Properties of Amines Questions in English

(C) $1$. Primary amine $(RNH_2)$ reacts with $C_6H_5SO_2Cl$ to form $N$-alkylbenzenesulfonamide $(RNHSO_2C_6H_5)$.
$2$. This product contains an acidic hydrogen attached to the nitrogen atom,so it dissolves in aqueous $KOH$ to form a water-soluble potassium salt,$RN(K^+)SO_2C_6H_5$.
$3$. Upon acidification of this solution,the potassium salt is converted back into the insoluble $N$-alkylbenzenesulfonamide $(RNHSO_2C_6H_5)$,which appears as a precipitate.

(C) The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom.
In $A$ (Pyridine),$B$ (Pyrrole),and $D$ (Aniline),the lone pair of electrons on the nitrogen atom is involved in resonance or is part of the aromatic system,making it less available for protonation.
In $C$ (Piperidine),the nitrogen atom is $sp^3$ hybridized and the lone pair is not involved in any resonance or aromatic system,making it highly available for protonation.
Therefore,Piperidine is the strongest base among the given options.

(A) The reaction described is the Hofmann bromamide degradation reaction.
$CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + 2KBr + K_2CO_3 + 2H_2O$
In this reaction,the carbonyl group $(C=O)$ of the amide is removed as a carbonate,resulting in a primary amine that contains one carbon atom less than the original amide.

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In aniline $(C_6H_5NH_2)$,$N$-methylaniline $(C_6H_5NHCH_3)$,and $o$-toluidine $(2-CH_3C_6H_4NH_2)$,the lone pair on the nitrogen atom is in conjugation with the benzene ring,which decreases its availability for protonation.
In benzylamine $(C_6H_5CH_2NH_2)$,the nitrogen atom is attached to a $CH_2$ group,not directly to the benzene ring. Therefore,the lone pair on the nitrogen atom is not involved in resonance with the benzene ring,making it more available for protonation.
Thus,benzylamine is the strongest base among the given options.

(C) The acetylation reaction involves the replacement of a hydrogen atom in the $-NH_2$ group with an acetyl group $(-COCH_3)$.
The increase in molecular weight per $-NH_2$ group is the difference between the mass of the acetyl group $(-COCH_3 = 12 + 16 + 12 + 3 = 43)$ and the hydrogen atom replaced $(H = 1)$.
Increase in mass per group $= 43 - 1 = 42$.
Total increase in molecular weight $= 390 - 180 = 210$.
Number of $-NH_2$ groups $= \frac{210}{42} = 5$.

(B) The correct option is $(B)$.
Unlike soaps,alkyl benzene sulphonates form $Ca^{2+}$ and $Mg^{2+}$ salts that are soluble in water.
Therefore,they do not form scum in hard water and remain effective as detergents.

(C) The compound $C_6H_5 - NH - NH - C_6H_5$ is $1,2$-diphenylhydrazine (hydrazobenzene).
It is a colourless crystalline solid.
In contrast,$C_6H_5 - N = N - C_6H_5$ (azobenzene) is orange-red,and $C_6H_5 - N = N(O) - C_6H_5$ (azoxybenzene) is yellow.

(B) Methylamine reacts with nitrous acid to produce methanol,nitrogen gas,and water.
$CH_3NH_2 + HNO_2 \rightarrow CH_3OH + N_2 + H_2O$
According to the stoichiometry of the reaction,$1 \ mol$ of methylamine produces $1 \ mol$ of $N_2$ gas.
At $NTP$,$1 \ mol$ of any ideal gas occupies $22.4 \ L$ volume.
Therefore,the reaction yields $22.4 \ L$ of $N_2$.

(C) To determine the basicity,we look at the conjugate acids of the given species:
$1$. The conjugate acid of $H_2O$ is $H_3O^+$.
$2$. The conjugate acid of $OH^-$ is $H_2O$.
$3$. The conjugate acid of $CH_3OH$ is $CH_3OH_2^+$.
$4$. The conjugate acid of $CH_3O^-$ is $CH_3OH$.
Stronger acids have weaker conjugate bases. The acidity order of the conjugate acids is $CH_3OH_2^+ > H_3O^+ > CH_3OH > H_2O$.
Therefore,the basicity order is the reverse: $H_2O < CH_3OH < OH^- < CH_3O^-$.
$CH_3O^-$ is a stronger base than $OH^-$ due to the electron-donating inductive effect $(+I)$ of the methyl group,which increases the electron density on the oxygen atom.

(A) Primary amines $(R-NH_2)$ react with acid anhydrides $((R'CO)_2O)$ to form amides. The reaction is an acylation reaction. The general equation is: $R-NH_2 + (R'CO)_2O \rightarrow R-NH-COR' + R'COOH$. The product formed is an amide.

(B) Schiff base is formed by the reaction of a primary amine with an aldehyde or ketone,resulting in the formation of an imine $(R-CH=N-R')$.
$1$. $C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O$ (Alcohol is formed).
$2$. $C_2H_5NH_2 + C_6H_5CHO \rightarrow C_6H_5CH=NC_2H_5 + H_2O$ (This is a Schiff base).
$3$. $C_2H_5NH_2 + NOCl \rightarrow C_2H_5Cl + N_2 + H_2O$ (Alkyl chloride is formed).
$4$. $C_2H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHC_2H_5 + HCl$ ($N$-ethylbenzenesulfonamide is formed).
Therefore,the product $b$ is a Schiff base.

(C) When an aliphatic primary amine reacts with $NaNO_2$ and $HCl$ at $0-5 \, ^\circ C$,it forms an unstable diazonium salt which decomposes to evolve $N_2$ gas.
The reaction is: $R-NH_2 + NaNO_2 + HCl$ $\rightarrow [R-N_2^+Cl^-]$ $\rightarrow R-OH + N_2 \uparrow + NaCl + H_2O$.

(C) The basicity of amines depends on the availability of the lone pair on the nitrogen atom.
$(1)$ $NH_3$: Ammonia has no electron-donating alkyl groups.
$(2)$ $CH_3CH_2NH_2$: Ethylamine is a primary amine with one electron-donating ethyl group,increasing electron density on $N$.
$(3)$ $(CH_3)_2NH$: Dimethylamine is a secondary amine with two electron-donating methyl groups,making it more basic than primary amines due to the $+I$ effect.
$(4)$ $CH_3CONH_2$: Acetamide is an amide where the lone pair on $N$ is involved in resonance with the carbonyl group $(C=O)$,making it the least basic.
Comparing the basicity: $(CH_3)_2NH > CH_3CH_2NH_2 > NH_3 > CH_3CONH_2$,which corresponds to $3 > 2 > 1 > 4$.

(C) Pure aniline is a colorless liquid. However,when exposed to air for a long time,it oxidizes and turns into a reddish-brown liquid.

(D) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $C_6H_5NH_2$,$p-NO_2-C_6H_4NH_2$,and $m-NO_2-C_6H_4NH_2$,the lone pair on nitrogen is involved in resonance with the benzene ring,which decreases its availability.
Specifically,the $-NO_2$ group is an electron-withdrawing group $(EWG)$ that further decreases basicity through $-I$ and $-M$ effects.
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,the lone pair on the nitrogen atom is not involved in resonance with the benzene ring,making it the most available for protonation.
Thus,$C_6H_5CH_2NH_2$ is the strongest base among the given options.

(D) In $CH_3NHCHO$,the lone pair of electrons on the nitrogen atom is involved in resonance with the adjacent carbonyl group $(C=O)$.
This delocalization significantly reduces the availability of the lone pair for protonation.
Although the $-NO_2$ group in $O_2NCH_2NH_2$ exerts a strong $-I$ effect,the resonance effect in the amide $(CH_3NHCHO)$ is more effective in reducing the basicity of the nitrogen atom.
Therefore,$CH_3NHCHO$ is the weakest base among the given options.

(C) Amines are basic in nature because they contain a lone pair of electrons on the nitrogen atom.
According to the $Lewis$ theory,a base is an electron pair donor. Since amines donate their lone pair,they act as $Lewis$ bases.
According to the $Bronsted-Lowry$ theory,a base is a proton acceptor. Amines accept a proton $(H^+)$ to form an ammonium ion $(R-NH_3^+)$,thus acting as $Bronsted$ bases.
Therefore,the basic character of amines is explained by both $Lewis$ and $Bronsted-Lowry$ theories.

(A) The given reaction sequence is:
$A$ $\xrightarrow{Sn/HCl} B$ $\xrightarrow{HNO_2, 0^{\circ}C} C$ $\xrightarrow{C_2H_5OH} C_6H_6$.
$1$. $A$ is nitrobenzene $(C_6H_5NO_2)$ containing the $-NO_2$ group.
$2$. Reduction of $A$ with $Sn/HCl$ gives $B$,which is aniline $(C_6H_5NH_2)$ containing the $-NH_2$ group.
$3$. Reaction of $B$ with $HNO_2$ at $0^{\circ}C$ (diazotization) gives $C$,which is benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ containing the $-N_2^+Cl^-$ group.
$4$. Reaction of $C$ with $C_2H_5OH$ reduces the diazonium salt to benzene $(C_6H_6)$.
Therefore,the functional groups are $-NO_2, -NH_2,$ and $-N_2^+Cl^-$.

(C) Aniline is highly activated towards electrophilic substitution due to the strong electron-donating effect of the $-NH_2$ group.
When aniline reacts with $Br_2$ in a polar solvent like water,it forms $2, 4, 6$-tribromoaniline.
However,when the reaction is carried out in a non-polar solvent like $CCl_4$ at low temperature,the activation of the benzene ring is reduced.
This allows for mono-substitution,resulting in a mixture of $o$-bromoaniline and $p$-bromoaniline.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $Aniline$ $(C_6H_5NH_2)$,the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring,which significantly reduces its availability for protonation.
In $Piperidine$,$Cyclohexylamine$,and $Methylamine$ $(CH_3NH_2)$,the lone pair is localized on the nitrogen atom and is not involved in resonance,making them much stronger bases than $Aniline$.
Therefore,$Aniline$ is the weakest Bronsted base among the given options.

(A) Aniline decomposes at its boiling point when heated directly. Therefore,it is purified by steam distillation. In this process,steam is passed through the mixture,which carries the aniline vapor to the condenser,where it is condensed back into a liquid state.

(A) The carbylamine reaction (also known as the isocyanide test) is a chemical test for the detection of primary amines.
In this reaction,a primary amine (aliphatic or aromatic) is heated with chloroform $(CHCl_3)$ and an alcoholic solution of potassium hydroxide $(KOH)$.
This results in the formation of an isocyanide (carbylamine),which has a foul smell.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH (\text{alc.}) \rightarrow R-NC + 3KCl + 3H_2O$.
Therefore,the required reagents are a primary amine and chloroform.

(C) 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),which has an offensive smell.
The balanced chemical equation is:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC + 3KCl + 3H_2O$
Here,$(a) = CH_3CH_2NC$ (Ethyl isocyanide or $C_2H_5NC$) and $(b) = 3KCl$.

(B) The reaction of primary amines with carbon disulfide $(CS_2)$ followed by treatment with mercuric chloride $(HgCl_2)$ is known as the Hofmann mustard oil reaction. The product formed is an alkyl isothiocyanate,which has the general formula $RNCS$. This compound is responsible for the characteristic smell of mustard oil.

(B) The reaction of a primary amine with carbon disulfide $(CS_2)$ followed by treatment with mercuric chloride $(HgCl_2)$ is known as the Hofmann mustard oil reaction.
For ethylamine $(CH_3CH_2NH_2)$,the reaction proceeds as follows:
$CH_3CH_2NH_2 + CS_2 \rightarrow CH_3CH_2NHCSSH$ (Dithiocarbamic acid)
$CH_3CH_2NHCSSH + HgCl_2 \rightarrow CH_3CH_2N=C=S + HgS + 2HCl$
The product formed is ethyl isothiocyanate $(CH_3CH_2NCS)$,which has a characteristic mustard-like smell.

(C) The conversion of an aliphatic primary amine $(C_2H_5NH_2)$ to an alkyl chloride $(C_2H_5Cl)$ is typically achieved using nitrosyl chloride $(NOCl)$.
This reaction is a specific method for replacing the amino group with a chlorine atom.
The reaction is: $C_2H_5NH_2 + NOCl \rightarrow C_2H_5Cl + N_2 + H_2O$.

(B) The Hofmann bromamide degradation reaction proceeds through the following steps:
$1$. Formation of $N$-bromamide: $R-CONH_2 + Br_2 + 2KOH \to R-CONHBr + KBr + H_2O$
$2$. Formation of isocyanate: $R-CONHBr + KOH \to R-NCO + KBr + H_2O$
$3$. Hydrolysis of isocyanate: $R-NCO + 2KOH \to R-NH_2 + K_2CO_3$
In this mechanism,$RCONHBr$ and $R-NCO$ are intermediates. $RCON$ and $RNC$ (isocyanide) are not formed as intermediates.

(D) $1$. Primary amines $(R-NH_2)$ have two hydrogen atoms attached to the nitrogen atom,allowing them to form intermolecular hydrogen bonds.
$2$. Tertiary amines $(R_3N)$ have no hydrogen atoms attached to the nitrogen atom,so they cannot form intermolecular hydrogen bonds.
$3$. Butylamine $(CH_3CH_2CH_2CH_2NH_2)$ is a primary amine because the nitrogen is attached to only one alkyl group.
$4$. Isopropylamine $(CH_3CH(NH_2)CH_3)$ is a primary amine because the nitrogen is attached to one carbon atom,which is bonded to two other methyl groups. Therefore,the statement that isopropylamine is a secondary amine is incorrect.

(D) This reaction is the $Carbylamine$ reaction,which is a characteristic test for primary amines.
When an aliphatic or aromatic primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,an isocyanide (carbylamine) is formed.
The balanced chemical equation is:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC + 3KCl + 3H_2O$.
Here,$(a)$ is $CH_3CH_2NC$ (which is $C_2H_5NC$) and $(b)$ is $3KCl$.

(C) During the acetylation of primary or secondary amines,the acetyl group $(CH_3CO-)$ replaces the hydrogen atom attached to the nitrogen atom.
The reaction is as follows:
$R-NH_2 + (CH_3CO)_2O \xrightarrow{\text{acetylation}} R-NHCOCH_3 + CH_3COOH$
Here,the $H$ atom on the $N$ atom is replaced by the $CH_3CO$ group.

(A) The basic strength of amines in the gaseous phase or non-polar solvents depends on the electron-donating 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 $NH_3 < CH_3NH_2 < (CH_3)_2NH$.

(D) In a strongly acidic medium,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$. The $-NH_3^+$ group is electron-withdrawing and acts as a meta-directing group in electrophilic aromatic substitution reactions.

(B) The reaction sequence is as follows:
$1$. $C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O$ (Product $A$ is $C_2H_5OH$)
$2$. $C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl + POCl_3 + HCl$ (Product $B$ is $C_2H_5Cl$)
$3$. $C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 + HCl$ (Product $C$ is $C_2H_5NH_2$)
Therefore,the final product $C$ is $C_2H_5NH_2$ (Ethylamine).

(D) Primary amines $(R-NH_2)$ do not undergo hydrolysis under standard conditions to yield these products. The hydrolysis of isocyanides $(R-NC)$ yields primary amines and formic acid,while the reduction of isocyanides yields secondary amines $(R-NH-CH_3)$. Given the options provided,none of the listed products are standard hydrolysis products of a primary amine.

(D) The basic strength of amines is inversely proportional to their $pK_b$ value.
Higher basicity corresponds to a higher $K_b$ value and a lower $pK_b$ value.
Among the given options,$NH_3$ is the weakest base because it lacks the electron-donating alkyl groups that increase the electron density on the nitrogen atom in amines.
Therefore,$NH_3$ has the highest $pK_b$ value.

(D) Primary $(1^o)$ aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which decompose to evolve $N_2$ gas: $RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O$.
Secondary $(2^o)$ amines react with $HNO_2$ to form $N$-nitrosamines,which are yellow oily liquids and do not evolve $N_2$ gas: $R_2NH + HNO_2 \rightarrow R_2N-N=O + H_2O$.
Tertiary $(3^o)$ aliphatic amines form water-soluble nitrite salts with $HNO_2$ and do not evolve $N_2$ gas: $R_3N + HNO_2 \rightarrow [R_3NH]^+NO_2^-$.
Therefore,both $2^o$ and $3^o$ amines do not evolve $N_2$ gas.

(B) The given reaction is the $Hofmann \ bromide \ degradation$ reaction.
$CH_3CH_2CONH_2 \xrightarrow[Br_2]{NaOH} CH_3CH_2NH_2 (A)$.
$A$ is $ethylamine$ $(CH_3CH_2NH_2)$,which is an aliphatic amine.
Aliphatic amines are basic in nature and turn red litmus blue.

(D) The mustard oil reaction is a characteristic test for primary amines $(R-NH_2)$ with $CS_2$ and $HgCl_2$,so statement $A$ is correct.
$1^o$ amines are basic and react with chloroplatinic acid $(H_2PtCl_6)$ to form insoluble salts of the type $(RNH_3)_2PtCl_6$,so statement $B$ is correct.
$1^o$ amines contain active hydrogen atoms attached to nitrogen,which react with sodium metal to release hydrogen gas $(H_2)$,so statement $C$ is correct.
$1^o$ amines do not undergo hydrolysis to form alcohols; instead,they react with nitrous acid $(HNO_2)$ to form diazonium salts (which are unstable and decompose to alcohols) or other products depending on the structure,but the direct hydrolysis of an amine to an alcohol is not a standard reaction. Thus,statement $D$ is incorrect.

(D) The hydrolysis of isocyanide $(R-NC)$ in the presence of an acid catalyst yields a primary amine and formic acid.
$R-NC + 2H_2O \xrightarrow{H^+} R-NH_2 + HCOOH$

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ with zinc $(Zn)$ and an alkali $(NaOH)$ leads to the formation of hydrazobenzene $(C_6H_5NH-NHC_6H_5)$.
The chemical reaction is:
$2C_6H_5NO_2 + 10[H] \xrightarrow{Zn/NaOH} C_6H_5NH-NHC_6H_5 + 4H_2O$

(C) The basicity of amines in aqueous solution depends on the inductive effect,solvation effect,and steric hindrance.
$C_6H_5NH_2$ (Aniline) is the least basic because the lone pair on nitrogen is involved in resonance with the benzene ring.
Among aliphatic amines,the order in aqueous solution is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$ due to the combined effect of $+I$ effect and solvation.
Therefore,the increasing order of basicity is $C_6H_5NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$.

(D) The boiling point of amines depends on the extent of hydrogen bonding. Primary $(1^{\circ})$ and secondary $(2^{\circ})$ amines have hydrogen atoms attached to the nitrogen atom,allowing for intermolecular hydrogen bonding. Tertiary $(3^{\circ})$ amines,such as $N,N-$dimethylmethanamine $(CH_3-N(CH_3)_2)$,do not have any hydrogen atoms attached to the nitrogen atom and therefore cannot form intermolecular hydrogen bonds. Thus,it has the lowest boiling point among the given options.

(D) Sodium metal reacts with compounds containing acidic hydrogen atoms to release hydrogen gas.
Among the given options,$CH_3CONH_2$ (acetamide) contains an acidic hydrogen atom attached to the nitrogen atom due to the electron-withdrawing effect of the adjacent carbonyl group.
The reaction is:
$CH_3CONH_2 + Na \to CH_3CONH^-Na^+ + \frac{1}{2}H_2$
Primary and secondary amines like $(CH_3)_2NH$,$CH_3NH_2$,and $C_6H_5NH_2$ are much less acidic and do not react with sodium metal under normal conditions to release $H_2$ gas.

(D) In aqueous solution,the basicity of aliphatic amines is influenced by inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
Aniline $(C_6H_5NH_2)$ is a much weaker base due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.
Therefore,dimethylamine $(CH_3)_2NH$ is the strongest base among the given options.

(C) Methylamine $(CH_3NH_2)$ is a primary $(1^{\circ})$ amine.
Due to the electron-donating inductive effect ($+I$ effect) of the methyl group,the electron density on the nitrogen atom increases,making the lone pair more available for donation.
Therefore,methylamine is a stronger base than ammonia $(NH_3)$.

(B) When aniline $(C_6H_5NH_2)$ reacts with $HNO_2$ at $273-278 \ K$,it forms benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
This diazonium salt undergoes a coupling reaction with an alkaline solution of $\beta$-naphthol to form a dye.
The product formed is $1$-phenylazo-$2$-naphthol,which is a red-colored dye (precipitate).

(C) In an aqueous solution,the order of basicity for methyl-substituted amines is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
This order is determined by the combined effect of inductive effect,solvation effect,and steric hindrance.
Dimethylamine $(CH_3)_2NH$ is the most basic among the given options.
Aniline is a weak base due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.

(B) In an aqueous solution,the basic strength of amines depends on the combined effect of the inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
Therefore,$CH_3NH_2$ is more basic than $(CH_3)_3N$ because $(CH_3)_3N$ experiences significant steric hindrance,which destabilizes the conjugate acid formed upon protonation.