Properties of Amines Questions in English

(N/A) $(i)$ Carbylamine reaction: Used to identify primary amines. Aliphatic and aromatic primary amines react with chloroform and ethanolic $KOH$ to form isocyanides (carbylamines) with unpleasant odors.
$R-NH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$
$(ii)$ Diazotisation: Aromatic primary amines react with nitrous acid $(NaNO_2 + HCl)$ at $273-278 \ K$ to form diazonium salts.
$(iii)$ Hofmann bromamide reaction: Amides react with $Br_2$ and $NaOH$ to produce a primary amine with one carbon atom less.
$R-CONH_2 + Br_2 + 4NaOH \to R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
$(iv)$ Coupling reaction: Arenediazonium salts react with phenol or aromatic amines to form colored azo compounds via electrophilic substitution.
$(v)$ Ammonolysis: Alkyl halides react with ethanolic ammonia to form amines via nucleophilic substitution,often resulting in a mixture of $1^o, 2^o, 3^o$ amines and quaternary salts.
$(vi)$ Acetylation: Introduction of an acetyl group into primary or secondary amines using acid chlorides or anhydrides to form amides.
$(vii)$ Gabriel phthalimide synthesis: $A$ method to prepare aliphatic primary amines by treating phthalimide with $KOH$,followed by alkyl halide and alkaline hydrolysis.

$(i)$ Aromatic primary amines react with nitrous acid (prepared in situ from $NaNO_2$ and a mineral acid such as $HCl$) at $273-278 \, K$ to form stable aromatic diazonium salts.
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \, K} [C_6H_5N_2^+Cl^-] + 2H_2O$
$(ii)$ Aliphatic primary amines react with nitrous acid to form unstable aliphatic diazonium salts,which decompose to produce alcohols and $HCl$ with the evolution of $N_2$ gas.
$R-NH_2 + HNO_2$ $\xrightarrow{NaNO_2 + HCl} [R-N_2^+ Cl^-]$ $\xrightarrow{H_2O} ROH + N_2 \uparrow + HCl$

(N/A) Two biologically active compounds,namely adrenaline and ephedrine,both containing a secondary amino group,are used to increase blood pressure.
Novocain,a synthetic amino compound,is used as an anaesthetic in dentistry.
Benadryl,a well-known antihistaminic drug,also contains a tertiary amino group.

(N/A) Like ammonia,the nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore $sp^{3}$ hybridised,and the geometry of amines is pyramidal.
Each of the three $sp^{3}$ hybridised orbitals of nitrogen overlaps with the orbitals of hydrogen or carbon,depending upon the composition of the amines.
The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Due to the presence of this unshared pair of electrons,the bond angle $C-N-E$ (where $E$ is $C$ or $H$) is less than $109.5^{\circ}$; for instance,it is $108^{\circ}$ in the case of trimethylamine.

(N/A) The lower aliphatic amines are gases with a fishy odour. Primary amines with three or more carbon atoms are liquid,and still higher ones are solid.
Aniline and other arylamines are usually colourless but become coloured on storage due to atmospheric oxidation.
Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules.
However,solubility decreases with an increase in the molar mass of amines due to the increase in the size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water.
Amines are soluble in organic solvents like alcohol,ether,and benzene.
Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between the nitrogen of one molecule and the hydrogen of another molecule.
This intermolecular association is more pronounced in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in primary amines.
Tertiary amines do not have intermolecular association due to the absence of hydrogen atoms available for hydrogen bond formation. Therefore,the order of boiling points of isomeric amines is as follows: Primary > Secondary > Tertiary.
Boiling points of amines,alcohols,and alkanes of almost the same molar mass:

Compound	Molar mass $(g/mol)$ and Boiling points $(K)$
$n-C_4H_9NH_2$	$73, 350.8$
$(C_2H_5)_2NH$	$73, 329.3$
$C_2H_5N(CH_3)_2$	$73, 310.8$
$C_2H_5CH(CH_3)_2$	$72, 300.3$
$n-C_4H_9OH$	$74, 390.3$

Amines,being basic in nature,react with acids to form salts.
Amine salts on treatment with a base like $NaOH$,regenerate the parent amine.
$RNH_3^+X^- + OH^- \longrightarrow RNH_2 + H_2O + X^-$
Amine salts are soluble in water but insoluble in organic solvents like ether.
This reaction is the basis for the separation of amines from non-basic organic compounds that are insoluble in water.
Amines have an unshared pair of electrons on the nitrogen atom,due to which they behave as a Lewis base.
$RNH_2 + H_2O \rightleftharpoons RNH_3^+ + OH^-$
$K = \frac{[RNH_3^+][OH^-]}{[RNH_2][H_2O]}$
$K[H_2O] = \frac{[RNH_3^+][OH^-]}{[RNH_2]}$
$K_b = \frac{[RNH_3^+][OH^-]}{[RNH_2]}$
$pK_b = -\log K_b$
Larger the value of $K_b$ or smaller the value of $pK_b$,stronger is the base.
The $pK_b$ value of ammonia is $4.75$. Aliphatic amines are stronger bases than ammonia due to the $+I$ effect of alkyl groups,leading to high electron density on the nitrogen atom.
Their $pK_b$ values lie in the range of $3$ to $4.22$.
On the other hand,aromatic amines are weaker bases than ammonia due to the electron-withdrawing nature of the aryl group.

(N/A) The basicity of amines is related to their structure. The basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine,the more basic the amine is.
Alkanamines versus ammonia: Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity:
$R-NH_2 + H^+ \rightleftharpoons R-NH_3^+$
$NH_3 + H^+ \rightleftharpoons NH_4^+$
Due to the electron-releasing nature of the alkyl group $(R)$,it pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid.
Moreover,the substituted ammonium ion formed from the amine gets stabilized due to the dispersal of the positive charge by the $+I$ effect of the alkyl group.
Hence,alkylamines are stronger bases than ammonia.
$(b)$ Comparison of $1^{\circ}, 2^{\circ}, 3^{\circ}$ alkylamines: The basic nature of aliphatic amines should increase with an increase in the number of alkyl groups. This trend is followed in the gaseous phase.
The order of basicity of amines in the gaseous phase follows the expected order: Tertiary amine $>$ Secondary amine $>$ Primary amine $>$ Ammonia $(3^{\circ} > 2^{\circ} > 1^{\circ} > NH_3)$.

(N/A) The basicity of amines is related to their structure. The basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine,the more basic the amine is.
Alkanamines versus ammonia: Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity:
$R-NH_2 + H^+ \rightleftharpoons R-NH_3^+$
$NH_3 + H^+ \rightleftharpoons NH_4^+$
Due to the electron-releasing nature of the alkyl group $(R)$,it pushes electrons towards the nitrogen atom and thus makes the unshared electron pair more available for sharing with the proton of the acid.
Moreover,the substituted ammonium ion formed from the amine gets stabilized due to the dispersal of the positive charge by the $+I$ effect of the alkyl group.
Hence,alkylamines are stronger bases than ammonia.
$(b)$ Comparison of $1^{\circ}, 2^{\circ}, 3^{\circ}$ alkylamines: The basic nature of aliphatic amines should increase with an increase in the number of alkyl groups due to the cumulative $+I$ effect. This trend is followed in the gaseous phase.
The order of basicity of amines in the gaseous phase follows the expected order: $3^{\circ} \text{ amine} > 2^{\circ} \text{ amine} > 1^{\circ} \text{ amine} > NH_3$.

(N/A) Aromatic amines,such as aniline,are much less basic than ammonia and aliphatic amines. This is because the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring.
In aniline,the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring,as shown by the resonance structures $(I)$ to $(V)$. This delocalization makes the lone pair less available for donation to a proton $(H^{+})$.
Furthermore,when aniline accepts a proton to form an anilinium ion,the positive charge on the nitrogen atom cannot be delocalized over the benzene ring. The anilinium ion has only two resonance structures $(I)$ and $(II)$,which are less stable than the resonance-stabilized aniline molecule.
Due to these factors,aromatic amines are weaker bases than ammonia and aliphatic amines.

(N/A) Aryl amines (e.g.,aniline) are much weaker bases than ammonia and alkyl amines.
In aryl amines,the $-NH_2$ group is directly attached to the benzene ring. The unshared electron pair on the nitrogen atom is in conjugation with the benzene ring,which makes it less available for protonation.
As shown in the resonance structures,the electron density on the nitrogen atom decreases due to the delocalization of the lone pair into the benzene ring.
Furthermore,the anilinium ion formed by accepting a proton is less stable than the aniline molecule itself because the resonance stabilization is lost upon protonation.
In contrast,in alkyl amines,the electron-releasing $+I$ effect of the alkyl group increases the electron density on the nitrogen atom,making the lone pair more available for protonation and stabilizing the resulting alkyl ammonium ion.

(N/A) Acetylation is a specific type of acylation reaction where an acetyl group $(CH_3CO-)$ is introduced into an organic compound,typically an amine or alcohol.
In the context of amines,aliphatic and aromatic primary and secondary amines react with acetylating agents like acetyl chloride $(CH_3COCl)$ or acetic anhydride $((CH_3CO)_2O)$ via a nucleophilic substitution reaction to form $N$-substituted amides.
The reaction is generally carried out in the presence of a base,such as pyridine,which acts as an acid scavenger to remove the $HCl$ formed,thereby shifting the equilibrium to the right.
Example of acetylation of ethanamine:
$C_2H_5NH_2 + CH_3COCl \xrightarrow{\text{Pyridine}} C_2H_5NHCOCH_3 + HCl$
(Ethanamine + Acetyl chloride $\rightarrow$ $N$-Ethylethanamide + $HCl$)

(N/A) Acylation: Aliphatic and aromatic primary and secondary amines react with acid chlorides,anhydrides,and esters via a nucleophilic substitution reaction. This process is known as acylation,and the resulting products are called amides.
The reaction is typically performed in the presence of a base stronger than the amine,such as pyridine. The base acts to remove the $HCl$ formed during the reaction,which shifts the equilibrium to the right,favoring the formation of the amide product.
$1.$ Acylation of Ethanamine: Ethanamine reacts with acetyl chloride in the presence of pyridine to form $N$-ethylethanamide.
$C_2H_5NH_2 + CH_3COCl \xrightarrow{\text{Pyridine}} C_2H_5NHCOCH_3 + HCl$
$2.$ Acylation of Aniline (Benzenamine): Aniline reacts with ethanoic anhydride in the presence of pyridine to form $N$-phenylethanamide (acetanilide).
$C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{Pyridine}} C_6H_5NHCOCH_3 + CH_3COOH$

(N/A) Carbylamine reaction: Aliphatic and aromatic primary amines,when heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,form isocyanides or carbylamines,which are substances with a foul smell.
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:
$R-NH_2 + CHCl_3 + 3 KOH \xrightarrow{\Delta} R-NC + 3 KCl + 3 H_2O$
Example with methanamine:
$CH_3NH_2 + CHCl_3 + 3 KOH \xrightarrow{\Delta} CH_3NC + 3 KCl + 3 H_2O$
(Methanamine $\rightarrow$ Methyl isocyanide)
Example with aniline:
$C_6H_5NH_2 + CHCl_3 + 3 KOH \xrightarrow{\Delta} C_6H_5NC + 3 KCl + 3 H_2O$
(Aniline $\rightarrow$ Phenyl isocyanide)

(N/A) Three classes of amines react differently with nitrous acid $(HNO_2)$,which is prepared in situ from a mineral acid and sodium nitrite $(NaNO_2)$.
$(i)$ Reaction of primary aliphatic amines with nitrous acid: Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts,which are unstable and decompose to liberate nitrogen gas $(N_2)$ quantitatively,forming alcohols $(R-OH)$.
$(ii)$ Reaction of primary aromatic amines with nitrous acid: Aromatic amines react with nitrous acid at low temperatures $(273-278 \ K)$ to form stable diazonium salts,which are important intermediates for the synthesis of various aromatic compounds.
$(iii)$ Secondary and tertiary amines react differently with $HNO_2$. Secondary amines form $N$-nitrosoamines (yellow oily liquids),while tertiary amines form salts that are soluble in water.

(N/A) Benzenesulphonyl chloride $(C_{6}H_{5}SO_{2}Cl)$,also known as Hinsberg's reagent,reacts with primary and secondary amines to form sulphonamides.
$(i)$ The reaction of benzenesulphonyl chloride with a primary amine $(R-NH_{2})$ yields $N$-alkylbenzenesulphonamide. The hydrogen atom attached to the nitrogen in the resulting sulphonamide is strongly acidic due to the presence of the strong electron-withdrawing sulphonyl group. Hence,it is soluble in alkali.
$(ii)$ In the reaction with a secondary amine $(R_{2}NH)$,an $N,N$-dialkylbenzenesulphonamide is formed. Since this product does not contain any hydrogen atom attached to the nitrogen atom,it is not acidic and is therefore insoluble in alkali.
$(iii)$ Tertiary amines $(R_{3}N)$ do not react with benzenesulphonyl chloride. This difference in reactivity is used to distinguish between primary,secondary,and tertiary amines and to separate them from a mixture. Note: $p$-toluenesulphonyl chloride is often used in place of benzenesulphonyl chloride.

Aniline reacts with bromine water at room temperature to give a white precipitate of $2,4,6$-tribromoaniline.
Reaction: $C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2(NH_2)Br_3(s) + 3HBr$
If we need to prepare a monosubstituted aniline derivative,this can be done by protecting the $-NH_2$ group by acetylation with acetic anhydride,then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.
The lone pair of electrons on the nitrogen of acetanilide interacts with the oxygen atom due to resonance as shown below:
$CH_3-CO-NH-C_6H_5 \leftrightarrow CH_3-C(O^-)=N^+-H-C_6H_5$
Hence,the lone pair of electrons on nitrogen is less available for donation to the benzene ring by resonance. Therefore,the activating effect of the $-NHCOCH_3$ group is less than that of the amino group,which prevents polysubstitution and allows for the formation of $p$-bromoacetanilide,which upon hydrolysis yields $4$-bromoaniline.

(N/A) Nitration: Direct nitration of aniline with concentrated $HNO_3$ and $H_2SO_4$ at $288 \ K$ yields a mixture of $p$-nitroaniline $(51\%)$,$m$-nitroaniline $(47\%)$,and $o$-nitroaniline $(2\%)$. The formation of the $m$-derivative is due to the protonation of the $-NH_2$ group in the strongly acidic medium to form the anilinium ion,which is meta-directing. To obtain $p$-nitroaniline as the major product,the $-NH_2$ group is protected by acetylation with acetic anhydride to form acetanilide,which is then nitrated and subsequently hydrolyzed.
Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate. Upon heating this salt with sulphuric acid at $453-473 \ K$,it undergoes rearrangement to produce $p$-aminobenzene sulphonic acid,commonly known as sulphanilic acid,as the major product. This exists as a zwitter ion.

(N/A) $C_6H_5N_2Cl$ is a colourless crystalline solid.
It is readily soluble in water and is stable in cold conditions,but it reacts with water when warmed.
It decomposes easily in the dry state.
$C_6H_5N_2BF_4$ (benzenediazonium fluoroborate) is water-insoluble and stable at room temperature.

(N/A) The chemical reactions of diazonium salts are broadly classified into two categories: $(A)$ reactions involving the displacement of nitrogen and $(B)$ reactions involving the retention of the diazo group.
$(a)$ Reactions involving displacement of nitrogen:
The diazonium group is an excellent leaving group and is substituted by other groups such as $Cl^-$,$Br^-$,$I^-$,$CN^-$,and $OH^-$. The nitrogen gas $(N_2)$ is evolved during these reactions.
$(i)$ Sandmeyer reaction:
Nucleophiles like $Cl^-$,$Br^-$,and $CN^-$ can be introduced into the benzene ring by treating the diazonium salt solution with the corresponding cuprous halide ($Cu_2Cl_2$,$Cu_2Br_2$) or cyanide $(CuCN)$ in the presence of the respective acid ($HCl$,$HBr$) or $KCN$.
$(ii)$ Gatterman reaction:
Chlorine or bromine can be introduced into the benzene ring by treating the diazonium salt solution with the corresponding halogen acid in the presence of copper powder. This is known as the Gatterman reaction.

(N/A) Diazonium salts serve as versatile intermediates for introducing various functional groups such as $-F, -Cl, -Br, -I, -CN, -OH,$ and $-NO_2$ into an aromatic ring.
$1.$ Aryl fluorides and iodides cannot be prepared via direct halogenation of benzene.
$2.$ The cyano group $(-CN)$ cannot be introduced through direct nucleophilic substitution of chlorine in chlorobenzene,whereas cyanobenzene is easily synthesized from diazonium salts.
$3.$ Diazonium salts allow for the synthesis of substituted aromatic compounds that are otherwise difficult or impossible to prepare through direct electrophilic substitution reactions on benzene or substituted benzene rings.

(N/A) Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives.
In the strongly acidic medium used for nitration,aniline is protonated to form the anilinium ion $(-NH_3^+)$,which is meta-directing.
This leads to the formation of a significant amount of meta-nitroaniline along with ortho and para derivatives.
By protecting the $-NH_2$ group through acetylation with acetic anhydride,the reactivity of the amino group is reduced.
This prevents the formation of the anilinium ion and allows for the controlled nitration,making the $p$-nitroacetanilide the major product,which can then be hydrolyzed to $p$-nitroaniline.

(C_6H_5CH_2OH) Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable aliphatic diazonium salts,which further decompose to form alcohols with the evolution of nitrogen gas $(N_2)$.
$C_6H_5CH_2NH_2 + HNO_2$ $\rightarrow [C_6H_5CH_2N_2^+Cl^-]$ $\xrightarrow{H_2O} C_6H_5CH_2OH + N_2 + HCl$
Thus,the major product is benzyl alcohol $(C_6H_5CH_2OH)$.

(A) The reaction proceeds in two steps:
$1$. Diazotization: $2\text{-Nitro-4-methylaniline}$ reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form the diazonium salt,$2\text{-nitro-4-methylbenzenediazonium chloride}$.
$2$. Deamination: The diazonium salt is then treated with $H_3PO_2$ and $H_2O$,which reduces the diazonium group to a hydrogen atom,resulting in $3\text{-methylnitrobenzene}$ (also known as $m\text{-nitrotoluene}$) as the final product $A$.

(N/A) Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at $273-278 \ K$.
Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid.
The conversion of primary aromatic amines into diazonium salts is known as diazotisation.
Due to its inherent instability,the diazonium salt is not generally stored and is used immediately after its preparation.
Benzene diazonium chloride is a colourless crystalline solid that is readily soluble in water.
It is stable in cold conditions but reacts with water when warmed and decomposes easily in the dry state.

(N/A) The activating effect of the $-NH_2$ group is due to the lone pair of electrons on the nitrogen atom,which is available for donation to the benzene ring through resonance,thereby increasing the electron density at the ortho and para positions.
When aniline is acetylated to form acetanilide $(C_6H_5NHCOCH_3)$,the lone pair of electrons on the nitrogen atom participates in resonance with the carbonyl group $(C=O)$. This delocalization of the lone pair towards the oxygen atom reduces its availability for donation to the benzene ring.
Consequently,the electron-donating power of the nitrogen atom towards the benzene ring is decreased,which reduces the activating effect of the group. This is represented by the following resonance structures:
$[:N-C(=O)-CH_3 \longleftrightarrow N^+=C-O^- -CH_3]$

(N/A) $MeNH_2$ is a stronger base than $MeOH$ because of the lower electronegativity of the nitrogen atom compared to the oxygen atom.
Nitrogen is less electronegative than oxygen,which means the lone pair of electrons on the nitrogen atom in $MeNH_2$ is held less tightly than the lone pair on the oxygen atom in $MeOH$.
Consequently,the nitrogen atom can more easily donate its lone pair of electrons to a proton,making $MeNH_2$ a stronger base.

(N/A) In the acylation reaction of amines,such as the reaction of aniline with acetyl chloride $(CH_3COCl)$,$HCl$ is produced as a side product.
Pyridine acts as a base in this reaction. Its primary role is to neutralize the $HCl$ produced during the reaction by forming pyridinium chloride $(C_5H_5NH^+Cl^-)$.
By removing $HCl$ from the reaction mixture,pyridine prevents the protonation of the amine reactant,which would otherwise render the amine non-nucleophilic and stop the reaction. Thus,it helps in driving the reaction to completion.

(A) The coupling reaction of aryldiazonium salts with aniline is carried out under mildly acidic conditions ($pH$ $4-5$).
In strongly basic conditions,the diazonium salt is converted into diazohydroxide or diazoate,which are not electrophilic and thus do not undergo coupling.
In highly acidic conditions,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$. The anilinium ion is not nucleophilic because the lone pair on the nitrogen atom is no longer available for the electrophilic attack on the diazonium salt.
Therefore,to maintain a sufficient concentration of nucleophilic aniline while keeping the diazonium salt electrophilic,the reaction is performed at a mild $pH$ of $4-5$.

(N/A) When aniline reacts with bromine in a non-polar solvent like $CS_2$ at low temperature,the reaction is controlled,and mono-substitution occurs.
This results in the formation of a mixture of $2-$bromoaniline and $4-$bromoaniline.
The chemical structures are as follows:
$2-$bromoaniline: $A$ benzene ring with an $-NH_2$ group at position $1$ and a $-Br$ group at position $2$.
$4-$bromoaniline: $A$ benzene ring with an $-NH_2$ group at position $1$ and a $-Br$ group at position $4$.

(N/A) The reaction of $1^o$ amine with $CH_3-X$ (alkylation) is difficult to stop at the $2^o$ amine stage because the product is more nucleophilic than the reactant.
To obtain only the $2^o$ amine,we can use the following sequence:
$RNH_2$ $\xrightarrow{CHCl_3 / KOH} RNC$ $\xrightarrow{H_2 / Pd} RNHCH_3$.
In this method,the $1^o$ amine is first converted to an isocyanide $(RNC)$ via the carbylamine reaction. Subsequent catalytic reduction of the isocyanide yields a secondary amine $(RNHCH_3)$ specifically,avoiding the formation of $3^o$ amines or quaternary ammonium salts.

(N/A) Aniline is a base that reacts with the acid $HCl$ to form a salt called anilinium chloride $(C_6H_5NH_3^+Cl^-)$.
This salt is an ionic compound,which makes it highly soluble in water.
The reaction is:
$C_6H_5NH_2 + HCl_{(aq)} \rightarrow C_6H_5NH_3^+Cl^-_{(aq)}$

(N/A) $(i)$ Toluene $\to$ $p$-toluidine:
First,toluene undergoes nitration with concentrated $HNO_3$ and $H_2SO_4$ to form $p$-nitrotoluene. Then,$p$-nitrotoluene is reduced using $Fe/HCl$ to yield $p$-toluidine.
$(ii)$ $p$-Toluidine diazonium chloride $\to$ $p$-toluic acid:
First,$p$-toluidine diazonium chloride reacts with $CuCN/KCN$ (Sandmeyer reaction) to form $p$-tolunitrile ($p$-methylbenzonitrile). Subsequently,the nitrile group is hydrolyzed with $H_2O/H^+$ to produce $p$-toluic acid.

(N/A) This reaction is an example of electrophilic aromatic substitution (coupling reaction).
In an alkaline medium,phenol exists as a phenoxide ion,which is more electron-rich than phenol and thus more reactive toward electrophilic attack.
The electrophile in this reaction is the aryldiazonium cation.
The reactivity of the diazonium cation depends on the electron-withdrawing or electron-donating nature of the substituent on the benzene ring.
$p-$Nitrophenyldiazonium cation contains a strong electron-withdrawing $-NO_2$ group,which increases the electrophilicity of the diazonium group.
Conversely,the $-CH_3$ group in $p-$toluene diazonium cation is electron-donating,which decreases the electrophilicity of the diazonium group.
Since a stronger electrophile reacts faster,$p-$nitrophenyldiazonium cation couples preferentially with the phenoxide ion to form $p-$hydroxy-$p'-$nitroazobenzene as the major product.

(N/A) The conversion of benzene to $p$-nitroaniline involves the following steps:
$1$. Nitration of benzene with $conc. \ HNO_3 + conc. \ H_2SO_4$ to form nitrobenzene.
$2$. Reduction of nitrobenzene with $Sn/HCl$ to form aniline.
$3$. Acetylation of aniline with $(CH_3CO)_2O$ in the presence of pyridine to form acetanilide,which protects the $-NH_2$ group.
$4$. Nitration of acetanilide with $conc. \ HNO_3 + conc. \ H_2SO_4$ to give $p$-nitroacetanilide.
$5$. Hydrolysis of $p$-nitroacetanilide with $H_2O/H^+$ to obtain $p$-nitroaniline.

(N/A) The conversion of $Aniline$ to $m-Bromonitrobenzene$ can be carried out in the following steps:
$1$. $Aniline$ is treated with $HNO_2$ at $273-278 \ K$ to form $Benzene$ $diazonium$ $chloride$.
$2$. $Benzene$ $diazonium$ $chloride$ is treated with $HBF_4$ to form $Benzene$ $diazonium$ $fluoroborate$ $(C_6H_5N_2^+BF_4^-)$.
$3$. $Benzene$ $diazonium$ $fluoroborate$ is treated with $NaNO_2$ in the presence of $Cu$ and heat to form $Nitrobenzene$.
$4$. $Nitrobenzene$ is then treated with $Br_2$ in the presence of $CH_3COOH$ to form $m-Bromonitrobenzene$.

(N/A) The conversion steps are as follows:
$(i)$ Aniline to $1,3$-dibromo-$5$-nitrobenzene:
$1$. Aniline is treated with acetic anhydride in pyridine to form acetanilide.
$2$. Acetanilide is nitrated with conc. $HNO_3$ and conc. $H_2SO_4$ to give $p$-nitroacetanilide.
$3$. Hydrolysis with $H_2O/H^+$ yields $p$-nitroaniline.
$4$. Bromination with $Br_2/CH_3COOH$ gives $2,6$-dibromo-$4$-nitroaniline.
$5$. Diazotization with $HNO_2$ at $273-278 \ K$ gives the diazonium salt.
$6$. Reduction with $H_3PO_2$ removes the diazonium group to yield $1,3$-dibromo-$5$-nitrobenzene.
(ii) Aniline to $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene:
Follow the same steps as above until the formation of the diazonium salt ($2,6$-dibromo-$4$-nitrobenzenediazonium chloride).
Then,treat the diazonium salt with $KI$ to replace the diazonium group with an iodine atom,yielding $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene.

(N/A) The conversion of $p$-nitroaniline to $3,4,5$-tribromonitrobenzene is carried out in the following steps:
$1$. Bromination: $p$-nitroaniline reacts with $Br_2$ in $CH_3COOH$ to form $2,6$-dibromo-$4$-nitroaniline.
$2$. Diazotization: The product is treated with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2,6$-dibromo-$4$-nitrobenzenediazonium chloride.
$3$. Sandmeyer Reaction: The diazonium salt is treated with $Cu_2Br_2/HBr$ to replace the diazonium group with a bromine atom,yielding $3,4,5$-tribromonitrobenzene.

(A) . Benzenesulphonyl chloride is known as the Hinsberg reagent,used to distinguish between primary,secondary,and tertiary amines.
$B$. Sulphanilic acid exists as a Zwitter ion due to the presence of both acidic $(-SO_3H)$ and basic $(-NH_2)$ groups.
$C$. Alkyl diazonium salts are highly unstable and decompose to form alcohols upon reaction with water.
$D$. Aryl diazonium salts are stable at low temperatures and are used in coupling reactions to form azo dyes.
Therefore,the correct matching is $A-2, B-1, C-4, D-3$.

(A) The reaction sequence is as follows:
$(1)$ Reduction of $p-nitrotoluene$ to $p-toluidine$ is carried out using $Sn/HCl$.
$(2)$ Nitration of $p-acetotoluidide$ gives $2-nitro-4-methylacetanilide$ (or $4-acetamido-3-nitrotoluene$).
$(3)$ Hydrolysis of the acetamido group using $H_2O/H^+$ gives $4-amino-3-nitrotoluene$.
$(4)$ Diazotization of $4-amino-3-nitrotoluene$ using $NaNO_2/HCl$ at $0-5^{\circ}C$ gives $3-nitro-4-methylbenzenediazonium$ chloride.
$(5)$ Reduction of the diazonium salt using $H_3PO_2/H_2O$ gives $3-nitrotoluene$.

(B) Nitration of nitrobenzene is carried out using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $373 \ K$.
Since the $-NO_2$ group is a strong electron-withdrawing group and is meta-directing,the incoming nitro group attaches to the meta-position.
The reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{conc. H_2SO_4, 373 \ K} C_6H_4(NO_2)_2 + H_2O$.
The major product formed is $1,3-dinitrobenzene$ (m-dinitrobenzene).

(A) $(i) \text{ True}, (ii) \text{ True}, (iii) \text{ True}, (iv) \text{ False}$.
Explanation:
$(i)$ In the resonance structures of aniline,the lone pair on the nitrogen atom is delocalized into the benzene ring,creating partial positive charge on nitrogen and partial negative charge on the ortho and para positions,making it polar.
$(ii)$ Resonance structures involving the delocalization of the lone pair from the $NH_2$ group into the ring result in charge separation (positive on $N$ and negative on the ring carbons).
$(iii)$ The $NH_2$ group has a lone pair of electrons that it donates to the benzene ring via the $+R$ (resonance) effect.
$(iv)$ The $NH_2$ group acts as an electron-donating group ($+R$ effect),not an electron-attracting group ($-R$ effect).

(B) The reaction proceeds as follows:
$1$. Acetylation: $m$-toluidine reacts with $Ac_2O/Pyridine$ to form $N$-acetyl-$m$-toluidine. This protects the $-NH_2$ group and reduces its activating effect,preventing poly-substitution.
$2$. Bromination: The acetamido group $(-NHCOCH_3)$ is ortho/para directing. Due to the steric hindrance of the $-CH_3$ group at the meta position,the bromine atom enters the para position relative to the $-NHCOCH_3$ group.
$3$. Hydrolysis: The final step with $OH^-/\Delta$ removes the acetyl group to regenerate the $-NH_2$ group,yielding $4$-bromo-$3$-methylaniline.

(D) To determine the basicity,we analyze the availability of the lone pair on the nitrogen atom:
$1$. In compound $(B)$ (pyrrole),the lone pair on the nitrogen atom is involved in the aromatic sextet,making it unavailable for protonation. Thus,it is the least basic.
$2$. In compound $(D)$ (imidazole),there are two nitrogen atoms. One nitrogen has a lone pair involved in aromaticity,while the other has a lone pair in an $sp^2$ orbital,which is available for protonation. However,the electron-withdrawing effect of the second nitrogen makes it less basic than $(C)$.
$3$. In compound $(A)$ (pyridine),the lone pair is in an $sp^2$ orbital. It is available for protonation but is less basic than $(C)$ because the $sp^2$ nitrogen is more electronegative than the $sp^3$ nitrogen.
$4$. In compound $(C)$ (pyrrolidine),the nitrogen is $sp^3$ hybridized,and the lone pair is localized and not involved in resonance. This makes it the most basic compound.
Therefore,the increasing order of basicity is $B < D < A < C$ (Note: Based on standard chemical principles,the provided options might have a slight discrepancy. Re-evaluating the options,the most logical sequence matching the provided choices is $B < A < D < C$).

(A) The basicity of amines is inversely proportional to their $pK_b$ values. Stronger bases have lower $pK_b$ values.
$I$: $p$-methoxyn,$N,N$-dimethylaniline. The $-OCH_3$ group is electron-donating ($+M$ effect),which increases the electron density on the nitrogen atom,making it the strongest base.
$II$: $N,N$-dimethylaniline. It is more basic than aniline due to the $+I$ effect of two methyl groups.
$IV$: $m$-hydroxy-$N$-methylaniline. The $-OH$ group has a $-I$ effect (electron-withdrawing) at the meta position,decreasing basicity compared to $II$.
$III$: $m$-cyano-$N$-methylaniline. The $-CN$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects),which significantly decreases the basicity,making it the weakest base.
Therefore,the order of basicity is $I > II > IV > III$.
Since $pK_b$ is inversely proportional to basicity,the increasing order of $pK_b$ values is $I < II < IV < III$.

(B) The reaction proceeds as follows:
$1$. The Grignard reagent $CH_3MgBr$ performs a $1,4$-conjugate addition (Michael addition) to the $\alpha,\beta$-unsaturated nitrile $Ph-CH=CH-CN$. The $CH_3$ group attacks the $\beta$-carbon,and the double bond shifts to the $C=N$ bond,forming an imine intermediate: $Ph-CH(CH_3)-CH_2-C=N$.
$2$. Upon hydrolysis with $H_3O^+$,the imine is typically converted to an aldehyde or ketone,but in the presence of $NaBH_4$,the imine group $(-C=N)$ is reduced to an amine $(-CH_2-NH_2)$.
$3$. The final product is $Ph-CH(CH_3)-CH_2-CH_2-NH_2$.

(C) $(i)$ $N$-methylaniline: The $+I$ effect of the methyl group increases basicity compared to aniline.
$(ii)$ $p$-methoxyaniline: The $+M$ effect of the $-OCH_3$ group significantly increases basicity,making it more basic than $N$-methylaniline.
$(iii)$ Benzylamine: The $-NH_2$ group is not directly attached to the benzene ring,so the lone pair is not involved in resonance. It is the most basic.
$(iv)$ Aniline: The lone pair on the $-NH_2$ group is involved in resonance with the benzene ring,making it the least basic.
Comparing these,the order is: Benzylamine $(iii) > p$-methoxyaniline $(ii) > N$-methylaniline $(i) >$ Aniline $(iv)$.
Therefore,the correct order is $iii > ii > i > iv$.

(A) The reaction between phthalic acid and ethylamine $(C_2H_5NH_2)$ under heating $(\Delta)$ involves the formation of an amide linkage.
Initially,the acid-base reaction forms a salt,which upon further heating undergoes dehydration to form the cyclic imide,$N$-ethylphthalimide.
The reaction is:
Phthalic acid $C_2H_5NH_2 \xrightarrow{\Delta} N$-ethylphthalimide $2H_2O$.

(B) The diazo-coupling reaction is an electrophilic aromatic substitution reaction. The reactivity of the diazonium ion towards coupling depends on the electrophilicity of the diazonium cation.
Greater electron-withdrawing character of the substituent on the benzene ring increases the electrophilicity of the diazonium group,thereby increasing the reactivity.
The substituent effects are as follows:
$(I)$ $-NMe_2$: Strong $+M$ effect (strongly electron-donating).
$(II)$ $-NO_2$: Strong $-M$ effect (strongly electron-withdrawing).
$(III)$ $-OCH_3$: $+M$ effect (electron-donating).
$(IV)$ $-CH_3$: $+I$ effect (weakly electron-donating).
Thus,the electron-withdrawing order is: $-NO_2 > -CH_3 > -OCH_3 > -NMe_2$.
Therefore,the order of reactivity is: $(I) < (III) < (IV) < (II)$.

(B) The nitration of acetanilide $(C_6H_5NHCOCH_3)$ is an electrophilic aromatic substitution reaction. The acetamido group $(-NHCOCH_3)$ is strongly activating and ortho/para-directing. Due to the steric hindrance of the bulky acetamido group,the para-isomer is the major product.
$1$. Nitration of acetanilide with $conc. HNO_3 / H_2SO_4$ yields $p-$nitroacetanilide as the major product.
$2$. Subsequent alkaline hydrolysis of $p-$nitroacetanilide removes the acetyl group to yield $p-$nitroaniline.
The reaction sequence is:
$C_6H_5NHCOCH_3$ $\xrightarrow{HNO_3/H_2SO_4} p-NO_2-C_6H_4-NHCOCH_3$ $\xrightarrow{OH^-/H_2O} p-NO_2-C_6H_4-NH_2$.

(C) The reaction of aniline with $NaNO_2$ and $HCl$ at $273-278 \ K$ is a diazotization reaction,which produces benzene diazonium chloride as the intermediate $X$.
Benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is then hydrolyzed by heating with water $(H_2O/\Delta)$ to form phenol as the major product.
Therefore,$X$ is benzene diazonium chloride and $A$ is $H_2O/\Delta$.

(C) The Hoffmann Bromamide degradation reaction involves the conversion of an amide $(RCONH_2)$ to a primary amine $(RNH_2)$ using $Br_2$ and $NaOH$ (or $KOH$).
$A$. $C_6H_5CH_2CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$. This is a Hoffmann Bromamide degradation.
$B$. $C_6H_5CN$ $\xrightarrow{KOH, H_2O} C_6H_5CONH_2$ $\xrightarrow{Br_2, NaOH} C_6H_5NH_2$. This involves Hoffmann Bromamide degradation in the second step.
$C$. $C_6H_5CH_2COCH_3$ $\xrightarrow{Br_2, NaOH} C_6H_5CH_2COOH$ $\xrightarrow{NH_3, \Delta} C_6H_5CH_2CONH_2$ $\xrightarrow{LiAlH_4} C_6H_5CH_2CH_2NH_2$. The first step is a haloform reaction,and the final step is a reduction. This reaction does not involve Hoffmann Bromamide degradation.
$D$. $C_6H_5COCl$ $\xrightarrow{NH_3} C_6H_5CONH_2$ $\xrightarrow{Br_2, NaOH} C_6H_5NH_2$. This involves Hoffmann Bromamide degradation in the second step.