Properties of Amines Questions in English

(D) The given reaction is the $Carbylamine$ reaction,which is a characteristic test for primary amines.
In this reaction,a primary amine $(CH_3CH_2NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine) and potassium chloride $(KCl)$ as a byproduct.
The balanced chemical equation is:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow CH_3CH_2NC + 3KCl + 3H_2O$
Comparing this with the given equation,$(a)$ is $C_2H_5NC$ (ethyl isocyanide) and $(b)$ is $3KCl$.

(D) When aniline is treated with concentrated sulfuric acid,it forms aniline hydrogen sulfate.
Heating aniline hydrogen sulfate at $453 - 473 \, K$ leads to a rearrangement reaction known as baking process.
This results in the formation of $p$-aminobenzenesulfonic acid,which is commonly known as sulfanilic acid.
The reaction is: $C_6H_5NH_2 + H_2SO_4$ $\rightarrow C_6H_5NH_3^+HSO_4^-$ $\xrightarrow{453-473 \, K} NH_2-C_6H_4-SO_3H + H_2O$.

(B) Aliphatic primary amines react with nitrous acid $(HNO_2)$ at low temperatures $(273-278 \ K)$ to form unstable aliphatic diazonium salts,which immediately decompose to evolve nitrogen gas and form alcohols.
The reaction is: $R-NH_2 + HNO_2 \xrightarrow{273-278 \ K} R-OH + N_2 + H_2O$.

(C) The reaction of a primary amine with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction is a test for primary amines,where the amine is converted into an isocyanide (carbylamine),which has a foul smell.
The reaction for $n$-propylamine $(CH_3CH_2CH_2NH_2)$ is:
$CH_3CH_2CH_2NH_2 + CHCl_3 + 3KOH \text{ (alc.)} \rightarrow CH_3CH_2CH_2NC + 3KCl + 3H_2O$.
Thus,the product $X$ is $n$-propyl isocyanide,which has the structure $CH_3CH_2CH_2NC$.

(D) Blue litmus turns red in the presence of an acidic substance.
$RNH_3^+Cl^-$ is a salt of a weak base (amine) and a strong acid $(HCl)$.
Due to the hydrolysis of the $RNH_3^+$ cation,the solution becomes acidic.
$RNH_3^+ + H_2O \rightleftharpoons RNH_2 + H_3O^+$.
Therefore,it turns blue litmus red.

(D) Amines are basic in nature and react with acids to form salts. Primary amines undergo the carbylamine reaction (isocyanide test) and the mustard oil reaction (specifically primary aliphatic amines with $CS_2$ and $HgCl_2$). However,amines do not undergo hydrolysis to form alcohols. Hydrolysis is a characteristic reaction of esters,amides,or alkyl halides,not amines.

(C) The reaction of aniline with nitrous acid $(HNO_2)$ in the presence of $HCl$ at $273-278 \, K$ $(0-5 \, ^\circ C)$ is known as the diazotization reaction.
In this reaction,aniline is converted into benzene diazonium chloride,which is a diazo compound.
The chemical equation is:
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \, K} [C_6H_5N_2]^+Cl^- + 2H_2O$
Thus,the correct option is $C$.

(A) When benzamide $(C_6H_5CONH_2)$ is reacted with phosphorus oxychloride $(POCl_3)$,it undergoes a dehydration reaction.
The $POCl_3$ acts as a dehydrating agent,removing a molecule of water from the amide group to form a nitrile group.
The reaction is: $C_6H_5CONH_2 + POCl_3 \rightarrow C_6H_5CN + H_2O$.
Thus,the product formed is benzonitrile.

(D) The reaction of a primary amine $(R-NH_2)$ with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction or Isocyanide test.
In this reaction,the amine group $(-NH_2)$ is converted into an isocyanide group $(-NC)$.
For aniline $(C_6H_5NH_2)$,the reaction is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$.
The product $C_6H_5NC$ is known as phenyl isonitrile or phenyl carbylamine.

(B) The reaction between benzene diazonium chloride and phenol in a weakly basic medium (typically $pH \approx 9-10$) is a coupling reaction.
In this reaction,the electrophilic diazonium ion attacks the electron-rich ortho or para position of the phenol ring.
Since the para position is sterically less hindered,the major product formed is $p-$hydroxyazobenzene,which is an orange-red dye.

(A) The reaction of aryl amines with nitrous acid at low temperatures $(0-5 \ ^{\circ}C)$ yields diazonium salts,not phenols. Phenols are formed only when the diazonium salt solution is warmed with water. Therefore,the statement that aryl amines react with nitrous acid to give phenols directly is incorrect.

(D) The reaction proceeds in two steps:
$1$. The first step is the Hofmann bromamide degradation reaction,where the amide $p-CH_3-C_6H_4-CONH_2$ reacts with $NaOH/Br_2$ to form the primary amine $p-CH_3-C_6H_4-NH_2$ ($p$-toluidine).
$2$. The second step is the acylation of the amine with benzoyl chloride $(C_6H_5COCl)$. The lone pair on the nitrogen atom of the amine attacks the carbonyl carbon of the benzoyl chloride,leading to the formation of an amide: $p-CH_3-C_6H_4-NH-CO-C_6H_5$ ($N$-(p-tolyl)benzamide).
Comparing this with the given options,option $D$ represents the correct structure.

(C) The reduction of $Nitrobenzene$ $(C_6H_5NO_2)$ in an alkaline medium can yield different products depending on the reducing agent used.
When $Nitrobenzene$ is reduced using $Alkaline \ sodium \ arsenite$ $(Na_3AsO_3)$ or $Alkaline \ glucose \ solution$,the major product obtained is $Azoxybenzene$ $(C_6H_5-N(O)=N-C_6H_5)$.
Therefore,both the reagents mentioned in options $A$ and $B$ are capable of producing $Azoxybenzene$ as the major product.

(D) Aryl amines react with $HNO_2$ at low temperatures $(0-5 \ ^{\circ}C)$ to form benzene diazonium salts,not phenol. Phenol is formed only when the diazonium salt is warmed with water.

(D) The reduction of isocyanides $(R-NC)$ with a reducing agent like $LiAlH_4$ or $H_2/Ni$ leads to the formation of secondary amines.
Specifically,the reduction of $t$-butyl isocyanide $((CH_3)_3C-NC)$ results in the formation of $N$-methyl-$t$-butylamine,which is a secondary amine.
The reaction is: $(CH_3)_3C-NC + 4[H] \rightarrow (CH_3)_3C-NH-CH_3$.
Since this product is not listed in the options,the correct choice is $D$.

(B) The reduction of a Schiff base $(R-CH=N-R')$ yields a secondary amine $(R-CH_2-NH-R')$.
Reaction: $R-CH=N-R' \xrightarrow{\text{Reduction}} R-CH_2-NH-R'$ ($2^\circ$ amine).

(B) Aniline is an aromatic amine where the lone pair on the nitrogen atom is involved in resonance with the benzene ring,reducing its availability for protonation.
$P$-nitroaniline has an electron-withdrawing $-NO_2$ group,which further decreases basicity.
Diphenylamine and triphenylamine have more phenyl groups attached to the nitrogen,increasing resonance and steric hindrance,which significantly decreases basicity.
Benzylamine $(C_6H_5CH_2NH_2)$ is an aliphatic amine where the lone pair on the nitrogen is not involved in resonance with the benzene ring because of the intervening $-CH_2-$ group.
Therefore,benzylamine is much more basic than aniline.

(B) Direct nitration of aniline $(C_6H_5NH_2)$ results in the formation of tarry oxidation products along with nitro derivatives because the $-NH_2$ group is strongly activating and ortho/para-directing.
To control the reaction and obtain the desired para-nitroaniline,the $-NH_2$ group is protected by acetylation using acetic anhydride or acetyl chloride to form acetanilide $(C_6H_5NHCOCH_3)$.
This reduces the activating effect of the group and prevents oxidation,allowing for controlled nitration.

(C) The reaction sequence is as follows:
$1$. $CH_3CH_2Br \xrightarrow[\Delta]{Aq. KOH} CH_3CH_2OH$ (Product $A$ is ethanol).
$2$. $CH_3CH_2OH \xrightarrow[\Delta]{KMnO_4/H^+} CH_3COOH$ (Product $B$ is acetic acid).
$3$. $CH_3COOH \xrightarrow[\Delta]{NH_3} CH_3CONH_2$ (Product $C$ is acetamide).
$4$. $CH_3CONH_2 \xrightarrow[alkali]{Br_2} CH_3NH_2$ (Product $D$ is methanamine via Hofmann bromamide degradation reaction).

(B) Methylamine $(CH_3NH_2)$ contains a lone pair of electrons on the nitrogen atom,which makes it a Lewis base.
When dissolved in water,it accepts a proton $(H^+)$ from water to form a methylammonium ion and a hydroxide ion $(OH^-)$.
$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$
Since it produces $OH^-$ ions in an aqueous solution,it acts as a base.

(C) Methylamine $(CH_3NH_2)$ contains a methyl group which is an electron-donating group due to the $+I$ effect.
This increases the electron density on the nitrogen atom,making the lone pair more available for donation.
Therefore,methylamine is more basic than ammonia $(NH_3)$.

(C) Primary aliphatic amines react with $HNO_2$ to form unstable diazonium salts which decompose to evolve $N_2$ gas.
Carbamide $(NH_2CONH_2)$ also reacts with $HNO_2$ to evolve $N_2$.
Secondary amines react with $HNO_2$ to form $N$-nitrosamines (yellow oily liquids),which do not evolve $N_2$ gas.
Alkanamides do not react with $HNO_2$ to evolve $N_2$ gas.

(A) The oxidation of primary aliphatic amines like ethylamine $(CH_3CH_2NH_2)$ with strong oxidizing agents such as acidic $KMnO_4$ leads to the formation of aldehydes or ketones depending on the structure.
For ethylamine,the oxidation proceeds through the formation of an imine intermediate which hydrolyzes to form acetaldehyde $(CH_3CHO)$.

(D) The ammonolysis of alkyl halides involves the nucleophilic substitution reaction where the halogen atom is replaced by an amino group.
$R-X + NH_3 \rightarrow R-NH_2 + HX$
Since the primary amine formed is also a nucleophile,it reacts further with the alkyl halide to form secondary and tertiary amines,and finally a quaternary ammonium salt.
$R-NH_2 + R-X \rightarrow R_2NH + HX$
$R_2NH + R-X \rightarrow R_3N + HX$
$R_3N + R-X \rightarrow R_4N^+X^-$
Thus,the reaction produces a mixture of primary,secondary,tertiary amines and quaternary ammonium salt.

(D) When aniline is treated with excess bromine water or liquid bromine,the $-NH_2$ group strongly activates the benzene ring.
This leads to the electrophilic substitution of bromine atoms at all available ortho and para positions.
As a result,$2, 4, 6$-tribromoaniline is formed as a white precipitate.

(C) Ethylamine $(C_2H_5NH_2)$ is soluble in water because it can form intermolecular hydrogen bonds with water molecules.
The lone pair on the nitrogen atom of the amine group interacts with the hydrogen atoms of water,facilitating dissolution.

(D) When a primary amine $(R-NH_2)$ reacts with a ketone $(R'_2C=O)$,it undergoes a condensation reaction to form an imine,which is commonly known as a Schiff base. The reaction involves the loss of a water molecule: $R'_2C=O + R-NH_2 \rightarrow R'_2C=N-R + H_2O$. Thus,the correct product is a Schiff base.

(D) Methylamine $(CH_3NH_2)$ has one methyl group with a $(+I)$ effect,which increases the electron density on the nitrogen atom.
Dimethylamine $((CH_3)_2NH)$ has two methyl groups,which provide a stronger $(+I)$ effect,making it more basic than methylamine.
Therefore,the statement that the basicity of dimethylamine is less than that of methylamine is incorrect.

(A) The Hofmann rearrangement (also known as the Hofmann degradation) involves the conversion of a primary amide to a primary amine with one carbon atom less.
During the mechanism,the migration of the alkyl or aryl group from the carbonyl carbon to the nitrogen atom occurs within the same molecule.
Therefore,the rearrangement is an $Intramolecular$ process.

(D) The $pK_b$ value is inversely proportional to the basic strength of the compound. $A$ higher $pK_b$ value indicates a weaker base. Among the given options,$NH_3$ is the weakest base compared to the alkyl amines ($R-NH_2$,$R_2NH$,$R_3C-NH_2$) due to the absence of the electron-donating inductive effect of alkyl groups. Therefore,$NH_3$ has the highest $pK_b$ value.

(D) Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which decompose to evolve nitrogen gas $(N_2)$.
Similarly,nitrosyl chloride $(NOCl)$ also reacts with primary amines to produce nitrogen gas.
Therefore,both $HNO_2$ and $NOCl$ produce nitrogen gas upon reaction with primary amines.

(A) The boiling point of amines depends on the extent of intermolecular hydrogen bonding.
Primary amines $(1^o)$ have two hydrogen atoms attached to the nitrogen,allowing for extensive hydrogen bonding.
Secondary amines $(2^o)$ have one hydrogen atom attached to the nitrogen,allowing for moderate hydrogen bonding.
Tertiary amines $(3^o)$ have no hydrogen atoms attached to the nitrogen,so they cannot form intermolecular hydrogen bonds.
Therefore,the order of boiling points is $1^o > 2^o > 3^o$.
$n$-Butylamine $(I)$ is a primary amine,diethylamine $(II)$ is a secondary amine,and $N,N$-dimethylethylamine $(III)$ is a tertiary amine.
Thus,the correct order is $III < II < I$.

(B) Schiff base is formed by the reaction of a primary amine with an aldehyde or ketone.
Aniline $(C_6H_5NH_2)$ reacts with benzaldehyde $(C_6H_5CHO)$ to form benzylidene aniline,which is a Schiff base.
The reaction is:
$C_6H_5NH_2 + OCH-C_6H_5 \xrightarrow{\Delta, -H_2O} C_6H_5N=CH-C_6H_5$.

(A) The reaction sequence is as follows:
$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.
$2$. Reduction of Nitrobenzene with $Sn/HCl$ gives Aniline $(B = C_6H_5NH_2)$.
$3$. Diazotization of Aniline with $NaNO_2/HCl$ at $0^{\circ}C$ gives Benzene diazonium chloride $(C = C_6H_5N_2^+Cl^-)$.
$4$. Hydrolysis of Benzene diazonium chloride with $H_2O/\Delta$ yields Phenol $(C_6H_5OH)$.
Thus,$A$ is Nitrobenzene,$B$ is Aniline,and $C$ is Benzene diazonium chloride.

(D) The reduction of nitrobenzene in an alkaline medium (using $Zn/NaOH$ or $Na_3AsO_3/NaOH$) proceeds through the formation of nitrosobenzene and phenylhydroxylamine. These intermediates then condense to form azoxybenzene,azobenzene,and finally hydrazobenzene. Nitrosobenzene is an intermediate product formed during the initial stages of reduction,but it is not the final product of the alkaline reduction process. However,among the given options,all are products formed at various stages of the reduction process. If the question implies the final stable product,the sequence is $Nitrobenzene$ $\rightarrow Azoxybenzene$ $\rightarrow Azobenzene$ $\rightarrow Hydrazobenzene$. Nitrosobenzene is a reactive intermediate. In many textbook contexts,$Nitrosobenzene$ is considered the starting point of the condensation sequence rather than a product of the alkaline reduction pathway itself.

(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 an alcoholic base $(KOH)$ to form an isocyanide (carbylamine),which has an unpleasant odor.
The reaction is: $CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O$.
The product $CH_3NC$ is methyl isocyanide,which can be represented as $CH_3-N^+ \equiv C^-$.
Therefore,the correct option is $A$.

(B) The reaction of a primary amine with chloroform and alcoholic $KOH$ is known as the carbylamine reaction.
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Aniline $(C_6H_5NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic $KOH$ to form phenyl isocyanide $(C_6H_5NC)$,which is also known as phenyl carbylamine.

(B) Aniline is purified by $Steam \ distillation$. This method is used for substances which are steam volatile and are immiscible with water.

(B) This reaction is known as the $\text{Carbylamine reaction}$.
It is a characteristic test for primary amines $(R-NH_2)$.
When a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it produces an isocyanide (or carbylamine),which has a foul smell.
The chemical equation is: $R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.

(C) The reaction of aniline with nitrous acid (generated in situ from $NaNO_2$ and $HCl$) at low temperatures $(273-278 \ K)$ is known as diazotization.
The reaction is: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.
Thus,the product $(X)$ is $C_6H_5N_2^+Cl^-$,which is benzene diazonium chloride.

(C) In an aqueous solution,the basicity of aliphatic amines depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
Dimethylamine $(CH_3)_2NH$ is a secondary amine and is the strongest base among the given options due to the optimal balance of the $+I$ effect and solvation.
Aniline is a very weak base due to the resonance stabilization of the lone pair on the nitrogen atom into the benzene ring.

(B) The Hofmann reaction (specifically the Hofmann carbylamine reaction) is used to identify primary amines,producing isocyanides $(R-NC)$. The Hofmann mustard oil reaction produces isothiocyanates $(R-N=C=S)$. The Hofmann bromamide degradation reaction produces primary amines $(RNH_2)$. Option $B$ $(RCH_2NH_2)$ is not a direct product of these specific Hofmann named reactions.

(B) The reactant is $N$-phenylbenzamide (acetanilide derivative).
In the presence of $conc. HNO_3$ and $conc. H_2SO_4$,nitration occurs.
The $-NH-CO-C_6H_5$ group is ortho/para directing due to the lone pair on the nitrogen atom,which can participate in resonance with the benzene ring attached to it.
However,due to the bulky nature of the $-NH-CO-C_6H_5$ group,steric hindrance makes the ortho position less favorable.
Therefore,the major product formed is the para-substituted isomer,which is $N$-($4$-nitrophenyl)benzamide.

(B) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom and the stability of the conjugate acid formed.
$1$. $C_2H_5NH_2$ (Ethylamine) is a stronger base than $CH_3NH_2$ (Methylamine) due to the greater $+I$ effect of the ethyl group compared to the methyl group.
$2$. Both aliphatic amines are stronger bases than $NH_3$ because the alkyl groups increase the electron density on the nitrogen atom.
$3$. $C_6H_5NH_2$ (Aniline) is the weakest base because the lone pair on the nitrogen atom is involved in resonance with the benzene ring,making it less available for protonation.
Therefore,the correct decreasing order is $C_2H_5NH_2 > CH_3NH_2 > NH_3 > C_6H_5NH_2$.

(A) $CH_3CH_2I \xrightarrow{NaCN} CH_3CH_2CN$ (Propanenitrile,$A$)
$CH_3CH_2CN \xrightarrow{OH^- \text{ (partial hydrolysis)}} CH_3CH_2CONH_2$ (Propanamide,$B$)
$CH_3CH_2CONH_2 \xrightarrow{Br_2/NaOH} CH_3CH_2NH_2$ (Ethanamine,$C$)
This is a Hoffmann bromamide degradation reaction.

(B) Primary amines $(R-NH_2)$ react with aldehydes $(R'-CHO)$ to form imines (also known as Schiff bases) with the elimination of a water molecule.
The general reaction is: $R-NH_2 + R'-CHO \rightarrow R-N=CH-R' + H_2O$.

(C) The reaction proceeds as follows:
$CH_3Br + KCN \to CH_3CN + KBr$
$X$ is $CH_3CN$ (acetonitrile or ethanenitrile).
Reduction of $CH_3CN$ using $Na / C_2H_5OH$ (Mendius reduction) yields $CH_3CH_2NH_2$.
$CH_3CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2NH_2$
Thus,$Y$ is $CH_3CH_2NH_2$ (ethanamine).