Properties of Amines Questions in English

(B) Sulphanilic acid exists as a zwitterion,which is a $Bipolar \ ionic$ structure.
Due to strong electrostatic forces of attraction between these ions,it exhibits a high melting point and is insoluble in organic solvents.

(A) The strength of a base is inversely proportional to the strength of its conjugate acid.
$1$. The conjugate acids of the given bases are: $C_2H_6$ (for $C_2H_5^-$),$C_2H_5COOH$ (for $C_2H_5COO^-$),$C_2H_5OH$ (for $C_2H_5O^-$),and $H_2O$ (for $OH^-$).
$2$. The order of acidity of these conjugate acids is: $C_2H_5COOH > C_2H_5OH > H_2O > C_2H_6$.
$3$. Since $C_2H_6$ is the weakest acid,its conjugate base $C_2H_5^-$ is the strongest base.

(A) The reaction of urea with nitrous acid $(HNO_2)$ is as follows:
$NH_2CONH_2 + 2HNO_2 \to CO_2 + 2N_2 + 3H_2O$
As shown in the balanced chemical equation,nitrogen gas $(N_2)$ is evolved along with carbon dioxide and water.

(D) The nucleophilicity of a species depends on its ability to donate an electron pair.
$CH_3O^-$ is a strong nucleophile because the negative charge is localized on the oxygen atom,making it highly available for donation.
In contrast,$C_6H_5O^-$ is less nucleophilic due to resonance stabilization of the negative charge on the benzene ring.
$RNH_2$ and $ROH$ are neutral species and are generally weaker nucleophiles compared to their corresponding alkoxide ions.

(A) The correct answer is $A$.
Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
Factors affecting nucleophilicity include charge,electronegativity,and solvent effects.
Comparing the given species:
$1$. $C_2H_5SH$ (Ethanethiol): Sulfur is less electronegative than oxygen and nitrogen,making its lone pair more available for donation.
$2$. $CH_3COO^-$ (Acetate ion): The negative charge is delocalized over two oxygen atoms via resonance,which decreases its nucleophilicity.
$3$. $CH_3NH_2$ (Methylamine): Nitrogen is more electronegative than sulfur,making its lone pair less available.
$4$. $NCCH_2^-$ (Cyanomethyl anion): The negative charge is stabilized by the strong electron-withdrawing $-I$ and $-M$ effect of the cyano $(-CN)$ group,significantly reducing its nucleophilicity.
Therefore,$C_2H_5SH$ acts as the strongest nucleophile among the given options due to the lower electronegativity of sulfur compared to oxygen and nitrogen,and the lack of resonance stabilization or strong electron-withdrawing groups.

(C) compound is optically active if it contains a chiral carbon atom (a carbon atom bonded to four different groups).
In $CH_3-CH(NH_2)-CH_2-CH_3$ (Secondary butyl amine),the $C-2$ carbon is bonded to a hydrogen atom,a methyl group,an amino group,and an ethyl group.
Since all four groups attached to the $C-2$ carbon are different,it is a chiral center.
Therefore,Secondary butyl amine is optically active.

(D) The basicity of amines is primarily due to the availability of a lone pair of electrons on the nitrogen atom for donation to a proton or Lewis acid.
In triethylamine,the lone pair on the nitrogen is localized and readily available for donation.
In pyridine,the nitrogen atom is $sp^2$ hybridized,and its lone pair is part of the aromatic $\pi$-electron system (delocalized) to satisfy $H$ückel's rule.
Because the lone pair is delocalized,it is significantly less available for protonation,making pyridine less basic than triethylamine.

(A) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $Benzylamine$ $(C_6H_5CH_2NH_2)$,the $-NH_2$ group is attached to an $sp^3$ hybridized carbon atom,not directly to the benzene ring. Thus,the lone pair is fully available.
In $Aniline$ $(C_6H_5NH_2)$,the lone pair is involved in resonance with the benzene ring,reducing its availability.
In $Acetanilide$ $(CH_3CONHC_6H_5)$,the lone pair is involved in resonance with the carbonyl group $(C=O)$,further reducing basicity.
In $p-nitroaniline$,the strong electron-withdrawing $-NO_2$ group decreases the electron density on the nitrogen atom,making it the least basic.
Therefore,$Benzylamine$ is the most basic compound.

(B) The reaction of a diazonium salt with cuprous chloride $(Cu_2Cl_2)$ in the presence of hydrochloric acid $(HCl)$ to form an aryl chloride is a classic example of the Sandmeyer's reaction.
In this reaction,the diazonium group $(-N_2^+)$ is replaced by a chlorine atom $(Cl)$ using a copper$(I)$ salt as a catalyst.

(C) The reaction of ammonia with an alkyl halide is a nucleophilic substitution reaction that proceeds until the quaternary ammonium salt is formed if the alkyl halide is in excess.
$NH_3 + C_2H_5Cl \rightarrow C_2H_5NH_2 + HCl$
$C_2H_5NH_2 + C_2H_5Cl \rightarrow (C_2H_5)_2NH + HCl$
$(C_2H_5)_2NH + C_2H_5Cl \rightarrow (C_2H_5)_3N + HCl$
$(C_2H_5)_3N + C_2H_5Cl \rightarrow [(C_2H_5)_4N]^+ Cl^-$
Since ethyl chloride is in excess,the final product is the quaternary ammonium salt,which is $Tetraethyl \ ammonium \ chloride$.

(C) This reaction is known as the $Carbylamine$ reaction.
When a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic $KOH$,an isocyanide (carbylamine) is formed.
The reaction is: $R-NH_2 + CHCl_3 + 3KOH \to R-NC + 3KCl + 3H_2O$.
The product,alkyl isocyanide $(R-NC)$,is characterized by an extremely offensive odour.

(B) The carbylamine reaction (also known as the Hofmann isocyanide test) is a diagnostic test for primary amines. $R-NH_2 + CHCl_3 + 3KOH (\text{alc.}) \rightarrow R-NC + 3KCl + 3H_2O$. In this reaction,a primary amine reacts with chloroform and alcoholic potassium hydroxide to produce an isocyanide (carbylamine),which has a foul,offensive smell.

(A) The reaction of primary amines (like aniline) with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ is known as the Carbylamine reaction.
This reaction produces a foul-smelling substance called phenyl isocyanide (or phenyl carbylamine).
The chemical equation is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Therefore,the correct option is $(A)$.

(B) The reaction is the carbylamine reaction: $C_2H_5NH_2 + CHCl_3 + 3KOH \to C_2H_5NC + 3KCl + 3H_2O$.
In this reaction,ethyl isocyanide $(C_2H_5NC)$ is formed,which is known for its characteristic unpleasant and offensive smell.

(B) 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 a highly offensive odour.
The balanced chemical equation is:
$CH_3NH_2 + CHCl_3 + 3KOH \to CH_3NC + 3KCl + 3H_2O$
Therefore,$X$ is $CHCl_3$.

(A) The reaction involves a nucleophilic substitution $(S_N2)$ where $OH^-$ acts as the nucleophile.
In $(CH_3)_4N^+I^-$,the quaternary ammonium salt acts as a good leaving group (trimethylamine,$(CH_3)_3N$) because the positive charge on the nitrogen atom makes the attached methyl group highly electrophilic.
$(CH_3)_4N^+I^- + OH^- \rightarrow CH_3OH + (CH_3)_3N + I^-$.
This reaction proceeds readily due to the high reactivity of the methyl group attached to the positively charged nitrogen atom.

(C) When ethylamine $(C_2H_5NH_2)$ reacts with nitrous acid $(HNO_2)$,it undergoes a diazotization reaction followed by the evolution of nitrogen gas to form ethyl alcohol $(C_2H_5OH)$.
The chemical equation is:
$C_2H_5NH_2 + HNO_2 \to C_2H_5OH + N_2 + H_2O$

(D) Benzylamine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine.
When primary aliphatic amines react with nitrous acid $(HNO_2)$,they form unstable aliphatic diazonium salts,which decompose to form alcohols with the evolution of nitrogen gas $(N_2)$.
The reaction is:
$C_6H_5CH_2NH_2 + HNO_2 \rightarrow C_6H_5CH_2OH + N_2 + H_2O$
Thus,the product formed is benzyl alcohol $(C_6H_5CH_2OH)$.

(B) The reaction between benzenediazonium chloride and phenol in a weakly basic medium is a coupling reaction.
In this reaction,the diazonium group $(-N=N-)$ replaces the hydrogen atom at the para-position of the phenol ring.
The product formed is $p$-hydroxyazobenzene,which is an orange-red dye.
The reaction is: $C_6H_5N_2Cl + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4OH + HCl$.

(A) Azo-dyes are prepared by the coupling reaction of a diazonium salt with an aromatic compound such as an amine or a phenol. The most common starting material for the preparation of the diazonium salt is $Aniline$ $(C_6H_5NH_2)$. The reaction involves the coupling of the diazonium cation with an electron-rich aromatic ring,such as $Phenol$ or $Aniline$,to form the azo group $(-N=N-)$. Since $Aniline$ is the primary precursor for the diazonium salt used in the synthesis of almost all azo-dyes,it is the correct answer.

(A) $1$. The reaction sequence starts with methanol $(CH_3OH)$,which reacts with $HI$ to form methyl iodide $(CH_3I)$.
$2$. $CH_3I$ reacts with $KCN$ to form methyl cyanide $(CH_3CN)$.
$3$. Reduction of methyl cyanide $(CH_3CN)$ using a reducing agent (like $LiAlH_4$ or $H_2/Ni$) yields ethylamine $(CH_3CH_2NH_2)$,which is $X$.
$4$. Ethylamine $(CH_3CH_2NH_2)$ reacts with nitrous acid $(HNO_2)$ to form ethyl alcohol $(CH_3CH_2OH)$,which is $Y$.

(D) Step $1$: Reduction of acetonitrile $(CH_3CN)$ with $Na/C_2H_5OH$ (Mendius reduction) gives ethylamine $(A)$: $CH_3CN + 4[H] \rightarrow CH_3CH_2NH_2$ $(A)$.
Step $2$: Reaction of ethylamine $(A)$ with nitrous acid $(HNO_2)$ gives ethanol $(B)$: $CH_3CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2OH + N_2 + H_2O$ $(B)$.
Step $3$: Dehydrogenation of ethanol $(B)$ over heated copper catalyst at $573 \ K$ gives acetaldehyde $(C)$: $CH_3CH_2OH \xrightarrow{Cu, 573 \ K} CH_3CHO + H_2$ $(C)$.

(A) Urea $(NH_2CONH_2)$ contains a carbonyl group attached to two amino groups. Due to the resonance effect,the lone pair on the nitrogen atoms is involved in delocalization,making it a weak base. It reacts with one equivalent of a strong acid like nitric acid $(HNO_3)$ to form urea nitrate $(NH_2CONH_2 \cdot HNO_3)$. Therefore,it acts as a monoacidic base.

(C) Ammonium cyanate isomerizes to urea upon heating: $NH_4CNO \xrightarrow{\Delta} NH_2CONH_2$.
On prolonged heating of urea,two molecules of urea condense with the elimination of an ammonia molecule $(NH_3)$ to form biuret:
$NH_2CONH_2 + NH_2CONH_2 \xrightarrow{\Delta} NH_2CONHCONH_2 + NH_3$.
Thus,the product obtained is biuret.

(B) When urea $(NH_2CONH_2)$ is heated slowly,it undergoes a condensation reaction to form biuret $(NH_2CONHCONH_2)$ with the evolution of ammonia $(NH_3)$ gas.
The chemical reaction is:
$2NH_2CONH_2 \xrightarrow{\Delta} NH_2CONHCONH_2 + NH_3$

(D) The correct order of basic strength is: $\text{cyclohexylamine} > \text{aniline} > \text{benzamide}$.
In $\text{cyclohexylamine}$,the nitrogen lone pair is localized.
In $\text{aniline}$,the lone pair is involved in resonance with the benzene ring,reducing basicity.
In $\text{benzamide}$,the lone pair is involved in resonance with both the benzene ring and the carbonyl group,significantly reducing basicity.
Therefore,the order given in option $D$ is incorrect.

(C) The conversion of an amide $(C_6H_5CONHCH_3)$ to an amine $(C_6H_5CH_2NHCH_3)$ involves the reduction of the carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
$LiAlH_4$ is a strong reducing agent that reduces amides to amines,but it typically reduces the carbonyl group to a methylene group $(C_6H_5CONHCH_3 \xrightarrow{LiAlH_4} C_6H_5CH_2NHCH_3)$.
However,the Clemmensen reduction $(Zn-Hg/HCl)$ is specifically used to reduce aldehydes and ketones to alkanes,but it is not the standard reagent for reducing amides to amines.
In the context of standard organic chemistry reagents,$LiAlH_4$ is the correct reagent for the reduction of amides to amines.

(D) Aniline is highly reactive and sensitive to oxidation. During nitration,the $-NH_2$ group is protonated in the acidic medium to form the anilinium ion $(-NH_3^+)$,which is meta-directing. To obtain ortho and para products and to protect the amino group from oxidation,it is acetylated using acetic anhydride ($CH_3CO)_2O$ to form acetanilide. After nitration,the acetyl group is removed by hydrolysis.

(C) In $1,3,5$-triaminobenzene,each amino group $(-NH_2)$ is attached directly to the benzene ring.
Since each nitrogen atom is bonded to only one carbon atom (the aromatic ring carbon),these are classified as primary $(1^o)$ amines.

(B) The molecular formula $C_4H_{11}N$ corresponds to a saturated amine.
Primary amines $(1^o)$ have the general structure $R-NH_2$.
The possible isomeric primary amines for the butyl group are:
$1$. $n$-butylamine: $CH_3CH_2CH_2CH_2NH_2$
$2$. Isobutylamine: $(CH_3)_2CHCH_2NH_2$
$3$. $sec$-butylamine: $CH_3CH_2CH(NH_2)CH_3$
$4$. $tert$-butylamine: $(CH_3)_3CNH_2$
Thus,there are $4$ isomeric primary amines.

(B) The reaction described is the $\text{Hofmann bromamide degradation reaction}$.
In this reaction,an amide is treated with bromine $(Br_2)$ in the presence of an aqueous or ethanolic solution of sodium hydroxide $(NaOH)$.
This process results in the degradation of the amide to a primary amine containing one carbon atom less than the original amide.
For example: $\text{CH}_3\text{CONH}_2$ $\xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$.
Thus,the amine formed has one less $C$ atom than the amide.

(D) Acetanilide is prepared by the acetylation of aniline.
When aniline $(C_6H_5NH_2)$ reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of an acid or base catalyst,the amino group of aniline undergoes acetylation to form acetanilide $(C_6H_5NHCOCH_3)$ and acetic acid $(CH_3COOH)$ as a byproduct.
The reaction is:
$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$
Therefore,the correct option is $(D)$.

(B) When nitro compounds (nitroalkanes or nitroarenes) are reduced in a neutral medium using $Zn$ and $NH_4Cl$,the reaction proceeds to form $N$-substituted hydroxylamines as the major product.
For example,the reduction of nitrobenzene with $Zn / NH_4Cl$ yields $N$-phenylhydroxylamine $(C_6H_5NHOH)$.
The general reaction is: $R-NO_2 + 4[H] \xrightarrow{Zn/NH_4Cl} R-NHOH + H_2O$.
Thus,the correct product is $R-NHOH$.

(B) Nitrosobenzene $(C_6H_5NO)$ is prepared by the oxidation of aniline $(C_6H_5NH_2)$ using Caro's acid $(H_2SO_5)$.
This reaction involves the controlled oxidation of the amino group to a nitroso group.

(A) Chloropicrin $(CCl_3NO_2)$ is prepared by the reaction of nitromethane with chlorine in the presence of sodium hydroxide.
The chemical equation is:
$CH_3NO_2 + 3Cl_2 + 3NaOH \to CCl_3NO_2 + 3NaCl + 3H_2O$.

(A) Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which further decompose to yield alcohols,nitrogen gas,and water.
For methyl amine: $CH_3NH_2 + HNO_2 \rightarrow CH_3OH + N_2 + H_2O$.
Aniline $(C_6H_5NH_2)$ forms a stable benzene diazonium salt at low temperatures $(0-5 \ ^\circ C)$.
Secondary amines form $N$-nitrosoamines,and tertiary amines form salts with $HNO_2$.

(B) The reaction of aniline $(C_6H_5NH_2)$ with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and dil. $HCl$,at a low temperature of $0^{\circ} - 5^{\circ}C$ is known as the diazotization reaction.
The chemical equation for this reaction is:
$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0^{\circ} - 5^{\circ}C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
The product formed is benzene diazonium chloride.

(B) The reaction sequence is as follows:
$1.$ Propanoic acid $(CH_3-CH_2-COOH)$ reacts with $SOCl_2$ to form propanoyl chloride $(X = CH_3-CH_2-COCl)$.
$2.$ Propanoyl chloride reacts with $NH_3$ to form propanamide $(Y = CH_3-CH_2-CONH_2)$.
$3.$ Propanamide undergoes Hoffmann bromamide degradation with $Br_2 + KOH$ to form ethylamine $(Z = CH_3-CH_2-NH_2)$.
The overall reaction is: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} CH_3-CH_2-COCl$ $\xrightarrow{NH_3} CH_3-CH_2-CONH_2$ $\xrightarrow{Br_2 + KOH} CH_3-CH_2-NH_2$.

(D) The reaction sequence is as follows:
$1$. $X$ is Aniline $(C_6H_5NH_2)$.
$2$. Bromination of Aniline with $Br_2$ water gives $2,4,6$-Tribromoaniline $(Y)$.
$3$. Treatment of $Y$ with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$ yields the diazonium salt,$2,4,6$-Tribromobenzenediazonium chloride $(Z)$.
$4$. Boiling $Z$ with ethanol $(C_2H_5OH)$ results in the reduction of the diazonium group to a hydrogen atom,producing $1,3,5$-Tribromobenzene.
Therefore,$X$ is Aniline.

(A) The given reaction is $C_6H_5NH_2 + CHCl_3 + 3KOH \to C_6H_5NC + 3KCl + 3H_2O$.
This reaction is known as the Carbylamine reaction.
It is a characteristic test for primary amines,where a primary amine reacts with chloroform and alcoholic $KOH$ to form an isocyanide (carbylamine),which has a foul smell.

(A) The reaction described is the $Carbylamine$ reaction,which is a characteristic test for primary amines ($R-NH_2$ or $Ar-NH_2$).
When aniline $(C_6H_5NH_2)$ reacts with chloroform $(CHCl_3)$ in the presence of an alcoholic base like potassium hydroxide $(KOH)$,it forms 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 $(A)$.

(D) When an aromatic primary amine (like aniline) is treated with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at a low temperature of $0-5^{\circ}C$ $(273-278 \ K)$,it undergoes a diazotization reaction.
This reaction results in the formation of an aromatic diazonium salt,such as benzene diazonium chloride.
The reaction is: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^{\circ}C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.

(C) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom and the stability of the conjugate acid formed.
$1$. $CH_3NH_2$ (Methylamine) is a $1^{\circ}$ aliphatic amine. The $+I$ effect of the methyl group increases the electron density on the nitrogen atom,making it more basic than ammonia.
$2$. $NH_3$ (Ammonia) has no alkyl group to increase electron density.
$3$. $C_6H_5NH_2$ (Aniline) is less basic because the lone pair on nitrogen is involved in resonance with the benzene ring.
$4$. $C_6H_5NHCH_3$ ($N$-methylaniline) is also less basic than aliphatic amines due to the resonance of the lone pair with the benzene ring.
Therefore,$CH_3NH_2$ is the strongest base among the given options.

(B) The given reaction is a Sandmeyer reaction where a diazonium salt reacts with $CuCN$ to replace the diazonium group $(-N_2^+)$ with a cyano group $(-CN)$.
In this case,the starting material is $p$-methylbenzenediazonium ion.
When it reacts with $CuCN$,the $-N_2^+$ group is replaced by $-CN$,resulting in $p$-tolunitrile ($4$-methylbenzonitrile).

(A) This reaction is known as the Hofmann mustard oil reaction.
When ethyl amine $(C_2H_5NH_2)$ is heated with carbon disulfide $(CS_2)$ in the presence of mercuric chloride $(HgCl_2)$,it forms ethyl isothiocyanate $(C_2H_5NCS)$.
The chemical equation is:
$C_2H_5NH_2 + CS_2 + HgCl_2 \to C_2H_5NCS + 2HCl + HgS$
Thus,the correct option is $A$.

(D) . $C_6H_5NH_2$ reacts with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$ to form benzene diazonium chloride $(C_6H_5N_2Cl)$.
This diazonium salt,upon hydrolysis with water $(H_2O)$,yields phenol $(C_6H_5OH)$ along with the evolution of $N_2$ gas and the formation of $HCl$.
The reaction sequence is: $C_6H_5NH_2$ $\xrightarrow{NaNO_2/HCl} C_6H_5N_2Cl$ $\xrightarrow{H_2O} C_6H_5OH + N_2 + HCl$.

(C) The reaction of aniline $(C_6H_5NH_2)$ with nitrous acid $(HNO_2)$,generated in situ from $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$ $(273-278 \ K)$,is known as the diazotization reaction.
The chemical equation is: $C_6H_5NH_2 + NaNO_2 + 2HCl \to C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.
Here,$X$ is $C_6H_5N_2Cl$,which is known as Benzenediazonium chloride.

(B) The reaction of aniline $(C_6H_5NH_2)$ with sodium nitrite $(NaNO_2)$ and hydrochloric acid $(HCl)$ at a low temperature of $0-5\,^{\circ}C$ is known as the diazotization reaction.
This reaction results in the formation of benzene diazonium chloride,which is a diazonium salt.
The chemical equation is: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,^{\circ}C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.

(A) The reaction of nitroalkanes with $Sn/HCl$ is a reduction reaction.
$CH_3NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3NH_2 + 2H_2O$.
Here,the product is methylamine $(CH_3NH_2)$.
Comparing $CH_3NH_2$ with $CH_3X$,we find that $X$ is $-NH_2$.

(C) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2/HCl$ at $0 - 5\ ^\circ C$ to form benzene diazonium chloride $(C_6H_5N_2^+ Cl^-)$,which is $X$.
Step $2$: Benzene diazonium chloride $(C_6H_5N_2^+ Cl^-)$ reacts with $HNO_2$ in the presence of $H_2O$ to form nitrobenzene $(C_6H_5NO_2)$,which is $Y$,along with the evolution of $N_2$ gas and formation of $HCl$.
Therefore,$X$ is $C_6H_5N_2^+ Cl^-$ and $Y$ is $C_6H_5NO_2$.