However,in the context of nitrogen-containing compounds,$N_{3}H$ is often associated with the release of ammonia under specific reducing conditions or in specific chemical contexts. Given the options,$N_{3}H$ is the only compound containing nitrogen that can be involved in pathways leading to $NH_{3}$ formation.

(A) Primary,secondary,and tertiary amines can be distinguished using Hinsberg's reagent,which is $Para-Toluene$ sulphonyl chloride $(CH_3C_6H_4SO_2Cl)$.
$1.$ Primary amines react with $Para-Toluene$ sulphonyl chloride to form an $N-alkyl$ sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in $NaOH$.
$2.$ Secondary amines react to form an $N,N-dialkyl$ sulphonamide,which lacks an acidic hydrogen atom,rendering it insoluble in $NaOH$.
$3.$ Tertiary amines do not react with $Para-Toluene$ sulphonyl chloride because they lack a hydrogen atom on the nitrogen.

(D) The reaction of an organic compound $A$ with benzene sulphonyl chloride (Hinsberg's reagent) to form a product $B$ that is soluble in $NaOH$ indicates that $A$ is a primary amine $(R-NH_2)$.
Primary amines react with benzene sulphonyl chloride to form $N$-alkylbenzene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in $NaOH$.
Secondary amines form $N,N$-dialkylbenzene sulphonamides,which lack an acidic hydrogen and are insoluble in $NaOH$.
Tertiary amines do not react with benzene sulphonyl chloride.
Among the given options:
$A$: $C_6H_5-N(CH_3)_2$ is a tertiary amine.
$B$: $C_6H_5-NH-CH_2-CH_3$ is a secondary amine.
$C$: $C_6H_5-CH_2-NH-CH_3$ is a secondary amine.
$D$: $C_6H_5-CH(CH_3)-NH_2$ is a primary amine.
Therefore,the correct compound is $C_6H_5-CH(CH_3)-NH_2$.

(D) The carbylamine test is given by $1^{\circ}$ amines. When aniline $(C_6H_5NH_2)$ is treated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it undergoes a reaction to form phenyl isocyanide $(C_6H_5NC)$,which is characterized by an extremely unpleasant odor.
The chemical equation is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$

(C) Ceric ammonium nitrate is used as a test for the identification of alcohols,which gives a characteristic red or yellow color.
$CHCl_3 / \text{alc. } KOH$ is used in the carbylamine test,which is specific for the identification of primary amines $(R-NH_2)$.

(C) $p$-toluenesulphonyl chloride $(CH_3C_6H_4SO_2Cl)$ is commonly known as Hinsberg's reagent.
It is used to distinguish between primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines.
$1$. Primary amines react with Hinsberg's reagent to form $N$-alkylbenzenesulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in alkali.
$2$. Secondary amines react with Hinsberg's reagent to form $N,N$-dialkylbenzenesulphonamide. This product does not contain any acidic hydrogen atom attached to the nitrogen,so it is insoluble in alkali.
$3$. Tertiary amines do not react with Hinsberg's reagent because they lack an acidic hydrogen atom.
Therefore,the statement that the product formed with a secondary amine is soluble in alkali is incorrect.

(B) Aniline $(C_6H_5NH_2)$ is an aromatic primary amine,while ethylamine $(C_2H_5NH_2)$ is an aliphatic primary amine.
When treated with $NaNO_2 / HCl$ at $0-5 \ ^\circ C$,aniline forms benzene diazonium chloride,which is stable at low temperatures and undergoes a coupling reaction with $2-$naphthol to form an azo dye (red color).
In contrast,ethylamine forms an unstable aliphatic diazonium salt that immediately decomposes to evolve $N_2$ gas and forms ethanol,thus it does not give the azo dye coupling reaction.
Therefore,the reaction with $NaNO_2 / HCl$ followed by treatment with $2-$naphthol can distinguish between them.

(A) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which has the chemical formula $K_2[HgI_4]$.

(B) Compound $P$ has the molecular formula $C_{14}H_{13}ON$.
Upon hydrolysis with $HCl$ and $\Delta$,it yields a carboxylic acid $(Q)$ and an amine $(R)$.
$Q$ gives effervescence with $NaHCO_3$,indicating it is a carboxylic acid.
$R$ reacts with Hinsberg's reagent $(PhSO_2Cl)$ to form a product that is soluble in $NaOH$,which confirms that $R$ is a primary amine ($1^{\circ}$ amine).
Based on the molecular formula $C_{14}H_{13}ON$,the compound $P$ is $N$-phenyl$-4-$methylbenzamide $(CH_3-C_6H_4-CONH-C_6H_5)$.
Hydrolysis of $N$-phenyl$-4-$methylbenzamide gives $4-$methylbenzoic acid $(Q)$ and aniline $(R)$.
Aniline $(R)$ reacts with Hinsberg's reagent to form $N$-phenylbenzenesulfonamide,which is soluble in $NaOH$ due to the acidic hydrogen on the nitrogen atom.

(D) The chemical formula for Nessler's reagent is $K_2[HgI_4]$.
It consists of potassium $(K)$,mercury $(Hg)$,and iodine $(I)$.
Therefore,nitrogen $(N)$ is the element not present in Nessler's reagent.

(D) $A.$ Alkaline solution of copper sulphate and sodium citrate is known as Benedict's solution,which is used to test aliphatic aldehydes. Thus,it tests compound $(III)$.
$B.$ Neutral $FeCl_3$ solution is used to test phenolic compounds. Thus,it tests compound $(IV)$.
$C.$ Alkaline chloroform solution (carbylamine test) is used to test primary amines. Thus,it tests compound $(II)$.
$D.$ Potassium iodide and sodium hypochlorite generate $I_2$ in situ,which is used for the iodoform test. This test is positive for compounds containing the $CH_3CO-$ or $CH_3CH(OH)-$ group. Thus,it tests compound $(I)$.

(D) Hinsberg's reagent $(C_6H_5SO_2Cl)$ reacts with primary $(1^{\circ})$ and secondary $(2^{\circ})$ amines to form sulfonamides. Tertiary $(3^{\circ})$ amines do not react with it.
Let us analyze the given compounds:
$1$. Benzenediazonium chloride: Does not react.
$2$. Dibenzamide: Does not react (amide).
$3$. Aniline ($1^{\circ}$ amine): Reacts.
$4$. $N$-phenylpiperidine ($3^{\circ}$ amine): Does not react.
$5$. $N$-methylbenzylamine ($2^{\circ}$ amine): Reacts.
$6$. $N,N$-dimethylaniline ($3^{\circ}$ amine): Does not react.
$7$. Ethylenediamine ($1^{\circ}$ amine): Reacts.
$8$. Piperidine ($2^{\circ}$ amine): Reacts.
$9$. Pyridine ($3^{\circ}$ amine): Does not react.
$10$. Propylamine ($1^{\circ}$ amine): Reacts.
$11$. Urea: Does not react.
The compounds that react are: Aniline,$N$-methylbenzylamine,Ethylenediamine,Piperidine,and Propylamine.
Total count = $5$.

(C) Hinsberg's reagent $(C_6H_5SO_2Cl)$ reacts with primary amines $(R-NH_2)$ to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen atom attached to the nitrogen. This acidic hydrogen makes the product soluble in aqueous $NaOH$.
Secondary amines $(R_2NH)$ form $N,N$-dialkylbenzenesulfonamide,which lacks an acidic hydrogen and is insoluble in $NaOH$.
Tertiary amines $(R_3N)$ do not react with Hinsberg's reagent.
Let us analyze the given compounds:
$1$. Aniline $(C_6H_5NH_2)$: Primary amine,forms soluble product.
$2$. $o$-methoxyaniline $(o-CH_3OC_6H_4NH_2)$: Primary amine,forms soluble product.
$3$. Ethanamine $(CH_3CH_2NH_2)$: Primary amine,forms soluble product.
$4$. $N$-phenyl-$p$-phenylenediamine $(C_6H_5NH-C_6H_4-NH_2)$: Contains a primary amine group,forms soluble product.
$5$. Cyclohexanamine $(C_6H_{11}NH_2)$: Primary amine,forms soluble product.
Other compounds like amides,ureas,and secondary/tertiary amines do not meet the criteria.
There are $5$ such primary amine compounds.

(A) $p$-Toluenesulfonyl chloride (Hinsberg reagent) reacts only with primary $(1^{\circ})$ and secondary $(2^{\circ})$ amines.
$C_6H_5NH_2$ ($1^{\circ}$ amine) reacts to form $N$-phenyl-$p$-toluenesulfonamide,which is soluble in aqueous $NaOH$ due to the acidic hydrogen on the nitrogen atom.
$(C_6H_5)_2NH$ ($2^{\circ}$ amine) reacts to form $N,N$-diphenyl-$p$-toluenesulfonamide,which is insoluble in aqueous $NaOH$ because it lacks an acidic hydrogen.
$(C_6H_5)_3N$ ($3^{\circ}$ amine) does not react with $p$-toluenesulfonyl chloride.
Therefore,Statement $I$ is false because not all compounds react,and Statement $II$ is false because the products are not all soluble in $NaOH$.