Properties of Amines Questions in English

(C) Secondary amines react with nitrous acid $(HONO)$ at low temperature to form $N$-nitrosoamines,which are yellow oily liquids.
$(CH_3CH_2)_2NH + HONO \to (CH_3CH_2)_2N-NO + H_2O$ (oily nitrosoamine).
Primary amines produce alcohols and nitrogen gas,while tertiary amines form water-soluble salts.

(A) Azo dyes are prepared by the coupling reaction of diazonium salts with aromatic compounds like phenols or amines.
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ and $HCl$ at $0-5\,^{\circ}C$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. This diazonium salt then undergoes a coupling reaction with an aromatic compound (e.g.,phenol) in a weakly alkaline medium to produce an azo dye.

(D) Primary amines $(R-NH_2)$ react with nitrous acid $(HNO_2)$ to form diazonium salts,which decompose to release $N_2$ gas.
Secondary amines $(R_2NH)$ react with $HNO_2$ to form $N$-nitrosoamines.
Tertiary amines $(R_3N)$ do not contain an $\alpha-H$ atom on the nitrogen-bearing carbon (or rather,they lack the hydrogen necessary to form the intermediate required for $N_2$ evolution).
Therefore,$(CH_3)_3N$ (a tertiary amine) does not react with nitrous acid to evolve $N_2$ gas.

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
In $CH_3CH_2NH_2$ (Ethylamine),the ethyl group is an electron-donating group ($+I$ effect),which increases the electron density on the nitrogen atom,making it more available for donation.
In $CH_3NH_2$ (Methylamine),the methyl group also has a $+I$ effect,but it is weaker than the ethyl group.
In $NH_2OH$ (Hydroxylamine),the $-OH$ group is electron-withdrawing ($-I$ effect),which decreases the basicity.
In $C_6H_5NH_2$ (Aniline),the lone pair on the nitrogen atom is involved in resonance with the benzene ring,making it the least basic.
Therefore,$Ethylamine$ is the most basic among the given options.

(C) The reaction of aniline with nitrous acid $(HNO_2)$ in the presence of hydrochloric acid $(HCl)$ at $0-5\,^oC$ $(273-278\,K)$ 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:
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{0\,^oC} C_6H_5N_2^+Cl^- + 2H_2O$
Thus,the correct option is $(C)$.

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ to nitrosobenzene $(C_6H_5NO)$ is an intermediate step in the reduction process. Nitrosobenzene is highly reactive and typically undergoes further reduction to phenylhydroxylamine $(C_6H_5NHOH)$ or aniline $(C_6H_5NH_2)$ under most standard reduction conditions. Therefore,it cannot be isolated as a stable product under these conditions.

(C) The oxidation of primary amines like ethylamine $(CH_3CH_2NH_2)$ with $KMnO_4$ proceeds through the formation of an imine intermediate.
$1$. Ethylamine is oxidized to acetaldimine $(CH_3CH=NH)$.
$2$. The imine intermediate undergoes hydrolysis to form acetaldehyde $(CH_3CHO)$,which is an aldehyde.
Reaction: $CH_3-CH_2-NH_2$ $\xrightarrow{[O], KMnO_4} CH_3-CH=NH$ $\xrightarrow{H_2O} CH_3-CHO + NH_3$.

(B) The correct answer is $B$.
Only primary aromatic amines undergo diazotisation to form stable diazonium salts at low temperatures $(0-5 \ ^\circ C)$.
Primary aliphatic amines form highly unstable diazonium salts that decompose immediately to form alcohols and nitrogen gas.

(B) The reaction between a primary amine $(R-NH_2)$ and an aldehyde $(R'-CHO)$ results in the formation of a Schiff base,which is also known as an aldimine.
The general reaction is: $R-NH_2 + R'-CHO \to R-N=CH-R' + H_2O$ (Aldimine).

(C) The reaction of acetamide $(CH_3CONH_2)$ with nitrous acid $(HNO_2)$ is a diazotization-like reaction that results in the evolution of nitrogen gas.
The chemical equation is: $CH_3CONH_2 + HNO_2 \to CH_3COOH + N_2 \uparrow + H_2O$.
Therefore,the gas evolved is $N_2$.

(A) The reduction of alkyl nitrites $(R-O-N=O)$ using a reducing agent like $Sn/HCl$ results in the formation of alcohol and ammonia.
The chemical reaction is: $R-O-N=O + 6[H] \xrightarrow{Sn/HCl} R-OH + NH_3 + H_2O$
Therefore,the product obtained is an alcohol.

(D) . Primary amines are basic in nature due to the presence of a lone pair of electrons on the nitrogen atom.
When they react with a strong acid like $HCl$,they undergo protonation to form an ammonium salt.
For example: $CH_3-CH_2-NH_2 + HCl \to CH_3CH_2-NH_3^+ Cl^-$ (Ethyl ammonium chloride).

(A) The basic strength order of the given amines in the aqueous phase is $((CH_3)_2NH) > (CH_3NH_2) > ((CH_3)_3N) > (NH_3)$.
Among the given options,ammonia $(NH_3)$ is the weakest base.
This is because the alkyl group $(CH_3)$ exerts a $+I$ (inductive) effect,which increases the electron density on the nitrogen atom,thereby increasing the basicity of amines compared to ammonia.

(B) The reaction of primary amines with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction is a characteristic test for primary amines.
When aniline $(C_6H_5NH_2)$ reacts with chloroform and alcoholic $KOH$,it produces phenyl isocyanide $(C_6H_5NC)$,which has a foul smell.
The chemical equation is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$.
Therefore,the correct option is $(B)$.

(C) The correct answer is $C$.
Tertiary nitroalkanes do not react with $HNO_2$ because they lack an $\alpha$-hydrogen atom,which is necessary for the formation of nitrolic acids or pseudonitroles.

(B) $CH_3NH_2$ is a primary aliphatic amine,which acts as a Lewis base due to the presence of a lone pair of electrons on the nitrogen atom.
When dissolved in water,it forms methylammonium hydroxide $(CH_3NH_3^+OH^-)$,which releases $OH^-$ ions in the solution.
Basic solutions turn red litmus paper blue.
Therefore,the correct option is $B$.

(C) The correct statement is that methyl amine is a stronger base than $NH_3$.
The presence of the methyl group $(-CH_3)$ increases the electron density on the nitrogen atom due to the $+I$ (inductive) effect.
This increased electron density makes the lone pair on the nitrogen atom more available for donation,thereby increasing the basic strength compared to $NH_3$.

(A) The mustard oil reaction involves the reaction of a primary amine with carbon disulfide $(CS_2)$ followed by treatment with mercuric chloride $(HgCl_2)$.
For ethylamine,the reaction is: $CH_3-CH_2-NH_2 + CS_2 \xrightarrow{HgCl_2} CH_3-CH_2-N=C=S + H_2S$.
The product formed is an alkyl isothiocyanate,which has a characteristic smell similar to mustard oil.

(D) The reduction of nitrosobenzene $(C_6H_5NO)$ proceeds as follows:
$1$. Reduction of nitrosobenzene gives phenylhydroxylamine $(C_6H_5NHOH)$.
$2$. Further reduction of phenylhydroxylamine gives aniline $(C_6H_5NH_2)$.
$3$. Nitrosobenzene can also undergo condensation to form azoxybenzene and azobenzene $(C_6H_5-N=N-C_6H_5)$.
$4$. Nitrobenzene $(C_6H_5NO_2)$ is an oxidation product of nitrosobenzene,not a reduction product.

(B) The basic strength of aniline is determined by the availability of the lone pair of electrons on the nitrogen atom for protonation.
Halogens (like $Cl$,$Br$,$I$) exert an electron-withdrawing effect due to their high electronegativity ($-I$ effect).
When a halogen atom is present in the benzene ring,it withdraws electron density from the ring through the inductive effect.
This reduces the electron density on the nitrogen atom,making the lone pair less available for donation to a proton.
Consequently,the basicity of the substituted aniline decreases compared to aniline.

(A) Primary nitroalkanes $(R-CH_2-NO_2)$ react with nitrous acid $(HNO_2)$ to form nitrolic acids $(R-C(=NOH)-NO_2)$.
These nitrolic acids are acidic in nature and dissolve in $NaOH$ to form a sodium salt,which results in a blood-red colored solution.

(C) Alkyl isothiocyanates $(R-N=C=S)$ are known as mustard oils because they possess a characteristic pungent,mustard-like smell.

(B) The reaction between a primary amine,chloroform,and alcoholic $KOH$ is known as the carbylamine reaction.
The chemical equation is: $CHCl_3 + C_2H_5NH_2 + 3KOH \to C_2H_5NC + 3KCl + 3H_2O$.
In this reaction,ethyl amine $(C_2H_5NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to produce ethyl isocyanide $(C_2H_5NC)$,which has a characteristic foul smell.

(A) Secondary amines react with nitrous acid $(HNO_2)$ at low temperature $(273-278 \ K)$ to form $N$-nitrosoamines,which are yellow oily liquids.
$(C_2H_5)_2NH + HONO \to (C_2H_5)_2N-N=O + H_2O$
Diethylamine is a secondary amine,hence it forms diethyl nitrosoamine.

(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 $C_2H_5NH_2$)
Thus,the end product $C$ is ethyl amine.

(D) The hydrolysis of alkyl isocyanide $(R-NC)$ yields a primary amine and formic acid.
The reaction is: $R-NC + 2H_2O \xrightarrow{H^+} R-NH_2 + HCOOH$.
Therefore,the correct option is $(D)$.

(D) The reaction of primary aliphatic amines with nitrous acid $(HNO_2)$ is complex.
Methylamine $(CH_3NH_2)$ reacts with $HNO_2$ to form methyl nitrite $(CH_3-O-N=O)$ and dimethyl ether $(CH_3-O-CH_3)$ along with nitrogen gas and water.
The reactions are as follows:
$CH_3NH_2 + HNO_2 \to CH_3-O-N=O + N_2 + H_2O$
$2CH_3NH_2 + 2HNO_2 \to CH_3-O-CH_3 + 2N_2 + 3H_2O$
Therefore,both products are formed.

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ with zinc dust and aqueous ammonium chloride $(NH_4Cl)$ is a controlled reduction process.
This reaction proceeds through the formation of nitrosobenzene as an intermediate and finally yields phenylhydroxylamine $(C_6H_5NHOH)$ as the major product.
The reaction can be represented as:
$C_6H_5NO_2 + 4[H] \xrightarrow{Zn/NH_4Cl} C_6H_5NHOH + H_2O$
Therefore,the correct option is $(D)$.

(A) The basicity 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 more basic than $CH_3NH_2$ (Methylamine) due to the greater $+I$ effect of the ethyl group,which increases electron density on the nitrogen atom.
$2$. $NH_3$ (Ammonia) is less basic than aliphatic amines because aliphatic amines have electron-donating alkyl groups that stabilize the positive charge on the nitrogen in the conjugate acid.
$3$. $C_6H_5NH_2$ (Aniline) is the least basic 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 $C_2H_5NH_2 > CH_3NH_2 > NH_3 > C_6H_5NH_2$.

(C) In aqueous solution,the basicity of aliphatic amines depends on the inductive effect,solvation effect,and steric hindrance.
For ethyl-substituted amines,the order of basicity is $(C_2H_5)_2NH > C_2H_5NH_2 > (C_2H_5)_3N$.
$NH_3$ is less basic than these aliphatic amines.
$C_6H_5NH_2$ (aniline) is the least basic due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.
Therefore,the correct increasing order is $C_6H_5NH_2 < NH_3 < (C_2H_5)_3N < C_2H_5NH_2 < (C_2H_5)_2NH$.

(B) The correct answer is $B$.
Aniline $(C_6H_5NH_2)$ acts as a base because the nitrogen atom has a lone pair of electrons available for donation to an acid.
In contrast,benzene and nitrobenzene do not have a lone pair on the ring atoms to act as a base,and phenol is acidic in nature.
Therefore,aniline exhibits the strongest basic behavior among the given compounds.

(D) When aniline is treated with excess of bromine water,the highly activating $-NH_2$ group directs the bromine atoms to all available ortho and para positions.
This results in the formation of a white precipitate of $2, 4, 6-$tribromoaniline.

(D) The reaction is known as the Carbylamine test,which is a characteristic test for primary amines.
$R-NH_2 + CHCl_3 + 3KOH \to R-NC + 3KCl + 3H_2O$
(where $R$ is an alkyl or aryl group).
The unpleasant smell is due to the formation of isocyanide (carbylamine).

(C) When an aliphatic primary amine $(R-NH_2)$ reacts with nitrous acid ($HNO_2$,generated in situ from $NaNO_2 + HCl$) at low temperatures $(0-5 \ ^\circ C)$,it forms an unstable alkyl diazonium salt which immediately decomposes to evolve nitrogen gas $(N_2)$ and forms an alcohol.
$R-NH_2 + NaNO_2 + HCl \to [R-N_2^+Cl^-] \to R-OH + N_2 \uparrow + NaCl + H_2O$.
In contrast,aromatic primary amines form stable diazonium salts at these temperatures,which do not evolve nitrogen gas unless heated. Therefore,the compound is an aliphatic primary amine.

(C) When aniline $(C_6H_5NH_2)$ reacts with an excess of alkyl halide $(R-X)$,it undergoes successive alkylation.
$1$. The primary amine $(C_6H_5NH_2)$ reacts to form a secondary amine $(C_6H_5NHR)$.
$2$. The secondary amine reacts further to form a tertiary amine $(C_6H_5NR_2)$.
$3$. Finally,the tertiary amine reacts with the excess alkyl halide to form a quaternary ammonium salt $([C_6H_5NR_3]^+X^-)$.
Therefore,the final product is a quaternary ammonium compound.

(C) Direct nitration of aniline with conc. $HNO_3$ and conc. $H_2SO_4$ is not possible because the amino group $(-NH_2)$ is highly sensitive to oxidation.
$HNO_3$ acts as a strong oxidizing agent and oxidizes the aniline ring to form a complex mixture of oxidation products,which appears as a black tarry mass.

(D) The correct answer is $D$.
In methyl amine $(CH_3NH_2)$,there is one electron-releasing methyl group,while in dimethyl amine $((CH_3)_2NH)$,there are two electron-releasing methyl groups.
These groups increase the electron density on the nitrogen atom through the inductive effect ($+I$ effect),making dimethyl amine more basic than methyl amine.
Therefore,the statement that dimethyl amine is less basic than methyl amine is incorrect.

(C) Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable aliphatic diazonium salts,which decompose to evolve nitrogen gas $(N_2)$ and form alcohols.
$CH_3NH_2 + HNO_2 \to [CH_3N_2^+Cl^-] \to CH_3OH + N_2 + H_2O$
Therefore,the correct option is $C$.

(C) Tertiary amines $(R_3N)$ react with nitrous acid $(HNO_2)$ to form trialkylammonium nitrite salts,which are often referred to as quaternary ammonium salts in the context of this specific reaction type.
$R_3N + HNO_2 \to [R_3NH]^+ [NO_2]^-$

(B) When an aliphatic primary amine $(R-NH_2)$ reacts with nitrous acid $(HNO_2)$ at low temperatures (cold conditions),it forms an unstable aliphatic diazonium salt,which immediately decomposes to evolve nitrogen gas $(N_2)$ and forms an alcohol $(R-OH)$.
The reaction is as follows:
$R-NH_2 + HNO_2$ $\xrightarrow{\text{cold}} [R-N_2^+Cl^-]$ $\rightarrow R-OH + N_2 + H_2O$

(C) The reaction of amines with acetyl chloride $(CH_3COCl)$ is known as acetylation.
This reaction requires the presence of at least one replaceable hydrogen atom attached to the nitrogen atom.
$1^o$ and $2^o$ amines contain replaceable hydrogen atoms and thus undergo acetylation.
$3^o$ amines,such as $(CH_3)_3N$,do not contain any replaceable hydrogen atom on the nitrogen,hence they cannot be acetylated.

(B) In an aqueous solution,the basicity of amines is determined by the combined effect of inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
$C_6H_5NH_2$ (aniline) is the least basic due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring.
Therefore,$(CH_3)_2NH$ is the most basic among the given options.

(A) The reduction of nitrobenzene $(C_6H_5NO_2)$ proceeds through several intermediates depending on the reaction conditions.
Nitrobenzene is first reduced to nitrosobenzene $(C_6H_5NO)$,which is an intermediate.
Nitrosobenzene is then further reduced to $N$-phenylhydroxylamine $(C_6H_5NHOH)$.
Among the given options,$C_6H_5N=O$ (nitrosobenzene) is a known intermediate in the reduction process.
Therefore,the correct option is $A$.

(C) When aniline is treated with concentrated $HNO_3$,it undergoes oxidation rather than electrophilic substitution. The strong oxidizing nature of concentrated $HNO_3$ leads to the formation of $p$-benzoquinone as the major product.

(A) In an aqueous solution,the basicity of aliphatic amines is determined by a combination of inductive effect,solvation effect,and steric hindrance.
For methyl substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
Secondary amines are the most basic due to the optimal balance of the inductive effect of two methyl groups and the solvation of the protonated cation.

(B) The reaction of $m$-dinitrobenzene with ammonium hydrosulfide $(NH_4HS)$ is a selective reduction reaction.
$NH_4HS$ acts as a selective reducing agent that reduces only one of the two nitro $(-NO_2)$ groups to an amino $(-NH_2)$ group in dinitro compounds.
Therefore,the reaction of $m$-dinitrobenzene with $NH_4HS$ yields $m$-nitroaniline as the major product ($70\%$ to $80\%$ yield).
The correct option is $(B)$.

(A) The basicity of aromatic amines depends on the availability of the lone pair of electrons on the nitrogen atom.
Electron-withdrawing groups $(EWG)$ decrease the electron density on the nitrogen atom,thereby reducing basicity.
Electron-donating groups $(EDG)$ increase the electron density on the nitrogen atom,thereby increasing basicity.
In $p-Nitroaniline$,the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which significantly decreases the electron density on the nitrogen atom,making it the least alkaline among the given options.

(A) The diazotisation reaction involves the conversion of aniline to a diazonium salt using $NaNO_2$ and $HCl$ at $0-5 \ ^{\circ}C$.
An excess of $HCl$ is used to prevent the coupling reaction between the formed diazonium salt and the unreacted aniline.
If free aniline is present,it can react with the diazonium salt to form an aminoazo compound (coupling product).
Therefore,$HCl$ protonates the free aniline to form anilinium ions $(C_6H_5NH_3^+)$,which are not reactive towards the diazonium salt,thus suppressing the coupling reaction.

(B) Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which subsequently decompose to yield alcohols,nitrogen gas,and water.
The reaction is: $R-NH_2 + HNO_2 \to R-OH + N_2 + H_2O$.