Properties of Amines Questions in English

(B) $1$. Benzoic acid reacts with concentrated $HNO_3$ and concentrated $H_2SO_4$ (nitration) to form $m$-nitrobenzoic acid $(X)$.
$2$. Reduction of $m$-nitrobenzoic acid $(X)$ with $Sn/HCl$ converts the $-NO_2$ group to an $-NH_2$ group,forming $m$-aminobenzoic acid $(Y)$.
$3$. $m$-Aminobenzoic acid $(Y)$ reacts with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form the diazonium salt,which then reacts with $Cu_2Cl_2/HCl$ (Sandmeyer reaction) to replace the diazonium group with a chlorine atom,resulting in $m$-chlorobenzoic acid $(Z)$.

(D) The reaction of a nitrile with $Na-Hg$ in the presence of $C_2H_5OH$ is a classic reduction method known as the $Mendius$ reduction.
In this reaction,the nitrile group $(-CN)$ is reduced to a primary amine group $(-CH_2NH_2)$.
Here,$p$-tolunitrile ($4$-methylbenzonitrile) reacts with $Na-Hg$ and $C_2H_5OH$ to produce $p$-methylbenzylamine as the major product.

(C) Diazonium salts are a group of organic compounds that have the common functional group $R-N_2^+ X^-$,where $R$ is an organic group (alkyl or aryl) and $X$ is an inorganic anion (e.g.,$Cl^-$,$Br^-$,$HSO_4^-$).
$(a)$ Aliphatic diazonium salts are highly unstable and decompose even at low temperatures.
$(b)$ Aniline can be converted into phenol by heating its diazonium salt with water.
$(c)$ Aliphatic diazonium salts decompose to liberate nitrogen gas quantitatively,which is often used for the estimation of aliphatic amines.
$(d)$ Aniline can be converted into fluorobenzene via the Balz-Schiemann reaction using its diazonium salt.
Therefore,statement $(c)$ is true.

(D) The reaction sequence is as follows:
$1$. $p$-toluidine reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form $p$-toluenediazonium chloride (Diazotisation).
$2$. The diazonium salt reacts with $HBF_4$ to form $p$-toluenediazonium fluoroborate,which upon heating decomposes to give $p$-fluorotoluene (Balz-Schiemann reaction).
$3$. $p$-fluorotoluene reacts with $NaNO_2$ and $Cu$ upon heating to replace the fluorine atom with a nitro group,yielding $p$-nitrotoluene.
Therefore,the final product $X$ is $p$-nitrotoluene.

(D) In the given reaction,$p$-toluidine reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form a diazonium salt,which is $p$-methylbenzenediazonium chloride.
This diazonium salt then undergoes an electrophilic aromatic substitution (coupling reaction) with $N$-phenylpyrrolidine in the presence of $H^+$.
The coupling reaction occurs at the para-position of the $N$-phenylpyrrolidine ring because the pyrrolidinyl group is a strong ortho/para-directing group and the para-position is sterically less hindered.
Thus,the major product is the para-coupled azo dye as shown in option $D$.

(B) The reaction sequence is as follows:
$1$. Acetylation of aniline with $(CH_3CO)_2O$ and pyridine protects the $-NH_2$ group,forming acetanilide.
$2$. Electrophilic aromatic substitution with $Br_2$ in $CH_3CO_2H$ occurs at the para-position due to the steric hindrance of the $-NHCOCH_3$ group,yielding $p$-bromoacetanilide.
$3$. Hydrolysis with $NaOH(aq)$ removes the acetyl group to regenerate $p$-bromoaniline.
$4$. Diazotization with $HNO_2$ at $0-5 \ ^\circ C$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.
$5$. The Sandmeyer reaction with $CuCl/HCl$ replaces the diazonium group with a chlorine atom,resulting in $1$-bromo-$4$-chlorobenzene.

(D) Benzylamine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine.
When it reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at low temperatures $(273-278 \ K)$,it forms an unstable diazonium salt.
This diazonium salt immediately undergoes hydrolysis to form benzyl alcohol $(C_6H_5CH_2OH)$ and nitrogen gas $(N_2)$.
The reaction is: $C_6H_5CH_2NH_2 + HNO_2 \rightarrow C_6H_5CH_2OH + N_2 + H_2O$.
Therefore,the major product is benzyl alcohol.

(B) The formation of benzonitrile from aniline involves two main steps:
$1$. Diazotization: Aniline reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: Benzene diazonium chloride reacts with $CuCN$ to form benzonitrile.
Therefore,the correct sequence is the reaction of aniline with $NaNO_2$ and $HCl$ at $273 \ K$ followed by the reaction with $CuCN$.

(A) $1$. The reaction of cyclohexylmagnesium chloride with $CO_2$ followed by acidic hydrolysis $(H^ )$ yields cyclohexanecarboxylic acid $(A)$.
$2$. Cyclohexanecarboxylic acid reacts with $NH_3$ upon heating $(\Delta)$ to form cyclohexanecarboxamide $(B)$.
$3$. The reaction of cyclohexanecarboxamide with $Br_2$ and $NaOH$ is the Hoffmann bromamide degradation reaction,which converts the amide into a primary amine with one less carbon atom.
$4$. Therefore,the product $C$ is cyclohexanamine.

(A) The $-NO_2$ group is a strong electron-withdrawing group that exerts both $-I$ (inductive) and $-M$ (mesomeric) effects.
These effects decrease the electron density on the nitrogen atom of the $-NH_2$ group,thereby reducing its basicity.
In compound $(i)$,there is no electron-withdrawing group,so it is the most basic.
In compound $(ii)$,the $-NO_2$ group is at the meta position,exerting only the $-I$ effect.
In compound $(iii)$,the $-NO_2$ group is at the para position,exerting both $-I$ and strong $-M$ effects,which significantly reduce the electron density on the nitrogen atom.
Therefore,the basicity order is $(i) > (ii) > (iii)$.

(D) The given compounds are:
$(i)$ $C_6H_5CH_2NH_2$ (Benzylamine)
(ii) $C_6H_5NHCH_3$ ($N$-methylaniline)
(iii) $C_6H_5N(CH_3)_2$ ($N,N$-dimethylaniline)
(iv) $C_6H_5NH_2$ (Aniline)
Benzylamine $(i)$ is an aliphatic amine where the lone pair on nitrogen is not involved in resonance with the benzene ring,making it the most basic.
Among the aryl amines (ii,iii,and iv),the basicity depends on the electron-donating effect of the methyl groups.
$N,N$-dimethylaniline (iii) has two electron-donating methyl groups,making it more basic than $N$-methylaniline (ii),which has one methyl group.
Aniline (iv) has no electron-donating methyl groups,making it the least basic.
Thus,the correct order of basicity is $i > iii > ii > iv$.

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In aromatic amines like $C_6H_5NH_2$,$C_6H_5NHCH_3$,and $C_6H_5N(CH_3)_2$,the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring,which decreases its availability for protonation.
In aliphatic amines like $CH_3NH_2$,there is no such resonance,and the electron-donating inductive effect of the alkyl group increases the electron density on the nitrogen atom,making it more available for protonation.
Therefore,$CH_3NH_2$ is a much stronger base than the given aromatic amines.

(B) The given reaction is the Hofmann bromamide degradation reaction.
In this reaction,a primary amide $(RCONH_2)$ reacts with bromine $(Br_2)$ and a strong base ($NaOH$ or $KOH$) to form a primary amine $(RNH_2)$ with one carbon atom less than the original amide.
The balanced chemical equation is:
$H_3CCONH_2 + Br_2 + 4NaOH \longrightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Comparing this with the given reaction,$Y$ is $CH_3NH_2$ (methylamine),which corresponds to option $B$.

(C) The reaction of aniline with $NaNO_2/HCl$ at $273-278 \ K$ produces benzene diazonium chloride.
When benzene diazonium chloride is treated with $KI$,it undergoes a substitution reaction to form iodobenzene as the final product $Z$.

(C) The reduction of nitrobenzene depends on the medium used:
$1$. In neutral medium $(Zn/NH_4Cl)$: Nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
$2$. In alkaline medium $(Zn + KOH/C_2H_5OH)$: Nitrobenzene undergoes reduction to form azoxybenzene,azobenzene,and finally hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(A) $1$. The $-NO_2$ group is a deactivating and meta-directing group. Therefore,when nitrobenzene reacts with $Cl_2$ in the presence of $Fe$ (a Lewis acid catalyst),electrophilic aromatic substitution occurs at the meta position to form $m$-chloronitrobenzene $(A)$.
$2$. The reduction of nitrobenzene with $LiAlH_4$ is a complex process. While $LiAlH_4$ is a strong reducing agent,its reaction with nitrobenzene typically leads to the formation of azobenzene $(C_6H_5-N=N-C_6H_5)$ as the major product $(B)$.

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) produces hydrazobenzene $(C_{12}H_{12}N_2)$ as the final product $X$.
Structure of hydrazobenzene: $Ph-NH-NH-Ph$.
Each phenyl ring $(C_6H_5)$ contains $6 \pi$ bonds (from the aromatic ring) and $12 \sigma$ bonds ($6$ $C-C$ and $6$ $C-H$).
For two phenyl rings,we have $12 \pi$ bonds and $24 \sigma$ bonds.
Additionally,the central $N-N$ linkage has $1 \sigma$ bond,and the two $N-H$ bonds contribute $2 \sigma$ bonds.
Total $\sigma$ bonds = $24 + 1 + 2 = 27$.
Total $\pi$ bonds = $3 + 3 = 6$ (from the two benzene rings).
Thus,$X$ contains $27 \sigma$ and $6 \pi$ bonds.

(A) The reduction of nitrobenzene $(C_6H_5NO_2)$ with $Zn$ dust and alcoholic $KOH$ is a specific chemical reaction that proceeds through several intermediates to finally yield hydrazobenzene $(C_6H_5NH-NHC_6H_5)$.
This reaction is a classic example of the reduction of nitro compounds in a basic medium.

(C) The reduction of nitrobenzene with lithium aluminium hydride $(LiAlH_4)$ in an alkaline medium or specific conditions leads to the formation of azobenzene as the major product.
The reaction is represented as:
$2 C_6H_5NO_2 \xrightarrow{LiAlH_4} C_6H_5-N=N-C_6H_5$ (Azobenzene).

(D) The reduction of nitrobenzene with zinc dust $(Zn)$ and aqueous ammonium chloride $(NH_4Cl)$ is a controlled reduction process.
This reaction yields $N$-phenylhydroxylamine as the major product.
The chemical equation is:
$C_6H_5NO_2 + 4[H] \xrightarrow{Zn, NH_4Cl} C_6H_5NHOH + H_2O$

(C) Amines react with alkyl halides in excess to undergo exhaustive alkylation,resulting in the formation of a quaternary ammonium salt.
Among the given options,$C_6H_5NH_2$ (aniline) is an amine,which reacts with excess methyl iodide $(CH_3I)$ to form a quaternary ammonium salt as shown below:
$C_6H_5NH_2 + 3CH_3I \rightarrow C_6H_5N^+(CH_3)_3I^- + 2HI$
Ethers ($C_2H_5OCH_3$ and $(CH_3)_2CHOC_2H_5$) do not form quaternary salts under these conditions,and nitro compounds $(C_6H_5NO_2)$ are not nucleophilic enough to undergo this reaction.

(B) Step $1$: Reduction of nitrobenzene $(C_6H_5NO_2)$ with $Sn/HCl$ gives aniline $(C_6H_5NH_2)$ as $X$.
$C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$
Step $2$: Reaction of aniline $(X)$ with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base (like $NaOH$) is known as benzoylation,which yields benzanilide $(C_6H_5NHCOC_6H_5)$ as $Y$.
$C_6H_5NH_2 + C_6H_5COCl \xrightarrow{NaOH} C_6H_5NHCOC_6H_5 + HCl$
Thus,$Y$ is benzanilide.

(B) The reaction involves the nucleophilic substitution of a sulfonyl chloride with a secondary amine. The nitrogen atom of dimethylamine acts as a nucleophile and attacks the sulfur atom of benzenesulfonyl chloride,leading to the displacement of the chloride ion. This reaction produces $N,N$-dimethylbenzenesulfonamide. Therefore,the reagent $P$ is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.