Preparation of Haloarenes Questions in English

(D) The reaction of benzene with chlorine in the presence of a Lewis acid like anhydrous $AlCl_3$ or $FeCl_3$ proceeds via electrophilic aromatic substitution to form chlorobenzene.
In the presence of direct sunlight ($UV$ light),the reaction proceeds via a free-radical addition mechanism to form benzene hexachloride ($BHC$ or $C_6H_6Cl_6$).
Therefore,$(i)$ requires anhydrous $AlCl_3$ and $(ii)$ requires direct sunlight.

(A) The given reaction is the Balz-Schiemann reaction. In this reaction,benzenediazonium tetrafluoroborate $(C_6H_5N_2^+BF_4^-)$ is heated,which leads to the decomposition of the diazonium salt to form fluorobenzene,nitrogen gas $(N_2)$,and boron trifluoride $(BF_3)$. The reaction is: $C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 + BF_3$. Therefore,the product $A$ is fluorobenzene.

(A) When a diazonium salt is treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$,it yields chlorobenzene.
This reaction is known as the Sandmeyer reaction.

(A) The $Raschig$ process is a commercial method for the production of chlorobenzene.
In this process,benzene reacts with hydrogen chloride $(HCl)$ and oxygen $(O_2)$ in the presence of a copper catalyst to form chlorobenzene $(C_6H_5Cl)$ and water $(H_2O)$.
The reaction is: $2C_6H_6 + 2HCl + O_2 \xrightarrow{CuCl_2} 2C_6H_5Cl + 2H_2O$.

(D) The preparation of chlorobenzene $(C_6H_5Cl)$ from aniline involves two main steps:
$1$. Diazotization: Aniline reacts with $HNO_2$ (prepared in situ from $NaNO_2 + HCl$) at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: The benzene diazonium chloride is then treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$ to yield chlorobenzene,with the evolution of nitrogen gas $(N_2)$.

(A) The reaction of silver benzoate with bromine in the presence of $CCl_4$ (Hunsdiecker reaction) leads to the formation of bromobenzene,carbon dioxide,and silver bromide.
$C_6H_5COOAg + Br_2 \xrightarrow{CCl_4} C_6H_5Br + CO_2 + AgBr$
Therefore,the correct product is bromobenzene.

(D) The conversion of benzene diazonium chloride $(C_6H_5N_2Cl)$ to bromobenzene $(C_6H_5Br)$ is achieved by treating it with hydrogen bromide $(HBr)$ in the presence of copper powder. This reaction is known as the Gattermann reaction. The reaction is: $C_6H_5N_2Cl + HBr \xrightarrow{Cu} C_6H_5Br + N_2 + HCl$.

(C) $1$. Nitration of toluene $(C_6H_5CH_3)$ gives a mixture of $o-$ and $p-$ nitrotoluene.
$2$. Reduction of these nitro compounds with $Sn + HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $o-$ and $p-$ toluidine.
$3$. Diazotization of $o-$ and $p-$ toluidine with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$ gives the corresponding diazonium salts.
$4$. Heating these diazonium salts with $Cu_2Br_2 + HBr$ (Sandmeyer reaction) replaces the $-N_2^+Cl^-$ group with $-Br$.
$5$. The final products are $o-$ and $p-$ bromotoluene.

(D) $1$. Nitration of toluene $(C_6H_5CH_3)$ gives a mixture of $o$-nitrotoluene and $p$-nitrotoluene.
$2$. Reduction of these nitro compounds using $Sn/HCl$ or $Fe/HCl$ yields $o$-toluidine and $p$-toluidine ($o$-methylaniline and $p$-methylaniline).
$3$. Diazotization of these amines with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ produces the corresponding diazonium salts ($o$-methylbenzenediazonium chloride and $p$-methylbenzenediazonium chloride).
$4$. Heating these diazonium salts with cuprous bromide $(CuBr/HBr)$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom,resulting in $o$-bromotoluene and $p$-bromotoluene.

(D) The preparation of chlorobenzene from aniline involves two main steps:
$1$. Diazotization: Aniline reacts with $HNO_2$ (formed in situ from $NaNO_2 + HCl$) at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: The benzene diazonium chloride is then treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$ to yield chlorobenzene.
Therefore,the sequence $HNO_2, HCl$ followed by $Cu_2Cl_2$ is the correct method.

(B) The reaction of a diazonium salt with $Cu_2Cl_2$ in the presence of $HCl$ to form chlorobenzene is known as the Sandmeyer reaction.
In this reaction,the diazonium group $(-N_2^+)$ is replaced by a chlorine atom.

(A) The reaction of a benzene diazonium salt with $Cu_2Cl_2$ in the presence of $HCl$ is known as the $Sandmeyer$ reaction. This reaction is used to prepare chlorobenzene from benzene diazonium chloride.

(B) When benzene reacts with chlorine in the presence of a Lewis acid catalyst like $Fe$ (or $FeCl_3$),it undergoes electrophilic aromatic substitution to form chlorobenzene and hydrogen chloride.
The reaction is: $C_6H_6 + Cl_2 \xrightarrow{Fe} C_6H_5Cl + HCl$.

(A) $1$. Nitration of $Toluene$ $(C_6H_5CH_3)$ with $HNO_3/H_2SO_4$ gives a mixture of $o-$ and $p-$ nitrotoluenes.
$2$. Reduction of these with $Sn/HCl$ converts the $-NO_2$ group into $-NH_2$ group,yielding $o-$ and $p-$ toluidines.
$3$. Diazotisation of these amines with $NaNO_2/HCl$ at $0-5 \ ^{\circ}C$ forms the corresponding diazonium salts.
$4$. Heating these diazonium salts with cuprous bromide $(CuBr)$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom.
$5$. Thus,the final product is a mixture of $o-$ bromotoluene and $p-$ bromotoluene.

(D) Fluorobenzene is prepared by the Balz-Schiemann reaction.
In this process,aniline is first converted to benzene diazonium chloride using $NaNO_2$ and $HCl$ at $0-5^{\circ}C$.
Then,it is treated with fluoroboric acid $(HBF_4)$ to form benzene diazonium tetrafluoroborate $(C_6H_5N_2^+BF_4^-)$.
Finally,heating this salt leads to the formation of fluorobenzene,$BF_3$,and $N_2$ gas.

(C) The conversion of benzene diazonium chloride to chlorobenzene using $CuCl/HCl$ is a classic example of the Sandmeyer reaction.
The chemical equation is: $C_6H_5N_2Cl \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2$
Therefore,the correct option is $C$.

(A) Aryl fluorides are prepared by the Balz-Schiemann reaction. In this reaction,arene diazonium chloride is treated with fluoroboric acid $(HBF_4)$ to form arene diazonium fluoroborate,which upon heating decomposes to give aryl fluoride,nitrogen gas,and boron trifluoride $(BF_3)$. The reaction is: $Ar-N_2^+Cl^- + HBF_4$ $\rightarrow Ar-N_2^+BF_4^- + HCl$ $\rightarrow Ar-F + N_2 + BF_3$.

(C) The conversion of benzene diazonium chloride to bromobenzene is achieved by treating it with copper powder $(Cu)$ and hydrogen bromide $(HBr)$.
This specific reaction is known as the Gattermann reaction.
The chemical transformation is represented as:
$C_6H_5N_2Cl \xrightarrow{Cu/HBr} C_6H_5Br + N_2 + HCl$

(B) Direct fluorination of benzene is highly exothermic and difficult to control,often leading to explosive reactions or polyfluorinated products.
The $Balz-Schiemann$ reaction is the most effective method for introducing a fluorine atom into the benzene ring.
In this process,an aniline is converted into a benzene diazonium tetrafluoroborate salt $(C_6H_5N_2^+BF_4^-)$ by reaction with $NaNO_2/HCl$ followed by $HBF_4$.
Upon heating,the salt decomposes to yield fluorobenzene,$N_2$ gas,and $BF_3$.

(B) The $Sandmeyer$ reaction is a well-known and efficient method for the synthesis of aryl halides from primary aromatic amines via diazonium salts.
It provides a high yield of aryl halides compared to other methods like the direct halogenation of benzene (which often leads to poly-substitution) or the reaction of phenol with $PCl_5$ (which gives poor yields of chlorobenzene).
Therefore,the $Sandmeyer$ reaction is the most suitable for obtaining a high yield of aryl halides.

(B) Fluorobenzene is prepared in the laboratory using the $Schiemann$ reaction.
First,aniline is converted to benzene diazonium chloride via diazotization.
Then,it is treated with fluoroboric acid $(HBF_4)$ to form benzene diazonium fluoroborate $(C_6H_5N_2BF_4)$.
Finally,heating this salt causes its decomposition to yield fluorobenzene,$N_2$,and $BF_3$.

(B) Industrially,chlorobenzene is prepared by the $Raschig$ process. In this process,benzene is reacted with $HCl$ and $O_2$ in the presence of a copper catalyst $(CuCl_2)$ at high temperature. The reaction is: $2C_6H_6 + 2HCl + O_2 \xrightarrow{CuCl_2} 2C_6H_5Cl + 2H_2O$.

(A) Aryl halides cannot be prepared by the methods used for alkyl halides because the $C-OH$ bond in phenols is much stronger than the $C-OH$ bond in alcohols.
$(a)$ Preparation from arenes by electrophilic substitution:
Method: Aryl chlorides and bromides are prepared by the reaction of arenes with chlorine $(Cl_2)$ or bromine $(Br_2)$ in the presence of a Lewis acid catalyst like iron or iron $(III)$ chloride.
Mechanism: This reaction proceeds via electrophilic aromatic substitution. The Lewis acid catalyst generates the electrophiles $Cl^+$ and $Br^+$.
Reactions: $C_6H_6 + X_2 \xrightarrow{Fe/FeX_3} C_6H_5X + HX$ (where $X = Cl, Br$).
The ortho and para isomers can be easily separated due to their large difference in melting points.
Iodination: The reaction is reversible,so an oxidizing agent like $HNO_3$ or $HIO_4$ is required to oxidize $HI$ formed during the reaction.
Fluorination: Fluorine is too reactive to be used directly for this method.
$(b)$ Preparation from amines by Sandmeyer reaction:
$(i)$ Diazotization: Primary aromatic amines dissolved or suspended in cold aqueous mineral acid are treated with sodium nitrite $(NaNO_2)$ to form diazonium salts $(ArN_2^+X^-)$.
$(ii)$ Sandmeyer reaction: The freshly prepared diazonium salt solution is treated with cuprous chloride $(Cu_2Cl_2)$ or cuprous bromide $(Cu_2Br_2)$ to replace the diazonium group with $-Cl$ or $-Br$.

(N/A) Aryl halides cannot be prepared by the method used for alkyl halides because the $C-OH$ bond in phenol is stronger than in alcohols.
$(a)$ Preparation of aryl halides from arenes by electrophilic substitution:
Method: Arenes react with chlorine $(Cl_2)$ or bromine $(Br_2)$ in the presence of a Lewis acid catalyst like iron or iron $(III)$ chloride to form aryl chlorides and aryl bromides,respectively.
Mechanism: This reaction proceeds via electrophilic aromatic substitution. The catalyst generates electrophiles ($Cl^+$ or $Br^+$).
Reaction: $C_6H_6 + X_2 \xrightarrow{Fe/FeX_3, \Delta} C_6H_5X + HX$ (where $X = Cl, Br$).
The ortho and para isomers can be easily separated due to the large difference in their melting points.
Iodination: Requires an oxidizing agent like $HNO_3$ or $HIO_4$ to oxidize $HI$ formed during the reaction.
Fluorination: Cannot be done by this method due to the extremely high reactivity of fluorine.
$(b)$ Preparation of aryl halides from amines via Sandmeyer reaction:
$(i)$ Diazotization: Primary aromatic amines dissolved or suspended in cold aqueous mineral acid are treated with sodium nitrite $(NaNO_2)$ to form diazonium salts $(ArN_2^+ X^-)$.
Reaction: $C_6H_5NH_2 + NaNO_2 + 2HX \xrightarrow{273-278 K} C_6H_5N_2^+ X^- + 2H_2O + NaX$.
$(ii)$ Sandmeyer reaction: The freshly prepared diazonium salt solution is treated with cuprous chloride $(Cu_2Cl_2)$ or cuprous bromide $(Cu_2Br_2)$ to replace the diazonium group $(N_2^+ X^-)$ with $-Cl$ or $-Br$.
Reaction: $C_6H_5N_2^+ X^- \xrightarrow{Cu_2X_2} C_6H_5X + N_2$.

(N/A) Iodination reactions are reversible in nature. The reaction is as follows:
$C_6H_6 + I_2 \rightleftharpoons C_6H_5I + HI$
To carry out the reaction in the forward direction,the $HI$ formed during the reaction must be removed. This is achieved by using an oxidizing agent such as $HIO_3$ (iodic acid),which oxidizes $HI$ back to $I_2$:
$5HI + HIO_3 \rightarrow 3I_2 + 3H_2O$
This prevents the reverse reaction and drives the equilibrium towards the formation of the aryl iodide.

(N/A) Lewis acids like $FeX_{3}$ $(X = Cl, Br)$ are used to generate the electrophile,the halonium ion $(X^{\oplus})$,from the halogen molecule $(X_{2})$.
The reaction is as follows:
$X-X + FeX_{3} \rightarrow [FeX_{4}]^{-} + X^{\oplus}$
The electrophile $X^{\oplus}$ then attacks the benzene ring to form a $\sigma$-complex,which subsequently loses a proton to form the aryl halide.
Overall reaction:
$C_{6}H_{6} + X_{2} \xrightarrow{FeX_{3}, \text{dark}} C_{6}H_{5}X + HX$ where $X = Cl, Br$.

(N/A) The compound $A$ with molecular formula $C_7H_8$ is toluene $(C_6H_5CH_3)$.
When toluene is treated with $Cl_2$ in the presence of $FeCl_3$ (a Lewis acid),electrophilic aromatic substitution occurs.
The $-CH_3$ group is ortho and para-directing.
Therefore,the reaction yields a mixture of $o$-chlorotoluene and $p$-chlorotoluene.
The reaction is as follows:
$C_6H_5CH_3 + Cl_2 \xrightarrow{FeCl_3} \text{o-Chlorotoluene} + \text{p-Chlorotoluene}$

In phenol,the lone pair of electrons on the oxygen atom is involved in resonance with the benzene ring.
This gives the $C-O$ bond a partial double bond character.
Due to this partial double bond character,the $C-O$ bond is much stronger and shorter than a pure single bond,making it difficult to break.
Therefore,the nucleophilic substitution reaction required to replace the $-OH$ group with a $-Cl$ atom cannot occur easily under these conditions.

(N/A) To obtain monobromobenzene from aniline,the following two-step process is used:
$1$. Diazotization: Aniline is treated with $NaNO_2$ and $HCl$ at a low temperature of $273-278 \ K$ to form benzene diazonium chloride.
$2$. Sandmeyer's Reaction: The benzene diazonium chloride is then treated with cuprous bromide $(Cu_2Br_2)$ in the presence of hydrobromic acid $(HBr)$ to yield bromobenzene.
The reaction sequence is:
$C_6H_5NH_2$ $\xrightarrow{NaNO_2 + HCl, 273-278 \ K} C_6H_5N_2^+Cl^-$ $\xrightarrow{Cu_2Br_2/HBr} C_6H_5Br$

(N/A) Haloarenes are prepared by the electrophilic substitution of arenes. Aryl chlorides and bromides are prepared by the chlorination and bromination of arenes in the presence of Lewis acid catalysts like iron or iron$(III)$ chloride $(FeCl_3)$.
The reaction of toluene with $X_2$ (where $X = Cl, Br$) in the presence of $Fe$ in the dark yields a mixture of $o$- and $p$-halotoluenes. These isomers can be separated due to the large difference in their melting points.
Direct iodination of arenes is reversible and requires the presence of an oxidizing agent like $HIO_3$ or $HNO_3$ to oxidize the $HI$ formed,thereby driving the reaction forward:
$5 HI + HIO_3 \rightarrow 3 I_2 + 3 H_2O$
Fluoroarenes are not prepared by this method because the high reactivity of fluorine is difficult to control.

(C) The reaction of benzene diazonium chloride $(ArN_2Cl)$ with copper powder $(Cu)$ in the presence of hydrochloric acid $(HCl)$ to form aryl chloride $(ArCl)$ is known as the Gattermann reaction.
Note: If $Cu_2Cl_2 / HCl$ were used instead of $Cu / HCl$,it would be the Sandmeyer reaction.

(N/A) $(i)$ Diazotization: When a primary aromatic amine,dissolved or suspended in cold aqueous mineral acid,is treated with sodium nitrite,a diazonium salt is formed. The reaction is known as diazotization.
$(ii)$ Sandmeyer's reaction: Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by $-Cl$ or $-Br$.
$(iii)$ Iodide product: Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.

(D) Chlorobenzene can be synthesized by the electrophilic aromatic substitution of benzene with chlorine in the presence of a Lewis acid catalyst like anhydrous $FeCl_{3}$.
The reaction is:
$C_{6}H_{6} + Cl_{2} \xrightarrow{\text{anhydrous } FeCl_{3}} C_{6}H_{5}Cl + HCl$
Alternatively,it can also be prepared from aniline via the Sandmeyer reaction (using $NaNO_{2}, HCl$ followed by $CuCl$),but option $D$ is a direct and standard method for the synthesis of chlorobenzene from benzene.

(D) Sandmeyer's reaction involves the treatment of benzene diazonium salts with $CuCl/HCl$ or $CuBr/HBr$ to yield chlorobenzene or bromobenzene,respectively.
Among the given structures:
$I$: Fluorobenzene (prepared by Balz-Schiemann reaction)
$II$: Chlorobenzene (prepared by Sandmeyer's reaction)
$III$: Bromobenzene (prepared by Sandmeyer's reaction)
$IV$: Iodobenzene (prepared by treatment with $KI$)
$V$: Astatobenzene (not prepared by Sandmeyer's reaction)
Thus,only $II$ and $III$ can be prepared by Sandmeyer's reaction.
The total number is $2$.

(C) In the Gattermann reaction,chlorine or bromine can be introduced into the arene ring by treating the arene diazonium salt solution with the corresponding halogen acid in the presence of copper powder.
Option $C$ represents the Gattermann reaction where $ArN_{2}^{+}X^{-}$ reacts with $Cu/HBr$ to form $ArBr$.

(C) $(C)$
Sandmeyer's reaction is the process where the diazonium group $(-N_{2}^{+}X^{-})$ is replaced by a halogen (like $-Cl$ or $-Br$) using the corresponding cuprous salt ($CuCl$ or $CuBr$) in the presence of its acid ($HCl$ or $HBr$).
The chemical equation is: $Ar-N_{2}^{+}X^{-} \xrightarrow{CuCl / HCl} Ar-Cl + N_{2} \uparrow$

(A) The reaction in which a diazonium group $(-N_2^+)$ is replaced by a halogen or a cyano group using copper$(I)$ salts ($CuCl/HCl$,$CuBr/HBr$,or $CuCN/KCN$) is known as the Sandmeyer reaction.
In this reaction,the diazonium salt is treated with the corresponding cuprous halide in the presence of its respective acid to yield aryl halides or aryl nitriles.

(D) The Balz-Schiemann reaction is a chemical reaction used to synthesize aryl fluorides from primary aromatic amines.
In this reaction,the arene diazonium salt is treated with fluoroboric acid $(HBF_{4})$ to form arene diazonium fluoroborate $(Ar-N_{2}^{+} BF_{4}^{-})$,which precipitates out.
Upon heating,this salt decomposes to yield aryl fluoride $(Ar-F)$,boron trifluoride $(BF_{3})$,and nitrogen gas $(N_{2})$.
The reaction is represented as:
$Ar-N_{2}^{+} Cl^{-} + HBF_{4} \rightarrow Ar-N_{2}^{+} BF_{4}^{-} + HCl$
$Ar-N_{2}^{+} BF_{4}^{-} \xrightarrow{\Delta} Ar-F + BF_{3} + N_{2}$

(A) The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper salts as reagents or catalysts.
Specifically,the reaction of benzene diazonium chloride with $CuCl/HCl$ or $CuBr/HBr$ or $CuCN/KCN$ leads to the formation of chlorobenzene,bromobenzene,or benzonitrile,respectively.
Among the given options,$CuCN / KCN$ is the reagent used to introduce a cyano group via the Sandmeyer reaction.

(B) When arene diazonium chloride $(Ar-N_2^+Cl^-)$ is treated with fluoroboric acid $(HBF_4)$,it forms arene diazonium fluoroborate $(Ar-N_2^+BF_4^-)$ as a precipitate.
Upon heating,this salt decomposes to yield fluoroarene $(Ar-F)$,boron trifluoride $(BF_3)$,and nitrogen gas $(N_2)$.
The reaction is: $Ar-N_2^+Cl^- + HBF_4 \rightarrow Ar-N_2^+BF_4^- + HCl$.
Then,$Ar-N_2^+BF_4^- \xrightarrow{\Delta} Ar-F + BF_3 + N_2$.

(B) Sandmeyer's reaction involves the conversion of benzene diazonium salts into aryl halides (chloro,bromo) or aryl cyanides using copper$(I)$ salts ($CuCl$,$CuBr$,$CuCN$).
Specifically:
$1$. $Ph-N_2^+Cl^- + CuCl \rightarrow Ph-Cl + N_2$
$2$. $Ph-N_2^+Cl^- + CuBr \rightarrow Ph-Br + N_2$
$3$. $Ph-N_2^+Cl^- + CuCN \rightarrow Ph-CN + N_2$
Iodobenzene $(Ph-I)$ is $NOT$ prepared by Sandmeyer's reaction because it is prepared by treating benzene diazonium salt with potassium iodide $(KI)$ solution,which does not require a copper catalyst.

(A) The reaction of benzene diazonium chloride with cuprous chloride $(CuCl)$ in the presence of hydrochloric acid $(HCl)$ to form chlorobenzene is a classic example of the Sandmeyer reaction.
In this reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a chlorine atom.

(D) The correct answer is $D$.
Electrophilic substitution of benzene with iodine is a reversible reaction,requiring an oxidizing agent (like $HNO_3$ or $HIO_4$) to remove $HI$ and drive the reaction forward; hence,it is not considered a direct preparation.
Fluorination is extremely exothermic and highly reactive,making it difficult to control for direct electrophilic substitution.
Therefore,aryl iodide and aryl fluoride cannot be prepared directly by simple electrophilic substitution.

(B) The reaction of a diazonium salt with fluoroboric acid $(HBF_4)$ followed by heating the resulting diazonium fluoroborate salt to yield an aryl fluoride is known as the Balz-Schiemann reaction.
The reaction sequence is:
$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- + HCl$
$C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 + BF_3$

(C) The Raschig process is a commercial method for the preparation of $chlorobenzene$ from $benzene$.
The reaction is given by:
$C_6H_6 + HCl + \frac{1}{2}O_2 \xrightarrow{CuCl_2, 500 \ K} C_6H_5Cl + H_2O$
Thus,$benzene$ is the raw material used in this process.

(B) The reaction of benzene diazonium chloride with $Cu_{2}Cl_{2} / HCl$ to produce chlorobenzene is a classic example of Sandmeyer's reaction.
In this reaction,the diazonium group $(-N_{2}^{+}Cl^{-})$ is replaced by a chlorine atom.

(C) The synthesis of bromobenzene from benzene diazonium salts is a classic example of the Sandmeyer reaction.
In this reaction,the benzene diazonium salt is treated with cuprous bromide $(Cu_2Br_2)$ in the presence of hydrobromic acid $(HBr)$.
The reaction proceeds as follows:
$C_6H_5N_2^+Cl^- + Cu_2Br_2 \xrightarrow{HBr} C_6H_5Br + N_2 + Cu_2Cl_2$
Thus,the presence of $Cu_2Br_2$ and $HBr$ is required.

(B) The reaction $ArN_2^+ Cl^- + Cu + HCl \longrightarrow ArCl + N_2 + CuCl$ is known as the Gattermann reaction.
In this reaction,the diazonium salt reacts with copper metal in the presence of $HCl$ to produce chlorobenzene.
This reaction is a modification of the Sandmeyer reaction,where copper powder is used instead of copper$(I)$ chloride $(CuCl)$ or copper$(I)$ bromide $(CuBr)$.

(A) In the Gattermann reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a chlorine atom $(-Cl)$ using copper powder $(Cu)$ in the presence of hydrochloric acid $(HCl)$.
Specifically,the nucleophilic species involved in the substitution is the chloride ion $(Cl^{\ominus})$,and the reagents used are $Cu$ powder and $HCl$.
Therefore,$\underline{X} = Cl^{\ominus}$ and $\underline{Y} = Cu / HCl$.