Properties of Amines Questions in English

(B) The correct order is $I > III > II > IV$.
$I$ (Benzylamine) is the most basic because the lone pair on the nitrogen atom is not delocalized by resonance with the benzene ring.
In $II$ (Aniline),$III$ ($p$-Toluidine),and $IV$ ($p$-Nitroaniline),the lone pair on the nitrogen atom is involved in resonance with the benzene ring,which significantly decreases their basicity.
Comparing $II$,$III$,and $IV$:
- $III$ is more basic than $II$ because the $-CH_3$ group exerts a $+I$ and $+H$ (hyperconjugation) effect,which increases the electron density on the nitrogen atom.
- $IV$ is the least basic because the $-NO_2$ group exerts a strong $-R$ (or $-M$) and $-I$ effect,which strongly withdraws electron density from the ring and the nitrogen atom.
Thus,the order is $I > III > II > IV$.

(C) $1$. The reaction of aniline with acetic anhydride leads to the formation of $N$-phenylacetamide (acetanilide) as product $X$. This step protects the amino group.
$2$. $N$-phenylacetamide undergoes electrophilic aromatic substitution (nitration) with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $15^{\circ}C$. The acetamido group $(-NHCOCH_3)$ is ortho/para directing,but due to steric hindrance,the para-isomer is the major product $Y$ ($p$-nitroacetanilide).
$3$. Finally,the hydrolysis of $p$-nitroacetanilide with aqueous $NaOH$ removes the acetyl group to yield $p$-nitroaniline as the final product $Z$.

(A) The reaction involves the nitration of acetanilide using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$.
In acetanilide,the $-NHCOCH_3$ group is ortho/para-directing due to the resonance effect of the lone pair on the nitrogen atom.
However,due to the steric hindrance caused by the bulky $-NHCOCH_3$ group,the para-position is more accessible than the ortho-position.
Therefore,the major product formed is $p$-nitroacetanilide.

(B) The correct option is $(b)$.
The conversion of benzene diazonium hydrogensulphate to benzene is a reduction reaction.
$C_{6}H_{5}N_{2}^{+}HSO_{4}^{-} + H_{3}PO_{2} + H_{2}O \rightarrow C_{6}H_{6} + N_{2} + H_{3}PO_{3} + H_{2}SO_{4}$
Phosphinic acid $(H_{3}PO_{2})$ acts as a reducing agent in the presence of water to reduce the diazonium group to a hydrogen atom,yielding benzene.

(C) The correct answer is $C$ $(III)$.
Step $1$: Aniline reacts with $NaNO_{2} +$ dil. $HCl$ at $0-5^{\circ} C$ to form benzene diazonium chloride $(C_{6}H_{5}N_{2}^{+}Cl^{-})$.
Step $2$: The benzene diazonium chloride then reacts with $CuCN$ (Sandmeyer reaction) to replace the diazonium group with a cyano group $(-CN)$,yielding cyanobenzene (benzonitrile) and releasing $N_{2}$ gas.
Structure $III$ represents cyanobenzene.

(A) When aniline reacts with excess bromine water $(Br_2 / H_2O)$,the $-NH_2$ group strongly activates the benzene ring towards electrophilic substitution.
This results in the substitution of bromine atoms at all available ortho and para positions,yielding $2,4,6-$tribromoaniline as the major white precipitate product.

(A) Aniline is a highly reactive species towards electrophilic substitution due to the strong activating effect of the $-NH_2$ group. To control the reaction and obtain a monobrominated product,the $-NH_2$ group is first protected by acetylation using $(CH_3CO)_2O$ in the presence of pyridine to form acetanilide.
Acetanilide then undergoes electrophilic bromination with $Br_2/CH_3CO_2H$ to yield $p$-bromoacetanilide as the major product $(X)$.
Finally,the acetyl protecting group is removed by hydrolysis using aqueous concentrated $NaOH$ to yield $p$-bromoaniline $(Y)$.

(A) The reaction of aniline $(C_6H_5NH_2)$ with acetic anhydride $(CH_3CO)_2O$ is an acetylation reaction.
In this reaction,the lone pair of electrons on the nitrogen atom of the aniline attacks the carbonyl carbon of the acetic anhydride.
This leads to the formation of $N$-phenylethanamide,commonly known as acetanilide $(C_6H_5NHCOCH_3)$,and acetic acid $(CH_3COOH)$ as a byproduct.
The reaction is:
$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$
Therefore,the major product is $N$-phenylethanamide.

(A) Benzene sulphonyl chloride,also known as Hinsberg's reagent,reacts with $1^{\circ}$ and $2^{\circ}$ amines to form sulphonamides.
The sulphonamide formed from a $1^{\circ}$ amine contains an acidic hydrogen atom attached to the nitrogen,making it soluble in alkali.
The sulphonamide formed from a $2^{\circ}$ amine does not contain any acidic hydrogen atom attached to the nitrogen,making it insoluble in alkali.
Given that compound $X (C_{7}H_{9}N)$ reacts with benzene sulphonyl chloride to form a product $Y$ that is insoluble in alkali,$X$ must be a $2^{\circ}$ amine.
Among the given options,$N$-methylaniline $(C_{6}H_{5}NHCH_{3})$ is a $2^{\circ}$ amine,which satisfies the given conditions.

(B) The correct option is $(B)$.
The reaction involves the removal of the diazonium group $(-N_2^+ Cl^-)$ from the benzene ring and its replacement with a hydrogen atom.
This transformation is achieved using hypophosphorous acid $(H_3PO_2)$ in the presence of water. The diazonium salt is reduced to the corresponding arene,while $H_3PO_2$ is oxidized to phosphorous acid $(H_3PO_3)$.
The reaction is represented as:
$Ar-N_2^+ Cl^- + H_3PO_2 + H_2O \rightarrow Ar-H + N_2 + H_3PO_3 + HCl$
This is a standard method for the deamination of aromatic amines via their diazonium salts.

(B) The correct option is $(B)$.
In the given compound,the most acidic proton is $N^b-H$. This is because the resulting conjugate base is resonance-stabilized by the adjacent carbonyl group $(C=O)$.
The most nucleophilic nitrogen is $N^c$. This is because the lone pair of electrons on this nitrogen is localized in an $sp^3$-hybrid orbital and is not involved in resonance,making it readily available for donation.

(A) The correct order of basicity is $IV < I < III < II$.
$1$. Compound $IV$ (pyrrole) is the least basic because the lone pair on the nitrogen atom is involved in the aromatic sextet,making it unavailable for protonation.
$2$. Compound $I$ ($p$-nitroaniline) is less basic than $III$ (pyridine) because the strong electron-withdrawing $-NO_2$ group decreases the electron density on the nitrogen atom through both $-I$ and $-M$ effects.
$3$. Compound $III$ (pyridine) has the nitrogen atom $sp^2$ hybridized,which is more electronegative than the $sp^3$ hybridized nitrogen in aliphatic amines,making it less basic than $II$.
$4$. Compound $II$ (cyclohexylamine) is the most basic as it is a primary aliphatic amine with no electron-withdrawing groups attached,and the nitrogen is $sp^3$ hybridized.

(B) Pyridine is basic in nature due to the presence of a free lone pair on the nitrogen atom. Its basicity is influenced by the substituents attached to the ring. Electron-withdrawing groups $(EWG)$ decrease the basicity of pyridine,while electron-releasing groups $(ERG)$ increase its basicity.
Substituents like $-N(CH_3)_2$ and $-CH_3$ are electron-releasing,but the $-N(CH_3)_2$ group is a much stronger electron-donating group due to the resonance effect (+$R$ effect) compared to the inductive effect (+$I$ effect) of the $-CH_3$ group.
Conversely,the $-Cl$ group is electron-withdrawing due to its strong inductive effect (-$I$ effect),which decreases the electron density on the nitrogen atom,making it the least basic.
The order of basicity is: $4\text{-Chloropyridine} < \text{Pyridine} < 4\text{-Methylpyridine} < 4\text{-(Dimethylamino)pyridine}$.
Therefore,$4\text{-(Dimethylamino)pyridine}$ is the most basic compound.

(C) The reaction between a carboxylic acid $(CH_3COOH)$ and an amine $(CH_3CH_2NH_2)$ at room temperature $(25^{\circ} C)$ is an acid-base reaction.
$CH_3COOH$ acts as a Bronsted-Lowry acid and $CH_3CH_2NH_2$ acts as a Bronsted-Lowry base.
The proton transfer occurs from the acid to the amine,resulting in the formation of an ammonium carboxylate salt.
$CH_3COOH + CH_3CH_2NH_2 \xrightarrow{25^{\circ} C} CH_3CH_2NH_3^+ \cdot CH_3COO^-$

(B) $1$. The reaction of nitrobenzene with $Sn/HCl$ is a reduction reaction that converts the nitro group $(-NO_2)$ into an amino group $(-NH_2)$,forming aniline $(X)$.
$2$. Aniline reacts with $NaNO_2/HCl$ at low temperatures $(0-5 \ ^\circ C)$ to undergo diazotization,forming benzenediazonium chloride.
$3$. The subsequent reaction with $CuBr$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom,resulting in the formation of bromobenzene $(Y)$.

(C) $1$. The reaction of an aliphatic primary amine with $HNO_2$ gives an alcohol with the evolution of $N_2$ gas.
$2$. The compound $Y$ reacts with $I_2 / NaOH$ (iodoform test) to give a sodium salt of a carboxylic acid,which indicates that $Y$ must be a secondary alcohol containing the $CH_3-CH(OH)-$ group.
$3$. Among the given options,$CH_3-CH(NH_2)-CH_2-CH_3$ (butan$-2-$amine) reacts with $HNO_2$ to form butan$-2-$ol $(CH_3-CH(OH)-CH_2-CH_3)$.
$4$. Butan$-2-$ol contains the $CH_3-CH(OH)-$ group,which gives a positive iodoform test with $I_2 / NaOH$ to produce sodium propionate and iodoform $(CHI_3)$.

(C) Statement $-I$ is correct. Pure aniline and other arylamines are generally colourless liquids or solids when freshly prepared.
Statement $-II$ is incorrect. Arylamines get coloured on storage due to atmospheric oxidation,not reduction. They form complex coloured products when exposed to air and light.

(C) The given reaction is a coupling reaction between a diazonium salt derivative and $1$-naphthylamine. The reaction proceeds as follows:
$1$-naphthylamine reacts with the diazonium species to form an azo dye.
This specific azo dye product is known to have a red colour.
Therefore,the correct option is $C$.

(A) In an aqueous medium,the basic strength of amines is determined by a combination of three factors: inductive effect,solvation effect,and steric hindrance.
$1$. The inductive effect ($+I$ effect) of methyl groups increases the electron density on the nitrogen atom,making it more basic.
$2$. The solvation effect stabilizes the protonated amine cation through hydrogen bonding with water molecules. More substituted cations are less stabilized due to steric hindrance.
$3$. For methyl-substituted amines,the combined effect results in the following order of basicity: $Me_2NH > MeNH_2 > Me_3N > NH_3$.

(D) The basic strength of amines is inversely proportional to their $pK_{b}$ values.
$1$. Aniline $(A)$ is the least basic due to resonance,so it has the highest $pK_{b}$ value: $9.38$ $(III)$.
$2$. Among aliphatic amines,the basicity order in aqueous solution is determined by inductive effect,solvation,and steric hindrance. For ethylamines,the order is: $N$-Ethylethanamine ($C$,secondary) > $N,N$-Diethylethanamine ($D$,tertiary) > Ethanamine ($B$,primary).
$3$. Corresponding $pK_{b}$ values are:
- $C$ ($N$-Ethylethanamine): $3.00$ $(II)$
- $D$ ($N,N$-Diethylethanamine): $3.25$ $(I)$
- $B$ (Ethanamine): $3.29$ $(IV)$
Thus,the correct matching is: $A-III, B-IV, C-II, D-I$.

(D) The reaction of an amide with $Br_2$ and an aqueous base $(KOH)$ is known as the $Hoffmann$ $Bromamide$ degradation reaction.
In this reaction,the amide is converted into a primary amine containing one carbon atom less than the original amide.
Propanamide is $CH_3CH_2CONH_2$ (a $3$-carbon amide).
Upon treatment with $Br_2 / KOH$,it undergoes degradation to form ethylamine $(CH_3CH_2NH_2)$,which is a $2$-carbon primary amine.

(D) The conversion of $p$-nitrotoluene $(X)$ to $3,5$-dibromotoluene $(Y)$ involves the following steps:
$1$. Reduction of the nitro group to an amino group: $p$-nitrotoluene is reduced using $Fe/H^{+}$ to form $p$-toluidine.
$2$. Bromination: $p$-toluidine is treated with $Br_2 (aq)$ to form $2,6$-dibromo-$4$-methylaniline due to the strong activating effect of the $-NH_2$ group.
$3$. Diazotization: The amino group is converted to a diazonium salt using $HNO_2$ at $0-5 \ ^{\circ}C$.
$4$. Deamination: The diazonium group is removed using $H_3PO_2$ and water to yield $3,5$-dibromotoluene $(Y)$.

(D) The reaction sequence is as follows:
$1$. Nitrobenzene $(C_6H_5NO_2)$ is reduced using $H_2/Pd$ in ethanol $(C_2H_5OH)$ to form aniline $(C_6H_5NH_2)$,which is compound $[A]$.
$2$. Aniline $(C_6H_5NH_2)$ reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of pyridine to undergo acetylation,forming $N$-phenylacetamide (acetanilide),$C_6H_5NHCOCH_3$,which is compound $[B]$.

(D) $1$. The molecular formula $C_4H_{11}N$ corresponds to a saturated amine.
$2$. The reaction with $HNO_2$ to release $N_2$ gas indicates that the compound is a primary aliphatic amine $(R-NH_2)$.
$3$. The reaction with $PhSO_2Cl$ (Hinsberg reagent) produces a sulfonamide that is soluble in $KOH$. This confirms that the amine is a primary amine,as the resulting sulfonamide contains an acidic hydrogen on the nitrogen atom.
$4$. The compound must be optically active. Among the primary amines with the formula $C_4H_{11}N$,butan$-2-$amine $(CH_3CH_2CH(NH_2)CH_3)$ contains a chiral carbon atom and is therefore optically active.
$5$. Butan$-1-$amine is achiral.

(B) Assertion $(A)$ is correct. The reaction of $\alpha$-halocarboxylic acids with aqueous $NH_3$ is a standard method for preparing $\alpha$-amino acids because the reaction is controlled and the product is stable. In contrast,the reaction of alkyl halides with $NH_3$ (ammonolysis) often leads to a mixture of primary,secondary,and tertiary amines,resulting in a low yield of the primary amine.
Reason $(R)$ is correct. Amino acids do exist as zwitter ions in aqueous medium due to the internal acid-base reaction between the $-COOH$ and $-NH_2$ groups.
However,the zwitter ion nature of amino acids is not the reason why $\alpha$-halocarboxylic acids give a better yield of amino acids compared to alkyl halides. The yield difference is due to the nature of the substitution reaction and the stability of the products formed.
Therefore,both $(A)$ and $(R)$ are correct,but $(R)$ is not the correct explanation of $(A)$.

(A) Step $1$: Aniline reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of pyridine to form acetanilide $(C_6H_5NHCOCH_3)$.
Step $2$: Acetanilide is then reduced by $LiAlH_4$ (a strong reducing agent) which reduces the amide group $(-NHCOCH_3)$ to an amine group $(-NHCH_2CH_3)$.
Step $3$: Hydrolysis $(H_2O)$ yields the final product,$N$-ethylaniline $(C_6H_5NHCH_2CH_3)$.

(B) The starting material is acetanilide.
$1$. Reaction with $Br_2/AcOH$: The $-NHCOCH_3$ group is strongly activating and ortho/para-directing. Due to the steric hindrance of the bulky acetamido group,the para-substitution is favored,yielding $4$-bromoacetanilide as the major product $A$.
$2$. Reaction with $LiAlH_4$: $LiAlH_4$ is a strong reducing agent that reduces the amide group $(-NHCOCH_3)$ to an amine group $(-NHCH_2CH_3)$. Thus,acetanilide is reduced to $N$-ethylaniline,which is the major product $B$.

(D) The reactions of benzenediazonium chloride $(C_6H_5N_2^ Cl^-)$ are as follows:
$A$. Reaction with aniline $(C_6H_5NH_2)$ leads to coupling reaction to form $p$-aminoazobenzene (Product $III$).
$B$. Reaction with $HBF_4$ followed by heating $(\Delta)$ is the Balz-Schiemann reaction,which forms fluorobenzene (Product $I$).
$C$. Reaction with $Cu/HCl$ is the Gattermann reaction,which forms chlorobenzene (Product $IV$).
$D$. Reaction with $CuCN/KCN$ is the Sandmeyer reaction,which forms benzonitrile (Product $II$).
Thus,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) Step $1$: Electrophilic aromatic substitution of $p$-nitrotoluene with $Br_2$ gives $2$-bromo-$4$-nitrotoluene.
Step $2$: Reduction of the nitro group using $H_2/Pd$ yields $2$-bromo-$4$-methylaniline.
Step $3$: Diazotization using $NaNO_2/HCl$ at $0^{\circ}C$ converts the amino group into a diazonium salt,forming $2$-bromo-$4$-methylbenzenediazonium chloride.
Step $4$: Reduction of the diazonium salt with $H_3PO_2$ replaces the $-N_2^+Cl^-$ group with a hydrogen atom,resulting in $2$-bromo-$1$-methylbenzene (also known as $o$-bromotoluene).

(D) $(P)$ Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. Among the given options,$2$-phenylethanamine $(C_6H_5CH_2CH_2NH_2)$ is an aliphatic primary amine.
$(Q)$ $2^{\circ}$-amines react with Hinsberg's reagent (benzenesulfonyl chloride) to form a sulfonamide that is insoluble in $NaOH$ because it lacks an acidic hydrogen on the nitrogen atom. $N$-ethylaniline $(C_6H_5NHC_2H_5)$ is a $2^{\circ}$-amine.
$(R)$ Aromatic primary amines react with nitrous acid $(HONO)$ at low temperature $(273-278 \ K)$ to form diazonium salts,which undergo coupling reactions with $\beta$-naphthol in alkaline medium to form a red azo dye. $o$-ethylaniline is an aromatic primary amine.
Therefore,the correct sequence is $(P) = 2$-phenylethanamine,$(Q) = N$-ethylaniline,$(R) = o$-ethylaniline. This corresponds to option $D$.

(A) $LiAlH_4$ is a strong reducing agent that reduces amides to amines and ketones to secondary alcohols.
In the given substrate,the amide group $(-CONH_2)$ is reduced to a primary amine $(-CH_2NH_2)$ and the ketone group $(>C=O)$ is reduced to a secondary alcohol $(-CH(OH)-)$.
The alkene double bond $(C=C)$ remains unaffected by $LiAlH_4$.
Therefore,the product '$X$' is the molecule containing both a primary amine and a secondary alcohol group,with the alkene intact.

(A) When $o$-phenylenediamine reacts with nitrous acid $(HNO_2)$,one of the amino groups is converted into a diazonium salt group $(-N_2^+)$.
Due to the proximity of the other amino group $(-NH_2)$ at the ortho position,an intramolecular nucleophilic attack occurs.
The lone pair on the nitrogen of the $-NH_2$ group attacks the diazonium nitrogen,leading to the formation of a cyclic compound known as benzotriazole.
The reaction proceeds as follows:
$o$-Phenylenediamine $\xrightarrow{HNO_2}$ $o$-amino-benzenediazonium ion $\xrightarrow{-H_2O, -H^+}$ Benzotriazole.

(B) The reaction of $N,N$-dimethylaniline with $NaNO_2$ and $HX$ (nitrous acid) at low temperature $(0-5^{\circ}C)$ is an electrophilic aromatic substitution reaction.
$1$. The electrophile generated is the nitrosonium ion,$NO^{+}$.
$2$. Since the $-N(CH_3)_2$ group is a strong ortho/para directing group,the electrophile $NO^{+}$ attacks the para position to form $p$-nitroso-$N,N$-dimethylaniline.
$3$. Option $A$ is correct as $NO^{+}$ is the electrophile.
$4$. Option $B$ is incorrect because the product is $p$-nitroso-$N,N$-dimethylaniline,not an $N$-nitroso ammonium compound (which would form with secondary amines).
$5$. Option $C$ is correct as the reaction is performed at low temperatures to stabilize the nitrous acid and prevent side reactions.
$6$. Option $D$ is correct as the major product is the $p$-nitroso derivative.

(D) The reaction sequence is as follows:
$1$. $A$ is $Aniline$ $(C_6H_5NH_2)$.
$2$. When $Aniline$ reacts with $Br_2$ in water (or $Br_2/CS_2$ at low temperature),it undergoes electrophilic substitution to form $2,4,6-tribromoaniline$ (Compound $B$).
$3$. $2,4,6-tribromoaniline$ reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form the diazonium salt,$2,4,6-tribromobenzenediazonium$ chloride (Compound $C$).
$4$. Finally,reduction of the diazonium salt with $H_3PO_2$ and water removes the $-N_2^+Cl^-$ group,yielding $1,3,5-tribromobenzene$.
Thus,compound $A$ is $Aniline$.

(B) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ in the presence of $HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is product '$A$'.
Step $2$: Benzenediazonium chloride undergoes a diazo coupling reaction with $N,N$-dimethylaniline in a weakly acidic medium. The diazonium group $( -N=N- )$ attacks the para-position of the $N,N$-dimethylaniline ring to form $N,N$-dimethyl$-4-$(phenylazo)aniline,which is product '$B$'.
The reaction is: $C_6H_5N_2^+Cl^- + C_6H_5N(CH_3)_2 \rightarrow C_6H_5-N=N-C_6H_4-N(CH_3)_2 + HCl$.

(D) Let us analyze each reaction:
$(A)$ $CH_3CN + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3CH_2NH_2$ (Ethylamine,a primary amine).
$(B)$ $CH_3CONH_2 + Br_2/KOH \rightarrow CH_3NH_2$ (Methylamine,a primary amine) via the Hoffmann bromamide degradation reaction.
$(C)$ $CH_3CONH_2 + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3CH_2NH_2$ (Ethylamine,a primary amine).
$(D)$ $CH_3NC + (i) LiAlH_4 / (ii) H_3O^{\oplus} \rightarrow CH_3NHCH_3$ (Dimethylamine,a secondary amine).
Thus,reaction $(D)$ does not yield a primary amine.

(D) Bronsted base is a substance that can accept a proton $(H^+)$. The strength of a base depends on the availability of the lone pair of electrons on the nitrogen atom.
$1$. In $A$ (Aniline),the lone pair on nitrogen is delocalized into the benzene ring due to resonance,making it less available for protonation.
$2$. In $B$ (Diphenylamine),the lone pair is delocalized into two benzene rings,making it even less basic than aniline.
$3$. In $C$ (Pyrrole),the lone pair on nitrogen is involved in the aromatic sextet (it is part of the $6\pi$ electron system),so it is not available for protonation.
$4$. In $D$ (Pyrrolidine),the nitrogen atom is $sp^3$ hybridized and the lone pair is localized (not involved in resonance). This makes the lone pair readily available for donation to a proton,making it the strongest base among the given options.

(B) When $1,4$-dichlorobutane reacts with excess ammonia,it undergoes nucleophilic substitution to form the ammonium salt $A$ $([H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-})$.
Treatment of this salt with a strong base like $NaOH$ releases the free diamine $B$ $(H_2N-(CH_2)_4-NH_2)$,along with water and sodium chloride.
The reaction is:
$Cl-(CH_2)_4-Cl + 2NH_3 \rightarrow [H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} (A)$
$[H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} + 2NaOH \rightarrow H_2N-(CH_2)_4-NH_2 (B) + 2H_2O + 2NaCl$

(C) The reaction sequence is as follows:
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ and $HCl$ at $0-5 \ ^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$. This is known as the diazotization reaction.
$2$. The benzenediazonium chloride then reacts with $Cu_2Cl_2$ (cuprous chloride) to form chlorobenzene $(C_6H_5Cl)$. This is a classic Sandmeyer reaction.
Therefore,the final product $A$ is chlorobenzene.

(D) The reaction shown is the diazotization of aniline followed by a coupling reaction to form an azo dye.
Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride. Thus,reagent $(A)$ is $NaNO_2 + HCl$ at $0-5^{\circ}C$.
Step $2$: Benzenediazonium chloride reacts with $\beta\text{-naphthol}$ in the presence of $NaOH$ to form a scarlet red dye (an azo dye). Thus,reagent $(B)$ is $\beta\text{-naphthol}$.
Therefore,the correct option is $(D)$.

(C) $1$. The reaction of benzene with $conc. HNO_3$ and $conc. H_2SO_4$ at $323-333 \ K$ gives nitrobenzene.
$2$. Reduction of nitrobenzene with $Sn/HCl$ gives aniline $(A)$.
$3$. Aniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ $(0-5^{\circ}C)$ to form benzenediazonium chloride.
$4$. Coupling of benzenediazonium chloride with phenol in a basic medium yields $p$-hydroxyazobenzene $(B)$.

(D) Statement $I$ is true: Aniline reacts with conc. $H_2SO_4$ to form anilinium hydrogen sulphate,which on heating at $453-473 \ K$ undergoes rearrangement to form $p$-aminobenzene sulphonic acid (sulphanilic acid). Since this molecule contains both $N$ and $S$,it gives a positive Lassaigne's test for both elements,resulting in the formation of $[Fe(SCN)]^{2+}$,which is blood red in colour.
Statement $II$ is true: Aniline is a Lewis base due to the lone pair on the nitrogen atom. It reacts with the Lewis acid $AlCl_3$ (used in Friedel-Crafts reactions) to form an acid-base adduct (salt). The nitrogen atom acquires a positive charge,which makes the $-NH_3^+$ group strongly electron-withdrawing (deactivating),thereby inhibiting the Friedel-Crafts reaction.

(C) Acetylation involves the replacement of a hydrogen atom $(-H)$ by an acetyl group $(-COCH_3)$.
The net change in molar mass for each amino group $(-NH_2)$ reacting is the replacement of one $H$ atom (mass $1$) with one $COCH_3$ group (mass $43$).
Gain in molecular weight per amino group $= 43 - 1 = 42 \ g \ mol^{-1}$.
Total increase in molecular weight $= 192 - 108 = 84 \ g \ mol^{-1}$.
Therefore,the number of amino groups in $(x) = \frac{84}{42} = 2$.

(A) The $NH_2$ group in Aniline is ortho and para directing and a powerful activating group because the lone pair on the nitrogen atom exhibits a strong $+M$ effect.
Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation) because the lone pair on the nitrogen atom of Aniline reacts with the Lewis acid $AlCl_3$ to form a salt (complex). This results in a positive charge on the nitrogen atom,which acts as a strong deactivating group for the benzene ring,thereby preventing the reaction.

(C) The reaction of ethylamine $(CH_3CH_2NH_2)$ with $NaNO_2 / HCl$ followed by water produces ethanol,nitrogen gas,and water:
$CH_3CH_2NH_2 + NaNO_2 + HCl \rightarrow CH_3CH_2OH + N_2(g) + NaCl + H_2O$
From the stoichiometry,$1 \ mole$ of ethylamine $(45 \ g)$ produces $1 \ mole$ of $N_2$ gas.
At $STP$,$1 \ mole$ of gas occupies $22.4 \ L$.
Given that $2.24 \ L$ of $N_2$ gas is evolved,the number of moles of $N_2$ is:
$n = \frac{2.24 \ L}{22.4 \ L/mol} = 0.1 \ mol$
Since $1 \ mole$ of ethylamine produces $1 \ mole$ of $N_2$,$0.1 \ mol$ of ethylamine is required.
The mass of $0.1 \ mol$ of ethylamine is:
$Mass = 0.1 \ mol \times 45 \ g/mol = 4.5 \ g$
We need to express $4.5 \ g$ as $X \times 10^{-1} \ g$,so $X = 45$.

(B) The reaction of aniline with bromine water is:
$C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr$
(Aniline) $\rightarrow$ ($2,4,6-$tribromoaniline)
Molar mass of aniline $(C_6H_7N)$ = $93 \ g/mol$.
Molar mass of $2,4,6-$tribromoaniline $(C_6H_4Br_3N)$ = $12 \times 6 + 1 \times 4 + 80 \times 3 + 14 = 72 + 4 + 240 + 14 = 330 \ g/mol$.
According to the stoichiometry,$93 \ g$ of aniline produces $330 \ g$ of $2,4,6-$tribromoaniline.
Therefore,$9.3 \ g$ of aniline should theoretically produce:
$\frac{330}{93} \times 9.3 = 33 \ g$ of $2,4,6-$tribromoaniline.
The actual mass obtained is $26.4 \ g$.
Percentage yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$
Percentage yield = $\frac{26.4}{33} \times 100 = 80 \ \%$.

(C) The reaction between aniline $(C_6H_5NH_2)$ and benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is an electrophilic substitution reaction (coupling reaction) that produces $p$-aminoazobenzene (aniline yellow).
The molar mass of aniline $(C_6H_5NH_2)$ is $93 \ g/mol$.
The number of moles of aniline given is $n = \frac{279 \ g}{93 \ g/mol} = 3 \ mol$.
Since the reaction is with one equivalent of benzenediazonium chloride,and assuming complete conversion,$3 \ moles$ of aniline will produce $3 \ moles$ of aniline yellow $(C_{12}H_{11}N_3)$.
The molar mass of aniline yellow $(C_{12}H_{11}N_3)$ is $(12 \times 12) + (11 \times 1) + (3 \times 14) = 144 + 11 + 42 = 197 \ g/mol$.
The mass of aniline yellow formed is $3 \ mol \times 197 \ g/mol = 591 \ g$.

(D) The basic strength depends on the availability of the lone pair on the nitrogen atom.
$1$. In Piperidine,the nitrogen is $sp^3$ hybridized and the lone pair is localized,making it the most basic.
$2$. In Pyridine,the nitrogen is $sp^2$ hybridized and the lone pair is localized in an $sp^2$ orbital,making it less basic than Piperidine.
$3$. In Pyrrole,the lone pair on the nitrogen is involved in the aromatic sextet (delocalized),making it the least basic.
Therefore,the correct order is: $\text{Piperidine} > \text{Pyridine} > \text{Pyrrole}$.

(D) Statement $I$ is true: Aniline is a Lewis base. It reacts with the Lewis acid catalyst $(AlCl_3)$ used in Friedel-Crafts reactions to form an anilinium ion-aluminium chloride salt. This deactivates the ring and prevents the reaction.
Statement $II$ is true: Gabriel phthalimide synthesis involves the nucleophilic substitution of an alkyl halide by the phthalimide anion. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions,so aniline cannot be prepared by this method.

(C) The given reaction is the $Carbylamine$ reaction (also known as the isocyanide test).
Primary amines (both aliphatic and aromatic) react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines),which have a foul smell.
The balanced chemical equation is:
$CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O$.
The nitrogen-containing compound formed is methyl isocyanide $(CH_3NC)$,which is represented as $CH_3-N \equiv C$ or $CH_3-N^{+} \equiv C^{-}$.