Properties of Amines Questions in English

(B) The conversion of benzamide $(C_6H_5CONH_2)$ to aniline $(C_6H_5NH_2)$ involves the removal of a carbonyl group,which is achieved by the Hoffmann bromamide degradation reaction using $NaOH + Br_2$. Thus,$X = NaOH + Br_2$.
The conversion of benzamide $(C_6H_5CONH_2)$ to benzylamine $(C_6H_5CH_2NH_2)$ involves the reduction of the amide group to an amine group,which is achieved using a strong reducing agent like $LiAlH_4$ followed by hydrolysis. Thus,$Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.
Therefore,$X = NaOH + Br_2$ and $Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.

(C) $1$. The reaction of $CH_3COOH$ with $P_2O_5/\Delta$ produces acetic anhydride $(M)$,which is $(CH_3CO)_2O$.
$2$. Acetic anhydride reacts with aniline $(C_6H_5NH_2)$ to form acetanilide $(N)$,which is $CH_3CONHC_6H_5$.
$3$. Acetanilide is then subjected to nitration using $HNO_3/H_2SO_4$ at $288 \ K$. The $-NHCOCH_3$ group is ortho/para directing,but due to steric hindrance,the para-isomer is the major product.
$4$. Finally,hydrolysis with $H^+/H_2O$ removes the acetyl group to yield $p-$nitroaniline as the major product '$R$'.

(B) The given reaction is the Hoffmann Bromamide degradation reaction.
In this reaction,an amide is treated with bromine and an aqueous or alcoholic solution of sodium hydroxide.
The product formed is a primary amine with one carbon atom less than the starting amide.
The balanced chemical equation is:
$CH_3CH_2CH_2CH_2CONH_2 + Br_2 + 4 NaOH \rightarrow CH_3CH_2CH_2CH_2NH_2 + Na_2CO_3 + 2 NaBr + 2 H_2O$
Thus,the products are $CH_3CH_2CH_2CH_2NH_2$,$Na_2CO_3$,$2 NaBr$,and $2 H_2O$.

(D) The reaction between benzene diazonium chloride and $o$-toluidine ($2$-methylaniline) at $273 \ K$ and $pH \ 4-5$ is an electrophilic aromatic substitution reaction known as coupling reaction.
In this reaction,the diazonium cation acts as an electrophile.
The $-NH_2$ group is a strong activating group and is ortho/para directing.
Due to the steric hindrance caused by the methyl group at the ortho position,the electrophile attacks the para position with respect to the $-NH_2$ group.
Therefore,the major product is $4$-amino-$3$-methylazobenzene.

(B) $1$. The reduction of propanenitrile $(CH_3CH_2CN)$ with $LiAlH_4$ gives propan$-1-$amine $(CH_3CH_2CH_2NH_2)$,which is compound $A$.
$2$. The reaction of a primary amine $(A)$ with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
$3$. This reaction produces an isocyanide or carbylamine $(CH_3CH_2CH_2NC)$,which is compound $B$,characterized by an offensive smell.
$4$. Therefore,the conversion of $A$ to $B$ is an example of the Carbylamine reaction.

(B) The reaction of primary amines with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the carbylamine reaction. This reaction produces isocyanides (carbylamines),which are characterized by a highly foul smell.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$
Among the given options,only benzylamine $(C_6H_5CH_2NH_2)$ is a primary amine. $N$-methylaniline is a secondary amine,$N,N$-dimethylaniline is a tertiary amine,and benzamide is an amide. Therefore,only benzylamine will undergo the carbylamine reaction to produce a foul-smelling substance.

(C) The reaction of primary amines (both aliphatic and aromatic) with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the carbylamine reaction.
In this reaction,aniline $(C_6H_5NH_2)$ reacts with $CHCl_3$ and $KOH$ to form phenyl isocyanide $(C_6H_5NC)$,which is also known as phenyl carbylamine.
The chemical equation is: $C_6H_5NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow C_6H_5NC + 3KCl + 3H_2O$.
Therefore,the correct product is phenyl isocyanide.

(D) The carbylamine test is a characteristic reaction of primary amines $(R-NH_2)$.
Secondary and tertiary amines do not give this test.
$A$. Hoffmann bromide degradation produces primary amines.
$B$. Gabriel phthalimide synthesis produces primary amines.
$C$. Reduction of nitriles with $LiAlH_4$ produces primary amines.
$D$. Reduction of tertiary amides with $LiAlH_4$ produces tertiary amines $(R_3N)$.
Since tertiary amines do not contain the $-NH_2$ group,they fail to give the carbylamine test.

(C) The conversion of $p-$methyl aniline to $p-$methyl benzoic acid involves the following steps:
$1$. Diazotization: $p-$methyl aniline reacts with $NaNO_2 + HCl$ at $273 \ K$ to form $p-$methyl benzene diazonium chloride.
$2$. Sandmeyer reaction: The diazonium salt reacts with $CuCN / KCN$ to form $p-$methyl benzonitrile.
$3$. Hydrolysis: The nitrile group is hydrolyzed using $H_3O^{+}$ to form $p-$methyl benzoic acid.
Thus,the correct sequence is $NaNO_2 + HCl / 273 \ K ; CuCN / KCN ; H_3O^{+}$.

(C) The reaction sequence is as follows:
$1$. Benzoic acid reacts with $NH_3$ followed by heating to form benzamide $(X)$.
$2$. Benzamide undergoes Hofmann bromamide degradation with $Br_2/NaOH$ to form aniline $(Y)$.
$3$. Aniline reacts with $CHCl_3/KOH$ (carbylamine reaction) to form phenyl isocyanide $(Z)$,which is characterized by a foul smell.

(C) The reaction sequence is as follows:
$1$. $N$-acylation of aniline with $(CH_3CO)_2O$ in the presence of pyridine yields acetanilide $(C_6H_5NHCOCH_3)$.
$2$. Nitration of acetanilide using concentrated $HNO_3$ and concentrated $H_2SO_4$ occurs primarily at the para-position due to the steric hindrance of the bulky $-NHCOCH_3$ group,forming $p$-nitroacetanilide.
$3$. Finally,acid-catalyzed hydrolysis $(H_2O/H^+)$ of $p$-nitroacetanilide removes the acetyl group to yield $p$-nitroaniline ($4$-nitroaniline) as the major product.

(A) $I$. Piperidine is the most basic. The $N$ lone pair is in an $sp^3$ hybrid orbital,there is no resonance,and due to the $+I$ effect of the alkyl group attached,it becomes more basic than $sp^3$ hybridized ammonia.
$II$. In pyridine,the lone pair is in an $sp^2$ hybridized orbital. The $sp^2$ hybrid is smaller than $sp^3$,so there is a stronger attraction to the nucleus,making it less basic.
$III$. $NH_3$ is more basic than aniline because in ammonia,the lone pair is localized and easily available for donation.
$IV$. The basicity of aniline is lower than that of ammonia because in aniline,the lone pair of electrons on $N$ is involved in resonance with the benzene ring.
Therefore,the correct order of basic strength is $I > III > II > IV$.

(A) The $-NH_2$ group of aniline is a strong electron-donating group due to the $+M$ (mesomeric) effect.
This effect increases the electron density at the ortho and para positions of the benzene ring,making them more nucleophilic and thus more susceptible to electrophilic attack.
During electrophilic aromatic substitution,the intermediate arenium ion (sigma complex) formed by the attack of an electrophile at the ortho or para positions is stabilized by the resonance donation of the lone pair of electrons from the nitrogen atom of the $-NH_2$ group.
Therefore,both the assertion and the reason are correct,and the reason correctly explains why the $-NH_2$ group is ortho,para directing.

(A) $1$. $CH_3CH_2OH + HCl \xrightarrow{ZnCl_2} CH_3CH_2Cl (A) + H_2O$.
$2$. $CH_3CH_2Cl + AgCN \rightarrow CH_3CH_2NC (B, \text{minor}) + CH_3CH_2CN (C, \text{major})$.
$3$. $B$ is ethyl isocyanide and $C$ is ethyl cyanide. They are functional isomers ($I$ is correct).
$4$. Reduction: $CH_3CH_2NC + 4[H] \rightarrow CH_3CH_2NHCH_3$ ($2^{\circ}$ amine) and $CH_3CH_2CN + 4[H] \rightarrow CH_3CH_2CH_2NH_2$ ($1^{\circ}$ amine). Thus,$II$ is incorrect.
$5$. Hydrolysis: $CH_3CH_2NC + 2H_2O \rightarrow CH_3CH_2NH_2 + HCOOH$ (formic acid). $CH_3CH_2CN + 2H_2O \rightarrow CH_3CH_2COOH + NH_3$. $CH_3CH_2COOH$ is $C_3H_6O_2$. Thus,$III$ is correct.
$6$. $C$ (nitrile) does not form isocyanate with $HgO$; $B$ (isocyanide) can be oxidized to isocyanate. Thus,$IV$ is incorrect.
Therefore,statements $I$ and $III$ are correct.

(A) The reaction of aniline with acetic anhydride $(CH_3CO)_2O$ leads to the formation of acetanilide $(A)$. This process is known as acetylation. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group due to the resonance of the lone pair of nitrogen with the carbonyl group. This reduces the electron density on the ring,preventing polybromination. When acetanilide $(A)$ is treated with $Br_2$ in $CH_3COOH$,the bulky acetamido group directs the incoming bromine atom primarily to the para position due to steric hindrance at the ortho position. Thus,the major product $B$ is $p$-bromoacetanilide.

(C) The reaction proceeds as follows:
$(i)$ $Sn/HCl$ reduces the $-NO_2$ group to an $-NH_2$ group,forming $p$-toluidine.
(ii) $Br_2$ $(1 \ eq.)$ in the presence of the strongly activating $-NH_2$ group leads to ortho-bromination,yielding $2$-bromo-$4$-methylaniline.
(iii) $NaNO_2/HCl$ at $273-278 \ K$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.
(iv) $Cu_2Br_2/HBr$ (Sandmeyer reaction) replaces the diazonium group with a $-Br$ atom,resulting in $1,2$-dibromo-$4$-methylbenzene (also known as $3,4$-dibromotoluene).

(C) $1$. The reduction of $4-$nitrotoluene with $Fe / HCl$ yields $4-$methylaniline ($p-$toluidine).
$2$. The $-NH_2$ group is a strongly activating and $ortho-para$ directing group.
$3$. The $-CH_3$ group is also an activating and $ortho-para$ directing group.
$4$. In $4-$methylaniline,the $ortho$ positions relative to the $-NH_2$ group are the $2$ and $6$ positions.
$5$. Since the $para$ position is already occupied by the $-CH_3$ group,electrophilic bromination with an excess of $Br_2$ occurs at both available $ortho$ positions ($2$ and $6$) relative to the $-NH_2$ group,resulting in $2,6-$dibromo-$4-$methylaniline.

(C) The basic strength of an anion is inversely proportional to the acidic strength of its conjugate acid.
First,identify the conjugate acids of the given bases:
$CH_{3}^{-}$'s conjugate acid is $CH_{4}$ $(pKa \approx 50)$.
$NH_{2}^{-}$'s conjugate acid is $NH_{3}$ $(pKa \approx 38)$.
$CN^{-}$'s conjugate acid is $HCN$ $(pKa \approx 9.2)$.
$HS^{-}$'s conjugate acid is $H_{2}S$ $(pKa \approx 7)$.
Since the acidic strength order is $CH_{4} < NH_{3} < HCN < H_{2}S$,the basic strength order of their conjugate bases is the reverse: $CH_{3}^{-} > NH_{2}^{-} > CN^{-} > HS^{-}$.

(D) When aniline is treated with conc. $H_{2}SO_{4}$,it first forms anilinium hydrogen sulphate.
Upon heating this salt at $200^{\circ}C$,it undergoes a rearrangement reaction (sulphonation) to form $p$-aminobenzenesulphonic acid,which is commonly known as sulphanilic acid.
The reaction proceeds through the formation of a zwitter ion.

(A AND C) The reaction of cyclobutyl amine with $HNO_{2}$ produces a diazonium salt,which is unstable and loses $N_{2}$ to form a cyclobutyl carbocation. This carbocation can undergo ring contraction to form a more stable cyclopropylmethyl carbocation. Both carbocations can then react with water to form their respective alcohols. Therefore,the products are cyclobutanol and cyclopropylmethanol. The reaction sequence is as follows:
$1. \text{Cyclobutylamine} + HNO_{2} \rightarrow \text{Cyclobutyl diazonium ion} + H_{2}O$
$2. \text{Cyclobutyl diazonium ion} \rightarrow \text{Cyclobutyl carbocation} + N_{2}$
$3. \text{Cyclobutyl carbocation} \xrightarrow{\text{Ring contraction}} \text{Cyclopropylmethyl carbocation}$
$4. \text{Cyclobutyl carbocation} + H_{2}O \rightarrow \text{Cyclobutanol} + H^{+}$
$5. \text{Cyclopropylmethyl carbocation} + H_{2}O \rightarrow \text{Cyclopropylmethanol} + H^{+}$

(D) The reduction of benzenediazonium chloride $(C_{6}H_{5}N_{2}^{+}Cl^{-})$ to phenylhydrazine $(C_{6}H_{5}NHNH_{2})$ is a standard chemical transformation.
It can be achieved using $SnCl_{2} / HCl$ (stannous chloride in hydrochloric acid) or $Na_{2}SO_{3}$ (sodium sulfite).
Therefore,both reagents are effective for this reduction.

(C) The basicity of the given compounds depends on the stability of the conjugate acid formed after protonation and the resonance stabilization of the lone pair on nitrogen atoms.
$(1)$ $CH_3CH_2NH_2$ (Ethylamine): $A$ primary aliphatic amine,basic due to the $+I$ effect of the ethyl group.
$(2)$ $CH_3CH=NH$ (Ethylideneimine): The nitrogen atom is $sp^2$ hybridized,making the lone pair less available for protonation compared to $sp^3$ hybridized nitrogen.
$(3)$ $CH_3C(=NH)NH_2$ (Acetamidine): The lone pair on the imine nitrogen is involved in resonance with the amino group,stabilizing the conjugate acid significantly.
$(4)$ $H_2NC(=NH)NH_2$ (Guanidine): This is the most basic compound because the positive charge on the conjugate acid is delocalized over three nitrogen atoms through resonance,providing maximum stability.
Thus,the order of basicity is $2 < 1 < 3 < 4$.

(B) The reaction of amines with benzene sulphonyl chloride (Hinsberg's reagent) is used to distinguish between primary,secondary,and tertiary amines.
$1$. Primary amines $(R-NH_2)$ form a sulfonamide that is soluble in $aq. NaOH$ due to the presence of an acidic hydrogen on the nitrogen atom.
$2$. Secondary amines $(R_2NH)$ form a sulfonamide that is insoluble in $aq. NaOH$ because they lack an acidic hydrogen on the nitrogen atom.
$3$. Tertiary amines $(R_3N)$ do not react with benzene sulphonyl chloride.
Since the given amine forms a white precipitate that is insoluble in $aq. NaOH$,it must be a secondary amine.
The molecular formula $C_3H_9N$ corresponds to a secondary amine,$N$-methylethanamine $(CH_3-NH-CH_2CH_3)$.

(A) The reaction shows the removal of the diazonium group $(-N_{2}^{+}Cl^{-})$ and its replacement with a hydrogen atom $(H)$.
This is a reductive deamination reaction.
$H_{3}PO_{2}$ (hypophosphorous acid) in the presence of water is the standard reagent used to reduce diazonium salts to the corresponding arene.
The reaction is: $Ar-N_{2}^{+}Cl^{-} + H_{3}PO_{2} + H_{2}O \rightarrow Ar-H + N_{2} + H_{3}PO_{3} + HCl$.
Therefore,the correct reagent is $H_{3}PO_{2}$.

(D) Direct nitration of aniline with concentrated $HNO_3$ and $H_2SO_4$ (nitrating mixture) leads to the formation of a significant amount of $m$-nitroaniline due to the protonation of the $-NH_2$ group to form the $-NH_3^+$ ion,which is meta-directing.
In acidic medium,aniline exists as anilinium ion $(C_6H_5NH_3^+)$,which is strongly deactivating and meta-directing.
Therefore,the major product obtained under these conditions is $m$-nitroaniline.

(D) The term $carbylamine$ is another name for an isocyanide or isonitrile,which has the general chemical formula $RNC$.
In the carbylamine reaction,primary amines react with chloroform $(CHCl_3)$ and an alcoholic base $(KOH)$ to form isocyanides $(RNC)$,which are characterized by their foul smell.

(B) In aniline $(C_6H_5NH_2)$,the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring through resonance.
This reduces the availability of the lone pair for protonation,making aniline a weaker base.
In contrast,in methylamine $(CH_3NH_2)$,the electron-donating methyl group increases the electron density on the nitrogen atom,making it more basic than aniline.

(B) Aniline is a primary amine (contains $-NH_{2}$ group).
When it is treated with chloroform $(CHCl_{3})$ under alkaline conditions (using $KOH$),it results in the formation of a bad-smelling compound known as phenyl isonitrile (or phenyl carbylamine).
The chemical reaction is:
$C_{6}H_{5}NH_{2} + CHCl_{3} + 3KOH(alc.) \rightarrow C_{6}H_{5}NC + 3KCl + 3H_{2}O$.
This reaction is known as the carbylamine reaction.

(C) The reaction is an electrophilic aromatic substitution (bromination) of $N$-phenylbenzamide using $Br_2$ and $FeBr_3$.
$N$-phenylbenzamide consists of two phenyl rings attached to an amide group.
The nitrogen atom of the amide group is directly attached to one of the phenyl rings. The lone pair on the nitrogen atom is delocalized into this ring,making it an activated ring towards electrophilic substitution.
Conversely,the other phenyl ring is attached to the carbonyl group $(C=O)$,which is an electron-withdrawing group,making that ring deactivated.
Therefore,the electrophilic substitution occurs on the activated ring at the ortho and para positions. Due to steric hindrance,the para-substituted product is the major product.
Thus,the major product is $N$-($4$-bromophenyl)benzamide.

(C) The reaction proceeds in two steps:
$1$. Nitrobenzene $(C_6H_5NO_2)$ is reduced by $NH_4Cl / Zn$ dust to form phenylhydroxylamine $(C_6H_5NHOH)$.
$2$. Phenylhydroxylamine is then oxidized by $H_2SO_4 / Na_2Cr_2O_7$ to form nitrosobenzene $(C_6H_5NO)$.

(D) The molecular formula $C_6H_7N$ and the given reactions indicate that $x$ is Aniline $(C_6H_5NH_2)$.
$1$. $x$ is a primary amine because it gives the carbylamine test (unpleasant smell of $y$,which is phenyl isocyanide,$C_6H_5NC$).
$2$. Reaction with benzenesulphonyl chloride (Hinsberg reagent): $C_6H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5-SO_2-NH-C_6H_5$ $(z)$.
$3$. Compound $z$ is $N$-phenylbenzenesulfonamide.
$4$. Different types of $H$ atoms in $z$:
- Ring $1$ (attached to $SO_2$): $3$ types (ortho,meta,para).
- Ring $2$ (attached to $NH$): $3$ types (ortho,meta,para).
- $H$ atom on Nitrogen: $1$ type.
Total different $H$ atoms = $3 + 3 + 1 = 7$.

(C) $1$. Empirical formula calculation:
$C = 61.01/12 = 5.08$,$H = 15.25/1 = 15.25$,$N = 23.74/14 = 1.69$.
Dividing by the smallest value $(1.69)$: $C = 3$,$H = 9$,$N = 1$.
The empirical formula is $C_3H_9N$.
$2$. Reaction with $HNO_2$:
Aliphatic primary amines react with $HNO_2$ to form unstable diazonium salts,which decompose to form alcohols $(Y)$ and evolve $N_2$ gas.
For $C_3H_9N$ (isopropylamine),the reaction is:
$(CH_3)_2CH-NH_2 + HNO_2 \rightarrow (CH_3)_2CH-OH (Y) + N_2 + H_2O$.
$3$. Oxidation and Iodoform test:
Compound '$Y$' is isopropyl alcohol (propan$-2-$ol).
Controlled oxidation of propan$-2-$ol gives acetone $(Z)$,which is $CH_3COCH_3$.
Acetone contains the $CH_3CO-$ group and thus gives a positive iodoform test.
Therefore,'$X$' is isopropylamine,$(CH_3)_2CH-NH_2$.

(D) $1$. In reaction $(A)$,nitration of benzanilide occurs primarily at the ortho and para positions of the aniline ring due to the activating effect of the $-NHCOPh$ group. The product shown in the image is incorrect.
$2$. In reaction $(B)$,aryl halides do not undergo nucleophilic substitution with potassium phthalimide under standard conditions,so no reaction occurs.
$3$. In reaction $(C)$,the carbylamine reaction produces an isocyanide,which upon reduction with $Sn/HCl$ yields a secondary amine ($N$-methylbenzylamine),not the primary amine shown.
$4$. In reaction $(D)$,the Hoffmann bromamide degradation of propanamide $(CH_3CH_2CONH_2)$ correctly yields ethylamine $(CH_3CH_2NH_2)$ along with the specified byproducts. This is a standard textbook reaction.

(A) The stability of benzene diazonium salts is influenced by the nature of the substituent attached to the benzene ring.
Electron-donating groups $(EDG)$ increase the stability of the diazonium cation by dispersing the positive charge,while electron-withdrawing groups $(EWG)$ decrease the stability by intensifying the positive charge.
In the given compounds:
$A$: $-OCH_3$ is a strong $EDG$ ($+M$ effect).
$C$: Benzene diazonium chloride (no substituent).
$D$: $-CN$ is an $EWG$ ($-M$ effect).
$B$: $-NO_2$ is a very strong $EWG$ ($-M$ effect).
Therefore,the stability order is $A > C > D > B$.

(B) The reaction is: $C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl$
Molar mass of aniline $(C_6H_5NH_2)$ = $6 \times 12 + 7 \times 1 + 14 = 93 \ g \ mol^{-1}$.
Molar mass of benzanilide $(C_6H_5NHCOC_6H_5)$ = $13 \times 12 + 11 \times 1 + 14 + 16 = 197 \ g \ mol^{-1}$.
Moles of aniline = $\frac{5.8 \ g}{93 \ g \ mol^{-1}} \approx 0.06237 \ mol$.
Theoretical moles of benzanilide = $0.06237 \ mol$.
Actual moles of benzanilide = $0.06237 \times 0.82 \approx 0.05114 \ mol$.
Mass of benzanilide = $0.05114 \ mol \times 197 \ g \ mol^{-1} \approx 10.07 \ g$.
Rounding to the nearest integer,the mass is $10 \ g$.

(C) The reaction proceeds in two steps:
$1$. Cyclohexylmethanamine reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base $(NaOH)$ to form an amide,$N$-(cyclohexylmethyl)benzamide,which is product $[A]$.
$2$. The amide $[A]$ is then reduced using lithium aluminium hydride $(LiAlH_4)$ followed by hydrolysis $(H_2O)$ to yield the corresponding amine,$N$-benzylcyclohexylmethanamine,which is the final product $[B]$.

(C) The reaction shown is the Hofmann bromamide degradation of $2$-methylbenzamide using $NaOBr$,which yields $2$-methylaniline (o-toluidine) as Product "$X$".
$2$-methylaniline is a primary aromatic amine $(Ar-NH_2)$.
$1$. It reacts with Hinsberg's reagent (benzenesulfonyl chloride) to form a sulfonamide,which is soluble in alkali.
$2$. It reacts with nitrous acid $(HNO_2)$ at $0-5^{\circ}C$ to form a diazonium salt,which gives a positive Azo dye test with $\beta$-naphthol.
$3$. It is a primary amine,and its $2^\circ$ amine isomer is $N$-methylaniline $(C_6H_5NHCH_3)$.
$4$. Primary aromatic amines are generally insoluble in aqueous $NaOH$ because they are basic in nature,not acidic.
Therefore,the correct statement is that it has one isomer of $2^\circ$ amine,which is $N$-methylaniline.

(A) Reactivity towards $HCl$ depends on the basicity of the amine.
Amines act as bases by donating the lone pair of electrons on the nitrogen atom to the $H^+$ ion.
Aliphatic amines are more basic than aromatic amines because the alkyl groups exert an electron-donating inductive effect ($+I$ effect),which increases the electron density on the nitrogen atom.
In aqueous solution,the basicity of aliphatic amines is determined by a combination of inductive effect,solvation effect,and steric hindrance.
For dimethylamine $((CH_3)_2NH)$,the combination of these factors makes it the most basic among the given options,thereby making it the most reactive towards dilute $HCl$.

(B) In the reduction of nitro compounds using iron scrap and $HCl$,iron reacts with $HCl$ to produce $FeCl_2$ and hydrogen gas.
The $FeCl_2$ formed undergoes hydrolysis in the presence of water to produce $Fe_3O_4$ and regenerate $HCl$.
This regenerated $HCl$ further reacts with more iron to continue the reduction process.
Therefore,only a catalytic amount of $HCl$ is required to initiate the reaction.
Both the assertion and the reason are correct,and the reason provides the correct explanation for the assertion.

(B) The carbylamine reaction (or isocyanide test) is a characteristic reaction of primary amines. In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and an alcoholic base (like $KOH$) to form an isocyanide (carbylamine),which has a foul,offensive smell.
The reaction for $4$-methylaniline ($p$-toluidine) is as follows:
$CH_3-C_6H_4-NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow CH_3-C_6H_4-NC + 3KCl + 3H_2O$
Here,$4$-methylaniline reacts with $CHCl_3$ and $KOH$ to produce $4$-methylphenyl isocyanide $(CH_3-C_6H_4-NC)$.