Properties of Haloarenes Questions in English

(B) Chlorobenzene undergoes electrophilic aromatic substitution (nitration) when treated with a mixture of concentrated $HNO_3$ and $H_2SO_4$.
Since the chlorine atom is ortho/para-directing,the reaction yields a mixture of $2-$chloronitrobenzene and $4-$chloronitrobenzene.

(D) The reaction for chlorobenzene to phenol is: $C_6H_5Cl \xrightarrow[(ii) dil. HCl]{(i) 6-8\% NaOH, 623 K, 300 atm} C_6H_5OH$.
The presence of the $-NO_2$ group at $o$- and $p$- positions withdraws electron density from the benzene ring due to its strong $-I$ and $-M$ effects.
This reduction in electron density facilitates the nucleophilic attack of $OH^-$ on the haloarene.
Furthermore,the intermediate carbanion formed is stabilized through resonance by the $-NO_2$ group at the $o$- and $p$- positions,which delocalizes the negative charge.

(B) The reaction involved is an electrophilic aromatic substitution.
In the presence of $FeCl_{3}$ (a Lewis acid) and in the dark,the halogen $(X_{2})$ reacts to generate an electrophile $(X^{+})$.
This electrophile attacks the benzene ring of toluene.
Since the methyl group $(-CH_{3})$ is an ortho/para-directing group due to its $+I$ effect and hyperconjugation,the electrophilic substitution occurs primarily at the ortho and para positions to yield ortho and para-halotoluene.

(C) Nucleophilic aromatic substitution reactions in haloarenes are facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions relative to the halogen atom.
These groups stabilize the carbanion intermediate formed during the reaction.
Among the given options,$2,4-$dinitrochlorobenzene has two $-NO_2$ groups at the ortho and para positions,which provide the maximum stabilization to the intermediate,making it the most reactive towards nucleophilic substitution with sodium methoxide.

(D) The reaction sequence is as follows:
$1$. Starting with $p$-nitro toluene $(P)$,bromination gives $2-$bromo$-4-$nitrotoluene.
$2$. Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $2-$bromo$-4-$aminotoluene $(Q)$.
$3$. Diazotization with $NaNO_2/HCl$ at $273-278 \ K$ followed by reduction with $H_3PO_2/H_2O$ removes the $-NH_2$ group,resulting in $o$-bromotoluene $(R)$.
$4$. Oxidation of the methyl group with $KMnO_4/OH^-$ gives $2-$bromobenzoic acid.
Thus,the starting compound $(P)$ is $p$-nitro toluene.

(C) The reaction of chlorobenzene with bromine in the presence of a Lewis acid like $Anhyd. AlBr_3$ is an example of electrophilic aromatic substitution.
In this reaction,$AlBr_3$ acts as a catalyst to generate the electrophile $Br^+$ from $Br_2$.
This electrophile then attacks the benzene ring at the ortho and para positions due to the activating effect of the chlorine atom,resulting in the substitution of a hydrogen atom by a bromine atom.

(B) The correct answer is $(B)$.
Nucleophilic aromatic substitution in haloarenes is facilitated by 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.
$2,4,6$-trinitrochlorobenzene has three electron-withdrawing $-NO_2$ groups at the ortho and para positions,which makes it highly reactive towards nucleophilic substitution.
Therefore,it can undergo hydrolysis even under mild conditions like warming with water or dilute aqueous $NaOH$.

(B) The presence of electron-withdrawing groups like $-NO_2$ at ortho and para positions increases the reactivity of haloarenes towards nucleophilic substitution. As the number of $-NO_2$ groups increases,the reaction conditions become milder.
$1.$ $1$-$Chloro$-$4$-$nitrobenzene$ requires $(i) \ NaOH, 443 \ K, (ii) \ H^{+}$ to form $4$-$nitrophenol$.
$2.$ $1$-$Chloro$-$2,4$-$dinitrobenzene$ requires $(i) \ NaOH, 368 \ K, (ii) \ H^{+}$ to form $2,4$-$dinitrophenol$.
$3.$ $1$-$Chloro$-$2,4,6$-$trinitrobenzene$ reacts with just $\text{Warm } H_2O$ to form $2,4,6$-$trinitrophenol$ (picric acid).

(B) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ and $HCl$ at $0-5^{\circ}C$ to form Benzenediazonium chloride $(X = C_6H_5N_2^+Cl^-)$.
Step $2$: Benzenediazonium chloride reacts with $CuCl/HCl$ (Sandmeyer reaction) to form Chlorobenzene $(Y = C_6H_5Cl)$.
Step $3$: Chlorobenzene reacts with $CH_3Cl$ and $Na$ in the presence of dry ether (Wurtz-Fittig reaction) to form Toluene $(Z = C_6H_5CH_3)$.

(C) Due to the $+R$ effect (resonance effect) of the $Cl$ atom with the benzene ring,the $C-Cl$ bond in chlorobenzene acquires partial double bond character.
This partial double bond character makes the $C-Cl$ bond in chlorobenzene shorter and stronger than the $C-Cl$ bond in methyl chloride $(CH_3-Cl)$,which exhibits only a single bond character.

(A) In haloarenes, the $C-Cl$ bond acquires partial double bond character due to resonance between the lone pair of electrons on the chlorine atom and the $\pi$-electron system of the benzene ring.
This partial double bond character results in a shorter bond length compared to the pure single $C-Cl$ bond found in haloalkanes.
The bond length of $C-Cl$ in chlorobenzene (haloarene) is approximately $169 \ pm$, while in chloromethane (haloalkane) it is approximately $177 \ pm$.
Therefore, the correct order is $169 \ pm$ and $177 \ pm$.

(B) Deactivating groups are those that withdraw electron density from the benzene ring through inductive or resonance effects, making the ring less reactive towards electrophilic substitution reactions.
Among the given groups:
$1$. $-Cl$: Deactivating (due to strong $-I$ effect).
$2$. $-SO_3H$: Deactivating (due to strong $-I$ and $-M$ effects).
$3$. $-OH$: Activating (due to $+M$ effect).
$4$. $-NHC_2H_5$: Activating (due to $+M$ effect).
$5$. $-COOCH_3$: Deactivating (due to $-I$ and $-M$ effects).
$6$. $-CH_3$: Activating (due to $+I$ and hyperconjugation).
Therefore, the deactivating groups are $-Cl, -SO_3H, \text{ and } -COOCH_3$.
The total number of deactivating groups is $3$.

(A) In reaction $A$,chlorobenzene $(I)$ is converted to ethylbenzene $(II)$. This is a $Wurtz-Fittig$ reaction where chlorobenzene reacts with ethyl chloride in the presence of sodium $(Na)$ and dry ether to form ethylbenzene.
In reaction $B$,chlorobenzene $(I)$ is converted to $4-chloroacetophenone$ $(III)$. This is a $Friedel-Crafts$ acylation reaction where chlorobenzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ to form $4-chloroacetophenone$ as the major product.
Therefore,the correct sequence is $Wurtz-Fittig$ and $Friedel-Crafts$.

(C) Aryl halides are less reactive towards nucleophilic substitution reactions due to the following reasons:
$1$. Resonance effect: The lone pair of electrons on the halogen atom participates in conjugation with the benzene ring,giving the $C-X$ bond a partial double bond character. This makes the bond stronger and shorter,making it difficult to break.
$2$. Hybridization of carbon: In aryl halides,the carbon atom attached to the halogen is $sp^2$ hybridized,which is more electronegative than the $sp^3$ hybridized carbon in alkyl halides. This increases the bond strength and reduces the polarity of the $C-X$ bond.
Therefore,statements $b$ and $d$ are correct.

(D) The given reaction is a Wurtz-Fittig reaction,which involves the coupling of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether to form an alkylbenzene.
The reaction is:
$C_6H_5Cl + CH_3CH_2CH_2CH_2Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-CH_2CH_2CH_2CH_3 + 2NaCl$
The product $X$ formed is $n$-butylbenzene (or simply butylbenzene).

(C) The Fittig reaction is a chemical reaction in which two aryl halides react with sodium metal in the presence of dry ether to form a diaryl compound (biphenyl).
The general reaction is:
$2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$
For example,when chlorobenzene reacts with sodium in the presence of dry ether,it produces biphenyl:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$
Therefore,the product of the Fittig reaction is biphenyl.

(C) The reaction sequence is as follows:
$1$. Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $X$,which is benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. Benzene diazonium chloride reacts with $Cu_2Br_2/HBr$ (Sandmeyer reaction) to form $Y$,which is bromobenzene $(C_6H_5Br)$.
$3$. Bromobenzene reacts with $Na$ in the presence of dry ether. Since only one type of aryl halide is used,this is the $Fittig$ reaction,which produces biphenyl $(Z)$.

(D) The reaction of benzene diazonium chloride with $Cu$ powder in the presence of $HCl$ is known as the $Gattermann$ reaction.
In this reaction,the diazonium group is replaced by a chlorine atom to form chlorobenzene.
Note that if $Cu_2Cl_2$ and $HCl$ were used instead of $Cu$ powder,it would be called the $Sandmeyer$ reaction.

(D) Nucleophilic aromatic substitution in haloarenes is facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions relative to the halogen atom.
These groups stabilize the carbanion intermediate formed during the reaction through the resonance effect.
As the number of electron-withdrawing $-NO_2$ groups increases,the electron density on the benzene ring decreases,making the carbon atom attached to the chlorine more susceptible to nucleophilic attack.
Therefore,the reactivity increases with the number of $-NO_2$ groups.
Among the given options,$1$-chloro-$2,4,6$-trinitrobenzene has the maximum number of $-NO_2$ groups (three),making it the most reactive towards nucleophilic substitution.

(C) Nucleophilic substitution reactions in aryl halides are difficult due to the partial double bond character of the $C-Cl$ bond caused by resonance.
Electron-withdrawing groups like $-NO_2$ at ortho or para positions increase the reactivity towards nucleophilic substitution by stabilizing the carbanion intermediate.
$p$-Nitrochlorobenzene and $m$-Nitrochlorobenzene are aryl halides,which are generally less reactive than alkyl halides like benzyl chloride and allyl chloride.
Among the given options,$m$-Nitrochlorobenzene is the least reactive because the $-NO_2$ group at the meta position does not effectively stabilize the negative charge of the intermediate carbanion through resonance,unlike the para position.
Therefore,$m$-Nitrochlorobenzene is the least reactive towards nucleophilic substitution.

(A) $1$. Conversion of chlorobenzene to phenol is done by Dow's process,which requires $NaOH$ at $623 \ K$ and $300 \ atm$ pressure,followed by acidification with $H^+$.
$2$. The major product of nitration of chlorobenzene is $p$-nitrochlorobenzene. This compound is more reactive towards nucleophilic substitution due to the electron-withdrawing $-NO_2$ group at the para position. It can be converted to $p$-nitrophenol using $NaOH$ at a milder condition of $443 \ K$,followed by acidification with $H^+$.
$3$. Thus,$A = (i) \ NaOH, 623 \ K, 300 \ atm \ (ii) \ H^+$ and $B = (i) \ NaOH, 443 \ K \ (ii) \ H^+$.

(D) The conversion of benzene diazonium chloride to diphenyl occurs in two steps:
$1$. The conversion of benzene diazonium chloride to chlorobenzene using $CuCl$ is known as the $Sandmeyer$ reaction.
$2$. The coupling of two aryl halide molecules (chlorobenzene) in the presence of sodium $(Na)$ and dry ether to form diphenyl is known as the $Fittig$ reaction.
Therefore,the correct sequence is $Sandmeyer$ and $Fittig$.

(B) $1$. The nitration of chlorobenzene with $conc. HNO_3$ and $conc. H_2SO_4$ yields $1$-chloro-$4$-nitrobenzene as the major product $(X)$ because the $-Cl$ group is ortho/para directing and the para-isomer is sterically favored.
$2$. The conversion of $1$-chloro-$4$-nitrobenzene to $4$-nitrophenol involves nucleophilic aromatic substitution.
$3$. The reaction requires strong nucleophilic conditions,which are provided by $(i) NaOH$ at $443 \ K$ followed by $(ii) H^+$ (acidification) to convert the phenoxide ion to the phenol.
$4$. Thus,$Y$ is $(i) NaOH, 443 \ K, (ii) H^+$ and $Z$ is $4$-nitrophenol.

(B) The Wurtz-Fittig reaction involves the reaction of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether to form an alkyl-substituted aromatic compound.
The general reaction is:
$Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$
In this reaction,the compound $X$ must be an aryl halide (such as chlorobenzene) to react with the alkyl halide $(R-X)$ to form the substituted benzene product.
Therefore,the correct compound $X$ is chlorobenzene.

(C) The reaction sequence is as follows:
$1$. Benzene reacts with $Cl_2$ in the presence of $FeCl_3$ (a Lewis acid) to form chlorobenzene $(X)$ via electrophilic aromatic substitution.
$2$. Chlorobenzene $(X)$ reacts with an alkyl halide $(R-X)$ in the presence of sodium metal and dry ether to form an alkylbenzene $(Y)$.
This specific reaction between an aryl halide and an alkyl halide in the presence of sodium metal and dry ether is known as the Wurtz-Fittig reaction.

(A) The nitration of chlorobenzene with a mixture of concentrated $HNO_3$ and $H_2SO_4$ yields a mixture of $1-$chloro$-2-$nitrobenzene (minor) and $1-$chloro$-4-$nitrobenzene (major) due to the ortho/para directing nature of the $-Cl$ group. Thus,Statement $I$ is correct.
The $-Cl$ atom is an electron-withdrawing group due to its $-I$ effect,which deactivates the benzene ring towards electrophilic aromatic substitution. Consequently,chlorobenzene reacts more slowly than benzene in nitration. Thus,Statement $II$ is also correct.

(A) 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 (biphenyl).
The general reaction is:
$2 Ar-X + 2 Na \xrightarrow{\text{dry ether}} Ar-Ar + 2 NaX$
For example,with chlorobenzene:
$2 C_6H_5Cl + 2 Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2 NaCl$

(D) The reaction of chlorobenzene with $Br_2$ in the presence of anhydrous $FeCl_3$ is an electrophilic aromatic substitution reaction (halogenation).
$Cl$ is an ortho/para-directing group due to its $+R$ effect,which increases electron density at the ortho and para positions.
Due to steric hindrance at the ortho position,the para-substituted product is formed as the major product.
Therefore,the major product is $1$-chloro-$4$-bromobenzene.

(D) Chlorobenzene is treated with aqueous sodium hydroxide at a temperature of $623 \ K$ and pressure of $300 \ atm$ to convert it into sodium phenoxide.
Acidification of sodium phenoxide yields phenol. This process is known as the Dow process.
The reaction is:
$C_6H_5Cl \xrightarrow[(ii) \ HCl]{(i) \ NaOH, \ 623 \ K, \ 300 \ atm} C_6H_5OH$

(A) The given reaction is the sulphonation of chlorobenzene.
Halogen atoms are ortho-para directing but deactivating groups due to the $-I$ effect.
In the presence of concentrated sulphuric acid $(H_2SO_4)$,chlorobenzene undergoes an aromatic electrophilic substitution reaction.
The electrophile involved is sulphur trioxide $(SO_3)$.
The reaction produces $2-$chlorobenzenesulphonic acid (minor product) and $4-$chlorobenzenesulphonic acid (major product).
Therefore,concentrated sulphuric acid is the best suitable reagent for this reaction.

(B) The reaction shown is the Wurtz-Fittig reaction,which involves the coupling of an aryl halide with an alkyl halide in the presence of sodium metal in dry ether.
In the first step,bromobenzene reacts with sodium to form phenylsodium $(C_6H_5Na)$ and sodium bromide $(NaBr)$.
In the second step,the phenylsodium intermediate reacts with methyl bromide $(CH_3Br)$ to form toluene $(C_6H_5CH_3)$ and sodium bromide $(NaBr)$.
The overall reaction is: $C_6H_5Br + 2Na + CH_3Br \rightarrow C_6H_5CH_3 + 2NaBr$.

(B) The reaction of benzyl chloride with $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$ is an electrophilic aromatic substitution reaction.
The $-CH_2Cl$ group is ortho/para directing due to its inductive effect and the nature of the benzene ring activation.
Therefore,the reaction produces a mixture of ortho-chlorobenzyl chloride and para-chlorobenzyl chloride.
In the given reaction,$A$ represents the ortho-substituted product,which is o-chlorobenzyl chloride.

(B) Due to the $+R$ effect of the halogen atom $X$,the $C-X$ bond of haloarene acquires partial double bond character.
This makes the $C-X$ bond shorter and stronger than the $C-X$ bond in haloalkanes,thereby making nucleophilic substitution difficult.
Therefore,statement $2$ is correct.

$A$. Correct: Due to resonance, the $C-Cl$ bond in chlorobenzene acquires partial double bond character, making it shorter $(169 \ pm)$ than the $C-Cl$ bond in chloromethane $(178 \ pm)$.
$B$. Correct: The partial double bond character in chlorobenzene makes the $C-Cl$ bond stronger and harder to break compared to the single $C-Cl$ bond in benzyl chloride.
$C$. Correct: Resonance in chlorobenzene involves the lone pair of $Cl$ delocalizing into the ring, creating partial double bond character.
$D$. Incorrect: The $Cl$ atom is $o/p$-directing, so chlorination of chlorobenzene yields $o$-dichlorobenzene and $p$-dichlorobenzene, not $m$-dichlorobenzene.
Therefore, statements $A, B,$ and $C$ are correct.

(B) The given reaction involves the conversion of an aryl halide to an alkylbenzene by reacting it with an alkyl halide $(RX)$ and sodium metal in the presence of dry ether. This specific reaction is known as the Wurtz-Fittig reaction. The general equation is: $Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$. Therefore,the reagent $Z$ is $RX / Na / \text{dry } (C_2H_5)_2O$.

(C) In reaction $I$,toluene reacts with $Cl_2$ in the presence of $UV$ light,which leads to free radical substitution at the side chain,producing benzyl chloride $(C_6H_5CH_2Cl)$.
In reaction $II$,toluene undergoes electrophilic aromatic substitution with $Cl_2$ in the presence of $Fe$ (a Lewis acid) in the dark. The $-CH_3$ group is ortho/para directing,so $o$-chlorotoluene and $p$-chlorotoluene are formed,with $p$-chlorotoluene being the major product due to steric hindrance at the ortho position.
In reaction $III$,$p$-toluidine ($4$-methylaniline) reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form a diazonium salt,which then reacts with $Cu/HCl$ (Sandmeyer reaction) to replace the diazonium group with a chlorine atom,yielding $p$-chlorotoluene as the major product.
Thus,$p$-chlorotoluene is the major product in reactions $II$ and $III$.

(D) The reaction proceeds as follows:
$1$. Electrophilic aromatic substitution (nitration) of chlorobenzene with $Conc. \ HNO_3$ and $Conc. \ H_2SO_4$ yields a mixture of $o$-nitrochlorobenzene (minor) and $p$-nitrochlorobenzene (major).
$2$. Nucleophilic aromatic substitution of the major product,$p$-nitrochlorobenzene,with $NaOH$ at $443 \ K$ followed by acidification $(H_3O^+)$ replaces the $-Cl$ group with an $-OH$ group.
$3$. The final major product is $p$-nitrophenol ($4$-nitrophenol).

(C) The nitration mixture consists of concentrated $HNO_3$ and concentrated $H_2SO_4$. The reaction between them is as follows:
$HNO_3 + 2H_2SO_4 \rightarrow NO_2^{+} + H_3O^{+} + 2HSO_4^{-}$.
From the reaction,it is evident that $NO_2^{+}$,$H_3O^{+}$,and $HSO_4^{-}$ are present in the mixture.
$SO_4^{2-}$ ions are not formed in this reaction.

(D) The starting material is chlorobenzene. Friedel-Crafts acylation with $CH_3COCl$ in the presence of anhydrous $AlCl_3$ gives $p$-chloroacetophenone as the major product '$Q$' because the $-Cl$ group is ortho/para directing.
$Q = p-Cl-C_6H_4-COCH_3$.
Clemmensen reduction of '$Q$' using $Zn-Hg$ and $conc. HCl$ reduces the carbonyl group $(-COCH_3)$ to an ethyl group $(-CH_2CH_3)$.
Thus,the final product '$P$' is $p$-chloroethylbenzene,which is $Cl-C_6H_4-CH_2CH_3$.
In $p$-chloroethylbenzene $(C_8H_9Cl)$:
- The carbons in the ethyl group are: $-CH_2-$ $(sp^3)$ and $-CH_3$ $(sp^3)$. Total $sp^3$ carbons = $2$.
- The carbons in the benzene ring are all $sp^2$. Total $sp^2$ carbons = $6$.
- The ratio of $sp^3$ carbons to $sp^2$ carbons = $2 : 6 = 1 : 3$.

(C) Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction.
It does not occur with strongly deactivated rings (like benzoic acid,which has a $-COOH$ group) or with compounds that form a complex with the Lewis acid catalyst (like aniline,which has a $-NH_2$ group that coordinates with $AlCl_3$).
Chlorobenzene is deactivated due to the $-I$ effect of the chlorine atom,making it very slow to react,but it is generally considered to undergo the reaction under specific conditions.
Anisole (methoxybenzene) is an activated ring due to the $+M$ effect of the $-OCH_3$ group and readily undergoes Friedel-Crafts alkylation.
Therefore,$b$ (Chlorobenzene) and $d$ (Anisole) are the compounds that can undergo this reaction.

(A) In the presence of $Fe$ in the dark,the reaction follows electrophilic aromatic substitution,which occurs at the ortho and para positions of the toluene ring.
Therefore,the products formed are $o-$chlorotoluene and $p-$chlorotoluene.
In the presence of $UV$ light with the same reagent,the $C(sp^3)-H$ bond breaks down and forms a free radical,resulting in side-chain chlorination.

(A) $1$. The first step is the nitration of benzene using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$,which produces nitrobenzene $(X)$.
$2$. The second step is the chlorination of nitrobenzene in the presence of a Lewis acid catalyst $(FeCl_3)$.
$3$. The nitro group $(-NO_2)$ is a strong electron-withdrawing group and is meta-directing.
$4$. Therefore,the incoming chlorine atom will attach to the meta-position relative to the nitro group,resulting in $1-$chloro$-3-$nitrobenzene (or $m-$chloronitrobenzene) as the major product $Y$.

(B) The reactivity towards $S_{N}1$ reaction depends on the stability of the carbocation formed in the rate-determining step.
In $C_6H_5Cl$,the $C-Cl$ bond has partial double bond character due to resonance,making it extremely difficult to break.
Furthermore,the resulting phenyl cation $(C_6H_5^+)$ is highly unstable because the positive charge is on an $sp$-hybridized carbon atom.
In contrast,the other options form relatively stable carbocations: $C_6H_5CH_2^+$ (benzyl),$CH_2=CH-CH_2^+$ (allyl),and $(C_6H_5)_2CH^+$ (diphenylmethyl).
Therefore,$C_6H_5Cl$ is the least reactive towards $S_{N}1$ reaction.

(B) Meta-directing groups are electron-withdrawing groups that decrease the electron density of the benzene ring,particularly at the ortho and para positions,thereby directing the incoming electrophile to the meta position.
In the given options,the set $-NO_2, -CHO, -SO_3H, -COR$ consists entirely of electron-withdrawing groups that are meta-directing.
$-NH_2, -NHR, -OCH_3$ and $-NHCOCH_3$ are ortho/para-directing groups due to their electron-donating resonance effect.