Properties of Haloarenes Questions in English

(D) The melting point of isomeric dihalobenzenes depends on their symmetry.
$p-$Dichlorobenzene is more symmetrical than the $o-$ and $m-$ isomers.
Due to this symmetry,the $p-$isomer fits better into the crystal lattice,leading to stronger intermolecular forces of attraction.
Therefore,$p-$dichlorobenzene has the highest melting point among the given options.

(B) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.
$1$. $p$-Dichlorobenzene $(IV)$: The two $C-Cl$ bond dipoles are equal and opposite, canceling each other out. Thus, $\mu = 0 \ D$.
$2$. Toluene $(I)$: The methyl group is electron-donating, creating a small dipole moment $(\mu \approx 0.4 \ D)$.
$3$. $m$-Dichlorobenzene $(II)$: The angle between the two $C-Cl$ bonds is $120^{\circ}$. The resultant dipole is $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(120^{\circ})} = \mu_1$ (where $\mu_1$ is the dipole of chlorobenzene, $\approx 1.5 \ D$).
$4$. $o$-Dichlorobenzene $(III)$: The angle between the two $C-Cl$ bonds is $60^{\circ}$. The resultant dipole is $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(60^{\circ})} \approx 2.5 \ D$.
Comparing these values: $0 (IV) < 0.4 (I) < 1.5 (II) < 2.5 (III)$.
Therefore, the increasing order is $IV < I < II < III$.

(B) Vinyl halides $(CH_2=CH-Cl)$ and aryl halides are least reactive towards nucleophilic substitution reactions.
This is due to the partial double bond character between the carbon and the halogen atom caused by resonance,which makes the $C-X$ bond stronger and shorter,making it difficult for the nucleophile to replace the halogen atom.

(D) Alkyl halides react with alcoholic $AgNO_3$ to form $AgCl$ (white precipitate) if the $C-Cl$ bond is easily broken to form a carbocation.
In $CH_2=CH-Cl$ (vinyl chloride),the $C-Cl$ bond acquires partial double bond character due to resonance.
This makes the $C-Cl$ bond stronger and shorter,preventing the formation of a carbocation.
Therefore,vinyl chloride does not react with alcoholic $AgNO_3$ to give a white precipitate.

(D) The given reaction is the nucleophilic substitution of $1-chloro-2,4-dinitrobenzene$ with dilute $NaOH$.
Since the benzene ring is substituted with strong electron-withdrawing $-NO_2$ groups at the ortho and para positions,the electron density on the ring is significantly reduced.
This makes the carbon atom attached to the chlorine atom highly electrophilic,facilitating the attack of the nucleophile $(OH^-)$.
This mechanism is known as $S_NAr$ (Nucleophilic Aromatic Substitution) or activated nucleophilic substitution,where the intermediate formed is a resonance-stabilized carbanion called a Meisenheimer complex.

(B) The reaction of $o-$methoxychlorobenzene with $NaNH_2$ proceeds via the benzyne mechanism.
In $o-$methoxychlorobenzene,the methoxy group $(-OCH_3)$ is an electron-donating group by resonance.
When $NaNH_2$ is added,it abstracts the ortho-hydrogen to form a benzyne intermediate.
The nucleophilic attack by the amide ion $(NH_2^-)$ on the benzyne intermediate is directed by the inductive effect of the methoxy group.
The methoxy group exerts an electron-withdrawing inductive effect ($-I$ effect) at the ortho position,which makes the meta-carbon more electrophilic.
Consequently,the nucleophile attacks the meta-position,leading to the formation of $m-$methoxyaniline as the major product.

(A) The starting material is $1$-bromo-$4$-chlorobenzene.
When treated with $1$ mole of $Mg$ in $THF$,the more reactive $C-Br$ bond undergoes oxidative insertion to form the Grignard reagent,$4$-chlorophenylmagnesium bromide $(X = Cl-C_6H_4-MgBr)$.
The $C-Cl$ bond is less reactive towards $Mg$ than the $C-Br$ bond,so it remains intact.
Subsequent reaction with $CO_2$ followed by acidic workup converts the Grignard reagent into a carboxylic acid group.
Thus,the product $Y$ is $4$-chlorobenzoic acid.

(A) The given reaction is an example of the Ullmann reaction.
In the Ullmann reaction,an aryl halide (like $p$-iodotoluene) reacts with copper powder upon heating to form a biaryl compound.
Two molecules of $p$-iodotoluene react with $Cu$ to form $4,4'$-dimethylbiphenyl and $CuI_2$ (or $CuI$ depending on stoichiometry).
The reaction is: $2 CH_3-C_6H_4-I + Cu \xrightarrow{\Delta} CH_3-C_6H_4-C_6H_4-CH_3 + CuI_2$.
Thus,the product is $4,4'$-dimethylbiphenyl.

(C) The reaction starts with $1$-bromo-$3$-chlorobenzene.
When treated with $Mg$ in dry ether $(Et_2O)$,the more reactive $C-Br$ bond undergoes Grignard reagent formation to yield $3$-chlorophenylmagnesium bromide as intermediate $(A)$.
$(A) = 3-Cl-C_6H_4-MgBr$.
Next,this Grignard reagent reacts with acetaldehyde $(CH_3CHO)$ followed by acidic hydrolysis $(aq. NH_4Cl)$.
The nucleophilic attack of the aryl group on the carbonyl carbon of $CH_3CHO$ forms an alkoxide,which upon protonation yields a secondary alcohol.
The final product $(X)$ is $1-(3-chlorophenyl)ethanol$.

(C) The reactivity of haloarenes towards nucleophilic substitution $(OH^-)$ increases with the presence of electron-withdrawing groups $(-NO_2)$ at the ortho and para positions relative to the halogen atom.
$II$: $2,4,6$-trinitrobromobenzene has three $-NO_2$ groups (two ortho,one para).
$IV$: $2,4$-dinitrobromobenzene has two $-NO_2$ groups (one ortho,one para).
$III$: $p$-nitrobromobenzene has one $-NO_2$ group (para).
$I$: $m$-nitrobromobenzene has one $-NO_2$ group (meta),which is less effective at stabilizing the carbanion intermediate compared to ortho/para positions.
Thus,the order of reactivity is $II > IV > III > I$.

(A) The reaction of chlorobenzene with aqueous $NaOH$ at high temperature $(300\,^oC)$ and high pressure is known as the $Dow$ process.
In this reaction,the chlorine atom is replaced by a hydroxyl group to form sodium phenoxide,which upon acidification gives phenol.
The reaction is: $C_6H_5Cl + 2NaOH \xrightarrow{300\,^oC, \text{pressure}} C_6H_5ONa + NaCl + H_2O$.
$C_6H_5ONa + H^+ \rightarrow C_6H_5OH$ (Phenol).

(C) The reaction of chlorobenzene with $NaOH$ at high temperature and pressure is known as the $Dow$ process.
In this reaction,the primary product is sodium phenoxide $(C_6H_5ONa)$,which upon acidification gives phenol.
However,a side reaction occurs where the formed phenoxide ion reacts with unreacted chlorobenzene via a nucleophilic substitution reaction to form diphenyl ether $(C_6H_5OC_6H_5)$.
This side reaction reduces the overall yield of the desired phenol product.

(B) Aryl halides are less reactive towards nucleophilic substitution reactions compared to alkyl halides primarily due to resonance stabilization.
In aryl halides,the lone pair of electrons on the halogen atom is in conjugation with the $\pi$-electrons of the benzene ring.
This results in the development of partial double bond character in the $C-X$ bond,making it shorter and stronger,which is difficult to break.
Additionally,the $sp^2$ hybridized carbon atom attached to the halogen is more electronegative,further strengthening the bond.

(A) The Ullmann reaction is a coupling reaction between two aryl halides in the presence of copper $(Cu)$ powder to form a biaryl compound.
For example,when iodobenzene is heated with copper powder,it yields diphenyl:
$2C_6H_5I + Cu \xrightarrow{\Delta} C_6H_5-C_6H_5 + CuI_2$
Thus,the Ullmann reaction is specifically used for the preparation of diphenyl (a biaryl).

(A) The Ullmann reaction involves the coupling of aryl halides with copper powder to form biaryls. Chlorobenzene is generally unreactive due to the strong $C-Cl$ bond and the electron-donating nature of the chlorine atom,which reduces the electrophilicity of the carbon atom.
However,the presence of strong electron-withdrawing groups like the nitro group $(-NO_2)$ at the $ortho$ $(o-)$ or $para$ $(p-)$ positions increases the electrophilicity of the carbon atom attached to the chlorine,thereby facilitating the reaction.
Therefore,the presence of the $-NO_2$ group makes the Ullmann reaction possible.

(A) The reaction of chlorobenzene with aqueous ammonia $(NH_3)$ in the presence of cuprous oxide $(Cu_2O)$ or cuprous chloride $(Cu_2Cl_2)$ at high temperature and pressure is a nucleophilic substitution reaction.
This reaction is known as the ammonolysis of aryl halides.
The reaction is: $C_6H_5Cl + 2NH_3 \xrightarrow{Cu_2Cl_2, \Delta, P} C_6H_5NH_2 + NH_4Cl$.
The product formed is aniline $(C_6H_5NH_2)$.

(D) The nucleophilic substitution reaction of haloarenes is facilitated by the presence of electron-withdrawing groups like $-NO_2$ at the ortho $(o-)$ and para $(p-)$ positions.
These groups withdraw electron density from the benzene ring through the inductive effect and resonance effect,thereby stabilizing the carbanion intermediate formed during the nucleophilic attack.
Since $2, 4-$ dinitrochlorobenzene has two $-NO_2$ groups at the $o-$ and $p-$ positions,the electron density on the ring is significantly reduced,making the carbon atom attached to chlorine more electrophilic and facilitating the nucleophilic substitution.

(B) The reaction of $p-$chlorotoluene with $KNH_2$ in liquid $NH_3$ proceeds via the $benzyne$ mechanism (elimination-addition reaction).
In $p-$chlorotoluene,the $Cl$ atom is at the $para$ position relative to the $CH_3$ group.
When $KNH_2$ removes a proton from the $ortho$ position relative to the $Cl$ atom,a $benzyne$ intermediate is formed.
Nucleophilic attack by $NH_2^-$ on this $benzyne$ intermediate can occur at either the $ortho$ or $meta$ position relative to the $CH_3$ group.
Due to the inductive effect of the $CH_3$ group,the $meta$ position is more favored for nucleophilic attack,leading to a mixture of $m-$toluidine and $p-$toluidine,with $m-$toluidine being the major product.

(A) Nucleophilic aromatic substitution reactions are facilitated by the presence of electron-withdrawing groups $(EWG)$ on the benzene ring,which stabilize the negatively charged intermediate (Meisenheimer complex).
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which significantly stabilizes the intermediate.
$2$. The $-OCH_3$ group is an electron-donating group ($+M$ effect),which destabilizes the intermediate.
$3$. The $-CH_3$ group is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the intermediate.
$4$. The $-Cl$ group is weakly electron-withdrawing ($-I$ effect) but also electron-donating ($+M$ effect),making it less effective than $-NO_2$ at stabilizing the intermediate.
Therefore,$p$-Nitrochlorobenzene is the most reactive towards nucleophilic substitution.

(A) The directing nature of a group depends on its electronic effect on the benzene ring.
$-NH_2$ is a strong electron-donating group due to the $+M$ effect,making it the strongest ortho-para directing group.
$-NO_2$ is a strong electron-withdrawing group due to the $-M$ and $-I$ effects,making it the strongest meta-directing group.
Therefore,the correct pair is $-NH_2$ and $-NO_2$.

(C) The assertion is correct because Friedel-Crafts reaction is indeed used to introduce an alkyl or acyl group into the benzene ring.
The reason is incorrect because benzene cannot be used as a solvent for the Friedel-Crafts alkylation of bromobenzene.
Bromobenzene is deactivated towards electrophilic aromatic substitution $(S_E)$ due to the electron-withdrawing inductive effect of the bromine atom.
Benzene is more reactive than bromobenzene towards electrophilic substitution.
Therefore,if benzene were used as a solvent,it would undergo alkylation in preference to bromobenzene,leading to a mixture of products.

(A) The nitro group $(-NO_2)$ is a strong electron-withdrawing group $(-EWG)$.
In nucleophilic aromatic substitution,the rate-determining step is the formation of a resonance-stabilized carbanion intermediate (Meisenheimer complex).
The $-NO_2$ group stabilizes this intermediate carbanion by dispersing the negative charge through resonance and inductive effects.
Therefore,the presence of a nitro group at ortho or para positions significantly facilitates nucleophilic substitution reactions in aryl halides.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) The reaction of chloral $(CCl_3CHO)$ with chlorobenzene in the presence of concentrated $H_2SO_4$ produces $DDT$ ($p,p'$-dichlorodiphenyltrichloroethane).
This reaction involves the condensation of two molecules of chlorobenzene with one molecule of chloral,resulting in the elimination of a water molecule.
The mechanism is an electrophilic aromatic substitution where the protonated chloral acts as an electrophile attacking the chlorobenzene ring.
Since both the assertion (that chloral reacts with chlorobenzene to form $DDT$) and the reason (that it is an electrophilic substitution reaction) are correct,and the reason correctly explains the mechanism of the reaction,option $A$ is the correct answer.

(C) The Assertion is true because $S_{N}2$ reactions involve the attack of a nucleophile from the side opposite to the leaving group,leading to the Walden inversion (inversion of configuration).
However,the Reason is false. Aryl halides are extremely unreactive towards nucleophilic substitution reactions under ordinary conditions due to the partial double bond character of the $C-X$ bond caused by resonance. Therefore,they do not undergo $S_{N}2$ reactions with $KOH$ to form alcohols.

(A) The Assertion is correct because the presence of an electron-withdrawing $-NO_2$ group at the $para-$position stabilizes the carbanion intermediate formed during nucleophilic aromatic substitution.
The Reason is also correct. Chlorobenzene typically requires harsh conditions and proceeds via an elimination-addition mechanism (benzyne intermediate) due to the lack of electron-withdrawing groups. In contrast,$4-$nitrochlorobenzene undergoes nucleophilic aromatic substitution via an addition-elimination mechanism (Meisenheimer complex) because the $-NO_2$ group strongly activates the ring towards nucleophilic attack.
Since the mechanism described in the Reason explains why $4-$nitrochlorobenzene is more reactive,the Reason is the correct explanation of the Assertion.

(C) The cyanide ion $(CN^{-})$ is indeed a strong nucleophile because the negative charge is present on the carbon atom,which is less electronegative than nitrogen,making it a good electron donor. Thus,the Assertion is correct.
However,chlorobenzene does not undergo nucleophilic substitution with $KCN$ under normal conditions because the $C-Cl$ bond in chlorobenzene acquires partial double bond character due to resonance,making it resistant to nucleophilic attack. Therefore,benzonitrile cannot be prepared by this method. Thus,the Reason is incorrect.

(D) The Assertion is false because aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions. This is due to the resonance effect,which gives the $C-Cl$ bond partial double bond character,making it shorter and stronger,and thus difficult to replace by nucleophiles.
The Reason is also false because $S_{N}2$ reactions proceed with inversion of configuration,not retention.

(B) The compound $A$ is $3$-methylanisole ($1$-methoxy-$3$-methylbenzene).
In this molecule,the $-OCH_3$ group is a strong ortho/para-directing group,while the $-CH_3$ group is a weakly ortho/para-directing group.
The $-OCH_3$ group dominates the directing effect.
Electrophilic substitution occurs at the positions ortho and para to the $-OCH_3$ group.
The positions ortho to $-OCH_3$ are $2$ and $6$. Position $2$ is sterically hindered by the $-CH_3$ group at position $3$.
The position para to $-OCH_3$ is position $4$.
Substitution at position $4$ is favored due to less steric hindrance compared to position $2$.
Thus,the major product is $4$-bromo-$1$-methoxy-$3$-methylbenzene.

(D) Friedel-Crafts reactions (alkylation and acylation) do not occur with nitrobenzene because the strong electron-withdrawing $-NO_2$ group deactivates the benzene ring towards electrophilic attack.
Furthermore,the lone pair on the nitrogen atom of the $-NO_2$ group coordinates with the Lewis acid catalyst (e.g.,$AlCl_3$),which further deactivates the ring.
Therefore,the Assertion is incorrect.
Since nitrobenzene is highly deactivated,it does not easily undergo electrophilic substitution reactions,making the Reason also incorrect.

(B) Nitrobenzene is used as a solvent in Friedel-Craft's reaction because it is highly resistant to electrophilic substitution due to the strong electron-withdrawing $-NO_2$ group,which deactivates the benzene ring.
The fusion of nitrobenzene with solid $KOH$ is a known chemical reaction that produces a mixture of $o-$ and $p-$ nitrophenols,although the yield is generally low.
Both statements are factually correct,but the reason does not explain why nitrobenzene is used as a solvent in Friedel-Craft's reaction.
Therefore,the correct option is $B$.

(B) Electrophilic substitution reaction involves the replacement of a hydrogen atom on an aromatic ring by an electrophile.
In the reaction of benzene with chlorine in the presence of a Lewis acid like anhydrous $AlCl_3$,the electrophile $Cl^+$ is generated,which attacks the benzene ring to form chlorobenzene.
This is a classic example of electrophilic aromatic substitution.
Reaction: $C_6H_6 + Cl_2 \xrightarrow{Anhyd. AlCl_3} C_6H_5Cl + HCl$.

(A) The rate of nucleophilic substitution reaction $(NSR)$ depends on the stability of the transition state or the nature of the $C-X$ bond.
$A$. $C_{6}H_{5}Cl$ is an aryl halide. In aryl halides,the $C-Cl$ bond acquires partial double bond character due to resonance,making it very strong and difficult to break. Thus,it undergoes hydrolysis at the slowest rate.
$B$. $CH_{3}CH_{2}Cl$ is a primary alkyl halide,which undergoes $S_{N}2$ reaction.
$C$. $CH_{2}=CH-CH_{2}Cl$ is an allylic halide,which is highly reactive towards $S_{N}1$ or $S_{N}2$ due to resonance stabilization of the carbocation or transition state.
$D$. $C_{6}H_{5}CH_{2}Cl$ is a benzyl chloride,which is highly reactive due to resonance stabilization of the benzyl carbocation.
Therefore,the correct option is $A$.

(A) Step $1$: Toluene $(C_{7}H_{8})$ reacts with $3Cl_{2}$ in the presence of heat to undergo side-chain chlorination,forming benzotrichloride $(C_{6}H_{5}CCl_{3})$ as product '$A$'.
Step $2$: The $-CCl_{3}$ group is strongly electron-withdrawing and meta-directing. Therefore,electrophilic substitution with $Br_{2}/Fe$ occurs at the meta position,forming $m-$bromobenzotrichloride as product '$B$'.
Step $3$: Reduction of the $-CCl_{3}$ group with $Zn/HCl$ converts it back into a methyl group $(-CH_{3})$,yielding $m-$bromotoluene as the final product '$C$'.
Thus,the correct option is $A$.

(A) $1$. The starting material is $p$-toluidine ($4$-methylaniline).
$2$. Reaction with acetic anhydride $(Ac_2O)$ performs acetylation of the amino group to form $N$-($4$-methylphenyl)acetamide (compound $A$).
$3$. The acetamido group $(-NHCOCH_3)$ is an ortho/para directing group. Since the para position is already occupied by the methyl group,electrophilic bromination with $Br_2$ in acetic acid $(AcOH)$ occurs at the ortho position relative to the acetamido group.
$4$. The major product $B$ is $N$-($2$-bromo$-4-$methylphenyl)acetamide.

(N/A) $(i)$ Benzene is first brominated with $Br_2$ in the presence of anhydrous $FeBr_3$ to form bromobenzene. Bromobenzene is then nitrated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 \ K$ to give a mixture of $o-$ and $p-nitrobromobenzene$. The $p-isomer$ is separated by fractional distillation.
$(ii)$ Benzene is first nitrated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 \ K$ to form nitrobenzene. Nitrobenzene is then chlorinated with $Cl_2$ in the presence of anhydrous $AlCl_3$ to give $m-nitrochlorobenzene$ (since the $-NO_2$ group is meta-directing).
$(iii)$ Benzene is first alkylated with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene. Toluene is then nitrated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 \ K$ to give a mixture of $o-$ and $p-nitrotoluene$. The $p-isomer$ is separated by fractional distillation.
$(iv)$ Benzene undergoes Friedel-Crafts acylation with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ to form acetophenone.

(N/A) Chlorine withdraws electrons through the inductive effect ($-I$ effect) and releases electrons through resonance ($+R$ effect).
Through the inductive effect,chlorine destabilizes the intermediate carbocation formed during the electrophilic substitution.
Through resonance,the halogen tends to stabilize the carbocation,and this effect is more pronounced at the $ortho$- and $para$- positions.
The inductive effect is stronger than the resonance effect,causing net electron withdrawal,which leads to overall deactivation of the benzene ring.
However,the resonance effect opposes the inductive effect specifically for the attack at the $ortho$- and $para$- positions,making the deactivation less significant at these positions compared to the $meta$- position.
Thus,the overall reactivity is controlled by the stronger inductive effect,while the orientation of the incoming electrophile is controlled by the resonance effect.

(N/A) $p-$Dichlorobenzene is more symmetrical than $o-$ and $m-$ isomers.
Due to this symmetry,it fits more closely than $o-$ and $m-$ isomers in the crystal lattice.
Therefore,more energy is required to break the crystal lattice of $p-$dichlorobenzene.
As a result,$p-$Dichlorobenzene has a higher melting point and lower solubility than $o-$ and $m-$ isomers.

(N/A) The conversion of $4-$nitrotoluene to $2-$bromobenzoic acid can be achieved through the following steps:
$1$. Bromination of $4-$nitrotoluene: $4-$nitrotoluene reacts with $Br_2$ to form $2-$bromo-$4-$nitrotoluene.
$2$. Reduction: The nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn/HCl$ to give $2-$bromo-$4-$methylaniline.
$3$. Diazotization: $2-$bromo-$4-$methylaniline is treated with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2-$bromo-$4-$methylbenzenediazonium chloride.
$4$. Deamination: The diazonium group is removed using $H_3PO_2$ and $H_2O$ to yield $2-$bromotoluene.
$5$. Oxidation: Finally,the methyl group $(-CH_3)$ is oxidized to a carboxylic acid group $(-COOH)$ using $KMnO_4/OH^-$ to obtain $2-$bromobenzoic acid.

(N/A) The reasons for the low reactivity of aryl halides towards nucleophilic substitution reactions are as follows:
$(a)$ Resonance Effect: In haloarenes, the electron pairs on the halogen atom $(X)$ are in conjugation with the $\pi$-electrons of the ring. Due to resonance, the $C-X$ bond acquires partial double bond character. This makes the cleavage of the $C-X$ bond difficult compared to haloalkanes, where the bond is purely single.
$(b)$ Difference in Hybridization of Carbon Atom in $C-X$ Bond: In haloalkanes, the carbon atom attached to the halogen is $sp^3$ hybridized, whereas in haloarenes, it is $sp^2$ hybridized. The $sp^2$ hybridized carbon has more $s$-character $(33\%)$ compared to $sp^3$ $(25\%)$, making it more electronegative. Consequently, it holds the electron pair of the $C-X$ bond more tightly, resulting in a shorter and stronger bond ($169 \text{ pm}$ in haloarenes vs $177 \text{ pm}$ in haloalkanes), which is harder to break.

(N/A) The nucleophilic substitution of haloarenes is difficult due to the partial double bond character of the $C-Cl$ bond. However,it can be facilitated by harsh conditions or by the presence of electron-withdrawing groups $(NO_2)$ at ortho and para positions.
$(i)$ Chlorobenzene to Phenol: Requires $NaOH_{(aq)}$,$623 \ K$,and $300 \ atm$ pressure followed by acidification $(H^+)$.
$(ii)$ $p-$Nitrochlorobenzene to $p-$Nitrophenol: Requires $NaOH$ at $443 \ K$ followed by acidification $(H^+)$.
$(iii)$ $2,4-$Dinitrochlorobenzene to $2,4-$Dinitrophenol: Requires $NaOH$ at $368 \ K$ followed by acidification $(H^+)$.

(N/A) In haloarenes,the presence of an electron-withdrawing group like $-NO_2$ facilitates the nucleophilic substitution of the $-Cl$ atom by an $-OH$ group. The presence of $-NO_2$ groups at ortho and para positions stabilizes the carbanion intermediate through resonance,thereby increasing the reactivity towards $S_NAr$ reactions.
Example:
$1$. Chlorobenzene $(C_6H_5Cl)$ requires harsh conditions ($623 \ K$ and $300 \ atm$ pressure) for the substitution of $-Cl$ by $-OH$.
$2$. $p$-Nitrochlorobenzene $(p-NO_2C_6H_4Cl)$ requires milder conditions $(443 \ K)$ for the same substitution.
$3$. As the number of $-NO_2$ groups increases at ortho/para positions,the reaction conditions become even milder:
- With two $-NO_2$ groups ($2$,$4$-dinitrochlorobenzene),the reaction occurs at $368 \ K$.
- With three $-NO_2$ groups ($2$,$4$,$6$-trinitrochlorobenzene),the reaction occurs simply by warming with water.

(N/A) The nitro group $(-NO_2)$ is a strong electron-withdrawing group due to its $-I$ and $-M$ (resonance) effects.
When present at the ortho or para positions,the $-NO_2$ group withdraws electron density from the benzene ring through resonance.
This creates a positive charge at the ortho and para positions,as shown in the resonance structures $(I)$ and $(II)$.
As a result,the carbon atom attached to the halogen becomes more electron-deficient (more positive),which facilitates the attack of a nucleophile (like $OH^-$).
Thus,the presence of $-NO_2$ groups at ortho and para positions makes the haloarene more susceptible to nucleophilic substitution.

(N/A) The presence of an electron-withdrawing group $(-NO_2)$ at the ortho or para position of haloarenes (like chlorobenzene) significantly increases the reactivity towards nucleophilic substitution.
This reaction proceeds via an addition-elimination mechanism,often referred to as nucleophilic aromatic substitution $(S_NAr)$.
$1$. In the first slow step,the nucleophile $(OH^-)$ attacks the carbon atom bearing the halogen,forming a resonance-stabilized carbanion intermediate (Meisenheimer complex).
$2$. The negative charge is delocalized onto the oxygen atoms of the $-NO_2$ group,which provides extra stability to the intermediate.
$3$. In the second fast step,the leaving group $(Cl^-)$ is eliminated,restoring the aromaticity of the ring to form the final product (nitrophenol).
This mechanism is facilitated by the $-NO_2$ group,which stabilizes the transition state and the intermediate carbanion through resonance.

(N/A) The nucleophilic substitution reaction of $m$-nitrochlorobenzene with $HO^-$ follows the addition-elimination mechanism (often referred to as $S_NAr$).
When the $-NO_2$ group is at the ortho or para position,the intermediate carbanion has a special resonance structure $(II)$ where the negative charge is delocalized onto the carbon atom attached to the $-NO_2$ group,allowing the nitro group to stabilize the negative charge through its $-I$ and $-M$ effects.
In the case of $m$-nitrochlorobenzene,none of the resonance structures of the intermediate carbanion place the negative charge on the carbon atom directly attached to the $-NO_2$ group. Therefore,the presence of the $-NO_2$ group at the meta position does not significantly enhance the reactivity of the haloarene towards nucleophilic substitution compared to chlorobenzene.

(N/A) Haloarenes undergo electrophilic substitution reactions like halogenation,nitration,sulfonation,and Friedel-Crafts reactions at the ortho and para positions because the halogen atom is an $o,p$-directing group. The resonance structures of halobenzene are shown in the image.
In these structures,the electron density is higher at the ortho and para positions due to the negative charge. Consequently,the electrophile attacks the ortho and para positions of the haloarene. The $-X$ group is $o,p$-directing,and electrophilic substitution occurs at these positions.
However,haloarenes are less reactive than benzene towards electrophilic substitution. The halogen atom $(X)$ exerts an electron-withdrawing inductive effect $(-I)$.
Due to this,the halogen atom pulls electrons from the benzene ring towards itself,resulting in lower electron density in the halobenzene ring compared to benzene. Thus,haloarenes are less reactive than benzene,making their electrophilic substitution reactions slower and requiring more vigorous conditions.

(N/A) Chlorobenzene undergoes electrophilic substitution reactions at the $o$- and $p$-positions due to the resonance effect of the chlorine atom,which increases electron density at these positions. The reactions are as follows:
$(i)$ Halogenation (Chlorination): Chlorobenzene reacts with $Cl_2$ in the presence of anhydrous $FeCl_3$ to form $o$-dichlorobenzene and $p$-dichlorobenzene.
$(ii)$ Nitration: Chlorobenzene reacts with concentrated $HNO_3$ and concentrated $H_2SO_4$ to form $1$-chloro-$4$-nitrobenzene (major) and $1$-chloro-$2$-nitrobenzene (minor).
$(iii)$ Sulphonation: Chlorobenzene reacts with concentrated $H_2SO_4$ upon heating to form $4$-chlorobenzenesulphonic acid (major) and $2$-chlorobenzenesulphonic acid (minor).

(N/A) The reactions of aryl halides with metals are categorized into three main types: $(a)$ Wurtz-Fittig reaction,$(b)$ Fittig reaction,and $(c)$ Grignard reaction.
$(a)$ Wurtz-Fittig reaction: When a mixture of an alkyl halide and an aryl halide is treated with sodium metal in dry ether,an alkylarene is formed. This is known as the Wurtz-Fittig reaction.
$(b)$ Fittig reaction: Aryl halides react with sodium metal in dry ether to give analogous compounds where two aryl groups are joined together. This is known as the Fittig reaction.
$(c)$ Grignard reaction: Aryl halides react with magnesium metal in the presence of dry ether to form aryl magnesium halides,which are known as Grignard reagents.

(N/A) There are three main reactions of aryl halides with metals: $(a)$ Wurtz-Fittig reaction,$(b)$ Fittig reaction,and $(c)$ Grignard reaction.
$(a)$ Wurtz-Fittig reaction: When a mixture of an alkyl halide and an aryl halide is treated with sodium metal in dry ether,an alkylarene is formed. This is known as the Wurtz-Fittig reaction.
Example: $C_6H_5X + RX + 2Na \xrightarrow{\text{dry ether}} C_6H_5R + 2NaX$
$(b)$ Fittig reaction: When aryl halides are treated with sodium metal in dry ether,two aryl groups join together to form diaryl compounds. This is known as the Fittig reaction.
Example: $2C_6H_5X + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaX$
$(c)$ Grignard reaction: Aryl halides react with magnesium metal in the presence of dry ether to form aryl magnesium halides (Grignard reagents).
Example: $C_6H_5X + Mg \xrightarrow{\text{dry ether}} C_6H_5MgX$

(B) $p$-dibromobenzene has a higher melting point than its $o$-isomer.
This is due to its symmetrical structure,which allows it to pack more efficiently into a crystal lattice,resulting in stronger intermolecular forces of attraction.