Mix Examples of Amines Questions in English

(C) $1$. Aniline reacts with concentrated $H_2SO_4$ to form $p$-aminobenzenesulfonic acid (sulfanilic acid) as the major product $(X)$ due to the protection of the amino group by protonation,which prevents it from being highly activating and directs the substitution to the para position.
$2$. When $(X)$ is treated with excess $Br_2/H_2O$,the highly activating $-NH_2$ group directs the bromination to the ortho positions. Due to the high reactivity of the intermediate,the $-SO_3H$ group is displaced by a bromine atom,a process known as ipso-substitution.
$3$. Thus,the final product $(Y)$ is $2,4,6$-tribromobenzenamine.

(A) $1$. Aniline $(Ph-NH_2)$ reacts with Caro's acid $(H_2SO_5)$ to form nitrosobenzene $(Ph-N=O)$ as product $(A)$.
$2$. Nitrosobenzene then undergoes a condensation reaction with benzyl cyanide $(Ph-CH_2CN)$ in the presence of a base $(EtO^-)$ and heat.
$3$. The base abstracts a proton from the $\alpha$-carbon of benzyl cyanide to form a carbanion $(Ph-CH^-(CN))$.
$4$. This carbanion attacks the electrophilic nitrogen of the nitroso group in $Ph-N=O$.
$5$. Subsequent steps involving proton transfer and elimination of water lead to the formation of the final product,$Ph-N=C(CN)-Ph$.

(C) Aspirin is acetylsalicylic acid,which contains a carboxylic acid group $(-COOH)$.
In the basic environment of the intestine,the carboxylic acid group undergoes an acid-base reaction with the base to form a carboxylate ion $(-COO^-)$.
The structure that exists in the intestine is the one where the carboxylic acid group is deprotonated to form the carboxylate anion,while the ester group remains intact.
Therefore,the correct structure is the one with the $-COO^-$ group.

(B) The conversion of $p$-chloronitrobenzene to $p$-chlorophenol involves the following steps:
$1$. Reduction of the nitro group $(-NO_2)$ to an amino group $(-NH_2)$ using $Fe/HCl$ to form $p$-chloroaniline.
$2$. Diazotization of $p$-chloroaniline using $NaNO_2$ and $H_2SO_4$ (or $HCl$) at $0-5^{\circ}C$ to form $p$-chlorobenzenediazonium salt.
$3$. Hydrolysis of the diazonium salt by heating with water $(H_2O, \Delta)$ to yield $p$-chlorophenol.
Therefore, the correct sequence is $1. Fe, HCl; 2. NaNO_2, H_2SO_4; 3. H_2O, \Delta$. Option $B$ matches this sequence (noting that the $NaOH$ step in the options is redundant or implied for neutralization if needed, but the core reagents for the transformation are $Fe/HCl$, $NaNO_2/H_2SO_4$, and $H_2O/\Delta$).

(D) The reaction of $o-$phenylenediamine with $NaNO_2$ and $HCl$ leads to the formation of a diazonium salt intermediate at one of the amino groups.
The lone pair on the adjacent amino group then attacks the diazonium nitrogen atom,resulting in the formation of a new $N-N$ bond within the same molecule.
Since this cyclization occurs within the same molecule,it is classified as an intramolecular $N-N$ coupling reaction,yielding benzotriazole.

(A) The reaction involves the nucleophilic substitution of aniline with $1$-bromo-$4$-chlorobutane.
First,the lone pair on the nitrogen atom of aniline attacks the carbon atom attached to the bromine atom,displacing the bromide ion to form $N$-($4$-chlorobutyl)aniline.
Subsequently,an intramolecular Friedel-Crafts alkylation occurs. The nitrogen atom directs the alkyl chain to the ortho position of the benzene ring.
The lone pair on the nitrogen facilitates the formation of a cyclic intermediate,which then undergoes cyclization to form $1,2,3,4$-tetrahydroquinoline.
Thus,the correct product $(A)$ is $1,2,3,4$-tetrahydroquinoline.

(B) The reaction involves the cleavage of the disulfide bond $(S-S)$ in bis($2$-nitrophenyl) disulfide by chlorine $(Cl_2)$.
The chlorine molecule acts as an electrophile,attacking the sulfur atom.
This leads to the formation of two molecules of $2-$nitrobenzenesulfenyl chloride $(o-NO_2C_6H_4SCl)$ as the product $(A)$.

(D) $1$. The reaction of aniline $(Ph-NH_2)$ with $2 \ mole$ of $CH_3Cl$ leads to the formation of $N,N$-dimethylaniline as product $(A)$.
$Ph-NH_2 + 2CH_3Cl \rightarrow Ph-N(CH_3)_2 + 2HCl$
$2$. $N,N$-dimethylaniline $(A)$ then undergoes an electrophilic aromatic substitution reaction with benzenediazonium chloride $(Ph-N_2^+ Cl^-)$.
$3$. The $-N(CH_3)_2$ group is a strong activating group and is ortho/para directing. Due to steric hindrance,the para-substitution is the major product.
$4$. The resulting product $(B)$ is $p$-dimethylaminoazobenzene,which is commonly known as Butter Yellow.
Structure of $(B)$: $Ph-N=N-C_6H_4-N(CH_3)_2$ (para-substituted).

(A) $p$-Toluidine reacts with benzene diazonium chloride to form a diazoamino compound $(p-CH_3-C_6H_4-NH-N=N-Ph)$.
This compound exists in tautomeric equilibrium with the aminoazo form $(p-CH_3-C_6H_4-N=N-NH-Ph)$.
On boiling with $aq. H_2SO_4$ (acidic hydrolysis),both tautomers undergo cleavage.
The diazoamino form cleaves to give $p$-toluidine and benzene diazonium salt (which further decomposes to phenol and nitrogen,but the question asks for the products of the cleavage of the formed compound).
Looking at the provided reaction scheme,the hydrolysis of the diazoamino form yields $p$-toluidine and toluene (via reduction/decomposition pathways in specific conditions),while the aminoazo form yields $p$-toluidine and $p$-cresol.
Counting the distinct organic products formed across the pathways shown in the reaction scheme: $p$-toluidine,toluene,and $p$-cresol are the distinct organic products.
Thus,the total number of products is $3$.

(C) Step $1$: Reaction of $1-chloro-4-nitrobenzene$ with $aq. NH_3$ undergoes nucleophilic aromatic substitution $(S_NAr)$ to form $4-nitroaniline$ as product $(A)$.
Step $2$: $4-nitroaniline$ reacts with $Br_2/H_2O$ to form $2,6-dibromo-4-nitroaniline$ as product $(B)$ due to the strong activating effect of the $-NH_2$ group.
Step $3$: The reaction of $(B)$ with $(i) NaNO_2 + HCl$ followed by $(ii) H_3PO_2$ involves diazotization of the $-NH_2$ group followed by deamination (reduction of the diazonium salt to a hydrogen atom).
Thus,the final product $(C)$ is $1,3-dibromo-5-nitrobenzene$.

(B) The reaction proceeds in two steps:
$1$. The first step is a nucleophilic aromatic substitution $(S_NAr)$ reaction where $1-chloro-4-nitrobenzene$ reacts with dimethylamine $(CH_3-NH-CH_3)$. The dimethylamino group replaces the chlorine atom at the para position,forming $N,N-dimethyl-4-nitroaniline$ as product $(A)$.
$2$. The second step is the reduction of the nitro group $(-NO_2)$ to an amino group $(-NH_2)$ using $Fe/HCl$. This converts $N,N-dimethyl-4-nitroaniline$ into $N,N-dimethyl-p-phenylenediamine$ (also known as $4-(dimethylamino)aniline$) as product $(B)$.

(C) The reaction sequence is as follows:
$1$. Naphthalene undergoes catalytic oxidation with $Air/V_2O_5$ at high temperature to form phthalic anhydride.
$2$. Phthalic anhydride reacts with $NH_3$ followed by heating to form phthalimide.
$3$. Phthalimide reacts with $KOBr$ (Hofmann bromamide degradation) to form $2$-aminobenzoic acid (anthranilic acid).
Thus,the final product $A$ is $2$-aminobenzoic acid.

(C) The reaction sequence is as follows:
$1$. $p$-Nitrotoluene reacts with $Cl_2/AlCl_3$ to give $2$-chloro-$4$-nitrotoluene $(A)$.
$2$. Reduction of $A$ with $Fe/HCl$ followed by $NaOH$ gives $2$-chloro-$4$-aminotoluene $(B)$.
$3$. Acetylation of $B$ with acetic anhydride (implied) gives $2$-chloro-$4$-acetamidotoluene $(C)$.
$4$. Bromination of $C$ with $Br_2/FeBr_3$ gives $2$-chloro-$4$-acetamido-$5$-bromotoluene $(D)$.
$5$. Hydrolysis of $D$ with $NaOH$ gives $2$-chloro-$4$-amino-$5$-bromotoluene $(E)$.
$6$. Diazotization of $E$ with $NaNO_2/H_2SO_4$ at $0-5^{\circ}C$ gives the diazonium salt $(F)$.
$7$. Reduction of $F$ with $H_3PO_2/H_2O$ removes the amino group to give $2$-chloro-$5$-bromotoluene $(G)$.

(A) $1$. The starting material is $1$-methyl-$4$-nitrobenzene. Bromination in the presence of $Fe$ gives $2$-bromo-$1$-methyl-$4$-nitrobenzene $(A)$.
$2$. Reduction of the nitro group with $Sn/HCl$ converts $(A)$ to $2$-bromo-$4$-methylaniline $(B)$.
$3$. Treatment with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ converts the amine group to a diazonium salt $(C)$.
$4$. Reduction of the diazonium salt with $H_3PO_2$ removes the $-N_2^+$ group,yielding $2$-bromotoluene $(D)$.
$5$. Oxidation of the methyl group with $KMnO_4/HO^-$ followed by acidification yields $2$-bromobenzoic acid $(E)$.

(D) $1$. The reaction of $1-fluoro-2, 4-dinitrobenzene$ with hydrazine $(NH_2-NH_2)$ proceeds via nucleophilic aromatic substitution to form $2, 4-dinitrophenylhydrazine$ $(A)$. Thus,statement $A$ is correct.
$2$. The reaction of $2, 4-dinitrophenylhydrazine$ $(A)$ with $butanone$ in the presence of $H^+$ and heat is a condensation reaction (dehydration) that forms a hydrazone $(B)$. Thus,statement $B$ is correct.
$3$. The product $B$ is a hydrazone,not an oxime (oximes are formed from hydroxylamine,$NH_2OH$). Thus,statement $D$ is incorrect.
$4$. The hydrazone formed $(B)$ contains a $C=N$ bond with different groups on the nitrogen and carbon atoms,allowing for geometrical isomerism ($syn$ and $anti$ forms). Thus,statement $C$ is correct.

(B) $1$. Reaction of $C_2H_5-Cl$ with $NaCN$ gives $C_2H_5CN$ $(C)$.
$2$. Partial hydrolysis of $C_2H_5CN$ with $H_2O_2/OH^-$ gives $C_2H_5CONH_2$ $(D)$.
$3$. Hofmann bromamide degradation of $C_2H_5CONH_2$ $(D)$ with $KOH/Br_2$ gives $C_2H_5NH_2$ $(E)$.
$4$. Reaction of $C_2H_5-Cl$ with $AgCN$ gives $C_2H_5NC$ $(A)$.
$5$. Reduction of $C_2H_5NC$ $(A)$ with $LiAlH_4$ gives $C_2H_5NHCH_3$ $(B)$.
$6$. $E$ is $C_2H_5NH_2$ (primary amine) and $B$ is $C_2H_5NHCH_3$ (secondary amine).
$7$. Both $E$ $(C_2H_7N)$ and $B$ $(C_3H_9N)$ are amines,but they differ by a $-CH_2-$ group,making them homologues.

(A) The Hinsberg reagent is benzene sulfonyl chloride $(C_6H_5SO_2Cl)$.
It reacts with $1^o$ amines to form a sulfonamide which is soluble in alkali.
It reacts with $2^o$ amines to form a sulfonamide which is insoluble in alkali.
It does not react with $3^o$ amines.
Thus,it is used to distinguish and separate $1^o, 2^o$ and $3^o$ amines.
Regarding volatility,$1^o$ amines have two $H$-atoms,$2^o$ have one,and $3^o$ have none,leading to intermolecular hydrogen bonding strength $1^o > 2^o > 3^o$,which makes $3^o$ amines more volatile. Thus,the order of volatility is $3^o > 2^o > 1^o$.

(D) The starting material is $3$-methylbutan-$2$-amine,which is a primary aliphatic amine.
When a primary aliphatic amine reacts with nitrous acid $(HNO_2)$,it forms a highly unstable diazonium salt.
This diazonium salt rapidly loses nitrogen gas $(N_2)$ to form a carbocation.
In this case,the initial carbocation formed is a secondary carbocation: $(CH_3)_2CH-CH^+(CH_3)$.
This secondary carbocation can undergo a $1,2$-hydride shift to form a more stable tertiary carbocation: $(CH_3)_2C^+-CH_2CH_3$.
Both the secondary and tertiary carbocations can be attacked by water $(H_2O)$ to form alcohols.
The secondary carbocation leads to $3$-methylbutan-$2$-ol,and the tertiary carbocation leads to $2$-methylbutan-$2$-ol.
Therefore,a mixture of products is formed,including $3$-methylbutan-$2$-ol and $2$-methylbutan-$2$-ol.
Since the question asks for the product $(A)$ and lists multiple possibilities,the correct answer is that all of these (or a mixture) can be formed.

(C) $1$. Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(A)$.
$2$. Benzenediazonium chloride $(A)$ on heating with $H_3O^+$ undergoes hydrolysis to form phenol $(B)$.
$3$. Phenol $(B)$ reacts with benzenediazonium chloride $(A)$ in a weakly alkaline medium to undergo an electrophilic aromatic substitution reaction (coupling reaction) to form $p$-hydroxyazobenzene as the final product.

(D) Primary aliphatic amines react with nitrous acid $(NaNO_2/HCl)$ to form unstable diazonium salts,which immediately decompose to form alcohols with the evolution of $N_2$ gas.
All the given compounds are primary amines:
$1$. $(CH_3)_3C-NH_2$ is a primary amine (tert-butylamine).
$2$. $(CH_3)_2CH-NH_2$ is a primary amine (isopropylamine).
$3$. $CH_3-CH_2-NH_2$ is a primary amine (ethylamine).
Since all of them are primary aliphatic amines,they all react with $NaNO_2/HCl$ to form their respective alcohols.

(D) $1$. The reaction of aniline $(C_6H_5NH_2)$ with $NaNO_2/HCl$ at $0^\circ-5^\circ C$ is a diazotization reaction,which produces benzenediazonium chloride $(A = C_6H_5N_2^+Cl^-)$.
$2$. The subsequent reaction of benzenediazonium chloride with $N,N$-dimethylaniline $(Ph-N(CH_3)_2)$ is an electrophilic aromatic substitution reaction known as a coupling reaction.
$3$. The diazonium ion acts as an electrophile and attacks the para-position of the $N,N$-dimethylaniline ring to form an azo dye.
$4$. The product $B$ is $p$-(dimethylamino)azobenzene,which is also known as $N,N$-dimethyl$-4-$phenylazoaniline.

(D) In the given reaction sequence:
$1.$ $CH_3CH_2CONH_2 \xrightarrow{Br_2 / KOH} CH_3CH_2NH_2 (X)$. This is the Hofmann bromamide degradation reaction,where an amide is converted to a primary amine with one less carbon atom.
$2.$ $CH_3CH_2NH_2 (X) \xrightarrow{C_6H_5COCl} C_6H_5CONHCH_2CH_3 (Y)$. The reaction of an amine with an acid chloride (like benzoyl chloride) in the presence of a base is known as the Schotten-Baumann reaction.
$3.$ $X$ (Ethylamine) is an aliphatic amine,while $Y$ ($N$-ethylbenzamide) is an amide. Amines are much more basic than amides because the lone pair of electrons on the nitrogen atom in an amide is delocalized through resonance with the carbonyl group $(C=O)$.
Therefore,all the given statements are correct.

(NONE) $1$. Aniline $(C_6H_5NH_2)$ is less soluble in water than methylamine $(CH_3NH_2)$ because the large hydrophobic phenyl group in aniline hinders hydrogen bonding with water. Thus,statement $A$ is incorrect.
$2$. Ethanol $(C_2H_5OH)$ forms stronger hydrogen bonds with water than ethylamine $(C_2H_5NH_2)$ because oxygen is more electronegative than nitrogen. Thus,statement $B$ is incorrect.
$3$. In chlorobenzene,the $C-Cl$ bond is attached to an $sp^2$ hybridized carbon,which is more electronegative,and the lone pairs on $Cl$ are in resonance with the benzene ring,reducing the dipole moment. In cyclohexyl chloride,the $C-Cl$ bond is attached to an $sp^3$ hybridized carbon,which is less electronegative,resulting in a higher dipole moment. Thus,statement $C$ is incorrect.
$4$. Since all statements $A$,$B$,and $C$ are incorrect,none of the options provided are correct. However,based on standard chemistry curriculum,if this is a multiple-choice question where one must be selected,there might be an error in the question framing. Given the options,none are factually correct.

(A) The reaction sequence is as follows:
$CH_3-CN$ $\xrightarrow{\text{Reduction}} CH_3-CH_2-NH_2$ $\xrightarrow{HNO_2} CH_3-CH_2-OH$
Here,$A$ is $CH_3CN$ (acetonitrile),which upon reduction gives $B$ ($CH_3CH_2NH_2$,ethylamine). Ethylamine reacts with nitrous acid $(HNO_2)$ to form $C_2H_5OH$ (ethanol).

(D) Amines exhibit various types of structural isomerism:
$1$. Positional isomerism: Due to the different positions of the amino group $(-NH_2)$ on the carbon chain.
$2$. Functional isomerism: Primary,secondary,and tertiary amines with the same molecular formula are functional isomers of each other.
$3$. Metamerism: Due to the different distribution of alkyl groups attached to the nitrogen atom in secondary and tertiary amines.
Therefore,amines exhibit all the above types of isomerism.

(D) . Formyl chloride $(HCOCl)$ is unstable at room temperature and decomposes into $CO$ and $HCl$.
$B$. Acetamide $(CH_3CONH_2)$ is amphoteric; it can be protonated by strong acids to form $CH_3CONH_3^+Cl^-$ (base) and deprotonated by strong bases (acid).
$C$. Reduction of acetamide with $LiAlH_4$ yields ethylamine $(CH_3CH_2NH_2)$.
Since all statements $A$,$B$,and $C$ are correct,the correct answer is $D$.

(A) $CH_3COOH \xrightarrow[\Delta]{NH_3} CH_3CONH_2 (A)$ (Ethanamide)
$CH_3CONH_2 \xrightarrow{Br_2/NaOH} CH_3NH_2 (B)$ (Methanamine) via Hofmann bromamide degradation.
$CH_3NH_2 \xrightarrow{NaNO_2/dil. HCl} CH_3OH (C)$ (Methanol) via diazotization followed by hydrolysis.

(D) All aliphatic primary amines react with nitrous acid $(NaNO_2 + HCl)$ to form unstable alkyl diazonium salts,which decompose to liberate $N_2$ gas.
$RNH_2 + HONO \longrightarrow R-OH + N_2 + H_2O$
Since all the given options ($C_2H_5NH_2$,$CH_3NH_2$,and $(CH_3)_2CHNH_2$) are aliphatic primary amines,they all will evolve $N_2$ gas.

(A) $(i), (ii)$ The structures and their $IUPAC$ names of different isomeric amines corresponding to the molecular formula,$C_{4}H_{11}N$ are:
$(a)$ $CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$: $\text{Butan-1-amine } (1^{\circ})$
$(b)$ $CH_{3}CH_{2}CH(NH_{2})CH_{3}$: $\text{Butan-2-amine } (1^{\circ})$
$(c)$ $(CH_{3})_{2}CHCH_{2}NH_{2}$: $\text{2-Methylpropan-1-amine } (1^{\circ})$
$(d)$ $(CH_{3})_{3}CNH_{2}$: $\text{2-Methylpropan-2-amine } (1^{\circ})$
$(e)$ $CH_{3}CH_{2}CH_{2}NHCH_{3}$: $\text{N-Methylpropan-1-amine } (2^{\circ})$
$(f)$ $CH_{3}CH_{2}NHCH_{2}CH_{3}$: $\text{N-Ethylethanamine } (2^{\circ})$
$(g)$ $CH_{3}CH(CH_{3})NHCH_{3}$: $\text{N-Methylpropan-2-amine } (2^{\circ})$
$(h)$ $CH_{3}CH_{2}N(CH_{3})_{2}$: $\text{N,N-Dimethylethanamine } (3^{\circ})$
$(iii)$ The pairs $(a)$ and $(b)$,and $(e)$ and $(g)$ exhibit position isomerism.
The pairs $(a)$ and $(c)$,$(a)$ and $(d)$,$(b)$ and $(c)$,and $(b)$ and $(d)$ exhibit chain isomerism.
The pairs $(e)$ and $(f)$,and $(f)$ and $(g)$ exhibit metamerism.
Primary,secondary,and tertiary amines exhibit functional isomerism with each other.

(N/A) $(i)$ Benzene $\xrightarrow{Conc. HNO_3 / Conc. H_2SO_4}$ Nitrobenzene $\xrightarrow{H_2/Pd, Ethanol}$ Aniline.
$(ii)$ Benzene $\xrightarrow{Conc. HNO_3 / Conc. H_2SO_4}$ Nitrobenzene $\xrightarrow{H_2/Pd, Ethanol}$ Aniline $\xrightarrow{2CH_3Cl}$ $N,N$-Dimethylaniline.
$(iii)$ $Cl-(CH_2)_4-Cl$ $\xrightarrow{Ethanolic NaCN}$ $NC-(CH_2)_4-CN$ $\xrightarrow{H_2/Ni}$ $H_2N-(CH_2)_6-NH_2$ (Hexane-$1,6$-diamine).

(N/A) $(i)$ Conversion of $3-$Methylaniline to $3-$nitrotoluene:
$1.$ $3-$Methylaniline reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form $3-$methylbenzenediazonium chloride.
$2.$ This diazonium salt reacts with $HBF_4$ to form $3-$methylbenzenediazonium tetrafluoroborate.
$3.$ Finally,treatment with $NaNO_2$ in the presence of $Cu$ and heat (Schiemann-like reaction variant) yields $3-$nitrotoluene.
$(ii)$ Conversion of Aniline to $1,3,5-$tribromobenzene:
$1.$ Aniline reacts with $Br_2/H_2O$ to form $2,4,6-$tribromoaniline.
$2.$ $2,4,6-$tribromoaniline is treated with $NaNO_2/HCl$ to form $2,4,6-$tribromobenzenediazonium chloride.
$3.$ Reduction of this diazonium salt with $H_3PO_2$ and $H_2O$ removes the diazonium group,yielding $1,3,5-$tribromobenzene.

(A) $(i)$ $1-$Methylethanamine ($1^{\circ}$ amine)
$(ii)$ Propan$-1-$amine ($1^{\circ}$ amine)
$(iii)$ $N-$Methylpropan$-2-$amine ($2^{\circ}$ amine)
$(iv)$ $2-$Methylpropan$-2-$amine ($1^{\circ}$ amine)
$(v)$ $N-$Methylbenzenamine or $N-$methylaniline ($2^{\circ}$ amine)
$(vi)$ $N-$Ethyl$-N-$methylethanamine ($3^{\circ}$ amine)
$(vii)$ $3-$Bromobenzenamine or $3-$bromoaniline ($1^{\circ}$ amine)

(N/A) $(i)$ The basic strength order is $C_{6}H_{5}NH_{2} < C_{6}H_{5}NHCH_{3} < C_{2}H_{5}NH_{2} < (C_{2}H_{5})_{2}NH$. Since $pK_{b} \propto 1/\text{basic strength}$,the decreasing order of $pK_{b}$ is: $C_{6}H_{5}NH_{2} > C_{6}H_{5}NHCH_{3} > C_{2}H_{5}NH_{2} > (C_{2}H_{5})_{2}NH$.
$(ii)$ Basic strength order: $C_{6}H_{5}NH_{2} < C_{6}H_{5}N(CH_{3})_{2} < CH_{3}NH_{2} < (C_{2}H_{5})_{2}NH$.
$(iii)$ $(a)$ $p-$Nitroaniline $ < $ Aniline $ < $ $p-$Toluidine. $(b)$ $C_{6}H_{5}NH_{2} < C_{6}H_{5}NHCH_{3} < C_{6}H_{5}CH_{2}NH_{2}$.
$(iv)$ In gas phase,basicity depends on $+I$ effect: $(C_{2}H_{5})_{3}N > (C_{2}H_{5})_{2}NH > C_{2}H_{5}NH_{2} > NH_{3}$.
$(v)$ Boiling point depends on $H$-bonding: $(CH_{3})_{2}NH < C_{2}H_{5}NH_{2} < C_{2}H_{5}OH$.
$(vi)$ Solubility depends on $H$-bonding and hydrophobic part: $C_{6}H_{5}NH_{2} < (C_{2}H_{5})_{2}NH < C_{2}H_{5}NH_{2}$.

(N/A) $(i)$ $\text{CH}_3\text{COOH}$ $\xrightarrow{\text{SOCl}_2} \text{CH}_3\text{COCl}$ $\xrightarrow{\text{NH}_3} \text{CH}_3\text{CONH}_2$ $\xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2$
$(ii)$ $\text{C}_5\text{H}_{11}\text{CN}$ $\xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{C}_5\text{H}_{11}\text{COOH}$ $\xrightarrow{\text{SOCl}_2} \text{C}_5\text{H}_{11}\text{COCl}$ $\xrightarrow{\text{NH}_3} \text{C}_5\text{H}_{11}\text{CONH}_2$ $\xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_5\text{H}_{11}\text{NH}_2$
$(iii)$ $\text{CH}_3\text{OH}$ $\xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl}$ $\xrightarrow{\text{KCN}} \text{CH}_3\text{CN}$ $\xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{COOH}$
$(iv)$ $\text{CH}_3\text{CH}_2\text{NH}_2$ $\xrightarrow{\text{NaNO}_2/\text{HCl}} \text{CH}_3\text{CH}_2\text{OH}$ $\xrightarrow{\text{KMnO}_4} \text{CH}_3\text{COOH}$ $\xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2$ $\xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2$
$(v)$ $\text{CH}_3\text{COOH}$ $\xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH}$ $\xrightarrow{\text{PCl}_5} \text{CH}_3\text{CH}_2\text{Cl}$ $\xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CN}$ $\xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{COOH}$
$(vi)$ $\text{CH}_3\text{NH}_2$ $\xrightarrow{\text{NaNO}_2/\text{HCl}} \text{CH}_3\text{OH}$ $\xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl}$ $\xrightarrow{\text{KCN}} \text{CH}_3\text{CN}$ $\xrightarrow{\text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{NH}_2$
$(vii)$ $\text{CH}_3\text{NO}_2$ $\xrightarrow{\text{Sn/HCl}} \text{CH}_3\text{NH}_2$ $\xrightarrow{\text{CHCl}_3/\text{KOH}} \text{CH}_3\text{NC}$ $\xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{NHCH}_3$
$(viii)$ $\text{CH}_3\text{CH}_2\text{COOH}$ $\xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CH}_2\text{CONH}_2$ $\xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{CH}_2\text{NH}_2$ $\xrightarrow{\text{NaNO}_2/\text{HCl}} \text{CH}_3\text{CH}_2\text{OH}$ $\xrightarrow{\text{KMnO}_4} \text{CH}_3\text{COOH}$

(N/A) $(i)$ Nitrobenzene $\xrightarrow{Sn/HCl}$ Aniline $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C}$ Benzenediazonium chloride $\xrightarrow{CuCN}$ Benzonitrile $\xrightarrow{H_3O^+}$ Benzoic acid.
$(ii)$ Benzene $\xrightarrow{HNO_3/H_2SO_4}$ Nitrobenzene $\xrightarrow{Br_2/FeBr_3}$ $m-$Bromonitrobenzene $\xrightarrow{Sn/HCl}$ $m-$Bromoaniline $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C}$ $m-$Bromobenzenediazonium chloride $\xrightarrow{H_2O, \Delta}$ $m-$Bromophenol.
$(iii)$ Benzoic acid $\xrightarrow{NH_3, \Delta}$ Benzamide $\xrightarrow{Br_2/KOH}$ Aniline.
$(iv)$ Aniline $\xrightarrow{Br_2/H_2O}$ $2,4,6-$Tribromoaniline $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C}$ $2,4,6-$Tribromobenzenediazonium chloride $\xrightarrow{HBF_4, \Delta}$ $2,4,6-$Tribromofluorobenzene.
$(v)$ Benzyl chloride $\xrightarrow{KCN}$ Phenylacetonitrile $\xrightarrow{LiAlH_4}$ $2-$Phenylethanamine.
$(vi)$ Chlorobenzene $\xrightarrow{HNO_3/H_2SO_4}$ $p-$Nitrochlorobenzene $\xrightarrow{Sn/HCl}$ $p-$Chloroaniline.
$(vii)$ Aniline $\xrightarrow{(CH_3CO)_2O/Pyridine}$ Acetanilide $\xrightarrow{Br_2/CH_3COOH}$ $p-$Bromoacetanilide $\xrightarrow{H_3O^+}$ $p-$Bromoaniline.
$(viii)$ Benzamide $\xrightarrow{LiAlH_4}$ Benzylamine $\xrightarrow{NaNO_2/HCl}$ Benzyl alcohol $\xrightarrow{PCl_5}$ Benzyl chloride $\xrightarrow{Mg/ether, then H_2O}$ Toluene.
$(ix)$ Aniline $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C}$ Benzenediazonium chloride $\xrightarrow{CuCN}$ Benzonitrile $\xrightarrow{LiAlH_4}$ Benzylamine $\xrightarrow{HNO_2}$ Benzyl alcohol.

(N/A) $(i)$ $CH_3CH_2I$ $\xrightarrow{NaCN} \underset{(A)}{CH_3CH_2CN}$ $\xrightarrow[Partial\,hydrolysis]{OH^{-}} \underset{(B)}{CH_3CONH_2}$ $\xrightarrow{NaOH+Br_2} \underset{(C)}{CH_3NH_2}$
$(ii)$ $C_6H_5N_2Cl$ $\xrightarrow{CuCN} \underset{(A)}{C_6H_5CN}$ $\xrightarrow{H_2O/H^{+}} \underset{(B)}{C_6H_5COOH}$ $\xrightarrow[\Delta]{NH_3} \underset{(C)}{C_6H_5CONH_2}$
$(iii)$ $CH_3CH_2Br$ $\xrightarrow{KCN} \underset{(A)}{CH_3CH_2CN}$ $\xrightarrow{LiAlH_4} \underset{(B)}{CH_3CH_2CH_2NH_2}$ $\xrightarrow[0\ ^oC]{HNO_2} \underset{(C)}{CH_3CH_2CH_2OH}$
$(iv)$ $C_6H_5NO_2$ $\xrightarrow{Fe/HCl} \underset{(A)}{C_6H_5NH_2}$ $\xrightarrow[273\ K]{NaNO_2+HCl} \underset{(B)}{C_6H_5N_2^+Cl^-}$ $\xrightarrow[\Delta]{H_2O/H^{+}} \underset{(C)}{C_6H_5OH}$
$(v)$ $CH_3COOH$ $\xrightarrow[\Delta]{NH_3} \underset{(A)}{CH_3CONH_2}$ $\xrightarrow{NaOBr} \underset{(B)}{CH_3NH_2}$ $\xrightarrow{NaNO_2/HCl} \underset{(C)}{CH_3OH}$
$(vi)$ $C_6H_5NO_2$ $\xrightarrow{Fe/HCl} \underset{(A)}{C_6H_5NH_2}$ $\xrightarrow[273\ K]{HNO_2} \underset{(B)}{C_6H_5N_2^+Cl^-}$ $\xrightarrow{C_6H_5OH} \underset{(C)}{p-C_6H_5-N=N-C_6H_4OH}$

(N/A) The reaction of compound $B$ with $Br_2$ and $KOH$ to form compound $C$ $(C_6H_7N)$ is a Hoffmann bromamide degradation reaction. This indicates that $C$ is an amine and $B$ is an amide. The amine with molecular formula $C_6H_7N$ is aniline $(C_6H_5NH_2)$,which has the $IUPAC$ name benzenamine.
Since $C$ is aniline,the corresponding amide $B$ must be benzamide $(C_6H_5CONH_2)$.
Benzamide is formed by heating benzoic acid $(A)$ with aqueous ammonia. Thus,compound $A$ is benzoic acid $(C_6H_5COOH)$.
The structures and names are:
$A$: Benzoic acid $(C_6H_5COOH)$
$B$: Benzamide $(C_6H_5CONH_2)$
$C$: Benzenamine or Aniline $(C_6H_5NH_2)$
The reaction sequence is:
$C_6H_5COOH$ $\xrightarrow{NH_3, \Delta} C_6H_5CONH_2$ $\xrightarrow{Br_2, KOH} C_6H_5NH_2$

(N/A) $(i)$ $\underset{\text{Aniline}}{C_{6}H_{5}NH_{2}} + CHCl_{3} + 3 \text{ alc. } KOH$ $\xrightarrow{\text{Carbylamine reaction}} 3H_{2}O + 3KCl + \underset{\text{Phenyl isocyanide}}{C_{6}H_{5}NC}$
$(ii)$ $\underset{\text{Benzenediazonium chloride}}{C_{6}H_{5}N_{2}Cl} + H_{3}PO_{2} + H_{2}O$ $\rightarrow \underset{\text{Benzene}}{C_{6}H_{6}} + N_{2} + H_{3}PO_{3} + HCl$
$(iii)$ $\underset{\text{Aniline}}{C_{6}H_{5}NH_{2}} + \text{conc. } H_{2}SO_{4}$ $\rightarrow \underset{\text{Anilinium hydrogen sulphate}}{C_{6}H_{5}NH_{3}^{+}HSO_{4}^{-}}$
$(iv)$ $\underset{\text{Benzenediazonium chloride}}{C_{6}H_{5}N_{2}Cl} + \underset{\text{Ethanol}}{C_{2}H_{5}OH}$ $\rightarrow \underset{\text{Benzene}}{C_{6}H_{6}} + \underset{\text{Ethanal}}{CH_{3}CHO} + N_{2} + HCl$
$(v)$ $C_{6}H_{5}NH_{2} + 3Br_{2(aq)} \rightarrow \text{2,4,6-Tribromoaniline} + 3HBr$
$(vi)$ $\underset{\text{Aniline}}{C_{6}H_{5}NH_{2}} + \underset{\text{acetic anhydride}}{(CH_{3}CO)_{2}O}$ $\rightarrow \underset{\text{N-Phenylethanamide}}{C_{6}H_{5}NHCOCH_{3}} + \underset{\text{acetic acid}}{CH_{3}COOH}$
$(vii)$ $\underset{\text{Benzenediazonium chloride}}{C_{6}H_{5}N_{2}Cl}$ $\xrightarrow[(ii) NaNO_{2}/Cu, \Delta ]{(i) HBF_{4}} \underset{\text{Nitrobenzene}}{C_{6}H_{5}NO_{2}} + N_{2} + NaBF_{4}$

(N/A) $(i)$ Amines undergo deprotonation to give amide ions (amide anions).
$R-NH_2 \to R-NH^- + H^+$
Similarly,alcohol loses a proton to give an alkoxide ion.
$R-OH \to R-O^- + H^+$
In an amide ion,the negative charge is on the $N$ atom,whereas in an alkoxide ion,the negative charge is on the $O$ atom. Since $O$ is more electronegative than $N$,$O$ can accommodate the negative charge more easily than $N$. As a result,the amide ion is less stable than the alkoxide ion. Hence,amines are less acidic than alcohols of comparable molecular masses.
$(ii)$ In a molecule of a tertiary amine,there are no $H$ atoms attached to the nitrogen atom,whereas in primary amines,two hydrogen atoms are present. Due to the presence of $N-H$ bonds,primary amines undergo extensive intermolecular hydrogen bonding.
As a result,extra energy is required to separate the molecules of primary amines. Hence,primary amines have higher boiling points than tertiary amines.
$(iii)$ Due to the $-R$ (resonance) effect of the benzene ring,the lone pair electrons on the $N$ atom are delocalized into the ring in aromatic amines. Therefore,the electrons on the $N$ atom in aromatic amines are less available for donation. This explains why aliphatic amines are stronger bases than aromatic amines.

(N/A) $1.$ Aniline to Picric acid:
Aniline is first converted to benzene diazonium chloride using $NaNO_2$ and $HCl$ at $273-278 \ K$. This is then hydrolyzed with water to form phenol. Phenol undergoes nitration with concentrated $HNO_3$ to yield $2,4,6-$trinitrophenol,commonly known as picric acid.
$2.$ Ethanal to Butan$-2-$ol:
Ethanal reacts with ethylmagnesium bromide (a Grignard reagent) to form an addition product. Subsequent hydrolysis of this product yields butan$-2-$ol.

(A-4, B-3, C-1, D-2) $A-4, B-3, C-1, D-2$
$A$. Ammonolysis: The process of cleavage of the $C-X$ bond by an ammonia molecule is known as ammonolysis. It involves the reaction of alkyl halides with $NH_3$.
$B$. Gabriel phthalimide synthesis: This involves the reaction of phthalimide with $KOH$ followed by reaction with $R-X$ to synthesize primary amines.
$C$. Hoffmann Bromamide reaction: This reaction is used to convert amides to amines with a lesser number of carbon atoms.
$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
$D$. Carbylamine reaction: This is a specific detection test for primary amines.
$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\text{Heat}} R-NC + 3KCl + 3H_2O$

(A) The reactions are as follows:
$1$. Ozonolysis of $'A'$ $(C_4H_8)$ gives $2 \ mols$ of acetaldehyde $(CH_3CHO)$,which indicates that $'A'$ is $CH_3-CH=CH-CH_3$ (but$-2-$ene).
$2$. $CH_3-CH=CH-CH_3 + HCl \rightarrow CH_3-CH_2-CHCl-CH_3$ (Compound $'B'$,$2$-chlorobutane).
$3$. $CH_3-CH_2-CHCl-CH_3 + NH_3 \rightarrow CH_3-CH_2-CH(NH_2)-CH_3$ (Compound $'C'$,butan-$2$-amine).
$4$. $CH_3-CH_2-CH(NH_2)-CH_3 + NaNO_2/HCl$ $\rightarrow [CH_3-CH_2-CH(N_2^+)-CH_3]$ $\xrightarrow{H_2O} CH_3-CH_2-CH(OH)-CH_3$ (Compound $'D'$,butan-$2$-ol,which is optically active).
The structures are:
$'A': CH_3-CH=CH-CH_3$
$'B': CH_3-CH_2-CHCl-CH_3$
$'C': CH_3-CH_2-CH(NH_2)-CH_3$
$'D': CH_3-CH_2-CH(OH)-CH_3$

(A) The substance $'A'$ is $C_6H_5NH_2$ (Aniline).
$1$. $C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+Cl^-$ (Anilinium chloride,$'B'$).
$2$. $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$ (Phenyl isocyanide,$'C'$).
$3$. $C_6H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5NHSO_2C_6H_5 + HCl$ ($N$-phenylbenzenesulphonamide,$'D'$).
$4$. $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$ (Benzenediazonium chloride,$'E'$).
$5$. $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4OH$ ($p$-hydroxyazobenzene,orange dye,$'F'$).

(B) The molecular formula $C_8H_{11}N$ corresponds to several isomers.
$1$. $A$ and $C$ undergo diazotization followed by hydrolysis and oxidation to form $R$ and $S$. This sequence $(Ar-NH_2$ $\rightarrow Ar-N_2^+$ $\rightarrow Ar-OH$ $\rightarrow Ar-COOH)$ indicates that $A$ and $C$ are primary aromatic amines with an alkyl group on the ring.
$2$. $A$ forms $R$ (salicylic acid derivative with intramolecular $H$-bonding),which implies $A$ is $o$-ethylaniline.
$3$. $C$ forms $S$ (p-hydroxybenzoic acid derivative with intermolecular $H$-bonding),which implies $C$ is $p$-ethylaniline.
$4$. $B$ is an isomer that does not follow this path. Based on the provided options and the reaction of $N$-methylbenzylamine with $Ph-SO_2Cl$ forming a solid sulfonamide,$B$ is identified as $N$-methylbenzylamine.
Thus,$A = o$-ethylaniline,$B = N$-methylbenzylamine,and $C = p$-ethylaniline.

(C) $1$. The reaction of aniline with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which produces benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. The reaction of benzene diazonium chloride with $KCN$ (in the presence of $CuCN$) yields benzonitrile $(C_6H_5CN)$. Thus,$(A)$ is benzonitrile.
$3$. The reduction of benzonitrile with $SnCl_2/HCl$ followed by hydrolysis $(H_3O^+)$ is the Stephen reduction,which converts the nitrile group $(-CN)$ into an aldehyde group $(-CHO)$. Thus,$(B)$ is benzaldehyde.
$4$. Therefore,$(A)$ is benzonitrile and $(B)$ is benzaldehyde.

(C) The reaction proceeds as follows:
$1$. Reaction with alcoholic $NH_3$ converts the acid chloride $(R-COCl)$ into an amide $(R-CONH_2)$: $CH_3-CH(CH_3)-CH_2-CH_2-COCl \xrightarrow{NH_3} CH_3-CH(CH_3)-CH_2-CH_2-CONH_2$.
$2$. The Hofmann bromamide degradation reaction $(NaOH, Br_2)$ converts the amide into a primary amine $(R-NH_2)$: $CH_3-CH(CH_3)-CH_2-CH_2-CONH_2 \xrightarrow{NaOH, Br_2} CH_3-CH(CH_3)-CH_2-CH_2-NH_2$.
$3$. Treatment with $NaNO_2/HCl$ at low temperature converts the primary aliphatic amine into a diazonium salt,which is unstable and decomposes in the presence of water $(H_2O)$ to form an alcohol: $CH_3-CH(CH_3)-CH_2-CH_2-NH_2$ $\xrightarrow{NaNO_2, HCl} [CH_3-CH(CH_3)-CH_2-CH_2-N_2^+Cl^-]$ $\xrightarrow{H_2O} CH_3-CH(CH_3)-CH_2-CH_2-OH$.
Thus,the major product is $CH_3-CH(CH_3)-CH_2-CH_2OH$.

(D) The transformation involves converting nitrobenzene to $3$-chlorophenol.
Step $1$: Chlorination of nitrobenzene using $Cl_{2} / FeCl_{3}$ gives $m$-chloronitrobenzene because the $-NO_{2}$ group is meta-directing.
Step $2$: Reduction of the nitro group using $Fe / HCl$ yields $m$-chloroaniline.
Step $3$: Diazotization of $m$-chloroaniline using $NaNO_{2} / HCl$ at $0^{\circ} C$ produces $m$-chlorobenzenediazonium chloride.
Step $4$: Hydrolysis of the diazonium salt using $H_{2}O / H^{+}$ results in the formation of $3$-chlorophenol.
Thus,the correct sequence is $i. Cl_{2}, FeCl_{3}; ii. Fe, HCl; iii. NaNO_{2}, HCl, 0^{\circ} C; iv. H_{2}O / H^{+}$.

(A) $1$. The compound $A$ contains $N$ and $Cl$ and forms an acidic solution in water,suggesting it is an amine hydrochloride salt $(R-NH_3^+Cl^-)$.
$2$. The molecular weight of $C_6H_5NH_3Cl$ is $77 + 14 + 3 + 35.5 = 129.5$,which matches the given $131 \pm 2$.
$3$. Treatment with $NaOH$ releases the free amine: $C_6H_5NH_3Cl + NaOH \rightarrow C_6H_5NH_2 + NaCl + H_2O$. The liquid $C_6H_5NH_2$ (aniline) contains $N$ but no $Cl$.
$4$. Aniline reacts with nitrous acid $(HNO_2)$ at $0-5^{\circ}C$ to form benzenediazonium chloride,which then couples with phenol in basic medium to form $p$-hydroxyazobenzene,an orange dye.
$5$. Thus,$A$ is $C_6H_5NH_3Cl$ (anilinium chloride).

(C) . Benzenesulphonyl chloride $\rightarrow$ $III$. Hinsberg reagent
$B$. Hoffmann bromamide reaction $\rightarrow$ $IV$. Known reaction of isocyanates
$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
Intermediate: $R-N=C=O$ (isocyanate)
$C$. Carbylamine reaction $\rightarrow$ $I$. Test for primary amine
$R-NH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O$
$D$. Hoffmann orientation $\rightarrow$ $II$. Anti Saytzeff (Formation of less substituted alkene as major product)
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(A) The given reaction is a coupling reaction of benzene diazonium chloride with a phenol or amine in a basic medium to form an azo dye.
The reaction of benzene diazonium chloride with $\beta$-naphthol in the presence of $NaOH$ (basic medium) yields an orange-red azo dye.
The chemical equation is:
$C_6H_5N_2^+Cl^- + \beta\text{-Naphthol} \xrightarrow{NaOH} \text{Orange-red dye}$