Properties of Amines Questions in English

(A) $(i)$ Aniline is highly reactive towards electrophilic substitution. When treated with $Br_2$ in $H_2O$,it undergoes rapid poly-substitution to form $2, 4, 6-$tribromoaniline as the major product $(X)$.
$(ii)$ To control the reactivity of the $-NH_2$ group,it is first acetylated using acetic anhydride $(CH_3CO)_2O$ to form acetanilide. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group,which restricts the bromination to the $p-$position,yielding $p-$bromoacetanilide as the major product $(Y)$.

(C) In aqueous solution,the basic strength of aliphatic amines depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
The order for methyl-substituted amines is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
Comparing this with $NH_3$ and aniline $(C_6H_5NH_2)$:
$1$. $(CH_3)_2NH$ $(V)$ is the most basic due to the $+I$ effect and favorable solvation.
$2$. $CH_3NH_2$ $(IV)$ is next.
$3$. $(CH_3)_3N$ $(II)$ is less basic than $CH_3NH_2$ and $(CH_3)_2NH$ due to steric hindrance in the aqueous phase.
$4$. $NH_3$ $(III)$ is less basic than the aliphatic amines.
$5$. $C_6H_5NH_2$ $(I)$ is the least basic because the lone pair on nitrogen is delocalized into the benzene ring.
Thus,the correct order is $V > IV > II > III > I$.

(C) The given reaction sequence is the Sandmeyer reaction.
Step $1$: The conversion of primary aromatic amine $(Ar-NH_2)$ to benzene diazonium chloride $(Ar-\overset{+}{N} \equiv NCl^-)$ is called diazotization,which is carried out using $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$. Thus,$X = NaNO_2-HCl$.
Step $2$: The conversion of benzene diazonium chloride to aryl chloride $(Ar-Cl)$ is carried out using cuprous chloride $(CuCl)$ or copper powder in $HCl$. This is the Sandmeyer reaction. Thus,$Y = Cu/HCl$.

(C) The reduction of nitrobenzene depends on the medium used:
$(i)$ In neutral medium,using $Zn$ dust and $NH_4Cl$ solution,nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
(ii) In alkaline medium,using $Zn$ and $KOH/C_2H_5OH$,the reduction proceeds further to form hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(A) The reaction between benzene diazonium chloride $(C_6H_5N_2Cl)$ and aniline $(C_6H_5NH_2)$ in the presence of a mild acid $(H^+)$ is a coupling reaction.
This is a type of electrophilic aromatic substitution reaction where the diazonium cation acts as an electrophile.
The electrophile attacks the electron-rich ring of aniline,primarily at the para-position due to steric hindrance at the ortho-position.
The product formed is $p$-aminoazobenzene $(C_6H_5-N=N-C_6H_4-NH_2)$.

(C) The preparation of $p-$methylbenzonitrile ($p-$tolunitrile) is best achieved via the Sandmeyer reaction starting from $p-$toluidine.
$1$. $p-$Toluidine reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form the $p-$methylbenzenediazonium chloride salt.
$2$. This diazonium salt then reacts with $CuCN/KCN$ to yield $p-$methylbenzonitrile.
- Option $(a)$ and $(b)$: Aryl halides like $4-$chlorotoluene do not undergo nucleophilic substitution with $NaCN$ or $AgCN$ under standard conditions due to the partial double bond character of the $C-Cl$ bond.
- Option $(d)$: This is the carbylamine reaction,which converts primary amines into isocyanides $(R-NC)$,not nitriles $(R-CN)$.

(D) $1$. The first step,$Benzamide \xrightarrow{Br_2 / NaOH} X$,is the Hoffmann bromamide degradation reaction,where $X$ is $Aniline$ $(C_6H_5NH_2)$.
$2$. The second step,$X \xrightarrow[\text{alc. } KOH]{CHCl_3} Y$,involves the reaction of a primary amine $(Aniline)$ with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.
$3$. This specific reaction is known as the Carbylamine reaction,which produces an isocyanide ($Y = Phenyl \ isocyanide$ or $C_6H_5NC$).
$4$. Therefore,the conversion of $X$ to $Y$ is the Carbylamine reaction.

(A) Step $1$: $C_2H_5Cl + KCN \rightarrow C_2H_5CN (X) + KCl$. Here,$X$ is propanenitrile $(CH_3CH_2CN)$.
Step $2$: $C_2H_5CN + 2H_2 \xrightarrow{\text{Catalyst}} CH_3CH_2CH_2NH_2 (Y)$. Here,$Y$ is propan$-1-$amine.
Step $3$: $CH_3CH_2CH_2NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3CH_2CH_2NC (Z) + 3KCl + 3H_2O$. This is the carbylamine reaction,which is a test for primary amines. The product $Z$ is propyl isocyanide $(CH_3CH_2CH_2NC)$.

(A) The reaction of primary amines with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines) is known as the carbylamine reaction.
This reaction is a characteristic test for primary amines.
Secondary and tertiary amines do not undergo this reaction.
In the given options:
$A$ is a secondary amine $((CH_3)_2CH-NH-CH_3)$,
$B$ is a primary amine $(C_6H_5-NH_2)$,
$C$ is a primary amine $(3\text{-bromo-4-methylaniline})$,
$D$ is a primary amine $(4\text{-bromoaniline})$.
Therefore,the secondary amine $(CH_3)_2CH-NH-CH_3$ will not react.

(C) The carbylamine reaction (also known as Hoffmann isocyanide synthesis) is a test used to identify primary amines. In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and an ethanolic base (like $KOH$) to form an isocyanide (carbylamine),which is characterized by a foul,offensive smell.
Only primary amines ($R-NH_2$ or $Ar-NH_2$) undergo this reaction.
Secondary amines $(R_2NH)$ and tertiary amines $(R_3N)$ do not possess the necessary hydrogen atoms on the nitrogen to form the isocyanide intermediate and thus do not give the carbylamine test.
In the given options:
$CH_3-CH_2-NH_2$ is a primary amine.
$C_6H_5-NH_2$ (aniline) is a primary amine.
$(CH_3)_3N$ is a tertiary amine.
$CH_3-CH_2-NH-CH_3$ is a secondary amine.
Since both $(CH_3)_3N$ and $CH_3-CH_2-NH-CH_3$ do not react,and typically such questions look for the tertiary amine or the most substituted one,$(CH_3)_3N$ is the standard answer for this type of question.

(A) The conversion of aniline to chlorobenzene is a two-step process known as the Sandmeyer reaction.
Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 / HCl$ at a low temperature of $273-278 \ K$ $(0-5 \ ^\circ C)$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: The benzene diazonium chloride is then treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$ to yield chlorobenzene $(C_6H_5Cl)$ and nitrogen gas $(N_2)$.
Therefore,the correct set of reagents is $NaNO_2 / HCl, 273-278 \ K ; Cu_2Cl_2 / HCl$.

(D) Primary aliphatic amines $(R-NH_2)$ react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which decompose in the presence of water to release nitrogen gas $(N_2 \uparrow)$.
$C_2 H_5 NH_2 + HNO_2$ $\xrightarrow{0-5^{\circ}C} [C_2 H_5 N_2^+ Cl^-]$ $\xrightarrow{H_2 O} C_2 H_5 OH + N_2 \uparrow + HCl$
Secondary amines form $N$-nitrosoamines (yellow oily liquids),and tertiary amines form salts with nitrous acid without evolving nitrogen gas.
Therefore,$C_2 H_5 NH_2$ (a primary amine) is the correct answer.

(B) $1$. The reaction of aniline with acetic anhydride in the presence of pyridine is an acylation reaction,which converts the $-NH_2$ group into an acetamido group $(-NHCOCH_3)$. The product $X$ is acetanilide.
$2$. Acetanilide undergoes electrophilic aromatic substitution (bromination) with $Br_2$ in $CH_3COOH$. The $-NHCOCH_3$ group is ortho/para directing. Due to steric hindrance,the para-isomer is the major product. The product $Y$ is $p$-bromoacetanilide.

(D) The given reaction is the Hoffmann bromamide degradation reaction. In this reaction,an amide $(R-CONH_2)$ reacts with bromine $(Br_2)$ in the presence of an aqueous base (like $NaOH$) to form a primary amine $(R-NH_2)$ with one carbon atom less than the starting amide.
In the given reactant,benzamide $(C_6H_5CONH_2)$,the carbonyl carbon is removed as a carbonate,and the phenyl group attaches directly to the nitrogen atom.
Therefore,the reaction is:
$C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
The product formed is aniline $(C_6H_5NH_2)$.

(D) $1$. The reaction of aniline $(C_6H_5NH_2)$ with acetic anhydride $((CH_3CO)_2O)$ in the presence of pyridine yields acetanilide $(C_6H_5NHCOCH_3)$,which is $X$.
$2$. Nitration of acetanilide with $HNO_3/H_2SO_4$ at $288 \ K$ gives $p$-nitroacetanilide $(p-NO_2-C_6H_4NHCOCH_3)$ as the major product,which is $Y$.
$3$. Hydrolysis of $p$-nitroacetanilide with $OH^-$ yields $p$-nitroaniline $(p-NO_2-C_6H_4NH_2)$,which is $Z$.
$4$. Thus,the correct sequence of structures corresponds to option $D$.

(A) The reaction of $1-$bromopropane $(CH_3CH_2CH_2Br)$ with ethanolic $KCN$ is a nucleophilic substitution reaction. $CN^-$ is an ambident nucleophile. The major product is the nitrile $(CH_3CH_2CH_2CN)$ due to attack by carbon,while the minor product is the isocyanide $(CH_3CH_2CH_2NC)$ due to attack by nitrogen.
Hydrolysis of the minor product,$CH_3CH_2CH_2NC$,yields a primary amine and formic acid:
$CH_3CH_2CH_2NC + 2H_2O \xrightarrow{H^+} CH_3CH_2CH_2NH_2 + HCOOH$
Thus,the product is $n-$propylamine $(CH_3CH_2CH_2NH_2)$.

(B) $1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ to form ammonium acetate,which upon heating undergoes dehydration to form acetamide $(A)$: $CH_3COOH + NH_3$ $\rightarrow CH_3COONH_4$ $\xrightarrow{\Delta} CH_3CONH_2 (A) + H_2O$.
$2$. Acetamide $(A)$ is reduced by $LiAlH_4$ to form ethylamine $(B)$: $CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2 (B)$.
$3$. Ethylamine $(B)$ reacts with chloroform $(CHCl_3)$ in the presence of $KOH$ (Carbylamine reaction) to form ethyl isocyanide $(C)$: $CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC (C) + 3KCl + 3H_2O$.
$4$. Therefore,$B$ is $CH_3CH_2NH_2$ and $C$ is $CH_3CH_2NC$.

(C) Intermolecular $H$-bonding occurs in molecules where a hydrogen atom is covalently bonded to a highly electronegative atom $(N, O, F)$ and is attracted to another electronegative atom in a neighboring molecule.
Acetamide $(CH_3CONH_2)$ contains an $N-H$ bond,which allows it to act as a hydrogen bond donor,and a $C=O$ group,which acts as a hydrogen bond acceptor. This enables the formation of intermolecular $H$-bonds between the $N-H$ of one molecule and the $C=O$ of another.
Ethyl acetate,methyl formate,and acetic anhydride lack $N-H$ or $O-H$ bonds,so they cannot form intermolecular $H$-bonds.
Thus,the correct option is $(C)$.

(C) The basicity of amines is inversely proportional to their $pK_{b}$ values. Stronger bases have lower $pK_{b}$ values,while weaker bases have higher $pK_{b}$ values.
$1$. Comparing aliphatic amines: In aqueous solution,the basicity order of methyl-substituted amines is secondary $(CH_{3}NHCH_{3})$ > primary $(CH_{3}NH_{2})$ > tertiary $((CH_{3})_{3}N)$. Thus,$(CH_{3})_{3}N$ is a weaker base than $CH_{3}NH_{2}$,meaning $(CH_{3})_{3}N$ has a higher $pK_{b}$ value than $CH_{3}NH_{2}$.
$2$. Comparing with benzylamines: Benzylamines ($C_{6}H_{5}CH_{2}NH_{2}$ and $C_{6}H_{5}CH_{2}NHCH_{3}$) are significantly less basic than aliphatic amines due to the electron-withdrawing $-I$ effect of the phenyl group.
$3$. Comparing benzylamines: $C_{6}H_{5}CH_{2}NHCH_{3}$ (secondary) is more basic than $C_{6}H_{5}CH_{2}NH_{2}$ (primary) due to the $+I$ effect of the additional methyl group,despite the steric hindrance.
Basicity order: $CH_{3}NH_{2} > (CH_{3})_{3}N > C_{6}H_{5}CH_{2}NHCH_{3} > C_{6}H_{5}CH_{2}NH_{2}$.
Decreasing order of $pK_{b}$ values (inverse of basicity): $C_{6}H_{5}CH_{2}NH_{2} (d) > C_{6}H_{5}CH_{2}NHCH_{3} (c) > (CH_{3})_{3}N (b) > CH_{3}NH_{2} (a)$.
Thus,the correct order is $d > c > b > a$.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$(A)$ $(C_2H_5)_2NH$ is a secondary amine with two electron-donating ethyl groups,which increase the electron density on $N$ via the $+I$ effect,making it the most basic.
$(B)$ $C_2H_5NH_2$ is a primary amine with one ethyl group,making it less basic than the secondary amine.
$(C)$ $NH_3$ has no alkyl groups to increase electron density,so it is less basic than aliphatic amines.
$(D)$ $C_6H_5NH_2$ (aniline) has the lone pair of electrons on $N$ involved in resonance with the benzene ring,significantly reducing its availability for protonation,making it the least basic.
Thus,the correct order of basicity is $(A) > (B) > (C) > (D)$.

(A) The basic strength of aliphatic amines in aqueous solution is determined by the combined effect of three factors:
$(i)$ Inductive effect: The $+I$ effect of alkyl groups increases electron density on the nitrogen atom,making tertiary amines potentially more basic.
$(ii)$ Solvation effect: Primary amines are stabilized more effectively by hydrogen bonding with water molecules,which favors their basicity.
$(iii)$ Steric hindrance: Bulky alkyl groups in tertiary amines hinder the approach of a proton to the nitrogen lone pair,reducing basicity.
Considering the interplay of these factors for methyl-substituted amines,the observed order of basicity in aqueous solution is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.

(C) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom. Electron-withdrawing groups (like $-NO_2$) decrease basicity by pulling electron density away from the nitrogen atom via inductive $(-I)$ and resonance $(-R)$ effects.
$(a)$ Aniline: The lone pair is involved in resonance with the benzene ring.
$(b)$ $p$-Nitroaniline: The $-NO_2$ group at the para position exerts a strong $-R$ effect,significantly reducing the basicity compared to aniline.
$(c)$ $2,4$-Dinitroaniline: Two $-NO_2$ groups exert a very strong electron-withdrawing effect,making it the least basic.
$(d)$ Ammonia $(NH_3)$: It does not have a benzene ring to delocalize the lone pair,making it more basic than the substituted anilines.
Therefore,the order of basic strength is: $d > a > b > c$.

(B) Step $1$: Toluene undergoes nitration with $HNO_3/H_2SO_4$ at $288 \ K$ to give $p$-nitrotoluene as the major product.
Step $2$: Reduction of $p$-nitrotoluene using $Sn/HCl$ gives $4$-methylaniline ($p$-toluidine).
Step $3$: The $-NH_2$ group is a strong activating group. Bromination of $4$-methylaniline with $Br_2/H_2O$ results in the substitution of bromine atoms at both ortho positions relative to the $-NH_2$ group,yielding $2,6$-dibromo-$4$-methylaniline as the major product.

(A) $Step$ $1$: $p$-toluidine ($4$-methylaniline) reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form $4$-methylbenzenediazonium chloride.
$Step$ $2$: The diazonium salt reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form $4$-chlorotoluene.
$Step$ $3$: $4$-chlorotoluene reacts with $Mg$ in dry ether to form $4$-methylphenylmagnesium chloride (a Grignard reagent).
$Step$ $4$: The Grignard reagent reacts with $D_3O^+$ to replace the $-MgCl$ group with a deuterium atom,resulting in $4$-deuterotoluene (also known as $4$-methylphenyl-deuteride).

(D) The reaction of benzene diazonium chloride $(C_6H_5N_2Cl)$ with ethanol $(CH_3CH_2OH)$ is a well-known reduction reaction.
Ethanol acts as a reducing agent and gets oxidized to acetaldehyde $(CH_3CHO)$,while benzene diazonium chloride is reduced to benzene $(C_6H_6)$,with the evolution of nitrogen gas $(N_2)$ and formation of hydrochloric acid $(HCl)$.
The reaction is: $C_6H_5N_2Cl + CH_3CH_2OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl$.
Therefore,option $D$ is the correctly matched pair.

(A) The conversion of an arenediazonium salt $(ArN_2^+Cl^-)$ to an arene $(ArH)$ is a reduction reaction. This can be achieved using ethanol $(CH_3CH_2OH)$ or hypophosphorous acid $(H_3PO_2)$ in the presence of water.
In the given reaction scheme,$X$ represents the reagent $CH_3CH_2OH$ and $Y$ represents the reagent $H_3PO_2/H_2O$,which are both standard reagents for the deamination of diazonium salts to arenes.
Therefore,the correct option is $A$.

(B) The Friedel-Crafts reaction involves an electrophilic aromatic substitution catalyzed by a Lewis acid (e.g.,$AlCl_3$).
Aniline $(C_6H_5NH_2)$ contains a lone pair of electrons on the nitrogen atom.
This lone pair reacts with the Lewis acid catalyst $(AlCl_3)$ to form a complex $(C_6H_5NH_2 \rightarrow AlCl_3)$.
This formation of the complex deactivates the benzene ring towards electrophilic substitution and makes the nitrogen atom positively charged,which strongly withdraws electrons from the ring.
Therefore,aniline does not undergo Friedel-Crafts reactions.

(D) The reaction sequence is as follows:
$1$. The starting material is $4$-amino-$3$-bromoethylbenzene. Treatment with $NaNO_2/HCl$ at $273-278 \ K$ performs diazotization of the $-NH_2$ group to form the diazonium salt,$X$ ($4$-ethyl-$2$-bromobenzenediazonium chloride).
$2$. Reaction of $X$ with ethanol $(C_2H_5OH)$ reduces the diazonium group to a hydrogen atom,yielding $Y$ ($1$-bromo-$2$-ethylbenzene).
$3$. Finally,oxidation of the ethyl group with alkaline $KMnO_4$ converts the benzylic carbon into a carboxylic acid group,resulting in $Z$ ($2$-bromobenzoic acid).
Thus,the correct structure for $Z$ is $2$-bromobenzoic acid,which corresponds to option $(D)$.

(A) The conversion of $3$-methylaniline to $3$-nitrotoluene involves the following steps:
$1.$ Diazotization: $3-CH_3-C_6H_4-NH_2 + NaNO_2 + HCl \xrightarrow{273-278 \ K} 3-CH_3-C_6H_4-N_2^+Cl^-$ (Reagent $X = NaNO_2, HCl$).
$2.$ Reaction with $HBF_4$: $3-CH_3-C_6H_4-N_2^+Cl^- + HBF_4 \rightarrow 3-CH_3-C_6H_4-N_2^+BF_4^-$ (Reagent $Y = HBF_4$).
$3.$ Replacement by nitro group: $3-CH_3-C_6H_4-N_2^+BF_4^- + NaNO_2 \xrightarrow{Cu, \Delta} 3-CH_3-C_6H_4-NO_2$ (Reagent $Z = NaNO_2, Cu, \Delta$).

(B) The reaction of arene diazonium fluoroborate $(Ar-N_2^+ BF_4^-)$ with sodium nitrite $(NaNO_2)$ in the presence of copper $(Cu)$ powder and heat is a method to introduce a nitro group into an aromatic ring,known as the Schiemann-like reaction or a variation of the Sandmeyer-type reaction for nitro compounds.
The reaction proceeds as follows:
$Ar-N_2^+ BF_4^- + NaNO_2 \xrightarrow{Cu, \Delta} ArNO_2 + N_2 + NaBF_4$
Here,$A = ArNO_2$,$B = N_2$,and $C = NaBF_4$.
Thus,the correct option is $B$.

(A) Schiff bases are formed by the reaction of a primary amine $(R-NH_2)$ with a carbonyl compound,such as an aldehyde or a ketone.
In this reaction,the amine attacks the carbonyl carbon to form a hemiaminal intermediate,which subsequently undergoes dehydration to produce an imine,commonly known as a Schiff base.
Therefore,$A$ represents an aldehyde or a ketone.

(B) $N$-ethylbenzene sulphonamide $(C_6H_5SO_2NHC_2H_5)$ is soluble in alkali because the hydrogen atom attached to the nitrogen atom is acidic due to the strong electron-withdrawing effect of the sulphonyl group $(-SO_2-)$.
This allows it to form a water-soluble salt with alkali.
$N,N$-diethylbenzene sulphonamide does not have an acidic hydrogen atom attached to the nitrogen,so it is insoluble in alkali.
Tertiary amines do not react with acid chlorides or Hinsberg's reagent $(C_6H_5SO_2Cl)$ because they lack an acidic hydrogen atom on the nitrogen.

(C) The given reaction sequence is:
$1$. $CH_3CONH_2 \xrightarrow{Br_2/NaOH(aq)} CH_3NH_2 (X)$
This is the Hofmann bromamide degradation reaction,where an amide is converted into a primary amine with one carbon atom less.
$2$. $CH_3NH_2 (X) \xrightarrow[(ii) H_2O]{(i) NaNO_2 + HCl} CH_3OH (Y)$
Primary aliphatic amines react with nitrous acid ($HNO_2$,generated in situ from $NaNO_2 + HCl$) to form unstable diazonium salts,which decompose in the presence of water to form alcohols.
Therefore,$Y$ is methanol $(CH_3OH)$.

(A) The reaction sequence is as follows:
$1$. Nitrobenzene $(C_6H_5NO_2)$ reacts with $Fe/HCl$ to undergo reduction,forming aniline $(C_6H_5NH_2)$,which is '$X$'.
$2$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is '$Y$'.
$3$. Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ reacts with $H_3PO_2$ and $H_2O$ to undergo reduction,resulting in the formation of benzene $(C_6H_6)$,which is '$Z$'.

(C) Statement $A$ is incorrect: Benzene diazonium chloride is stable only at low temperatures $(273-278 \ K)$. At $285 \ K$,it is unstable and decomposes.
Statement $B$ is correct: The $-NHCOCH_3$ group in $N$-phenylethanamide is electron-withdrawing due to resonance with the carbonyl group,making the ring less electron-rich compared to the $-NH_2$ group in aniline. Thus,it is less reactive towards electrophilic substitution like nitration.
Statement $C$ is incorrect: Hinsberg reagent is benzenesulphonyl chloride $(C_6H_5SO_2Cl)$,not $p-CH_3C_6H_4COCl$.
Therefore,only statement $B$ is correct.

(B) $1$. The starting material is $4$-amino-$3$-bromoethylbenzene. Treatment with $NaNO_2/HCl$ at $273 \ K$ converts the $-NH_2$ group into a diazonium salt $(-N_2^+Cl^-)$,forming compound $A$.
$2$. Treatment of the diazonium salt with $H_3PO_2/H_2O$ reduces the diazonium group to a hydrogen atom,resulting in $1$-bromo-$2$-ethylbenzene (compound $B$).
$3$. Finally,oxidation of the ethyl group $(-CH_2CH_3)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ converts the alkyl side chain into a carboxylic acid group $(-COOH)$. The product $C$ is $2$-bromobenzoic acid.

(B) $1$. The starting material is $4$-methylbenzamide. Treatment with $NaOH/Br_2$ is the Hoffmann bromamide degradation reaction,which converts an amide into a primary amine with one less carbon atom. Thus,$X$ is $p$-toluidine ($4$-methylaniline).
$2$. The reaction of $p$-toluidine with benzoyl chloride $(C_6H_5COCl)$ is an acylation reaction. The lone pair on the nitrogen atom of the amine attacks the carbonyl carbon of the acid chloride,resulting in the formation of an amide. The product $Y$ is $N-(4-methylphenyl)benzamide$.

(D) secondary amine reacts with Hinsberg reagent ($p-$toluene sulphonyl chloride) to form a corresponding sulphonamide that is insoluble in aqueous $NaOH$. Thus,'$X$' must be a secondary amine. The reaction of a secondary amine $(R_1R_2NH)$ with benzoyl chloride $(C_6H_5COCl)$ is a benzoylation reaction (Schotten-Baumann reaction),which yields an $N,N-$disubstituted benzamide. Given the options,the secondary amine '$X$' is $N-$methylethanamine $(CH_3CH_2NHCH_3)$. The reaction is: $CH_3CH_2NHCH_3 + C_6H_5COCl \rightarrow CH_3CH_2N(CH_3)COC_6H_5 + HCl$.

(B) In $RNH_2$,the $N$ atom is $sp^3$ hybridized. In $RN=CHR^1$,the $N$ atom is $sp^2$ hybridized,whereas in $RC \equiv N$,the $N$ atom is $sp$ hybridized.
The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom. As the $s$-character of the hybrid orbital increases,the electronegativity of the nitrogen atom increases,making the lone pair less available for donation.
The $s$-character increases in the order: $sp^3 < sp^2 < sp$.
Therefore,the electronegativity of the $N$ atom increases in the order: $sp^3 < sp^2 < sp$.
Consequently,the basicity decreases as the $s$-character increases.
The order of decreasing basicity is: $RNH_2 (III) > RN=CHR^1 (I) > RC \equiv N (II)$.