Properties of Haloarenes Questions in English

(C) The given reaction sequence is as follows $:$
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$. Thus,$[A]$ is aniline.
$2$. Benzene diazonium chloride reacts with $KI$ to form iodobenzene $(C_6H_5I)$. Thus,$[B]$ is iodobenzene.
$3$. Iodobenzene reacts with $2Na$ in the presence of dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$. Thus,$[C]$ is biphenyl.
Therefore,the correct sequence is $[A] = \text{Aniline}, [B] = \text{Iodobenzene}, [C] = \text{Biphenyl}$.

(A) The reaction of benzenediazonium chloride with ethanol $(EtOH)$ is a deamination reaction,not an etherification reaction.
In this reaction,the diazonium group is replaced by a hydrogen atom,resulting in the formation of benzene,not phenetole $(Ph-OEt)$.
Therefore,the reaction shown in option $A$ is incorrect.
Options $B$,$C$,and $D$ represent standard reactions of diazonium salts:
$B$: Reduction with $H_3PO_2$ gives benzene.
$C$: Reaction with $KI$ gives iodobenzene.
$D$: Sandmeyer reaction with $CuCN$ gives benzonitrile.

(B) The reaction of bromobenzene with $SO_3/H_2SO_4$ (sulphonation) primarily yields the para-substituted product due to steric hindrance at the ortho position,making $A$ = $4$-bromobenzenesulphonic acid.
Next,the reaction of $4$-bromobenzenesulphonic acid with $Br_2/Fe$ (electrophilic aromatic substitution) occurs. The $-SO_3H$ group is meta-directing,while the $-Br$ group is ortho/para-directing. The position ortho to the $-Br$ group and meta to the $-SO_3H$ group is the most favored site for the incoming electrophile,leading to $B$ = $2,4$-dibromobenzenesulphonic acid.

(A) The nucleophilic aromatic substitution reaction of aryl halides with $NaOH$ depends on the presence of electron-withdrawing groups $(-NO_2)$ on the ring.
$A$. Chlorobenzene requires harsh conditions: $(a)$ $NaOH, 623 \ K, 300 \ atm$; $(b)$ $H_3O^+$ $(IV)$.
$B$. $1$-Chloro-$4$-nitrobenzene has one $-NO_2$ group,requiring milder conditions: $(a)$ $NaOH, 443 \ K$; $(b)$ $H_3O^+$ $(III)$.
$C$. $1$-Chloro-$2,4$-dinitrobenzene has two $-NO_2$ groups,requiring even milder conditions: $(a)$ $NaOH, 368 \ K$; $(b)$ $H_3O^+$ $(II)$.
$D$. $1$-Chloro-$2,4,6$-trinitrobenzene has three $-NO_2$ groups,which activate the ring significantly,allowing substitution with just warm $H_2O$ $(I)$.
Thus,the correct match is $A-IV, B-III, C-II, D-I$.

(A) The conversion of ethylbenzene to $p$-bromostyrene involves three steps:
$1$. Electrophilic aromatic substitution: Treatment of ethylbenzene with $Br_2 / Fe$ (or $FeBr_3$) leads to bromination at the para position due to the ortho/para-directing nature of the ethyl group.
$2$. Free radical halogenation: Treatment with $Cl_2, \Delta$ (or $h\nu$) results in the chlorination of the benzylic position of the ethyl group.
$3$. Dehydrohalogenation: Treatment with alcoholic $KOH$ (alc. $KOH$) causes the elimination of $HCl$ from the side chain to form the vinyl group (styrene derivative).
Thus,the correct sequence is $Br_2 / Fe ; Cl_2, \Delta ; \text{alc. } KOH$.

(A) The rate of Electrophilic Substitution Reaction $(ESR)$ depends on the electron density of the benzene ring. Electron-donating groups $(EDG)$ increase the rate,while electron-withdrawing groups $(EWG)$ decrease the rate.
$(a)$ Toluene: $-CH_3$ is an activating group ($+I$ and hyperconjugation effect),which increases electron density.
$(b)$ Nitrobenzene: $-NO_2$ is a strongly deactivating group ($-I$ and $-M$ effect),which significantly decreases electron density.
$(c)$ Benzene: Reference compound.
$(d)$ Chlorobenzene: $-Cl$ is a deactivating group ($-I$ effect dominates over $+M$ effect),which decreases electron density,but less than $-NO_2$.
Therefore,the order of reactivity is: Toluene $(a)$ > Benzene $(c)$ > Chlorobenzene $(d)$ > Nitrobenzene $(b)$.
Thus,the correct order is $a > c > d > b$.

(B) The reaction between $2$ moles of chlorobenzene and $1$ mole of chloral $(CCl_3CHO)$ in the presence of concentrated $H_2SO_4$ is a condensation reaction.
This reaction produces $1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane$,commonly known as $DDT$.
The chemical equation is: $2C_6H_5Cl + CCl_3CHO \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$.

(A) Let's analyze each reaction:
$(A)$ The reaction of benzene diazonium salt with $Cu_2X_2/HX$ is the Sandmeyer reaction,which is a nucleophilic substitution reaction,not an Electrophilic Substitution Reaction $(ESR)$.
$(B)$ Toluene with $X_2/Fe$ in the dark undergoes electrophilic aromatic substitution $(ESR)$ on the ring.
$(C)$ $p$-Nitroethylbenzene is highly deactivated due to the $-NO_2$ group,making it extremely difficult to undergo Friedel-Crafts alkylation or halogenation $(ESR)$.
$(D)$ $p$-Hydroxybenzyl alcohol with $HCl$ undergoes nucleophilic substitution $(NSR)$ at the benzylic position to form $p$-hydroxybenzyl chloride.
Option $(A)$ is clearly incorrect as it represents a nucleophilic substitution,not an $ESR$.

(D) The rate of nucleophilic aromatic substitution is directly proportional to the presence of electron-withdrawing groups $(EWG)$ on the aromatic ring.
These groups stabilize the negatively charged intermediate (Meisenheimer complex) formed during the reaction.
The nitro group $(-NO_2)$ is a strong electron-withdrawing group.
Compound $(D)$ contains two $-NO_2$ groups at the ortho and para positions relative to the leaving group $X$,which provides the maximum stabilization to the intermediate,making it the most reactive towards nucleophilic substitution.

(C) In aryl halides,the halogen atom is attached to an $sp^2$ hybridized carbon atom,making the $C-X$ bond shorter and stronger than in alkyl halides.
Nucleophilic substitution in aryl halides is facilitated by the presence of strong electron-withdrawing groups (like $-NO_2$) at the $ortho$ and $para$ positions due to the stabilization of the carbanion intermediate.
In $o$-nitrochlorobenzene,$p$-nitrochlorobenzene,and $2, 4, 6$-trinitrochlorobenzene,the $-NO_2$ groups are at positions that allow for resonance stabilization of the intermediate.
However,in $m$-nitrochlorobenzene,the $-NO_2$ group at the $meta$ position does not provide resonance stabilization to the carbanion intermediate formed during nucleophilic attack.
Therefore,$m$-nitrochlorobenzene has the most difficulty in breaking the $C-X$ bond compared to the other options.

(D) The reaction shown is the nucleophilic aromatic substitution of $2,4,6-$trinitrochlorobenzene with water (hydrolysis) to form $2,4,6-$trinitrophenol (picric acid).
Because of the presence of three strongly electron-withdrawing nitro $(-NO_2)$ groups at the ortho and para positions,the chlorine atom in $2,4,6-$trinitrochlorobenzene becomes highly susceptible to nucleophilic attack by water,even under mild conditions like warming with water.

(D) The bromination of anisole $(C_6H_5OCH_3)$ with bromine $(Br_2)$ in acetic acid $(CH_3COOH)$ is an electrophilic aromatic substitution reaction.
Anisole is an ortho/para directing group due to the $+M$ effect of the methoxy $(-OCH_3)$ group.
Due to steric hindrance at the ortho position,the para-isomer is the major product.
According to standard experimental data,the reaction yields $90 \%$ of $p$-bromoanisole and $10 \%$ of $o$-bromoanisole.
Therefore,the correct percentage of $p$-bromoanisole formed is $90 \%$.

(B) All the given groups are $o-$ and $p-$ directing.
However,in the $-OH$ group,the lone pair of electrons on the $O$-atom participates in resonance,which significantly increases the electron density at the $o-$ and $p-$ positions of the benzene ring.
This makes the $-OH$ group a strongly activating and strongly $o-$ and $p-$ directing group compared to the others.

(C) The reaction involving an aryl halide $(Ar-X)$ and an alkyl halide $(R-X)$ in the presence of sodium metal $(Na)$ in dry ether is known as the $Wurtz-Fittig$ reaction.
The general equation is: $Ar-X + 2Na + R-X \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.

(D) The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal and dry ether to form a higher alkane.
Option $A$,$B$,and $C$ involve alkyl halides and are examples of the Wurtz reaction.
Option $D$ involves bromobenzene (an aryl halide) reacting with sodium to form biphenyl. This is known as the Fittig reaction,not the Wurtz reaction.

(C) The reaction of chlorobenzene with fused $NaOH$ at high temperature $(623 \ K)$ and high pressure $(300 \ atm)$ is known as the Dow process.
The reaction proceeds as follows:
$C_6H_5Cl + 2NaOH \rightarrow C_6H_5ONa + NaCl + H_2O$
In this reaction,chlorobenzene reacts with fused $NaOH$ to form sodium phenoxide $(C_6H_5ONa)$ as the intermediate product.
Subsequently,acidification of sodium phenoxide yields phenol.
Therefore,the intermediate compound formed is sodium phenoxide.

(B) Chlorobenzene undergoes electrophilic aromatic substitution (chlorination) in the presence of anhydrous $FeCl_3$ as a Lewis acid catalyst.
The chlorine atom on the benzene ring is ortho- and para-directing due to the resonance effect.
However,the para-isomer ($1,4-$dichlorobenzene) is the major product due to less steric hindrance compared to the ortho-isomer ($1,2-$dichlorobenzene).

(D) The reaction between two molecules of an aryl halide (like chlorobenzene) with metallic sodium in the presence of dry ether to form a diaryl (like diphenyl) is known as the Fittig reaction.
The reaction is:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$
Note: The Wurtz–Fittig reaction involves one molecule of alkyl halide and one molecule of aryl halide. Since this reaction involves two molecules of aryl halide,it is specifically the Fittig reaction.

(C) The reaction of two molecules of an aryl halide (like chlorobenzene) with metallic sodium in the presence of dry ether to form a diaryl (biphenyl) is known as the $Fittig$ reaction.
The general equation is: $2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$.
Therefore,the correct option is $C$.

(C) The given reaction is $C_{6}H_{6} + CH_{3}Cl \xrightarrow{\text{Anhy. } AlCl_{3}} C_{6}H_{5}CH_{3} + HCl$.
This is an electrophilic aromatic substitution reaction where a methyl group is introduced into the benzene ring using an alkyl halide in the presence of a Lewis acid catalyst $(AlCl_{3})$.
This reaction is known as the Friedel-Crafts alkylation reaction.

(C) Chlorobenzene undergoes electrophilic aromatic substitution (nitration) when treated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$.
Because the chlorine atom is ortho/para-directing due to the resonance effect,the nitration occurs at both the $ortho$ and $para$ positions.
Therefore,the reaction yields a mixture of $1-$chloro$-2-$nitrobenzene ($ortho$-isomer) and $1-$chloro$-4-$nitrobenzene ($para$-isomer).

(D) The reaction shown is the nucleophilic aromatic substitution of a chlorobenzene derivative with water (hydrolysis) to form a phenol derivative.
The product formed is $2,4,6-$trinitrophenol,commonly known as picric acid.
This reaction proceeds easily under mild conditions (warm water) because the three electron-withdrawing $-NO_2$ groups at the ortho and para positions significantly activate the benzene ring towards nucleophilic attack by $H_2O$.
Therefore,the substrate '$S$' must be $2,4,6-$trinitrochlorobenzene.

(D) The reaction involving the cleavage of the $C-X$ bond in haloarenes is Nucleophilic Aromatic Substitution $(S_NAr)$.
This reaction is facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions,which stabilize the carbanion intermediate formed during the reaction.
Compound $I$ has one $-NO_2$ group at the para position.
Compound $II$ has two $-NO_2$ groups (one at ortho and one at para position).
Compound $III$ has three $-NO_2$ groups (two at ortho and one at para position).
As the number of electron-withdrawing groups increases,the reactivity towards nucleophilic substitution increases.
Therefore,the correct order of reactivity is $III > II > I$.

(B) The reaction between an aryl halide $(Ar-X)$ and an alkyl halide $(R-X)$ in the presence of sodium metal $(Na)$ and dry ether is known as the $Wurtz-Fittig$ reaction.
The general equation is: $Ar-X + 2Na + R-X \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.
This reaction is used to prepare alkyl-substituted aromatic compounds.

(B) The Fittig reaction is a coupling reaction where two aryl halides react with sodium metal in the presence of dry ether to form a diaryl compound.
Option $B$ represents the reaction of two molecules of chlorobenzene $(C_6H_5Cl)$ with sodium to form biphenyl $(C_6H_5-C_6H_5)$,which is the definition of the Fittig reaction.
Option $A$ is the Wurtz reaction.
Option $C$ is the Wurtz-Fittig reaction.
Option $D$ is the Wurtz reaction.

(C) The reaction of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether is known as the $Wurtz-Fittig$ reaction.
In this reaction,the alkyl group from the alkyl halide replaces the halogen atom on the aromatic ring.
The general equation is: $Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.

(D) The reaction involves an aryl halide (bromobenzene) and an alkyl halide (bromomethane) reacting with sodium metal in the presence of dry ether to form an alkylbenzene (toluene). This specific reaction is known as the $Wurtz-Fittig$ reaction. The general equation is: $C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaBr$.

(D) Nucleophilic aromatic substitution in aryl halides is facilitated by the presence of electron-withdrawing groups $(EWG)$ at the ortho and para positions relative to the halogen atom.
These groups stabilize the carbanion intermediate (Meisenheimer complex) formed during the reaction through resonance and inductive effects.
As the number of electron-withdrawing nitro groups $(-NO_2)$ increases,the electron density on the benzene ring decreases,making the carbon atom attached to the chlorine more electrophilic and thus more susceptible to nucleophilic attack.
Therefore,$2,4,6-$trinitrochlorobenzene has the highest number of electron-withdrawing groups,making it the most reactive towards nucleophilic substitution.

(C) The reaction of two molecules of an aryl halide with sodium metal in the presence of dry ether to form a diaryl (biphenyl) is known as the $Fittig$ reaction.
$2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$
Here,$Ar$ represents an aryl group and $X$ represents a halogen atom.

(B) The ease of breaking the $C-X$ bond in haloarenes depends on the electron density at the carbon atom attached to the halogen.
Electron-withdrawing groups (like $-NO_2$) at the ortho and para positions increase the reactivity of the $C-X$ bond towards nucleophilic substitution by stabilizing the carbanion intermediate.
In $m-$Nitrochlorobenzene,the $-NO_2$ group is at the meta position,which does not exert a strong electron-withdrawing effect on the carbon attached to the chlorine through resonance.
Therefore,$m-$Nitrochlorobenzene has the least activation for nucleophilic substitution,making the $C-X$ bond the most difficult to break among the given options.

(B) The reaction is a Friedel-Crafts acylation of $A$ with acetyl chloride in the presence of anhydrous $AlCl_3$.
The products formed are $1-\text{chloroacetophenone}$ (ortho-substituted) and $4-\text{chloroacetophenone}$ (para-substituted).
Since the products contain a chlorine atom attached to the benzene ring,the starting material $A$ must be $Chlorobenzene$.
Chlorobenzene is an ortho/para directing group due to the $+M$ effect of the chlorine atom,which explains the formation of the observed products.

(C) Chlorobenzene undergoes electrophilic aromatic substitution when treated with a nitrating mixture (concentrated $HNO_3$ and concentrated $H_2SO_4$).
Since the chlorine atom is ortho/para-directing due to the resonance effect,the nitration occurs at both the ortho and para positions.
Therefore,the reaction yields a mixture of $1-$chloro$-2-$nitrobenzene (ortho-isomer) and $1-$chloro$-4-$nitrobenzene (para-isomer).

(D) The Fittig reaction involves the coupling of two aryl halide molecules in the presence of sodium metal and dry ether to form a diaryl compound.
Option $A$ represents the Wurtz-Fittig reaction.
Options $B$ and $C$ represent the Wurtz reaction.
Option $D$ represents the Fittig reaction: $2C_6H_5Br + 2Na \xrightarrow[\text{dry ether}]{} C_6H_5-C_6H_5 + 2NaBr$.

(B) The reaction is a Friedel-Crafts acylation of a substituted benzene ring.
Since the products are $1-\text{chloroacetophenone}$ (ortho-isomer) and $4-\text{chloroacetophenone}$ (para-isomer),the starting material '$A$' must be chlorobenzene.
Chlorobenzene undergoes electrophilic aromatic substitution where the chlorine atom is ortho/para-directing,leading to the formation of these specific products.

(D) The reaction of chlorobenzene with chlorine in the presence of anhydrous $FeCl_3$ is an electrophilic aromatic substitution reaction (chlorination).
The $-Cl$ group attached to the benzene ring is ortho/para-directing due to the resonance effect.
Due to steric hindrance at the ortho position,the para-isomer is formed as the major product.
Therefore,$1,4$-dichlorobenzene is the major product.

(B) The reactivity of haloarenes towards nucleophilic substitution depends on the presence of electron-withdrawing groups $(EWG)$ at the $ortho$ and $para$ positions.
These groups stabilize the carbanion intermediate formed during the reaction.
An $EWG$ at the $meta$ position has practically no effect on the reactivity because the negative charge in the intermediate does not reside on the carbon atom attached to the $meta$ position.
Therefore,$m-$nitrochlorobenzene has the least reactivity,making it the most difficult to break the $C-Cl$ bond among the given options.

(A) The reactivity of haloarenes towards nucleophilic substitution reactions increases with the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions.
These groups stabilize the carbanion intermediate formed during the reaction by withdrawing electron density through the inductive and resonance effects.
Compound $I$ has one $-NO_2$ group at the para position.
Compound $II$ has two $-NO_2$ groups (one at ortho and one at para position).
Compound $III$ has three $-NO_2$ groups (two at ortho and one at para position).
Therefore,the reactivity order is $I < II < III$.

(C) The reaction of chlorobenzene with conc. $HNO_3$ in the presence of conc. $H_2SO_4$ is an electrophilic aromatic substitution reaction known as nitration.
Chlorine is an ortho/para-directing group due to the resonance effect.
Therefore,the nitration of chlorobenzene yields a mixture of $1-$chloro$-2-$nitrobenzene (ortho-isomer) and $1-$chloro$-4-$nitrobenzene (para-isomer).
The reaction is: $C_6H_5Cl + HNO_3 \xrightarrow{conc. H_2SO_4, \Delta} C_6H_4(Cl)(NO_2) \text{ (ortho + para isomers)}$.

(C) The reactivity of haloarenes towards nucleophilic substitution increases with the presence of electron-withdrawing groups (like $-NO_2$) at the $ortho$ and $para$ positions due to the stabilization of the carbanion intermediate.
Electron-withdrawing groups at the $meta$ position have a negligible effect on the reactivity compared to $ortho$ and $para$ positions.
Therefore,among the given options,$m-$nitrochlorobenzene is the least reactive,making it the most difficult to break the $C-X$ bond during a nucleophilic substitution reaction.

(A) The reaction of bromobenzene with methyl bromide in the presence of sodium metal and dry ether is known as the Wurtz-Fittig reaction.
In this reaction,an aryl halide reacts with an alkyl halide to form an alkylbenzene.
The chemical equation is: $C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaBr$.
Thus,the correct option is $A$.

(A) The $Wurtz-Fittig$ reaction involves the coupling of an aryl halide $(C_6H_5Cl)$ and an alkyl halide $(CH_3Cl)$ in the presence of sodium metal in dry ether to form an alkylbenzene $(C_6H_5-CH_3)$.
The reaction is: $C_6H_5Cl + CH_3Cl + 2 \ Na \xrightarrow[dry]{ether} C_6H_5-CH_3 + 2 \ NaCl$.
Option $A$ represents this reaction.

(A) The reaction of two moles of an aryl halide with sodium metal in the presence of dry ether to form a diaryl compound (biphenyl) is known as the $Fittig$ reaction.
The general reaction is:
$2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$
Where $Ar$ is an aryl group and $X$ is a halogen.

(C) The reaction of two molecules of aryl halides (like chlorobenzene) with sodium metal in the presence of dry ether to form diaryl compounds (like biphenyl) is known as the Fittig reaction.
The chemical equation is:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$
Therefore,the correct option is $C$.

(B) Ethylbenzene reacts with dilute nitric acid under specific conditions,but typically,dilute nitric acid is not a strong enough nitrating agent to perform electrophilic aromatic substitution on the benzene ring of ethylbenzene effectively. However,if the reaction is considered in the context of standard organic chemistry problems,the alkyl group $(-CH_2CH_3)$ is an ortho/para directing group. Therefore,the expected products of nitration would be a mixture of $o$-nitroethylbenzene and $p$-nitroethylbenzene. Given the options,both $A$ and $B$ are valid isomers,but usually,$p$-nitroethylbenzene is the major product due to steric hindrance at the ortho position. If the question implies a specific reaction condition leading to a single major product,$p$-nitroethylbenzene is the standard answer.

(D) The reaction of $Chlorobenzene$ with $Cl_2$ in the presence of anhydrous $FeCl_3$ is an electrophilic aromatic substitution reaction (chlorination).
$Chlorine$ atom on the benzene ring is ortho/para-directing due to the resonance effect.
However,the para-isomer is the major product due to less steric hindrance compared to the ortho-isomer.
Therefore,$1,4-dichlorobenzene$ is the major product.

(B) The reaction is an electrophilic aromatic substitution,specifically a Friedel-Crafts alkylation.
Chlorobenzene reacts with chloromethane in the presence of anhydrous $AlCl_3$ to form $1-\text{chloro}-2-\text{methylbenzene}$ $(o-\text{chlorotoluene})$ and $1-\text{chloro}-4-\text{methylbenzene}$ $(p-\text{chlorotoluene})$.
Since the product is a mixture of $1-\text{chlorotoluene}$ (ortho-isomer) and $4-\text{chlorotoluene}$ (para-isomer),the reactant $A$ must be $Chlorobenzene$ because the chlorine atom on the benzene ring is ortho/para-directing.

(B) The reaction is a Friedel-Crafts acylation of an aromatic compound '$A$' with acetyl chloride in the presence of anhydrous $AlCl_3$.
Since the products formed are $1-$chloroacetophenone (ortho-substituted) and $4-$chloroacetophenone (para-substituted),the starting material '$A$' must be chlorobenzene.
The chlorine atom on the benzene ring is ortho/para-directing,which explains the formation of these specific isomers.

(B) The given reaction is a Friedel-Crafts alkylation reaction.
In this reaction,$A$ reacts with chloromethane $(CH_3Cl)$ in the presence of anhydrous $AlCl_3$ to form $2$-chlorotoluene and $4$-chlorotoluene.
Since the products are ortho and para substituted chlorotoluenes,the reactant $A$ must be chlorobenzene $(C_6H_5Cl)$.
Chlorine is an ortho/para directing group,which directs the incoming methyl group to the ortho and para positions.
Therefore,$A$ is chlorobenzene.

(D) The given reaction is a Friedel-Crafts acylation reaction.
In this reaction,chlorobenzene reacts with an acylating agent in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$.
The reaction is:
$\text{Chlorobenzene} + CH_3COCl$ $\xrightarrow[anhydrous]{AlCl_3} 2-\text{chloroacetophenone} + 4-\text{chloroacetophenone}$
Here,the reactant $(A)$ is acetyl chloride $(CH_3COCl)$.