Mix Examples of Amines Questions in English

(B) The molecular formula $C_7H_9N$ corresponds to a degree of unsaturation of $4$,which is consistent with a benzene ring ($4$ degrees of unsaturation).
There are $5$ isomers that contain a benzene ring:
$1$. Benzylamine $(C_6H_5CH_2NH_2)$
$2$. $o$-Methylaniline ($o$-toluidine)
$3$. $m$-Methylaniline ($m$-toluidine)
$4$. $p$-Methylaniline ($p$-toluidine)
$5$. $N$-Methylaniline $(C_6H_5NHCH_3)$
Thus,the correct option is $B$.

(D) Ammonia acts as a nucleophile and reacts with alkyl halides $(RX)$ in a nucleophilic substitution reaction $(S_N2)$.
$RX + NH_3 \rightarrow RNH_2 + HX$
$RNH_2 + RX \rightarrow R_2NH + HX$
$R_2NH + RX \rightarrow R_3N + HX$
$R_3N + RX \rightarrow R_4N^+X^-$
Since the reaction continues until all hydrogen atoms on the nitrogen are replaced by alkyl groups,the final product is a mixture of primary $(1^o)$,secondary $(2^o)$,and tertiary $(3^o)$ amines,along with quaternary ammonium salts.

(B) The reaction described is the carbylamine test,which is characteristic of primary amines.
The reaction is: $C_6H_5NH_2 + 3 KOH + CHCl_3 \to C_6H_5NC + 3 KCl + 3 H_2O$.
Here,$(Y)$ is $CHCl_3$ (chloroform).
Chloroform $(Y)$ is prepared by the haloform reaction using ethanol $(Z)$ and chlorine in the presence of slaked lime $(Ca(OH)_2)$:
$C_2H_5OH + Cl_2$ $\xrightarrow{Ca(OH)_2} CH_3CHO$ $\xrightarrow{Cl_2} CCl_3CHO$ $\xrightarrow{Ca(OH)_2} CHCl_3$.
Thus,the compound $(Z)$ is $C_2H_5OH$ (ethanol).

(D) $C_3H_9N$ can form all three types of amines.
$1^o$ amine: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine)
$2^o$ amine: $CH_3CH_2NHCH_3$ ($N$-methylethanamine)
$3^o$ amine: $(CH_3)_3N$ ($N$,$N$-dimethylmethanamine)

(C) Ethylamine $(C_2H_5NH_2)$ can be prepared by the ammonolysis of alkyl halides and the catalytic amination of alcohols.
$1$. Action of $NH_3$ on ethyl iodide: $C_2H_5I + NH_3 \to C_2H_5NH_2 + HI$
$2$. Action of $NH_3$ on ethyl alcohol: $C_2H_5OH + NH_3 \xrightarrow[\Delta]{Al_2O_3} C_2H_5NH_2 + H_2O$
Therefore,both $(a)$ and $(b)$ are correct methods.

(D) The reaction of $methyl$ $iodide$ $(CH_3I)$ with $ammonia$ $(NH_3)$ is a nucleophilic substitution reaction known as $Hofmann$ $ammonolysis$.
Since the primary amine $(CH_3NH_2)$ formed is more nucleophilic than $ammonia$,it further reacts with $CH_3I$ to form secondary and tertiary amines.
Therefore,the reaction proceeds as follows:
$CH_3I + NH_3 \rightarrow CH_3NH_2 + HI$
$CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$
$(CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$
Thus,a mixture of primary,secondary,and tertiary amines is obtained.

(C) The reaction $RCN + 2H_2O \xrightarrow{H^{+}} RCOOH + NH_3$ represents the acid-catalyzed hydrolysis of a nitrile,which yields a carboxylic acid and ammonia,not an amine.
$A$. $RX + NH_3 \rightarrow RNH_2$ (Ammonolysis of alkyl halides yields amines).
$B$. $RCH = NOH + 4[H] \xrightarrow{Na/C_2H_5OH} RCH_2NH_2 + H_2O$ (Reduction of oximes yields primary amines).
$D$. $RCONH_2 + 4[H] \xrightarrow{LiAlH_4} RCH_2NH_2 + H_2O$ (Reduction of amides yields primary amines).

(C) . $CH_3CONH_2 \xrightarrow{KOH, Br_2} CH_3NH_2$ (primary amine,Hofmann bromamide degradation).
$B$. $CH_3CN \xrightarrow{LiAlH_4} CH_3CH_2NH_2$ (primary amine).
$C$. $CH_3NC \xrightarrow{LiAlH_4} CH_3NHCH_3$ (secondary amine).
$D$. $CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2$ (primary amine).
Therefore,option $C$ does not yield a primary amine.

(B) The reaction of an amide with $Br_2$ and $KOH$ is the $Hofmann$ bromamide degradation reaction,which produces a primary amine.
$C_6H_5CONH_2 + Br_2 + 4KOH \to C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O$
The product $C_6H_5NH_2$ (aniline) is an oily liquid.
When aniline is treated with acetic anhydride,it undergoes acetylation to form acetanilide $(CH_3CONHC_6H_5)$,which is a well-known antipyretic drug.
$C_6H_5NH_2 + (CH_3CO)_2O \to CH_3CONHC_6H_5 + CH_3COOH$
Therefore,the starting compound is benzamide.

(C) The $Wurtz$ reaction is used to prepare alkanes from alkyl halides.
$2 R - X + 2 Na \xrightarrow{\text{Dry ether}} R - R + 2 NaX$
Here,$R$ is an alkyl group and $X$ is a halide.
$Hinsberg$ method is used for the separation of amines.
$Hofmann$ method and $Curtius$ reaction are used for the synthesis of amines.

(A) $CH_3CH_2Br \xrightarrow{AgCN} CH_3CH_2NC (A)$
$CH_3CH_2NC (A) \xrightarrow{H_3O^+} HCOOH + CH_3CH_2NH_2 (B)$
$CH_3CH_2NH_2 (B) \xrightarrow{CHCl_3/KOH} CH_3CH_2NC (A)$
$CH_3CH_2NC (A) \xrightarrow{Reduction} CH_3CH_2NHCH_3 (C)$
Therefore,$A$ is ethyl isocyanide,$B$ is ethylamine,and $C$ is $N$-methylethanamine.

(A) This is the Hofmann bromamide degradation reaction,which converts an amide to an amine with one less carbon atom.
$(b)$ This is the decarboxylation reaction using soda lime,which removes $CO_2$ and results in an alkane with one less carbon atom.
$(c)$ This is the Hunsdiecker reaction,which converts a silver salt of a carboxylic acid to an alkyl bromide with one less carbon atom.
Therefore,all three reactions result in a decrease in the number of carbon atoms.

(A) $1.$ Aniline is used in the preparation of azo dyes via diazotization.
$2.$ Nitrobenzene is often used as a solvent in Friedel-Crafts reactions.
$3.$ Sulphanilamide is a well-known sulpha drug.
$4.$ Trinitrotoluene $(TNT)$ is widely used as an explosive.
Therefore,the correct matching is $1-a, 2-c, 3-b, 4-d$.

(D) The reaction of alcohols with ammonia in the presence of a catalyst like $Al_2O_3$ or $ThO_2$ at $300-400^{\circ}C$ is a catalytic ammonolysis process.
This reaction does not stop at the primary amine stage.
The primary amine formed $(CH_3NH_2)$ further reacts with the alcohol to form secondary amine $((CH_3)_2NH)$ and tertiary amine $((CH_3)_3N)$.
Therefore,the final product is a mixture of primary,secondary,and tertiary amines.

(B) Hydrolysis of nitriles $(R-CN)$ gives carboxylic acids and ammonia.
Hydrolysis of amides $(R-CONH_2)$ gives carboxylic acids and ammonia.
Hydrolysis of acetanilide $(CH_3CONHC_6H_5)$ gives acetic acid and aniline (a primary amine).
Hydrolysis of phenyl isocyanide $(C_6H_5NC)$ gives a primary amine $(C_6H_5NH_2)$ and formic acid $(HCOOH)$.
However,the question asks which does not produce an amine. Actually,all listed compounds can produce amines or are related to amine chemistry. Re-evaluating: Acetonitrile $(CH_3CN)$ on hydrolysis produces acetic acid and ammonia,not an amine. Therefore,$B$ is the correct answer.

(A) The reaction of aniline $(C_6H_5NH_2)$ with methyl iodide $(CH_3I)$ is an alkylation reaction.
Step $1$: Aniline reacts with $CH_3I$ to form $N$-methylaniline (a secondary amine).
Step $2$: $N$-methylaniline reacts with another molecule of $CH_3I$ to form $N,N$-dimethylaniline (a tertiary amine).
Therefore,the reaction sequence $Aniline$ $\xrightarrow{CH_3I}$ $\xrightarrow{CH_3I}$ leads to the formation of a tertiary amine.

(D) The reaction of $B$ with $Br_2$ and $KOH$ 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 is $CH_3CH_2NH_2$ (ethanamine),the amide $B$ must be $CH_3CH_2CONH_2$ (propanamide).
Since $A$ reacts with $NH_3$ on heating to form propanamide $(B)$,$A$ must be propanoic acid $(CH_3CH_2COOH)$.
The reaction sequence is: $CH_3CH_2COOH + NH_3$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (B)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$.

(C) The alkaline hydrolysis of ethyl isocyanate $(C_2H_5 - N = C = O)$ yields ethylamine $(C_2H_5NH_2)$.
$C_2H_5 - N = C = O + 2KOH \rightarrow C_2H_5NH_2 + K_2CO_3$.
When ethylamine $(C_2H_5NH_2)$ reacts with $t$-butyl magnesium bromide $((CH_3)_3CMgBr)$,the Grignard reagent acts as a base and abstracts the acidic proton from the amine group.
$C_2H_5NH_2 + (CH_3)_3CMgBr \rightarrow (CH_3)_3CH + C_2H_5NHMgBr$.
The product formed is isobutane $((CH_3)_3CH)$.

(B) $R-CN \xrightarrow{\text{reduction}} R-CH_2NH_2$ ($1^o$ amine) $(a)$
$R-CN$ $\xrightarrow[(ii) H_2O]{(i) CH_3MgBr} R-C(CH_3)=NMgBr$ $\xrightarrow{H_2O} R-C(=O)CH_3$ (methyl ketone) $(b)$
$R-NC \xrightarrow{\text{hydrolysis}} R-NH_2 + HCOOH$ ($1^o$ amine) $(c)$
$R-NH_2 \xrightarrow{HNO_2} R-OH$ (alcohol) $(d)$
Thus,the correct set is: $(a) = 1^o$ amine,$(b) = \text{methyl ketone}, (c) = 1^o$ amine,$(d) = \text{alcohol}$.

(C) Methyl isocyanate is produced by the reaction of methylamine with phosgene.
$CH_3NH_2 + COCl_2$ $\xrightarrow{-HCl} [CH_3NH-CO-Cl]$ $\xrightarrow{\Delta, -HCl} CH_3-N=C=O$

(B) Step $1$: The reaction of benzene with $CH_3CH_2COCl$ in the presence of $AlCl_3$ is a Friedel-Crafts acylation reaction. The product $X$ is propiophenone $(C_6H_5COCH_2CH_3)$.
Step $2$: The reaction of $X$ $(C_6H_5COCH_2CH_3)$ with $CH_3NH_2$ followed by reduction with $H_2, Ni$ is a reductive amination process. The ketone group $(C=O)$ reacts with the amine $(CH_3NH_2)$ to form an imine intermediate,which is then reduced to an amine. The final product is $N$-methyl$-1-$phenylpropan$-1-$amine,represented as $C_6H_5-CH(NHCH_3)-CH_2-CH_3$.

(D) The molecular formula $C_3H_9N$ corresponds to the general formula $C_nH_{2n+3}N$,which represents saturated amines.
It can form various isomers:
$1$. Primary $(1^\circ)$ amines: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine) and $CH_3CH(NH_2)CH_3$ (propan$-2-$amine).
$2$. Secondary $(2^\circ)$ amine: $CH_3CH_2NHCH_3$ ($N$-methylethanamine).
$3$. Tertiary $(3^\circ)$ amine: $(CH_3)_3N$ ($N$,$N$-dimethylmethanamine).
Thus,it represents all three types of amines.

(C) The molecular formula $C_3H_9N$ corresponds to the general formula $C_nH_{2n+3}N$,which indicates a saturated amine.
Possible isomers are:
$1$. Primary amines $(1^{\circ})$: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine) and $CH_3CH(NH_2)CH_3$ (propan$-2-$amine).
$2$. Secondary amines $(2^{\circ})$: $CH_3CH_2NHCH_3$ ($N$-methylethanamine).
$3$. Tertiary amines $(3^{\circ})$: $(CH_3)_3N$ ($N$,$N$-dimethylmethanamine).
Total number of isomers = $2 + 1 + 1 = 4$.

(D) Aniline is a weak base. $C_6H_5NH_2$ reacts with mineral acids like $HCl$ to form anilinium chloride,which is a salt. This process is essentially a salt formation reaction,not hydrolysis. Therefore,none of the reagents listed are used for the hydrolysis of aniline,as aniline does not undergo hydrolysis under these conditions.

(C) $A. \ CH_3(CH_2)_3NH_2$ is a primary amine,which gives the carbylamine test with $KOH$ (alc.) and $CHCl_3$ to produce a foul-smelling isocyanide $(ii)$.
$B. \ CH_3C\equiv CH$ is a terminal alkyne,which reacts with ammoniacal $AgNO_3$ to form a white precipitate of silver acetylide $(iii)$.
$C. \ CH_3CH_2COOCH_3$ is an ester,which undergoes alkaline hydrolysis to form an alcohol and a carboxylate salt $(i)$.
$D. \ CH_3CH(OH)CH_3$ is a secondary alcohol,which reacts with Lucas reagent $(ZnCl_2 + conc. HCl)$ to produce cloudiness within $5-10 \ minutes$ $(iv)$.
Thus,the correct matching is $A-(ii), B-(iii), C-(i), D-(iv)$.

(B) The reaction sequence is as follows:
$1$. Compound $A$ reacts with $Sn/HCl$ (a reducing agent) to form an amine. This indicates that $A$ is a nitro compound,specifically nitrobenzene $(C_6H_5NO_2)$.
$2$. Nitrobenzene $(C_6H_5NO_2)$ is reduced to aniline $(C_6H_5NH_2)$.
$3$. Aniline reacts with $HNO_2$ (nitrous acid) at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is the unstable compound $B$.
$4$. Benzenediazonium chloride $(B)$ undergoes a coupling reaction with phenol $(C_6H_5OH)$ in a basic medium to form $p$-hydroxyazobenzene $(C_6H_5-N=N-C_6H_4OH)$,which is the coloured compound $C$ with molecular formula $C_{12}H_{10}N_2O$.
Therefore,compound $A$ is nitrobenzene.

(B) $1$. The reaction of aniline $(A)$ with $NaNO_2$ and $HCl$ at $0-5^{\circ}C$ (cold conditions) produces benzene diazonium chloride $(B)$.
$2$. Benzene diazonium chloride $(B)$ undergoes an electrophilic aromatic substitution (coupling reaction) with $N,N$-dimethylaniline.
$3$. The coupling occurs at the para-position of the $N,N$-dimethylaniline ring to form $p$-dimethylaminoazobenzene,which is a yellow-coloured dye known as 'Butter yellow' $(C)$.
$4$. The structure of $C$ is $Ph-N=N-C_6H_4-N(CH_3)_2$.

(C) $CH_3-CH_2-CH_2-CONH_2$ (Butanamide) undergoes Hofmann bromamide (chloramide) degradation with $Ca(OH)_2$ and $Cl_2$ to form $CH_3-CH_2-CH_2-NH_2$ (Propan$-1-$amine) as $X$.
Reaction of $X$ with $HNO_2$ forms a primary carbocation $CH_3-CH_2-CH_2^+$,which rearranges via a $1,2$-hydride shift to a more stable secondary carbocation $CH_3-CH^+ -CH_3$.
Nucleophilic attack by water on this carbocation yields Propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ as the major product $Z$.

(C) For the molecular formula $C_3H_9N$,the possible isomers are:
$1$. Primary amines $(1^{\circ})$: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine) and $CH_3CH(NH_2)CH_3$ (propan$-2-$amine). Total = $2$.
$2$. Secondary amine $(2^{\circ})$: $CH_3CH_2NHCH_3$ ($N$-methylethanamine). Total = $1$.
$3$. Tertiary amine $(3^{\circ})$: $(CH_3)_3N$ ($N,N$-dimethylmethanamine). Total = $1$.
Thus,the number of primary,secondary,and tertiary amines are $2, 1, 1$ respectively.

(C) $1$. Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride $(A)$.
$2$. Benzene diazonium chloride $(A)$ on boiling with water gives phenol $(B)$.
$3$. Phenol $(B)$ reacts with $Br_2/H_2O$ to form $2,4,6-$tribromophenol $(C)$.
$4$. $2,4,6-$tribromophenol $(C)$ on treatment with $Zn$ dust undergoes reduction to form $1,3,5-$tribromobenzene $(D)$.

(B) The reaction sequence involves the Hofmann Bromamide Degradation followed by the Carbylamine reaction.
Step $1$: $C_2H_5-CONH_2 \xrightarrow{Br_2 / KOH} C_2H_5-NH_2$ (Ethylamine).
Step $2$: $C_2H_5-NH_2 \xrightarrow{CHCl_3 / KOH} C_2H_5-NC$ (Ethyl isocyanide).
Therefore,the major product is $C_2H_5-NC$.

(D) $1$. The reaction starts with cyclohexanone.
$2$. Addition of $HCN$ to cyclohexanone in an alkaline medium gives the cyanohydrin,$A$,which is $1-hydroxycyclohexanecarbonitrile$.
$3$. Reduction of the nitrile group $(-CN)$ with $LiAlH_4$ converts it into a primary amine $(-CH_2NH_2)$,yielding $B$,which is $1-(aminomethyl)cyclohexan-1-ol$.
$4$. Treatment of a primary aliphatic amine with $NaNO_2/HCl$ (nitrous acid) at low temperature leads to the formation of a diazonium salt,which is highly unstable and decomposes to form a carbocation. This carbocation then reacts with water to form an alcohol.
$5$. The product $C$ is $1-(hydroxymethyl)cyclohexan-1-ol$.

(B) $1$. The reaction sequence starts with $(A)$,which is $p$-methylbenzamide $(CH_3-C_6H_4-CONH_2)$.
$2$. Hofmann bromamide degradation: $(A) \xrightarrow{KOBr} (B)$. This converts the amide group $(-CONH_2)$ to a primary amine group $(-NH_2)$. Thus,$(B)$ is $p$-toluidine $(CH_3-C_6H_4-NH_2)$.
$3$. Carbylamine reaction: $(B) \xrightarrow{CHCl_3/KOH} (C)$. Primary amines react with chloroform and alcoholic $KOH$ to form isocyanides (isonitriles). Thus,$(C)$ is $p$-methylphenyl isocyanide $(CH_3-C_6H_4-NC)$.
$4$. Reduction: $(C) \xrightarrow{LiAlH_4} \text{Product}$. Isocyanides are reduced to secondary amines $(R-NH-CH_3)$. Therefore,the final product is $N$-methyl-$p$-toluidine $(CH_3-C_6H_4-NHCH_3)$.
$5$. The question asks for the structure of $(C)$,which is $p$-methylphenyl isocyanide.

(C) The distinction between primary $(1^{\circ})$ and secondary $(2^{\circ})$ amines is as follows:
$(i)$ Hinsberg test: $1^{\circ}$ amines react with benzenesulfonyl chloride to form a sulfonamide soluble in alkali,while $2^{\circ}$ amines form a sulfonamide insoluble in alkali.
$(ii)$ Carbyl amine test: Only $1^{\circ}$ amines give this test (foul-smelling isocyanide),while $2^{\circ}$ amines do not.
$(iii)$ Hofmann mustard oil test: Only $1^{\circ}$ amines give this test,while $2^{\circ}$ amines do not.
$(iv)$ $HNO_2$ test: $1^{\circ}$ aliphatic amines react with $HNO_2$ to evolve $N_2$ gas,whereas $2^{\circ}$ amines form $N$-nitrosamines (yellow oily liquids).
Since all four tests can distinguish between $1^{\circ}$ and $2^{\circ}$ amines,the correct option is $C$.

(B) $1$. The reaction starts with aniline $(C_6H_5NH_2)$.
$2$. Treatment with $NaNO_2 + HCl$ at $0-5^{\circ}C$ performs diazotization to form benzene diazonium chloride $(A)$,which is $C_6H_5N_2^+Cl^-$.
$3$. Reaction of $A$ with $CuCN/KCN$ (Sandmeyer reaction) replaces the diazonium group with a cyano group to form benzonitrile $(B)$,which is $C_6H_5CN$.
$4$. Acidic hydrolysis of benzonitrile $(B)$ with $H_2O/H^+$ converts the cyano group $(-CN)$ into a carboxylic acid group $(-COOH)$,resulting in benzoic acid $(C)$,which is $C_6H_5COOH$.

(B) $1.$ Primary Amines $(R-NH_2)$:
$CH_3CH_2CH_2CH_2NH_2$ ($n$-butylamine)
$CH_3CH_2CH(NH_2)CH_3$ ($sec$-butylamine)
$(CH_3)_2CHCH_2NH_2$ (isobutylamine)
$(CH_3)_3CNH_2$ ($tert$-butylamine)
$2.$ Secondary Amines $(R-NH-R')$:
$CH_3CH_2CH_2NHCH_3$ ($N$-methylpropan$-1-$amine)
$(CH_3)_2CHNHCH_3$ ($N$-methylpropan$-2-$amine)
$CH_3CH_2NHCH_2CH_3$ ($N$-ethylethanamine)
$3.$ Tertiary Amines $(R-N(R')-R'')$:
$CH_3CH_2N(CH_3)_2$ ($N,N$-dimethylethanamine)
Total structural isomers = $4$ (primary) + $3$ (secondary) + $1$ (tertiary) = $8$.

(D) The basic strength of amines depends on the inductive effect,solvation effect,and steric hindrance.
$(a)$ For methyl-substituted amines in a polar protic solvent,the order is $Me_2NH > MeNH_2 > Me_3N > NH_3$.
$(b)$ For ethyl-substituted amines in a polar protic solvent,the order is $Et_2NH > Et_3N > EtNH_2 > NH_3$.
$(c)$ In the gas phase,the solvation effect is absent,and the basicity is determined by the inductive effect of alkyl groups,leading to the order $3^o > 2^o > 1^o > NH_3$ (i.e.,$Me_3N > Me_2NH > MeNH_2 > NH_3$).
Since all these statements are correct,the correct option is $D$.

(C) The reactivity of an amine towards dilute hydrochloric acid $(HCl)$ depends on its basicity. The more basic the amine,the more readily it will accept a proton $(H^+)$ from the acid to form a salt.
$1$. $C_6H_5CH_2NH_2$ (Benzylamine) is a primary aliphatic amine,which is relatively basic.
$2$. $C_6H_5CONH_2$ (Benzamide) is an amide,where the lone pair on nitrogen is involved in resonance with the carbonyl group,making it very weakly basic.
$3$. $C_6H_5NH_2$ (Aniline) is an aromatic amine,where the lone pair on nitrogen is involved in resonance with the benzene ring,making it weakly basic.
$4$. $4-aminopyridine$ has two nitrogen atoms. The nitrogen in the ring (pyridine-like) is basic,and the amino group $(NH_2)$ can donate its lone pair into the ring,significantly increasing the electron density on the ring nitrogen. Upon protonation at the ring nitrogen,the resulting cation is stabilized by resonance,which includes an aromatic form. This makes it significantly more basic than the other options.

(A) The $d$ and $l$-amine mixture is a racemic mixture.
When a racemic mixture reacts with an optically active compound (like $l$-acid),it forms a mixture of diastereomers.
This is because the $d$-amine reacts with $l$-acid to form a $(d,l)$-salt,and the $l$-amine reacts with $l$-acid to form an $(l,l)$-salt.
These two salts are diastereomers of each other.

(D) The reaction of amines with methyl iodide is a nucleophilic substitution reaction $(S_N2)$.
For the same nucleophilic atom (nitrogen),the nucleophilicity is directly proportional to the basicity of the amine.
Structure $(A)$ is triethylamine,which is a flexible tertiary amine.
Structure $(B)$ is quinuclidine,a rigid bicyclic amine where the lone pair is more available due to less steric hindrance compared to $(A)$ and $(C)$.
Structure $(C)$ is $DABCO$ ($1$,$4$-diazabicyclo[$2.2$.$2$]octane),where the lone pair on one nitrogen is involved in resonance/inductive effects of the other nitrogen,reducing its basicity and nucleophilicity compared to $(B)$.
Thus,the order of basicity and nucleophilicity is $B > A > C$.

(C) Aniline $(C_6H_5NH_2)$ undergoes acetylation to form acetanilide $(C_6H_5NHCOCH_3)$ when reacted with acetylating agents like acetyl chloride $(CH_3COCl)$,acetic anhydride $((CH_3CO)_2O)$,or phenyl acetate $(CH_3COOC_6H_5)$.
However,reaction with acetaldehyde $(CH_3CHO)$ leads to the formation of a Schiff base (imine) via nucleophilic addition followed by dehydration:
$C_6H_5NH_2 + CH_3CHO \to C_6H_5N=CHCH_3 + H_2O$
Therefore,acetaldehyde does not produce acetanilide.

(B) The compounds are:
$(1)$ Aniline $(C_6H_5NH_2)$: The lone pair on nitrogen is involved in resonance with the benzene ring,making it a weak base.
$(2)$ Benzylamine $(C_6H_5CH_2NH_2)$: The lone pair on nitrogen is not involved in resonance,making it the strongest base among the given compounds.
$(3)$ $o$-Nitroaniline $(o-NO_2-C_6H_4NH_2)$: The $-NO_2$ group is a strong electron-withdrawing group ($-M$ and $-I$ effect). Additionally,the ortho effect further decreases its basicity.
$(4)$ Benzamide $(C_6H_5CONH_2)$: The lone pair on nitrogen is involved in resonance with the carbonyl group $(C=O)$,which is a very strong electron-withdrawing group,making it the weakest base.
Comparing the basicity:
$(4)$ is the weakest due to strong resonance with the carbonyl group.
$(3)$ is weaker than $(1)$ due to the electron-withdrawing $-NO_2$ group and ortho effect.
$(1)$ is weaker than $(2)$ because $(2)$ is an aliphatic amine where the lone pair is localized.
Thus,the order of increasing basic strength is $4 < 3 < 1 < 2$.

(C) $1$. The nitrile $X$ is $CH_{3}CN$ (acetonitrile or ethanenitrile).
$2$. Reduction of $CH_{3}CN$ with $LiAlH_{4}$ gives $CH_{3}CH_{2}NH_{2}$ (ethanamine),which is $Y$ $(C_{2}H_{7}N)$.
$3$. Acidic hydrolysis of $CH_{3}CN$ gives $CH_{3}COOH$ (ethanoic acid),which is $Z$.
$4$. When $Y$ (an amine,$CH_{3}CH_{2}NH_{2}$) and $Z$ (a carboxylic acid,$CH_{3}COOH$) are mixed,an acid-base reaction occurs to form a salt: $CH_{3}COOH + CH_{3}CH_{2}NH_{2} \rightarrow CH_{3}COO^{-}CH_{3}CH_{2}NH_{3}^{+}$.
$5$. Thus,the product is $(CH_{3}COO^{-})(CH_{3}CH_{2}NH_{3}^{+})$.

(B) The Hoffmann bromamide degradation reaction converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one carbon atom less.
In this specific case,the starting material is $\alpha$-hydroxy amide,$Ph-CH(OH)-CONH_2$.
Upon treatment with $Br_2/KOH$,the $-CONH_2$ group is converted into an $-NH_2$ group attached to the $\alpha$-carbon.
This yields the intermediate $\alpha$-hydroxy amine,$Ph-CH(OH)-NH_2$.
This intermediate is unstable because it has an $-OH$ and an $-NH_2$ group on the same carbon atom (a geminal amino alcohol).
Such compounds readily eliminate ammonia $(NH_3)$ to form a carbonyl compound.
Therefore,$Ph-CH(OH)-NH_2 \to Ph-CHO + NH_3$.
The final product is benzaldehyde $(Ph-CHO)$.

(B) The reaction is a Beckmann rearrangement of benzophenone oxime in the presence of $H_2SO_4$ and $H_2^{18}O$.
In the Beckmann rearrangement,the oxime undergoes protonation followed by the migration of a phenyl group to the nitrogen atom,resulting in the formation of a nitrilium ion.
This nitrilium ion is then attacked by the labeled water molecule $(H_2^{18}O)$.
After the loss of a proton,an imidic acid intermediate is formed,which tautomerizes to the final amide product,$N$-phenylbenzamide,containing the $^{18}O$ isotope in the carbonyl group: $Ph-NH-C(^{18}O)-Ph$.

(B) The reaction of $N$-bromosuccinimide $(NBS)$ with $KOBr$ (which provides $Br_2$ and $OH^-$) involves the hydrolysis of the imide ring. The $OH^-$ attacks one of the carbonyl carbons,leading to the opening of the succinimide ring. This results in the formation of $4$-amino$-4-$oxobutanoate (or its protonated form depending on pH),which is represented by the structure in option $B$.

(A) The reaction is an intramolecular Curtius rearrangement.
$1$. Upon heating,the acyl azide undergoes loss of $N_2$ to form an acyl nitrene or directly rearranges to an isocyanate intermediate.
$2$. The ortho-amino group then performs a nucleophilic attack on the electrophilic carbon of the isocyanate group.
$3$. This cyclization results in the formation of a cyclic urea derivative,specifically $1,3$-dihydro-$2H$-benzimidazol-$2$-one.

(D) The Hoffmann rearrangement involves the following steps:
$1$. Formation of $N$-bromamide by the reaction of amide with $Br_2$ and $NaOH$.
$2$. Formation of an acyl nitrene or direct rearrangement of the $N$-bromoamide anion to an isocyanate.
$3$. The isocyanate is then hydrolyzed to the amine.
Looking at the provided mechanism in the solution image,the species involved are:
$(a)$ $N$-bromocyclohexanecarboxamide
$(b)$ The conjugate base of the $N$-bromoamide
$(c)$ Cyclohexyl isocyanate
Since all these species are intermediates or products in the mechanism of the Hoffmann rearrangement,option $D$ is the correct answer.

(B) $1$. The reaction of benzene with phthalic anhydride in the presence of $AlCl_3$ (Friedel-Crafts acylation) gives $o$-benzoylbenzoic acid as product $(A)$.
$2$. Reduction of $(A)$ with $Zn-Hg/HCl$ (Clemmensen reduction) reduces the carbonyl group of the ketone to a methylene group,yielding $o$-benzylbenzoic acid as product $(B)$.
$3$. Treatment of $(B)$ with $SOCl_2$ converts the $-COOH$ group to $-COCl$. Subsequent intramolecular Friedel-Crafts acylation in the presence of $AlCl_3$ followed by hydrolysis yields anthrone as the final product $(C)$.