Properties of Amines Questions in English

(D) In an aqueous medium,the basicity of amines is determined by a combination of three factors: inductive effect $(+I)$,solvation effect (hydrogen bonding),and steric hindrance.
For ethyl-substituted amines,the secondary amine $(C_2H_5)_2NH$ provides the optimal balance between the electron-donating inductive effect of the two ethyl groups and the stability provided by solvation.
Therefore,$(C_2H_5)_2NH$ is the strongest base among the given options in an aqueous medium.

(B) Nucleophilicity is generally related to the basicity of the species. Stronger bases are better nucleophiles when the attacking atom is the same.
$1$. The order of basicity is $RO^{-} > HO^{-} > RCOO^{-} > ROH > H_2O$.
$2$. Since $RO^{-}$ is a stronger base than $HO^{-}$,it is a better nucleophile.
$3$. $RCOO^{-}$ is a weaker base due to resonance stabilization compared to $HO^{-}$.
$4$. Neutral molecules like $ROH$ and $H_2O$ are weaker nucleophiles than their conjugate bases.
Therefore,the correct order of nucleophilicity is $RO^{-} > HO^{-} > RCOO^{-} > ROH > H_2O$.

(B) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
In $C_6H_5NH_2$ (aniline),the lone pair is delocalized into the benzene ring due to resonance,reducing its availability.
In $O_2N-C_6H_4-NH_2$,the $-NO_2$ group is a strong electron-withdrawing group,which further decreases the electron density on nitrogen.
In $C_6H_5NHCOCH_3$,the lone pair is involved in resonance with the carbonyl group $(C=O)$,making it less available.
In $C_6H_{11}NH_2$ (cyclohexylamine),the nitrogen is attached to an $sp^3$ hybridized carbon atom. There is no resonance,and the alkyl group shows a $+I$ effect,which increases the electron density on the nitrogen atom,making it the strongest base among the given options.

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
In $C_6H_5NH_2$ (aniline),the lone pair is involved in resonance with the benzene ring,reducing its availability.
In $p-NO_2-C_6H_4NH_2$ and $m-NO_2-C_6H_4NH_2$,the $-NO_2$ group is a strong electron-withdrawing group $(EWG)$ that further decreases the electron density on the nitrogen atom,making them less basic than aniline.
In $C_6H_5CH_2NH_2$ (benzylamine),the $-NH_2$ group is attached to a $CH_2$ group,not directly to the benzene ring. Therefore,there is no resonance with the benzene ring,and the lone pair is fully available for protonation.
Thus,$C_6H_4CH_2NH_2$ is the most basic compound among the given options.

(A) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.
$1$. In $CH_3NH_2$,the methyl group $(CH_3)$ is an electron-donating group ($+I$ effect),which increases the electron density on the nitrogen atom,making it more basic.
$2$. In $NH_3$,there is no substituent effect.
$3$. In $ClNH_2$,the chlorine atom $(Cl)$ is an electron-withdrawing group ($-I$ effect),which decreases the electron density on the nitrogen atom,making it the least basic.
Therefore,the increasing order of basic strength is: $ClNH_2 < NH_3 < CH_3NH_2$.

(D) To determine the strongest base,we look at the availability of the lone pair on the nitrogen atom.
In $C_6H_5NH_2$ (aniline) and $C_6H_5NHCH_3$ ($N$-methylaniline),the lone pair on the nitrogen is in conjugation with the benzene ring,which significantly reduces its basicity.
In $o$-toluidine $(C_6H_4(CH_3)NH_2)$,the ortho-methyl group causes steric hindrance,further reducing basicity.
In $C_6H_5CH_2NH_2$ (benzylamine),the nitrogen atom is not directly attached to the benzene ring. The lone pair is not involved in resonance with the aromatic ring,making it much more available for protonation compared to the other options. Therefore,$C_6H_5CH_2NH_2$ is the strongest base.

(C) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$1$. $NH_3$ has no alkyl group to increase electron density on the nitrogen atom.
$2$. $CH_3NH_2$ ($1^{\circ}$ amine) has one methyl group which shows a $+I$ effect,increasing the electron density on the nitrogen atom,making it more basic than $NH_3$.
$3$. $(CH_3)_2NH$ ($2^{\circ}$ amine) has two methyl groups,providing a stronger $+I$ effect than $CH_3NH_2$,further increasing the electron density on the nitrogen atom.
Therefore,the correct order of increasing basic strength is $NH_3 < CH_3NH_2 < (CH_3)_2NH$.

(D) The basic strength depends on the availability of the lone pair of electrons on the nitrogen atom.
$I$ (Pyrrole): The lone pair on nitrogen is involved in aromaticity (delocalized),making it the least basic.
$II$ (Piperidine): The lone pair is on an $sp^3$ hybridized nitrogen and is localized,making it the most basic.
$III$ (Pyridine): The lone pair is on an $sp^2$ hybridized nitrogen and is localized,but $sp^2$ nitrogen is more electronegative than $sp^3$ nitrogen,making it less basic than $II$.
$IV$ (Aniline): The lone pair is delocalized into the benzene ring through resonance,making it more basic than $I$ but less basic than $III$.
Therefore,the correct decreasing order of basic strength is $II > III > IV > I$.

(C) The acidity of amines depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
In these compounds,the acidity is primarily determined by the electron-donating effect (inductive effect) of the alkyl groups attached to the nitrogen atom.
Alkyl groups are electron-releasing ($+I$ effect),which destabilizes the conjugate base (amide ion) by increasing electron density on the nitrogen atom.
Therefore,the compound with the fewest or least electron-donating alkyl groups will be the most acidic.
Comparing the options:
$NH_3$ has no alkyl groups.
$CH_3NH_2$ has one methyl group.
$CH_3CH_2NH_2$ has one ethyl group (more $+I$ effect than methyl).
$(CH_3)_2CHNH_2$ has two methyl groups.
Thus,$NH_3$ is the most acidic among the given options.

(C) In an aqueous medium,the basicity of methyl-substituted amines is determined by a combination of inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
$1$. $(CH_3)_2NH$ (dimethylamine) is the most basic due to the combined effect of the $+I$ effect of two methyl groups and sufficient solvation.
$2$. $CH_3NH_2$ (methylamine) is next.
$3$. $(CH_3)_3N$ (trimethylamine) is less basic than expected due to steric hindrance which prevents effective solvation of the protonated cation.
$4$. $NH_3$ (ammonia) is the least basic among these.

(B) The reaction of primary aliphatic amines with nitrous acid $(HNO_2)$ produces unstable diazonium salts,which decompose to form alcohols,nitrogen gas,and water.
The chemical equation is:
$C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O$
Thus,ethylamine reacts with nitrous acid to give ethyl alcohol and nitrogen gas.

(A) The reaction proceeds in two steps:
$1$. The ketone reacts with methylamine $(CH_3NH_2)$ to form an imine intermediate $(R_2C=NCH_3)$.
$2$. The imine is then reduced by lithium aluminum hydride $(LiAlH_4)$ followed by hydrolysis $(H_2O)$ to yield a secondary amine,specifically $N$-methyl$-2-$propanamine.

(C) Schiff base is formed by the reaction of a primary amine with an aldehyde or a ketone. The general reaction is $R-NH_2 + R'-C(=O)-R'' \rightarrow R-N=C(R')R'' + H_2O$. In this case,aniline $(C_6H_5NH_2)$ reacts with a carbonyl compound like acetone $(CH_3COCH_3)$ or benzaldehyde $(C_6H_5CHO)$ to form a Schiff base. Based on the provided image,the reaction shown is between acetone and aniline to form a Schiff base. Therefore,the correct option is acetone.

(D) This reaction is known as the Hofmann bromamide degradation reaction.
In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.
$CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
Propionamide $(CH_3CH_2CONH_2)$ reacts to form ethyl amine $(CH_3CH_2NH_2)$.

(C) Primary amides contain two $N-H$ bonds,which allow them to form the maximum number of intermolecular hydrogen bonds compared to secondary or tertiary amides. Due to this extensive hydrogen bonding,primary amides exhibit the highest melting points.

(C) This reaction is known as the $Hofmann$ bromamide degradation reaction.
When acetamide $(CH_3CONH_2)$ reacts with $Br_2$ and $KOH$ (or $NaOH$),it undergoes degradation to form methylamine $(CH_3NH_2)$.
The chemical equation is: $CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$.

(D) Acetamide $(CH_3CONH_2)$ is an amide. Amides exhibit amphoteric character because the lone pair on the nitrogen atom can be protonated by strong acids,and the hydrogen atoms attached to the nitrogen can be removed by strong bases.

(B) This reaction is known as the $Schotten-Baumann$ reaction.
$C_6H_5NH_2 + ClCOC_6H_5 \xrightarrow{NaOH} C_6H_5NHCOC_6H_5 + HCl$
In this reaction,the nucleophilic attack of the aniline nitrogen on the carbonyl carbon of benzoyl chloride leads to the formation of $N$-phenylbenzamide,commonly known as $Benzanilide$.

(C) The correct answer is $(C)$.
Amides exhibit high melting and boiling points due to strong intermolecular hydrogen bonding.
In primary amides $(R-CONH_2)$,there are two hydrogen atoms attached to the nitrogen atom,which allows for extensive intermolecular hydrogen bonding.
In secondary amides $(R-CONHR)$,there is only one hydrogen atom available for hydrogen bonding.
Tertiary amides $(R-CONR_2)$ lack hydrogen atoms attached to the nitrogen atom,preventing them from forming intermolecular hydrogen bonds with each other.
Therefore,the order of melting/boiling points is: $\text{Primary amides} > \text{Secondary amides} > \text{Tertiary amides} > \text{Amines}$.

(B) In amines,the nitrogen atom possesses a lone pair of electrons.
According to the Lewis theory,a species that can donate an electron pair is a Lewis base.
Since amines can donate their lone pair of electrons,they behave as Lewis bases.

(B) The reaction of a primary amine with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction produces an isocyanide (carbylamine),which has a foul smell.
The general chemical equation is: $R-NH_2 + CHCl_3 + 3KOH_{(alc)} \to R-NC + 3KCl + 3H_2O$.

(B) The $Carbylamine$ reaction is used for the preparation of isocyanides (or carbylamines).
In this reaction,a primary amine (aliphatic or aromatic) is heated with chloroform $(CHCl_3)$ and an alcoholic potassium hydroxide $(KOH)$ solution to form an isocyanide (carbylamine),which has a foul smell.
The reaction for the preparation of phenyl isocyanide from aniline is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Therefore,the correct option is $(B)$.

(C) In amines,the nitrogen atom is bonded to three other atoms (either $H$ or $R$ groups) and possesses one lone pair of electrons.
Using the formula for steric number: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $N$ in amines: $\text{Steric Number} = \frac{1}{2} [5 + 3 - 0 + 0] = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(C) The carbylamine reaction involves heating a primary amine with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$.
This reaction produces an alkyl isocyanide $(RNC)$,which is characterized by a highly offensive or foul smell.
The chemical equation is: $RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O$.
Therefore,the correct option is $(C)$.

(C) $C_2H_5NH_2 + HNO_2 \to C_2H_5OH (A) + N_2 + H_2O$
$C_2H_5OH + PCl_5 \to C_2H_5Cl (B) + POCl_3 + HCl$
$C_2H_5Cl + KCN \to C_2H_5CN (C) + KCl$
The final product $C$ is $C_2H_5CN$,which is named $Propane \ nitrile$.

(C) The basic strength of aromatic amines depends on the electron density on the nitrogen atom.
Electron-donating groups (like $-CH_3$) increase the electron density on the nitrogen atom,thereby increasing the basic strength.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density on the nitrogen atom,thereby decreasing the basic strength.
In the given compounds:
$(I)$ is Aniline.
$(II)$ is $p$-Nitroaniline,where $-NO_2$ is a strong electron-withdrawing group,making it the least basic.
$(III)$ is $p$-Toluidine,where $-CH_3$ is an electron-donating group,making it the most basic.
Therefore,the increasing order of basic strength is: $II < I < III$.

(C) In arylamines,the lone pair of electrons on the nitrogen atom is involved in resonance with the aromatic ring.
This delocalization makes the lone pair less available for protonation,thereby reducing the basicity of arylamines compared to alkylamines,where the alkyl group exerts a $+I$ effect that increases electron density on the nitrogen atom.

(C) The reaction of aniline $(C_6H_5NH_2)$ with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base like $NaOH$ is known as the Schotten-Baumann reaction.
This process involves the benzoylation of the amine to form an amide ($N$-phenylbenzamide).

(C) The electrolytic reduction of nitrobenzene $(C_6H_5NO_2)$ in a strongly acidic medium proceeds through the formation of phenylhydroxylamine $(C_6H_5NHOH)$.
In the presence of a strong acid $(H^+)$,phenylhydroxylamine undergoes a Bamberger rearrangement to form $p-$aminophenol $(H_2N-C_6H_4-OH)$.

(A) The reaction shown is a coupling reaction between benzenediazonium chloride and aniline.
This reaction is an electrophilic aromatic substitution where the diazonium cation acts as the electrophile.
Since the $-NH_2$ group is a strong ortho/para-directing group,the electrophile attacks the para-position to the $-NH_2$ group,yielding $p$-aminoazobenzene,which is a yellow dye.

(B) The diazonium salt containing an aryl group directly linked to the nitrogen atom $(C_6H_5N_2^+X^-)$ is the most stable.
This stability is due to the resonance stabilization between the benzene ring and the diazonium group $(-N_2^+)$,which delocalizes the positive charge over the benzene ring.
Aliphatic diazonium salts are highly unstable and decompose readily to form carbocations and nitrogen gas.
The resonance structures of the benzene diazonium ion are shown below:
(The image provided shows the resonance structures where the positive charge is delocalized into the ortho and para positions of the benzene ring.)

(A) The given reaction involves the removal of the diazonium group $(-N_2^+Cl^-)$ from the benzene ring,which is a deamination reaction.
The reagent $H_3PO_2$ (hypophosphorous acid) in the presence of water $(H_2O)$ is a standard reducing agent used to replace the diazonium group with a hydrogen atom.
Therefore,the conversion of $4-nitro-3-bromobenzenediazonium \ chloride$ to $1-bromo-3-nitrobenzene$ is achieved using $H_3PO_2$ and $H_2O$.

(A) The hydrolysis of an alkyl isocyanide $(R-NC)$ yields a primary amine $(R-NH_2)$ and formic acid $(HCOOH)$.
Here,the compound is $CH_3CH_2CH_2NC$ (propyl isocyanide).
Hydrolysis: $CH_3CH_2CH_2NC + 2H_2O \rightarrow CH_3CH_2CH_2NH_2 + HCOOH$.
$1$. $CH_3CH_2CH_2NH_2$ (propylamine) reacts with $NaNO_2/HCl$ to form $CH_3CH_2CH_2OH$ (propan$-1-$ol),which does not give the iodoform test.
$2$. $HCOOH$ (formic acid) reduces both Tollens reagent and Fehling's solution.
Thus,the compound is $CH_3CH_2CH_2NC$.

(D) The reaction of secondary aliphatic amines with nitrous acid $(HNO_2)$ produces $N$-nitrosamines,which are yellow oily compounds. The reaction is: $(CH_3)_2NH + HNO_2 \rightarrow (CH_3)_2N-N=O + H_2O$.
Option $D$ is incorrect because the reaction of dimethylamine with nitrous acid (generated from $NaNO_2 + HCl$) forms $N$-nitrosodimethylamine,not the structure implied by the notation in the option. Furthermore,the reaction shown in the option is chemically inaccurate regarding the products and stoichiometry for the given reactants.

(A) The molecular formula $(C_3H_9N)$ corresponds to a primary amine.
When treated with nitrous acid $(HNO_2)$,primary aliphatic amines evolve $N_2$ gas and form an alcohol.
The carbylamine reaction (warming with $CHCl_3$ and $KOH$) is a characteristic test for primary amines,forming an isocyanide $(C)$.
Reduction of the isocyanide $(CH_3)_2CH-NC$ gives isopropylmethylamine $(CH_3)_2CH-NH-CH_3$.
Therefore,$(A)$ is isopropylamine,$(CH_3)_2CH-NH_2$.

(A) The reduction of nitrobenzene with $Zn$ in the presence of $NH_4Cl$ is a controlled reduction process.
Nitrobenzene $(C_6H_5NO_2)$ reacts with $Zn/NH_4Cl$ to form $N$-phenylhydroxylamine $(C_6H_5NHOH)$.
Therefore,the correct product is phenylhydroxylamine.

(B) The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $C_6H_5-CH_2NH_2$ (benzylamine),the lone pair on the nitrogen atom is localised because the $-NH_2$ group is attached to an $sp^3$ hybridized carbon atom,which is not directly attached to the benzene ring.
In the other compounds ($p$-nitroaniline,acetanilide,and aniline),the lone pair on the nitrogen atom is delocalized over the benzene ring due to resonance,making it less available for donation.
Therefore,benzylamine is the most basic compound among the given options.

(B) $1$. The reaction of aniline with $NaNO_{2}/HCl$ at $273-278 \ K$ results in the formation of benzenediazonium chloride $(X)$.
$2$. Benzenediazonium chloride $(X)$ then undergoes an electrophilic aromatic substitution reaction (coupling reaction) with $N,N-dimethylaniline$ in a weakly acidic medium.
$3$. The coupling occurs at the para-position of $N,N-dimethylaniline$ to form $p-(dimethylamino)azobenzene$ $(Y)$,which is a yellow-orange coloured dye.
$4$. The structure of $Y$ is $C_6H_5-N=N-C_6H_4-N(CH_3)_2$.

(C) $(i)$ The presence of an electron-withdrawing substituent decreases basicity,while the presence of an electron-releasing substituent like $-CH_{3}, -C_{2}H_{5},$ etc.,increases basicity.
$(ii)$ $HNO_{2}$ converts the $-NH_{2}$ group of aliphatic amines into $-OH$ (alcohols),whereas it converts aromatic amines into diazonium salts at low temperatures $(0-5\,^{\circ}C)$.
$(iii)$ Alkyl amines are more basic than ammonia due to the electron-releasing effect of the alkyl group,while aryl amines are less basic than ammonia due to the electron-withdrawing nature of the phenyl ring.
$(iv)$ The reaction for alkyl amines: $R-NH_{2} \xrightarrow{HNO_{2}} R-OH + N_{2} + H_{2}O$.
$(v)$ The reaction for aryl amines: $C_{6}H_{5}NH_{2} \xrightarrow{NaNO_{2} + HCl, 273-278\,K} C_{6}H_{5}N_{2}^{+}Cl^{-}$.
Therefore,the statement that aryl amines react with nitrous acid to produce phenols is false,as they form diazonium salts under these conditions.

(A) $N$-methylaniline is a secondary amine. When it reacts with nitrous acid $(HNO_2)$,generated in situ by the reaction of $NaNO_2$ and $HCl$,it undergoes $N$-nitrosation to form $N$-nitroso-$N$-methylaniline. The reaction is as follows:
$C_6H_5NHCH_3 + NaNO_2 + HCl \rightarrow C_6H_5N(NO)CH_3 + NaCl + H_2O$
This product is a yellow oily liquid.

(A) $CH_{3}NC + 4[H] \xrightarrow{LiAlH_{4}} CH_{3}NHCH_{3}$
$CH_{3}NC$ (Methyl isocyanide) on reduction with lithium aluminium hydride $(LiAlH_{4})$ yields $CH_{3}NHCH_{3}$ (Dimethylamine),which is a secondary $(2^{\circ})$ amine.
In contrast,reduction of alkyl cyanides $(RCN)$ with $LiAlH_{4}$ yields primary $(1^{\circ})$ amines $(RCH_{2}NH_{2})$.
Reduction of acetamide $(CH_{3}CONH_{2})$ yields ethylamine $(CH_{3}CH_{2}NH_{2})$,a primary amine.
Reduction of nitroethane $(CH_{3}CH_{2}NO_{2})$ yields ethylamine $(CH_{3}CH_{2}NH_{2})$,a primary amine.

(A) Benzylamine,$C_{6}H_{5}CH_{2}NH_{2}$,is more basic than aniline because the benzyl group $(C_{6}H_{5}CH_{2}-)$ is an electron-donating group due to the $+I$ effect.
This increases the electron density on the nitrogen atom of the $-NH_{2}$ group,making the lone pair more available for donation.
In contrast,phenyl and nitro groups are electron-withdrawing groups ($-I$ or $-M$ effects),which decrease the electron density on the nitrogen atom.
Therefore,diphenylamine,triphenylamine,and $p-$nitroaniline are less basic than aniline.

(D) Aromatic amines are less basic than aliphatic amines.
Among aliphatic amines,the order of basicity in aqueous solution is $2^{o} > 1^{o} > 3^{o}$ due to a combination of inductive effect,solvation effect,and steric hindrance.
In $3^{o}$ amines (Trimethylamine),the steric hindrance of three alkyl groups makes the approach and bonding of a proton relatively difficult.
Aniline is much less basic than aliphatic amines because the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring.
Therefore,Dimethylamine ($2^{o}$ aliphatic amine) is the strongest base among the given choices.
$\therefore$ The correct order of basic strength is $\text{Dimethylamine} > \text{Methylamine} > \text{Trimethylamine} > \text{Aniline}$.

(A) The given reaction is the carbylamine reaction,which is a characteristic test for primary amines.
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine) and potassium chloride $(KCl)$.
The balanced chemical equation is:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow C_2H_5NC + 3KCl + 3H_2O$
Here,$(A)$ is $C_2H_5NC$ (ethyl isocyanide) and $(B)$ is $3KCl$.

(C) The reaction of aniline with $NaNO_2$ and $HCl$ at $278 \ K$ $(0-5 \ ^\circ C)$ is known as diazotization,which produces benzene diazonium chloride $(A)$.
Reaction: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.
When benzene diazonium chloride $(A)$ is treated with fluoroboric acid $(HBF_4)$,it forms benzene diazonium fluoroborate,which upon heating decomposes to give fluorobenzene $(B)$.
Reaction: $C_6H_5N_2^+Cl^- + HBF_4$ $\rightarrow C_6H_5N_2^+BF_4^- + HCl$ $\xrightarrow{\Delta} C_6H_5F + N_2 + BF_3$.
Thus,$A$ is benzene diazonium chloride and $B$ is fluorobenzene.

(B) During acetylation,one $H$ atom (atomic mass $1 \, u$) of an $-NH_2$ group is replaced by an acetyl group $(CH_3CO-)$ with a molecular mass of $43 \, u$.
The reaction is: $-NH_2 + CH_3COCl \longrightarrow -NHCOCH_3 + HCl$.
This indicates that the acylation of each $-NH_2$ group increases the molecular mass of the compound by $(43 - 1) = 42 \, u$.
Given that the initial molecular mass is $180 \, u$ and the final molecular mass is $390 \, u$,the total increase in mass is $390 - 180 = 210 \, u$.
Therefore,the number of $-NH_2$ groups present is $\frac{210}{42} = 5$.

(D) This reaction is known as the $Carbylamine$ test,which is used to identify primary amines.
When an aliphatic primary amine $(R-NH_2)$ is heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,it forms an alkyl isocyanide $(R-NC)$,which has a foul smell.
The chemical equation is: $R-NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow R-NC + 3KCl + 3H_2O$.
This reaction involves the formation of a dichlorocarbene intermediate $(:CCl_2)$.

(A) The basic strength of amines in aqueous solution depends on the combined effect of inductive effect,solvation effect,and steric hindrance. For methyl-substituted amines,the order of basic strength is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > C_6H_5NH_2$.
Since $pK_b = -\log K_b$,a higher $K_b$ value corresponds to a smaller $pK_b$ value.
Therefore,$(CH_3)_2NH$ (dimethylamine) has the highest basic strength and consequently the smallest $pK_b$ value.
$C_6H_5NH_2$ (aniline) is the least basic due to the resonance of the lone pair with the benzene ring.

(A) The reaction proceeds in two steps:
$1$. The reaction of $p$-toluidine ($4$-methylaniline) with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which forms the diazonium salt,$4$-methylbenzenediazonium chloride $(D)$.
$2$. The reaction of the diazonium salt $(D)$ with $CuCN/KCN$ is a Sandmeyer reaction,where the diazonium group $(-N_2^+Cl^-)$ is replaced by a cyano group $(-CN)$ to form $4$-methylbenzonitrile $(E)$.

(B) The $Hofmann$ bromamide degradation reaction is represented by the following chemical equation:
$RCONH_2 + 4NaOH + Br_2 \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
From the stoichiometry of the balanced equation,it is clear that for every $1$ mole of amine $(RNH_2)$ produced,$4$ moles of $NaOH$ and $1$ mole of $Br_2$ are consumed.