Properties of Amines Questions in English

(A) The reaction is the Hofmann bromamide degradation,where the amide nitrogen is retained in the amine product.
$1$. For $X$: Benzoic acid reacts with $ND_3$ and $\Delta$ to form benzamide $(C_6H_5COND_2)$. Subsequent reaction with $KOH$ and $Br_2$ yields $C_6H_5ND_2$ $(X)$,as the nitrogen atom of the amide is retained in the amine.
$2$. For $Y$: Benzoic acid reacts with $NH_3$ and $\Delta$ to form benzamide $(C_6H_5CONH_2)$. Subsequent reaction with $KOD$ and $Br_2$ yields $C_6H_5NH_2$ $(Y)$,as the nitrogen atom of the amide is retained in the amine.
Thus,$X$ is $C_6H_5ND_2$ and $Y$ is $C_6H_5NH_2$.

(A) Amines act as Lewis bases due to the presence of a lone pair of electrons on the nitrogen atom.
$(a)$ $CH_3CH_2NH_2 + HI \rightarrow [CH_3CH_2NH_3]^+ I^-$. The product is $CH_3CH_2NH_3^+ I^-$,which is named Ethylammonium iodide.
$(b)$ $(CH_3)_3N + HBr \rightarrow [(CH_3)_3NH]^+ Br^-$. The product is $(CH_3)_3NH^+ Br^-$,which is named Trimethylammonium bromide.
Therefore,the correct option is $(a)$ Ethylammonium iodide and $(b)$ Trimethylammonium bromide.

(D) The basicity of substituted $N,N$-dimethylanilines depends on the electron-donating or electron-withdrawing nature of the substituent and steric effects.
In $4$-methyl-$N,N$-dimethylaniline,the $-CH_3$ group at the para position exerts a $+I$ effect and hyperconjugation,which increases the electron density on the nitrogen atom,making it more available for protonation.
In $2$-methyl-$N,N$-dimethylaniline,the ortho-substituent causes steric hindrance,which inhibits the resonance of the lone pair of the nitrogen with the benzene ring (ortho effect),but the steric hindrance also makes it difficult for the proton to approach the nitrogen,generally reducing its basicity compared to the para-isomer.
$3$-methyl-$N,N$-dimethylaniline has a $+I$ effect but lacks the strong resonance stabilization of the para-isomer.
Therefore,$4$-methyl-$N,N$-dimethylaniline is the strongest base among the given options.

(B) Nitrogen $2$ is the most basic because its lone pair is not involved in resonance,and it is a tertiary amine with electron-donating alkyl groups (inductive effect).
Nitrogen $3$ is the least basic because its lone pair is in resonance with the $C=O$ double bond of the amide group,significantly reducing its availability.
Nitrogen $1$ is less basic than $2$ because its lone pair is involved in resonance with the indole ring system (aromatic system).
Therefore,the correct order of basicity is $2 > 1 > 3$.

(A) The reaction of $2-$aminocyclohexane$-1-$ol with nitrous acid $(HNO_2)$ is a deamination reaction involving the formation of a diazonium salt intermediate.
Since the amino group is adjacent to a hydroxyl group,the diazonium ion formed is unstable and undergoes a Tiffeneau-Demjanov rearrangement or ring expansion/contraction depending on the structure.
However,for $2-$aminocyclohexane$-1-$ol,the primary product is the ring-contracted product,which is $Cyclopentane \text{ } carboxaldehyde$.

(D) To determine the basicity,we analyze the availability of the lone pair on each nitrogen atom:
$(3)$ is a tertiary aliphatic amine. The lone pair is localized and available for donation,making it the most basic.
$(4)$ is an aniline derivative $(Ph-NH-)$. The lone pair is in conjugation with the benzene ring,reducing its availability,but it is more basic than amides.
$(1)$ is an amide nitrogen $(R-NH-CO-CH_3)$. The lone pair is involved in resonance with the carbonyl group $(C=O)$,making it significantly less basic.
$(2)$ is a primary amide nitrogen $(R-CO-NH_2)$. The lone pair is involved in resonance with the carbonyl group,and it is even less basic due to the strong electron-withdrawing effect of the carbonyl group in a primary amide structure.
Comparing these,the order of basicity is $(3) > (4) > (1) > (2)$.

(D) The hydrolysis of alkyl isocyanides with dilute mineral acids yields primary amines and formic acid.
The reaction is as follows:
$R-CH_2-NC + 2H_2O \xrightarrow{H^+} R-CH_2-NH_2 + HCOOH$

(A) The reaction of primary aliphatic amines with nitrous acid $(HNO_2)$ produces an unstable diazonium salt,which immediately decomposes to form a carbocation. This carbocation then reacts with water to form an alcohol as the major product.
$CH_3NH_2 + HNO_2 \rightarrow [CH_3N_2^+] + H_2O$
$[CH_3N_2^+] + H_2O \rightarrow CH_3OH + N_2 + H^+$

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In aromatic amines,the lone pair is involved in resonance with the benzene ring,which decreases its availability.
However,the presence of electron-donating groups (like alkyl groups) on the nitrogen atom increases the electron density on the nitrogen,thereby increasing its basicity.
Comparing the given options:
$C_6H_5-NH_2$ (Aniline) has no alkyl groups.
$C_6H_5-NHCH_3$ has one methyl group.
$C_6H_5-N(CH_3)_2$ has two methyl groups.
$C_6H_5-N(C_2H_5)_2$ has two ethyl groups.
Ethyl groups are better electron-donating groups than methyl groups due to the $+I$ effect.
Therefore,$C_6H_5-N(C_2H_5)_2$ is the most basic compound among the given options.

(B) The reaction $ArN_2^{\oplus} + PhOH \rightarrow Ar-OH$ is incorrect as written. The coupling reaction between a diazonium salt and phenol $(PhOH)$ produces an azo dye,specifically $p$-hydroxyazobenzene $(Ar-N=N-C_6H_4-OH)$,not a phenol $(Ar-OH)$.
Option $A$ is the Sandmeyer reaction,which is valid.
Option $C$ is the reduction of diazonium salt using hypophosphorous acid $(H_3PO_2)$,which is valid.
Option $D$ is the standard method for preparing aryl iodides,which is valid.

(D) The preparation of $m$-bromotoluene cannot be done by direct bromination of toluene because the $-CH_3$ group is ortho/para directing.
To obtain the meta-isomer,we use the following sequence:
$1$. Acetylation of $p$-toluidine to form aceto-$p$-toluidine.
$2$. Bromination of aceto-$p$-toluidine,where the $-NHCOCH_3$ group is ortho/para directing,directing the bromine to the position meta to the methyl group.
$3$. Hydrolysis of the amide to remove the acetyl group.
$4$. Deamination (diazotization followed by reduction) to remove the $-NH_2$ group,leaving the bromine at the meta position relative to the methyl group.

(B) In an acidic medium,aniline $(C_6H_5NH_2)$ gets protonated to form the anilinium ion $(C_6H_5NH_3^+)$.
The $-NH_3^+$ group is meta-directing because it is strongly electron-withdrawing due to the positive charge on the nitrogen atom.
Consequently,the electrophilic substitution (chlorination) occurs primarily at the meta-position.
Therefore,the major product formed is $m-$chloroaniline.

(D) Chlorination of aromatic compounds is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron-donating ability of the substituent group attached to the benzene ring. Stronger electron-donating groups increase the electron density in the ring,making it more reactive towards electrophiles.
The electron-donating power of the groups is as follows:
$-N(CH_3)_2 > -NHCH_3 > -NH_2 > -OH$.
Among the given options,the dimethylamino group $(-N(CH_3)_2)$ is the strongest electron-donating group due to the $+I$ effect of the two methyl groups,which further increases the electron density on the nitrogen atom and consequently on the benzene ring. Therefore,$N,N$-dimethylaniline $(C_6H_5N(CH_3)_2)$ undergoes chlorination at the fastest rate.

(A) Nitration is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring. Groups that donate electrons to the ring (activating groups) increase the rate of nitration,while groups that withdraw electrons (deactivating groups) decrease the rate.
$1$. $C_6H_5NH_2$ (Aniline): The $-NH_2$ group has a lone pair of electrons on the nitrogen atom,which it donates to the benzene ring through resonance ($+R$ effect). This strongly activates the ring towards electrophilic substitution.
$2$. $C_6H_5N^+(CH_3)_3$: The $-N^+(CH_3)_3$ group is a strong electron-withdrawing group due to its positive charge ($-I$ effect),which deactivates the ring.
$3$. $C_6H_5CH_2N^+(CH_3)_3$ and $C_6H_5CH_2CH_2N^+(CH_3)_3$: These also contain electron-withdrawing groups ($-I$ effect) attached to the ring,which deactivate the ring.
Therefore,$C_6H_5NH_2$ is the most activated and undergoes nitration fastest.

(A) Sulphanilic acid $(p-H_2NC_6H_4SO_3H)$ contains a strongly activating amino group $(-NH_2)$ and a sulfonic acid group $(-SO_3H)$.
In the presence of water,the amino group is protonated to form the zwitterion $(H_3N^+C_6H_4SO_3^-)$.
However,the $-NH_2$ group is highly activating and directs electrophilic substitution to the ortho positions.
When treated with excess bromine water,the $-SO_3H$ group is displaced by bromine due to the high reactivity of the intermediate,leading to the formation of $2,4,6-tribromoaniline$.
Therefore,the final product is a tribromo product.

(A) In $meta$-nitroaniline,the $-NH_2$ group is a strong ortho/para-directing group,while the $-NO_2$ group is a meta-directing group.
Both groups direct the incoming electrophile $(Cl^+)$ to the same position,which is the position ortho to the $-NH_2$ group and para to the $-NO_2$ group (position $4$).
Therefore,the major product formed is $4$-chloro-$3$-nitroaniline.

(A) $1$. The reaction of the acid chloride $(CH_3-C(H)(C_6H_4-D)-COCl)$ with $NH_3$ yields the corresponding amide,$CH_3-C(H)(C_6H_4-D)-CONH_2$. This step does not involve the chiral center,so the configuration is retained.
$2$. The subsequent reaction with $OH^-/Br_2$ is the Hoffmann Bromamide degradation. In this reaction,the carbonyl group $(CO)$ is removed,and the $NH_2$ group attaches directly to the chiral carbon atom.
$3$. The Hoffmann Bromamide degradation proceeds with complete retention of configuration at the chiral center because the migrating group does not break its bond with the chiral carbon until the new bond with nitrogen is formed.
$4$. Therefore,the final product is $CH_3-C(H)(C_6H_4-D)-NH_2$ with the same configuration as the starting material.

(D) The starting material is $4$-methylacetanilide. The acetamido group $(-NHCOCH_3)$ is a strong activating group due to the $+M$ effect of the nitrogen lone pair,making it ortho/para directing. The methyl group $(-CH_3)$ is also activating and ortho/para directing.
In $4$-methylacetanilide,the para position relative to the acetamido group is occupied by the methyl group. Therefore,electrophilic aromatic substitution (bromination) occurs at the ortho position relative to the acetamido group.
Since the methyl group is at the para position,the ortho position relative to the acetamido group is also the meta position relative to the methyl group. The acetamido group is a stronger activator than the methyl group,so it directs the incoming electrophile $(Br^+)$. The product is $2$-bromo-$4$-methylacetanilide.

(C) The chemical equation for the carbylamine reaction is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.
Here,the number of moles of $KOH$ $(X_1)$ is $3$ and the number of moles of $CHCl_3$ $(X_2)$ is $1$.
Therefore,$X_1 - X_2 = 3 - 1 = 2$.

(C) The conversion of $R-NH_2$ to $R-NH-CH_3$ involves two steps:
$1$. Carbylamine reaction: $R-NH_2 + CHCl_3 + 3KOH \to R-NC + 3KCl + 3H_2O$
$2$. Reduction: $R-NC + 4[H] \xrightarrow{LiAlH_4} R-NH-CH_3$.
This sequence effectively adds a methyl group to the nitrogen atom of a primary amine to form a secondary amine.

(B) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$1$. $CH_3CH_2CH_2NH_2$ (n-propylamine) is an aliphatic amine,where the lone pair is readily available for protonation.
$2$. $(CH_3)_3N$ (trimethylamine) is also an aliphatic amine with high electron density on nitrogen due to the inductive effect of three methyl groups.
$3$. Aniline $(C_6H_5NH_2)$ is less basic because the lone pair on nitrogen is involved in resonance with the benzene ring.
$4$. $2-$aminoacetophenone has an $-NH_2$ group attached to a benzene ring which also has an electron-withdrawing acetyl group $(-COCH_3)$ at the ortho position. The $-COCH_3$ group exerts a strong $-I$ and $-M$ effect,which significantly reduces the electron density on the nitrogen atom,making it the least basic among the given options.

(C) Ethyl amine $(C_2H_5NH_2)$ contains an acidic hydrogen atom attached to the nitrogen atom.
When it reacts with an active metal like sodium $(Na)$,the sodium replaces the hydrogen atom to form sodium ethylamide $(C_2H_5NHNa)$ and releases hydrogen gas $(H_2)$.
The chemical equation is:
$2C_2H_5NH_2 + 2Na \to 2C_2H_5NHNa + H_2 \uparrow$
Therefore,the gas evolved is $H_2$.

(C) The reaction between aniline $(C_6H_5NH_2)$ and benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is an electrophilic aromatic substitution reaction known as coupling reaction.
In this reaction,the diazonium ion acts as an electrophile and attacks the electron-rich ring of aniline.
Since the $-NH_2$ group is strongly activating and ortho/para-directing,the electrophile preferentially attacks the para-position due to steric hindrance at the ortho-position.
Thus,the major product formed is $p$-aminoazobenzene,which is a yellow-orange coloured dye.

(A) The reaction of an aliphatic primary amine with nitrous acid $(HNO_2)$ produces a diazonium salt,which is highly unstable and decomposes to form a carbocation.
In this case,the starting material is $Ph-CH_2-CH_2-CH(NH_2)-COOH$.
Upon treatment with $HNO_2$,the $-NH_2$ group is replaced by a diazonium group $(-N_2^+)$,which leaves as $N_2$ gas,generating a carbocation at the $\alpha$-position relative to the $-COOH$ group.
This carbocation is then attacked by water (the nucleophile) to form the corresponding $\alpha$-hydroxy acid.
The configuration at the chiral center is generally retained or undergoes partial inversion depending on the mechanism,but the primary product is the $\alpha$-hydroxy acid,which is $2$-hydroxy-$4$-phenylbutanoic acid.

(C) The reaction of primary aliphatic amines with nitrous acid $(NaNO_2 + HCl)$ produces unstable diazonium salts,which decompose to form carbocations. These carbocations then react with water to form alcohols as the major product.
In option $C$,the primary aliphatic amine (cyclopentyl-ethylamine) reacts with $NaNO_2 + HCl$ to form a diazonium salt,which subsequently yields the corresponding alcohol.
Option $A$ is a tertiary amine,which forms a salt.
Option $B$ is an aromatic primary amine,which forms a stable diazonium salt at $0-5^{\circ}C$.
Option $D$ is a secondary aromatic amine,which forms an $N$-nitroso compound.

(B) The molecular formula $C_3H_9N$ corresponds to the following four structural isomers of amines:
$1$. $CH_3CH_2CH_2NH_2$ (propan$-1-$amine)
$2$. $CH_3CH(NH_2)CH_3$ (propan$-2-$amine)
$3$. $CH_3CH_2NHCH_3$ ($N$-methylethanamine)
$4$. $(CH_3)_3N$ ($N,N$-dimethylmethanamine)
These represent primary,secondary,and tertiary amines. Thus,there are $4$ structural isomers.
Hence,option $B$ is correct.

(A) Step $1$: Aniline $(Ph-NH_2)$ reacts with $HNO_2$ at $0-5 \ ^\circ C$ to form a diazonium salt,but in the presence of $Br_2$,the reaction proceeds to form $2,4,6-tribromoaniline$ first,which then undergoes diazotization to form $2,4,6-tribromobenzenediazonium$ chloride $(A)$.
Step $2$: When $2,4,6-tribromobenzenediazonium$ chloride $(A)$ is warmed with water,it undergoes hydrolysis to form $2,4,6-tribromophenol$ $(B)$.
Note: Given the options provided,the intended reaction sequence likely refers to the formation of phenol if the $Br_2$ was not present,or the question implies the final product of the diazotization and hydrolysis sequence. However,based on the standard reaction of aniline with $HNO_2$ followed by $H_2O/warm$,the product is $Phenol$.

(D) The reaction of benzenediazonium chloride $(PhN_2Cl)$ with water $(H_2O)$ upon heating $(\Delta)$ is a standard hydrolysis reaction.
In this reaction,the diazonium group $(-N_2Cl)$ is replaced by a hydroxyl group $(-OH)$,resulting in the formation of phenol $(PhOH)$ and the release of nitrogen gas $(N_2)$ and hydrochloric acid $(HCl)$.
The reaction is:
$PhN_2Cl H_2O \xrightarrow{\Delta} PhOH N_2 HCl$
Therefore,the product is phenol.

(A) To determine the basicity,we analyze the availability of the lone pair on the nitrogen atom:
$I$ (Pyridine): The lone pair is in an $sp^2$ orbital,perpendicular to the aromatic $\pi$ system,making it available for protonation.
$II$ (Pyrrole): The lone pair is involved in the $6\pi$ aromatic system,making it unavailable for protonation. It is very weakly basic.
$III$ ($3$-Pyrroline): The lone pair is on a nitrogen atom in a non-aromatic ring. It is more basic than pyrrole but less basic than aliphatic amines.
$IV$ (Guanidine): The lone pair is highly stabilized by resonance across three nitrogen atoms,but the resulting conjugate acid is extremely stable due to resonance,making guanidine one of the strongest organic bases.
Comparing the basicity order: $IV > III > I > II$.
Therefore,the most basic compound is $IV$ and the least basic compound is $II$.

(B) In ethylamine $(I)$,the $+I$ effect of the ethyl group increases the electron density on the nitrogen atom,making it the strongest base.
In benzylamine $(II)$,the phenyl group exerts a $-I$ effect,which decreases the electron density on the nitrogen atom compared to ethylamine,but the lone pair is not involved in resonance with the ring.
In aniline $(III)$,the lone pair on the nitrogen atom is delocalized into the benzene ring due to resonance,which significantly reduces the availability of the lone pair for protonation,making it the weakest base.
Therefore,the correct order of basic strength is $I > II > III$.

(D) The basic strength of aromatic amines depends on the availability of the lone pair on the nitrogen atom. Electron-withdrawing groups $(EWG)$ like $-NO_2$ decrease basicity by pulling electron density away from the nitrogen atom through inductive and resonance effects. Electron-donating groups $(EDG)$ like $-CH_3$ (methyl) increase basicity.
$P$: Contains one $-NO_2$ group and two $-CH_3$ groups. The $-CH_3$ groups are electron-donating,which increases the electron density on the nitrogen atom,making it the most basic among the three.
$Q$: Contains three $-NO_2$ groups (two ortho,one para). These are strong electron-withdrawing groups that significantly reduce the electron density on the nitrogen atom,making it the least basic.
$R$: Contains one $-NO_2$ group at the para position. It is less basic than $P$ (due to the absence of electron-donating groups) but more basic than $Q$ (which has three electron-withdrawing groups).
Therefore,the order of basic strength is $Q < R < P$.

(B) The reaction is an electrophilic aromatic substitution (nitration) of $N$-phenylbenzamide (benzanilide).
The $-NHCOC_6H_5$ group is an ortho/para-directing group because the nitrogen atom has a lone pair of electrons that can be delocalized into the benzene ring.
Due to the steric hindrance caused by the bulky $-NHCOC_6H_5$ group,the para-position is more accessible than the ortho-position.
Therefore,the major product is the para-substituted isomer,which is $N$-($4$-nitrophenyl)benzamide.

(D) To determine the basicity,we look at the availability of the lone pair of electrons on the nitrogen atom for protonation.
$A$. In Pyrrole,the lone pair on the nitrogen atom is involved in the aromatic sextet,making it unavailable for protonation. Thus,it is very weakly basic.
$B$. In Aniline,the lone pair on the nitrogen atom is in conjugation with the benzene ring,which reduces its availability for protonation.
$C$. In Pyridine,the lone pair is in an $sp^2$ hybridized orbital. Although it is not involved in the aromatic system,the $sp^2$ character makes it less basic than aliphatic amines.
$D$. In Piperidine,the nitrogen atom is $sp^3$ hybridized and there is no resonance or aromaticity to delocalize the lone pair. Therefore,the lone pair is highly available for protonation,making Piperidine the most basic among the given options.

(C) The basicity of nitrogen atoms depends on the availability of the lone pair.
In $1,3-dimethyl-1,2,3,4-tetrahydroquinoxaline$,the amino group at the $5$-position is conjugated with the benzene ring,reducing its basicity.
The nitrogen at the $1$-position is an aniline-like nitrogen (conjugated with the ring),while the nitrogen at the $3$-position is an aliphatic tertiary amine.
Aliphatic amines are more basic than aromatic/conjugated amines.
Therefore,the nitrogen at the $3$-position is the most basic and will be protonated first to form the stable conjugate acid.

(C) The acidic strength depends on the stability of the conjugate base and the electronegativity of the atom bearing the positive charge.
$A$ is an oxonium ion $(R-OH^{+}-R)$,which is a very strong acid because oxygen is highly electronegative and carries a positive charge.
$D$ is an ammonium ion $(R-NH^{+}-CH_3)$,which is also a strong acid but less acidic than the oxonium ion because nitrogen is less electronegative than oxygen.
$C$ is an alcohol $(R-OH)$,which is a weak acid.
$B$ is an amine $(R-NH_2)$,which is a very weak acid (essentially basic).
Therefore,the decreasing order of acidic strength is $A > D > C > B$.

(C) To determine the basic strength,we analyze the availability of the lone pair on the nitrogen atom:
$1$. $(Q)$ is $N$-methylpiperidine,a tertiary aliphatic amine where the nitrogen is $sp^3$ hybridized and the lone pair is localized,making it the most basic.
$2$. $(P)$ is pyridine,where the nitrogen is $sp^2$ hybridized. The lone pair is in an $sp^2$ orbital,which is more electronegative than an $sp^3$ orbital,making it less basic than $(Q)$.
$3$. $(S)$ is aniline,where the lone pair on the nitrogen is involved in resonance with the benzene ring,significantly reducing its availability for protonation.
$4$. $(R)$ is $\delta$-valerolactam,an amide. The lone pair on the nitrogen is delocalized into the carbonyl group $(C=O)$ via resonance,making it the least basic among the given compounds.
Therefore,the order of basic strength is $(Q) > (P) > (S) > (R)$.

(C) In the presence of concentrated $H_2SO_4$,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$.
This anilinium ion is meta-directing because the $-NH_3^+$ group is strongly electron-withdrawing due to the positive charge on the nitrogen atom.
Consequently,the nitration of aniline in an acidic medium yields a significant amount of meta-nitroaniline $(47\%)$ along with ortho $(2\%)$ and para $(51\%)$ products.
Therefore,the correct distribution is $I$ (para) $\to 51\%$,$II$ (meta) $\to 47\%$,and $III$ (ortho) $\to 2\%$.

(A) The reaction sequence is as follows:
$CH_3CN$ $\xrightarrow{\text{Reduction}} CH_3CH_2NH_2$ $\xrightarrow{HNO_2} C_2H_5OH + N_2 + H_2O$
Here,$A$ is $CH_3CN$ (acetonitrile or ethanenitrile).
Reduction of $CH_3CN$ gives ethylamine $(CH_3CH_2NH_2)$,which on reaction with nitrous acid $(HNO_2)$ yields ethanol $(C_2H_5OH)$.

(A) To determine the basic strength,we look at the availability of the lone pair on the nitrogen atom:
$1$. In $Piperidine$,the nitrogen atom is $sp^3$ hybridized and its lone pair is localized,making it the most basic.
$2$. In $Pyridine$,the nitrogen atom is $sp^2$ hybridized. The lone pair is in an $sp^2$ orbital,which is more electronegative than an $sp^3$ orbital,making it less basic than $Piperidine$.
$3$. In $Pyrrole$,the lone pair on the nitrogen atom is involved in the aromatic sextet (delocalized). Therefore,it is not available for protonation,making it the least basic.
Thus,the correct order of basic strength is $Piperidine > Pyridine > Pyrrole$.

(B) To determine the $pK_b$ order,we first determine the basicity order. Higher basicity corresponds to lower $pK_b$ values.
$I$: Vinylamine $(CH_2=CH-NH_2)$. The lone pair on $N$ is in conjugation with the double bond,reducing its availability.
$II$: $1,2,3,4$-Tetrahydroquinoline. The lone pair on $N$ is not involved in resonance with the benzene ring due to the saturated carbons,making it a relatively strong base.
$III$: Pyrrolidine. It is a saturated secondary amine,making it the most basic among the given compounds.
$IV$: Pyrrole. The lone pair on $N$ is involved in the aromatic sextet,making it extremely weakly basic.
Basicity order: $III > II > I > IV$.
Since $pK_b = -\log(K_b)$,the order of $pK_b$ is the inverse of the basicity order: $IV > I > II > III$.

(B) $1$. The reduction of phenyl isocyanide $(Ph-NC)$ with $LiAlH_4$ yields $N$-methylaniline $(Ph-NH-CH_3)$ as product $A$.
$2$. $N$-methylaniline is a secondary aromatic amine.
$3$. Secondary amines react with nitrous acid $(NaNO_2 + HCl)$ at low temperatures $(0-5 \ ^\circ C)$ to form $N$-nitroso compounds.
$4$. Therefore,the reaction of $N$-methylaniline with $NaNO_2 + HCl$ gives $N$-nitroso-$N$-methylaniline as product $B$.

(B) The reaction involves the following steps:
$1$. The reaction of $3$-methylbutan-$2$-one (an aldehyde/ketone) with methylamine $(CH_3NH_2)$ forms an imine intermediate $(R_2C=NCH_3)$.
$2$. The subsequent reduction with $LiAlH_4$ reduces the $C=N$ double bond to a $C-NH$ single bond,resulting in a secondary amine.
$3$. The starting material is $3$-methylbutan-$2$-one,which is $(CH_3)_2CH-C(=O)-CH_3$.
$4$. The reaction with $CH_3NH_2$ forms the imine $(CH_3)_2CH-C(=NCH_3)-CH_3$.
$5$. Reduction with $LiAlH_4$ gives the amine $(CH_3)_2CH-CH(NHCH_3)-CH_3$.

(A) The $pK_b$ value is inversely proportional to the basic strength of the compound. Higher $pK_b$ means lower basic strength.
Among the given options,$NH_3$ is the weakest base because alkyl groups are electron-donating ($+I$ effect),which increases the electron density on the nitrogen atom,thereby increasing basicity.
In the gas phase,the order of basicity is $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$.
In aqueous solution,the order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ due to solvation effects and steric hindrance.
In both cases,$NH_3$ remains the weakest base,and therefore,it has the highest $pK_b$ value.

(D) The reaction between benzenediazonium chloride and aniline in a weakly acidic medium $(pH \approx 4-5)$ is an electrophilic aromatic substitution reaction known as coupling reaction.
In this reaction,the diazonium cation acts as an electrophile and attacks the electron-rich ring of aniline.
Since the $-NH_2$ group is a strong ortho/para directing group,and due to steric hindrance at the ortho position,the electrophile preferentially attacks the para position of aniline to form $p$-aminoazobenzene,which is a yellow dye.

(D) The reaction is a Hofmann elimination of a quaternary ammonium hydroxide.
In Hofmann elimination,the base $(OH^{-})$ abstracts a proton from the $\beta$-carbon that is least sterically hindered.
The possible $\beta$-hydrogens are on the cyclohexyl ring,the ethyl group,and the $n$-butyl group.
The transition state for the formation of the least substituted alkene is favored due to steric hindrance.
Among the alkyl groups attached to the nitrogen,the ethyl group has the most accessible $\beta$-hydrogens.
Therefore,the elimination of the ethyl group leads to the formation of ethene $(CH_2=CH_2)$ as the major product.

(B) The reaction between an acid chloride $(CH_3COCl)$ and a primary amine $(CH_3NH_2)$ is the most effective method for preparing amides.
Acid chlorides are highly reactive towards nucleophilic substitution by amines, providing high yields of the $N$-substituted amide at room temperature.

(D) The reaction of a primary aliphatic amine with $NaNO_2 + HCl$ (nitrous acid) forms an unstable diazonium salt.
This diazonium salt quickly decomposes to form a carbocation.
In this case,the $2$-methylcyclohexyl cation is formed.
This secondary carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation ($1$-methylcyclohexyl cation).
Finally,the nucleophilic attack by $H_2O$ on the tertiary carbocation yields $1$-methylcyclohexanol as the major product.

(C) The reaction sequence is as follows:
$1$. The starting material (not explicitly shown but implied as $4-$nitro$-2,6-$dibromoaniline or similar derivative) undergoes diazotization followed by Sandmeyer reaction with $CuCl/HCl$ to replace the amino group with a chlorine atom,forming compound $I$ $(1-chloro-2,6-dibromo-4-nitrobenzene)$.
$2$. Reduction of the nitro group in compound $I$ using $Sn/HCl$ yields compound $II$ $(4-amino-3,5-dibromochlorobenzene)$.
$3$. Diazotization of compound $II$ with $HNO_2$ at $0-5\,^oC$ followed by treatment with $H_3PO_2$ and $H_2O$ (reductive deamination) removes the amino group,resulting in compound $III$ $(1-chloro-2,6-dibromobenzene)$.

(A) The reaction sequence is as follows:
$1$. $p$-Toluidine reacts with acetic anhydride in the presence of pyridine to form $N$-(p-tolyl)acetamide (compound $A$).
$2$. Compound $A$ undergoes electrophilic aromatic substitution with $Br_2$ in $CH_3COOH$. The acetamido group $(-NHCOCH_3)$ is ortho/para directing. Since the para position is occupied by a methyl group,bromine enters the ortho position relative to the acetamido group to form $2$-bromo-$4$-methylacetanilide (compound $B$).
$3$. Finally,acid-catalyzed hydrolysis $(H^{\oplus}/H_2O)$ of the acetamido group in compound $B$ yields $2$-bromo-$4$-methylaniline (compound $C$).

(D) Hofmann bromamide degradation of $CH_3CONH_2$ yields methylamine $(CH_3NH_2)$,a primary amine.
$(b)$ Curtius rearrangement of $CH_3COCl$ with $NaN_3$ followed by hydrolysis yields methylamine $(CH_3NH_2)$,a primary amine.
$(c)$ Reduction of $CH_3CH_2CONH_2$ with $LiAlH_4$ yields ethylamine $(CH_3CH_2NH_2)$,a primary amine.
$(d)$ Reduction of methyl isocyanide $(CH_3NC)$ with $LiAlH_4$ yields dimethylamine $(CH_3NHCH_3)$,which is a secondary amine.
Therefore,the correct option is $(d)$.