Properties of Amines Questions in English

(B) The rate of electrophilic aromatic substitution depends on the electron density of the aromatic ring. Groups with a $+M$ (mesomeric) effect increase the electron density,thereby activating the ring towards electrophilic attack.
$A$: Tetralin has only alkyl groups,which are weakly activating.
$B$: $1,2,3,4$-Tetrahydroquinoline has an $-NH-$ group directly attached to the benzene ring. The lone pair on the nitrogen atom participates in resonance ($+M$ effect),significantly increasing the electron density of the ring.
$C$: $1,2,3,4$-Tetrahydroisoquinoline has the nitrogen atom separated from the ring by a $CH_2$ group,so it does not exert a direct $+M$ effect on the ring.
$D$: $3,4$-Dihydroquinolin-$2(1H)$-one has an amide group where the nitrogen lone pair is involved in resonance with the carbonyl group ($-M$ effect),which deactivates the ring.
Therefore,$1,2,3,4$-Tetrahydroquinoline $(B)$ undergoes bromination at the fastest rate due to the strong $+M$ effect of the $-NH-$ group.

(C) The given reactant is a tertiary amine,$(Ph-CH_2)_3N$.
Tertiary aliphatic amines react with nitrous acid $(HNO_2)$ generated from $NaNO_2/HCl$ to form a salt,which is a nitrosoammonium salt.
The reaction is: $(Ph-CH_2)_3N + HNO_2 \rightarrow (Ph-CH_2)_3N^{+}-N=O \, OH^{-}$.
Thus,the product $(A)$ is $(Ph-CH_2)_3N^{+}-N=O$.

(D) $p$-aminophenol contains both an amino group $(-NH_2)$ and a hydroxyl group $(-OH)$.
In the presence of a base like pyridine,the amino group is more nucleophilic than the hydroxyl group.
Therefore,the acetylation reaction with one equivalent of acetyl chloride occurs selectively at the $-NH_2$ group to form $N$-($4$-hydroxyphenyl)acetamide (paracetamol).
The correct option is $D$.

(B) $1$. In the presence of an acid $(H^{+})$,triphenylmethanol $(Ph_3COH)$ undergoes protonation followed by the loss of water to form a stable triphenylmethyl carbocation $(Ph_3C^{+})$.
$2$. Aniline $(PhNH_2)$ acts as a nucleophile. The amino group $(-NH_2)$ is a strongly activating group and is ortho/para-directing.
$3$. Due to the significant steric hindrance provided by the bulky triphenylmethyl group $(Ph_3C^{+})$,the electrophilic aromatic substitution occurs primarily at the para-position.
$4$. Therefore,the major product is $p$-tritylaniline.

(B) The reaction sequence is as follows:
$1$. Aniline reacts with acetic anhydride $(CH_3CO)_2O$ to form acetanilide $(A)$ (protection of the $-NH_2$ group).
$2$. Acetanilide undergoes nitration with $HNO_3/H_2SO_4$ to give $p$-nitroacetanilide $(B)$ as the major product due to the steric hindrance and the activating effect of the $-NHCOCH_3$ group.
$3$. Finally,hydrolysis of $(B)$ with $H^+/H_2O$ removes the acetyl group to yield $p$-nitroaniline $(C)$ as the final product.

(B) The reaction proceeds in two steps:
$1$. Reduction of $p$-nitrophenol with $H_2/Pd$ gives $p$-aminophenol as product $(A)$.
$2$. $p$-aminophenol reacts with acetic anhydride. Since the amino group $(-NH_2)$ is more nucleophilic than the hydroxyl group $(-OH)$,the nitrogen atom attacks the carbonyl carbon of acetic anhydride to form an amide linkage.
This results in the formation of $N-(4-hydroxyphenyl)acetamide$,commonly known as Paracetamol.

(B) The rate of electrophilic aromatic substitution depends on the electron density of the aromatic ring.
Groups that donate electrons via resonance ($+M$ effect) significantly increase the electron density,making the ring more reactive.
In $1,2,3,4$-tetrahydroquinoline (option $B$),the nitrogen atom has a lone pair of electrons that can participate in resonance with the benzene ring,exerting a strong $+M$ effect.
In $1,2,3,4$-tetrahydroisoquinoline (option $C$),the nitrogen is also attached to the ring,but the resonance effect is slightly less effective compared to the position in $B$ due to the connectivity.
Tetralin (option $A$) has only alkyl groups,which are weakly activating.
$3,4$-dihydroquinolin-$2(1H)$-one (option $D$) has an amide group where the lone pair on nitrogen is involved in resonance with the carbonyl group $(-C=O)$,reducing its availability for the aromatic ring,thus making it the least reactive.
Therefore,$1,2,3,4$-tetrahydroquinoline is the most reactive.

(B) The reactivity of diazonium ions towards diazo-coupling depends on the electrophilicity of the diazonium group $(-N_2^+)$.
Electron-withdrawing groups $(EWG)$ increase the electrophilicity of the diazonium ion,thereby increasing its reactivity towards coupling.
Electron-donating groups $(EDG)$ decrease the electrophilicity,thereby decreasing its reactivity.
The substituents at the para-position are:
$(I)$ $-NMe_2$ (Strong $EDG$)
$(II)$ $-NO_2$ (Strong $EWG$)
$(III)$ $-OCH_3$ (Moderate $EDG$)
$(IV)$ $-CH_3$ (Weak $EDG$)
The order of electron-withdrawing power is: $-NO_2 > -CH_3 > -OCH_3 > -NMe_2$.
Therefore,the order of reactivity is: $(I) < (III) < (IV) < (II)$.

(C) The reaction is a Friedel-Crafts alkylation of $N$-phenylbenzamide (benzanilide) with $CH_3Cl$ in the presence of $AlCl_3$.
In $N$-phenylbenzamide,the $-NHCOPh$ group is attached to the benzene ring. The $-NHCOPh$ group is a deactivating group because the lone pair on the nitrogen atom is involved in resonance with the carbonyl group $(-C=O)$,making it less available for resonance with the benzene ring.
However,the $-NHCOPh$ group is still ortho/para-directing due to the presence of the lone pair on the nitrogen atom.
Between the ortho and para positions,the para position is sterically less hindered. Therefore,the major product is the para-substituted isomer,which is $4$-methyl-$N$-phenylbenzamide.
Thus,the correct option is $(C)$.

(B) The given reaction involves the removal of the diazonium group $(-N_2^{+}Cl^{-})$ from the benzene ring and its replacement with a hydrogen atom.
This is a reductive deamination reaction of a diazonium salt.
$H_3PO_2$ (hypophosphorous acid) in the presence of water is a standard reagent used to reduce diazonium salts to the corresponding arenes.
The reaction is:
$Ar-N_2^{+}Cl^{-} + H_3PO_2 + H_2O \rightarrow Ar-H + N_2 + H_3PO_3 + HCl$
Therefore,the correct reagent is $H_3PO_2$.

(B) The ease of diazotisation depends on the basicity of the amine and the stability of the resulting diazonium salt.
$ (A) $ is an aliphatic amine,which forms highly unstable diazonium salts that decompose immediately. Thus,it has the lowest tendency for diazotisation.
Among aromatic amines,electron-donating groups increase the electron density on the nitrogen atom,increasing basicity and facilitating diazotisation. Electron-withdrawing groups decrease electron density,making diazotisation more difficult.
$ (B) $ is aniline.
$ (C) $ has an $ -OCOCH_3 $ group at the meta position,which is electron-donating by resonance but electron-withdrawing by induction. Overall,it is slightly activating or neutral.
$ (D) $ has an $ -COCH_3 $ group at the ortho position,which is a strong electron-withdrawing group,significantly reducing the basicity of the amine and making diazotisation difficult.
Therefore,the increasing order of diazotisation is $ (A) < (D) < (B) < (C) $.

(A) The reaction of $p$-aminobenzenesulfonic acid with $HNO_2$ produces a diazonium salt intermediate.
In the presence of acetic anhydride or similar conditions (often implied in such reaction schemes),the diazonium group can form a diazo-ester linkage.
Product $A$ is the diazo-ester formed from the reaction.
When this intermediate reacts with aniline $(C_6H_5NH_2)$,it undergoes a coupling reaction to form an amino-azo compound $(B)$,specifically $p$-aminoazobenzene-sulfonic acid derivative.

(B) When the given mixture is shaken with $1 \, M \, HCl$,the amine gets protonated to form a salt,which is a cation $\left( RNH_2CH_3^{\oplus} \right)$.
This salt is ionic and therefore dissolves in the aqueous layer $(H_2O)$ rather than the organic solvent.
Among the given compounds,$(II)$ is an amine ($N$-methylcyclohexylamine),which reacts with $HCl$ to form a water-soluble salt.
The other compounds (sulfide,ether,and ketone) do not react with $1 \, M \, HCl$ to form ionic species and remain in the organic layer.
Therefore,the correct option is $(II)$.

(A) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
$(II)$ Pyrrole: The lone pair on nitrogen is involved in the aromatic sextet,making it non-basic.
$(I)$ Aniline: The lone pair on nitrogen is delocalized into the benzene ring due to resonance,reducing its availability for protonation.
$(IV)$ Cyclohexylamine: This is a primary aliphatic amine. The lone pair is localized on the nitrogen atom,making it more basic than aniline.
$(III)$ $N$-Methylpiperidine: This is a tertiary aliphatic amine. It is more basic than primary aliphatic amines due to the electron-donating inductive effect $(+I)$ of the alkyl groups,which increases the electron density on the nitrogen atom.
Therefore,the increasing order of basic strength is: $II < I < IV < III$.

(C) The reaction described is the Balz-Schiemann reaction.
In this reaction,an aromatic diazonium salt is treated with fluoroboric acid $(HBF_4)$ to form an insoluble diazonium fluoroborate salt $(ArN_2^ BF_4^-)$.
Upon heating this salt,it undergoes thermal decomposition to yield an aryl fluoride,nitrogen gas $(N_2)$,and boron trifluoride $(BF_3)$.
Therefore,the condition required for the conversion of the diazonium fluoroborate to the aryl fluoride is simply heating $(\Delta)$.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
Aliphatic amines (like $CH_3NH_2$) are more basic than aromatic amines because the lone pair on the nitrogen atom in aromatic amines is involved in resonance with the benzene ring,making it less available for protonation.
Among aromatic amines,electron-donating groups (like $-OCH_3$) increase the electron density on the nitrogen atom,thereby increasing basicity.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density on the nitrogen atom,thereby decreasing basicity.
Therefore,the order of increasing basicity is: $p$-nitroaniline < aniline < $p$-methoxyaniline < $CH_3NH_2$.

(B) The reduction of $benzene-diazonium$ chloride $(C_6H_5N_2Cl)$ with $Zn/HCl$ is a strong reduction process.
It leads to the formation of $phenylhydrazine$ $(C_6H_5NHNH_2)$ as the final product.
Note: While mild reducing agents like $SnCl_2/HCl$ or $Na_2SO_3$ can produce $phenylhydrazine$,strong reduction with $Zn/HCl$ also yields $phenylhydrazine$ as the primary product.

(B) When $Methyl \ amine$ $(CH_3NH_2)$ reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$,it forms an unstable diazonium salt $(CH_3N_2^+Cl^-)$.
This salt immediately undergoes hydrolysis to produce $Methyl \ alcohol$ $(CH_3OH)$,nitrogen gas $(N_2)$,and water $(H_2O)$.
The overall reaction is: $CH_3NH_2 + NaNO_2 + HCl \rightarrow CH_3OH + N_2 + H_2O$.

(C) In the gaseous phase,the basicity of amines is determined solely by the inductive effect of the alkyl groups.
Alkyl groups are electron-donating ($+I$ effect),which increases the electron density on the nitrogen atom,making the lone pair more available for donation.
As the number of alkyl groups increases,the electron density on the nitrogen atom increases.
Therefore,the order of basicity in the gaseous phase is $3^o > 2^o > 1^o > NH_3$.

(B) The carbylamine reaction (also known as the isocyanide test) involves the reaction of a primary amine $(RNH_2)$ with chloroform $(CHCl_3)$ in the presence of a base like $KOH$.
In the first step of the mechanism,chloroform undergoes $\alpha$-elimination in the presence of a base to form dichlorocarbene $(:CCl_2)$,which acts as the reactive intermediate.
This dichlorocarbene then undergoes nucleophilic attack by the nitrogen atom of the primary amine to eventually yield the isocyanide $(RNC)$.
Therefore,the correct intermediate is a carbene.

(B) Benzylamine $(C_6H_5CH_2NH_2)$ is the most basic compound among the given options.
In benzylamine,the lone pair of electrons on the nitrogen atom is localized because it is not in conjugation with the benzene ring.
In contrast,in aniline $(C_6H_5NH_2)$,$p-$nitroaniline,and acetanilide $(CH_3CONHC_6H_5)$,the lone pair of electrons on the nitrogen atom is delocalized due to resonance with the benzene ring or the carbonyl group,which significantly decreases their basicity.

(B) The compounds are: $(I)$ Piperidine,$(II)$ Pyridine,$(III)$ Morpholine,and $(IV)$ Pyrrole.
In $(I)$,the nitrogen atom is $sp^3$ hybridized,and the lone pair is localized,making it highly basic.
In $(III)$,the nitrogen atom is $sp^3$ hybridized,but the presence of an electronegative oxygen atom exerts an electron-withdrawing $-I$ effect,reducing its basicity compared to $(I)$.
In $(II)$,the nitrogen atom is $sp^2$ hybridized,making the lone pair less available for protonation than in $sp^3$ hybridized amines.
In $(IV)$,the lone pair on the nitrogen atom is involved in the aromatic sextet ($6\pi$ electrons),making it unavailable for protonation,thus it is the least basic.
Therefore,the correct order of basicity is $(I) > (III) > (II) > (IV)$.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$I$ (Piperidine): The nitrogen atom is $sp^3$ hybridized,and the lone pair is not involved in resonance. It is the most basic.
$II$ (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.
$III$ (Morpholine): Similar to piperidine,the nitrogen is $sp^3$ hybridized,but the presence of an electronegative oxygen atom exerts an inductive effect ($-I$ effect),which reduces the electron density on the nitrogen atom,making it less basic than piperidine but more basic than pyridine.
$IV$ (Pyrrole): The nitrogen atom is $sp^2$ hybridized,and its lone pair is involved in the aromatic sextet (delocalized). Thus,it is the least basic.
Therefore,the order of basicity is $I > III > II > IV$.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
In compound $(III)$ (piperidine),the nitrogen atom is $sp^3$ hybridized,and the lone pair is localized,making it the most basic.
In compound $(I)$ (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 $(III)$.
In compound $(II)$ (pyrrole),the lone pair of electrons on the nitrogen atom participates in the aromatic sextet (delocalization). Therefore,it is not available for protonation,making it the least basic.
Thus,the decreasing order of basicity is $III > I > II$.

(D) The reactant is $2$-amino-$1$-phenylpropan-$1$-ol. It contains both an amino group $(-NH_2)$ and a hydroxyl group $(-OH)$.
When treated with acetic anhydride $((CH_3CO)_2O)$ in the presence of pyridine at room temperature,acetylation occurs.
Since the nucleophilicity of the amino group $(-NH_2)$ is significantly higher than that of the hydroxyl group $(-OH)$,the amino group undergoes acetylation preferentially to form an amide.
Therefore,the major product is the $N$-acetylated derivative,where the $-NH_2$ group is converted to $-NHCOCH_3$ while the $-OH$ group remains unchanged.

(B) The basicity of amines depends on the availability of the lone pair on the nitrogen atom and the stability of the resulting conjugate acid.
$(i)$ $PhNHCH_3$ $(4)$ is an aromatic amine where the lone pair on nitrogen is delocalized into the benzene ring,making it the least basic.
$(ii)$ Among aliphatic amines,basicity is influenced by the inductive effect $(+I)$,solvation,and steric hindrance.
$(iii)$ Diethylamine $(2)$ is a $2^\circ$ amine,which is more basic than Ethylamine $(1)$ ($1^\circ$ amine) due to the $+I$ effect of two ethyl groups.
$(iv)$ Trimethylamine $(3)$ is a $3^\circ$ amine. Due to significant steric hindrance and reduced solvation of its conjugate acid in water,it is less basic than Ethylamine $(1)$.
Thus,the increasing order of basicity is: $(4) < (3) < (1) < (2)$.

(D) The reaction of a primary aliphatic amine with $NaNO_2$ and $HCl$ at $0-5^{\circ}C$ produces an unstable diazonium salt.
This diazonium salt undergoes rapid loss of $N_2$ gas to form a primary carbocation.
Since the primary carbocation is unstable,it undergoes a ring expansion rearrangement to form a more stable secondary carbocation.
Specifically,the $5$-membered ring expands to a $6$-membered ring to form a decalin system.
Finally,the secondary carbocation reacts with water to form the major product,which is decalin-$2$-ol.

(C) The reaction of a compound with $Br_2/NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.
Given that the product $C_3H_9N$ gives a positive carbylamine test,it must be a primary aliphatic amine $(CH_3CH_2CH_2NH_2)$.
Since the product has $3$ carbon atoms,the starting amide must have $4$ carbon atoms.
Among the options,$CH_3CH_2CH_2CONH_2$ (butanamide) is the only primary amide with $4$ carbon atoms that would yield $CH_3CH_2CH_2NH_2$ upon reaction with $Br_2/NaOH$.

(A) The reaction of these compounds with alkyl halide is a nucleophilic substitution reaction where the nitrogen atom acts as the nucleophile.
Reactivity depends on the availability of the lone pair on the nitrogen atom.
$(a)$ Benzamide: The lone pair on nitrogen is involved in resonance with the carbonyl group $(C=O)$,making it very weakly nucleophilic.
$(b)$ Phthalimide: The lone pair on nitrogen is involved in resonance with two carbonyl groups,making it the least nucleophilic.
$(c)$ $2$-Cyanoaniline: The $-CN$ group is electron-withdrawing by both $-I$ and $-M$ effects,which reduces the electron density on the nitrogen atom.
$(d)$ Aniline: The lone pair on nitrogen is involved in resonance with the benzene ring,but it is more nucleophilic than the others because it lacks the strong electron-withdrawing carbonyl groups present in $(a)$ and $(b)$ or the strong $-I$ effect of the $-CN$ group in $(c)$.
Thus,the order of nucleophilicity (and reactivity) is $b < a < c < d$.

(D) The reaction proceeds as follows:
$1$. Treatment of the primary amine with $NaNO_2/H^+$ leads to the formation of a diazonium salt,which is unstable and undergoes hydrolysis to form a primary alcohol: $Ar-(CH_2)_3-OH$.
$2$. Oxidation with $CrO_3/H^+$ (Jones reagent) oxidizes the primary alcohol to a carboxylic acid: $Ar-(CH_2)_2-COOH$.
$3$. Treatment with conc. $H_2SO_4$ and heat causes an intramolecular Friedel-Crafts acylation. During this process,the ester group (acetoxy) is hydrolyzed to a phenolic $-OH$ group under acidic conditions. The resulting carboxylic acid cyclizes onto the benzene ring to form a five-membered ketone ring,yielding $5-hydroxy-1-indanone$.

(C) The basic strength of amines in the gaseous phase or non-polar solvents is determined by the inductive effect ($+I$ effect) of the alkyl groups attached to the nitrogen atom.
As the number of alkyl groups increases,the electron density on the nitrogen atom increases due to the $+I$ effect,making the lone pair more available for donation.
Therefore,the order of basic strength is $(C_2H_5)_2NH > C_2H_5NH_2 > NH_3$.

(D) The reaction sequence is as follows:
$1$. The starting material is $2$-amino-$3$-cyano-benzaldehyde.
$2$. Treatment with $CHCl_3/KOH$ (carbylamine reaction) converts the $-NH_2$ group into an isocyanide group $(-NC)$.
$3$. Subsequent reduction with $Pd/C/H_2$ reduces the isocyanide $(-NC)$ to a methylamino group $(-NHCH_3)$,the aldehyde group $(-CHO)$ to a secondary alcohol $(-CH(OH)CH_3)$,and the nitrile group $(-CN)$ to a primary amine $(-CH_2NH_2)$.

(A) $1$. Aniline reacts with $NaNO_2$ and $HCl$ at $0\,^oC$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. When this diazonium salt is added to a mixture of aniline and phenol in dilute $HCl$,the medium is acidic.
$3$. In an acidic medium,the amino group of aniline is protonated to form an anilinium ion $(-NH_3^+)$,which is deactivating and does not undergo coupling.
$4$. However,the diazonium salt reacts with aniline to form a diazoamino compound $(C_6H_5-N=N-NH-C_6H_5)$ in a slightly acidic or neutral medium,but in the presence of dilute $HCl$,the reaction with aniline is suppressed.
$5$. The coupling reaction with phenol occurs in a basic medium. Since the medium is acidic,the coupling reaction is generally slow or inhibited for both.
$6$. However,among the given options,the formation of the diazoamino compound $(C_6H_5-N=N-NH-C_6H_5)$ is the characteristic reaction of benzenediazonium chloride with aniline in the presence of dilute $HCl$ (where the aniline is not fully protonated or the reaction is kinetically favored).
$7$. Therefore,the major product formed is the diazoamino compound.

(D) Amines are more nucleophilic than alcohols.
Therefore, the $-NH_2$ group reacts preferentially with ethyl formate $(HCOOC_2H_5)$ to form the formamide derivative $CH_3-CH(OH)-CH_2-CH_2-NH-CHO$.

(A) Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
For charged species,the nucleophilicity is generally higher than for neutral species.
Among the given species,$H_2O$ is a neutral molecule and is the weakest nucleophile.
Comparing the anions,$CH_3SO_3^-$ is a very weak nucleophile because the negative charge is delocalized over three oxygen atoms,making it a stable conjugate base of a strong acid.
$CH_3CO_2^-$ has the negative charge delocalized over two oxygen atoms,making it a stronger nucleophile than $CH_3SO_3^-$.
$OH^-$ is the strongest nucleophile among these as the charge is localized on a single oxygen atom.
Thus,the increasing order is $(b) < (c) < (a) < (d)$.

(A) The $pK_b$ value is inversely proportional to the basic strength of the compound.
Basic strength depends on the availability of the lone pair on the nitrogen atom.
Electron-donating groups $(EDG)$ increase basicity (decrease $pK_b$),while electron-withdrawing groups $(EWG)$ decrease basicity (increase $pK_b$).
The substituents on the phenyl ring are: $(A) -F$ ($EWG$ by inductive effect),$(B) -OCH_3$ ($EDG$ by resonance),$(C) -NO_2$ (strong $EWG$),$(D) -CH_3$ ($EDG$ by hyperconjugation).
Basicity order: $(B) > (D) > (A) > (C)$.
Since $pK_b$ is inversely proportional to basicity,the increasing order of $pK_b$ is: $(C) < (A) < (D) < (B)$.

(B) When benzene diazonium chloride reacts with aniline in the presence of dilute $HCl$ at a low temperature $(273-278 \ K)$,an electrophilic substitution reaction occurs at the para-position of the aniline ring. This process is known as coupling reaction,which yields $p$-aminoazobenzene,a yellow-colored dye.

(B) $CH_3NC$ (methyl isocyanide) on reduction with $LiAlH_4$ gives $CH_3NHCH_3$ (dimethylamine),which is a secondary amine.
$CH_3CONH_2$ (acetamide) reduces to $CH_3CH_2NH_2$ (ethylamine),$CH_3CN$ (acetonitrile) reduces to $CH_3CH_2NH_2$ (ethylamine),and $C_6H_5NO_2$ (nitrobenzene) reduces to $C_6H_5NH_2$ (aniline). All of these are primary amines.

(D) The reaction between benzenediazonium chloride and aniline in a mildly acidic medium $(pH \approx 4-5)$ 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 para-position of the aniline ring.
Since the $-NH_2$ group is a strong ortho/para-directing group,the para-position is sterically favored over the ortho-position.
Thus,the major product formed is $p$-aminoazobenzene,which is a yellow-colored dye.

(B) Let's analyze each option:
$A$: $o$-Nitrobenzoic acid is more acidic than benzoic acid due to the ortho effect,which is correct.
$B$: $o$-Toluidine is less basic than aniline due to the Steric Inhibition of Protonation $(SIP)$ effect,which decreases the basic strength. Thus,the given order is incorrect.
$C$: $(CH_3)_3C-O^-$ is less stable than $CH_3-CH_2-O^-$ because the electron-donating inductive effect $(+I)$ of the three methyl groups in the tert-butoxide ion increases the electron density on the oxygen atom,making it less stable. The order $(CH_3)_3C-O^- < CH_3-CH_2-O^-$ is correct for stability. Wait,the question asks for the incorrect order. Let's re-evaluate.
$D$: Methylenecyclohexane is less stable than $1-$methylcyclohexene due to the lack of hyperconjugation and resonance stabilization of the double bond. Therefore,it has a higher heat of hydrogenation. This order is correct.
Since $B$ shows $o$-Toluidine $>$ Aniline,which is incorrect due to the $SIP$ effect,$B$ is the correct answer.

(B) The given reaction is a two-step process:
$1$. The first step uses $KOBr(aq)$,which is the reagent for the Hofmann bromamide degradation reaction. This converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one carbon atom less.
$CH_3-CH(CH_3)-CH_2-CONH_2 \xrightarrow{KOBr} CH_3-CH(CH_3)-CH_2-NH_2$ (This is $X$,isobutylamine).
$2$. The second step involves the reaction of a primary aliphatic amine with nitrous acid $(NaNO_2/HCl)$,which forms an unstable diazonium salt that decomposes to form a carbocation,eventually leading to the formation of an alcohol.
$CH_3-CH(CH_3)-CH_2-NH_2$ $\xrightarrow{NaNO_2/HCl} [CH_3-CH(CH_3)-CH_2-N_2^+]$ $\rightarrow CH_3-CH(CH_3)-CH_2^+$ $\rightarrow CH_3-CH(CH_3)-CH_2OH$ (This is $Y$,isobutyl alcohol).
Thus,the correct option is $B$.

(B) Step $1$: Oxidation of toluene with $KMnO_4/H^+$ gives benzoic acid $(A = C_6H_5COOH)$.
Step $2$: Reaction of benzoic acid with $SOCl_2$ gives benzoyl chloride $(B = C_6H_5COCl)$.
Step $3$: Reaction of benzoyl chloride with $NH_3$ gives benzamide $(C = C_6H_5CONH_2)$.
Step $4$: Reaction of benzamide with $KOH/Br_2$ is the Hofmann bromamide degradation reaction,which converts an amide to a primary amine with one carbon atom less. Thus,benzamide is converted to aniline $(D = C_6H_5NH_2)$.

(A) Primary,secondary,and tertiary amines with the same molecular formula are functional isomers. $CH_3-CH_2-CH_2-NH_2$ (a $1^\circ$ amine) and $CH_3-CH_2-NH-CH_3$ (a $2^\circ$ amine) both have the molecular formula $C_3H_9N$,thus they are functional isomers.
$(b)$ Homologues differ by a $-CH_2-$ group. These are not homologues.
$(c)$ In the gaseous phase,basicity increases with the number of alkyl groups due to the $+I$ effect: $CH_3-N(CH_3)-CH_3 > CH_3-NH-CH_3 > CH_3-NH_2$.
$(d)$ Boiling point depends on hydrogen bonding. $CH_3-NH_2$ has two $H$-atoms on $N$,$CH_3-NH-CH_3$ has one,and $CH_3-N(CH_3)-CH_3$ has none. Thus,the order is $CH_3-NH_2 > CH_3-NH-CH_3 > CH_3-N(CH_3)-CH_3$.

(A) $1$. $C_6H_5NH_2$ (Aniline) reacts with $HCl + NaNO_2$ at $0-5\,^{\circ}C$ to form $C_6H_5N_2^+Cl^-$ (Benzenediazonium chloride),which is $A$.
$2$. $A$ reacts with $CuCN$ (Sandmeyer reaction) to form $C_6H_5CN$ (Benzonitrile),which is $B$.
$3$. $B$ undergoes catalytic hydrogenation with $H_2/Ni$ to form $C_6H_5CH_2NH_2$ (Benzylamine),which is $C$.
$4$. $C$ reacts with $HNO_2$ (Nitrous acid) to form $C_6H_5CH_2OH$ (Benzyl alcohol),which is $D$.

(B) The reaction sequence is as follows:
$1.$ Carboxylic acid reacts with methylamine to form an ammonium salt: $CH_3-CH(CH_3)-COOH + CH_3-NH_2 \rightarrow CH_3-CH(CH_3)-COO^- NH_3^+CH_3$ ('$A$').
$2.$ Heating the salt results in dehydration to form an $N$-substituted amide: $CH_3-CH(CH_3)-COO^- NH_3^+CH_3 \xrightarrow{\Delta} CH_3-CH(CH_3)-CONH-CH_3$ ('$B$').
$3.$ Reduction of the amide with $LiAlH_4$ yields a secondary amine: $CH_3-CH(CH_3)-CONH-CH_3 \xrightarrow{LiAlH_4} CH_3-CH(CH_3)-CH_2-NH-CH_3$ ('$C$').
Since the nitrogen in '$C$' is bonded to two carbon atoms,it is a $2^o$ amine.

(B) The reaction of a primary aliphatic amine with nitrous acid $(HNO_2)$ produces an unstable diazonium salt,which rapidly decomposes to form a carbocation.
$CH_3CH_2CH_2CH_2NH_2 + HNO_2$ $\rightarrow [CH_3CH_2CH_2CH_2N_2^+]$ $\rightarrow CH_3CH_2CH_2CH_2^+ + N_2 + H_2O$
This primary carbocation can undergo rearrangement (hydride shift) to form a more stable secondary carbocation.
$CH_3CH_2CH_2CH_2^+ \rightarrow CH_3CH_2CH^+CH_3$
The secondary carbocation then reacts with water to form the major product,butan-$2$-ol.
$CH_3CH_2CH^+CH_3 + H_2O \rightarrow CH_3CH_2CH(OH)CH_3$
Therefore,the major product is butan-$2$-ol.

(A) The reaction of primary aliphatic amines with nitrous acid $(HNO_2)$ produces an unstable diazonium salt,which immediately decomposes in the presence of water to form an alcohol,nitrogen gas,and a proton.
The reaction is: $CH_3-NH_2 + HNO_2$ $\rightarrow [CH_3-N_2^+ Cl^-]$ $\xrightarrow{H_2O} CH_3-OH + N_2 + HCl$.
Therefore,the major product is methanol $(CH_3-OH)$.

(B) secondary amine has the general structure $R-NH-R'$.
For $C_5H_{13}N$,the sum of carbon atoms in $R$ and $R'$ must be $5$.
The possible combinations of $(R, R')$ are:
$1$. $(CH_3, C_4H_9)$: $C_4H_9$ has $4$ isomers ($n$-butyl,isobutyl,sec-butyl,tert-butyl). Total = $4$.
$2$. $(C_2H_5, C_3H_7)$: $C_3H_7$ has $2$ isomers ($n$-propyl,isopropyl). Total = $2$.
Summing these up: $4 + 2 = 6$.
Therefore,the total number of secondary amines is $6$.

(B) The reaction sequence is as follows:
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$,which is $X$.
$2$. Benzene diazonium chloride $(X)$ reacts with hypophosphorous acid $(H_3PO_2)$ and water to undergo reduction,yielding benzene $(C_6H_6)$ as the final product $Y$.
Therefore,$Y$ is benzene.