Properties of Haloarenes Questions in English

(D) $p$-dichlorobenzene has the highest melting point due to its symmetrical structure,which leads to more efficient packing in the crystal lattice compared to the ortho and meta isomers.

(B) The dipole moment $(\mu)$ depends on the vector sum of individual bond dipoles.
$(IV)$ $p$-dichlorobenzene: The two $C-Cl$ bond dipoles are equal and opposite, so $\mu = 0$.
$(I)$ Toluene: The methyl group is electron-donating, creating a small dipole moment.
$(II)$ $m$-dichlorobenzene: The angle between the two $C-Cl$ bonds is $120^{\circ}$, resulting in a net dipole moment.
$(III)$ $o$-dichlorobenzene: The angle between the two $C-Cl$ bonds is $60^{\circ}$, resulting in the largest net dipole moment due to the smaller angle.
Thus, the correct order is: $IV < I < II < III$.

(B) In $m$-Dinitrobenzene,the two $-NO_2$ groups are at the $1$ and $3$ positions. Both groups are strongly deactivating and meta-directing. The position $5$ is meta to both $-NO_2$ groups,making it the only equivalent position available for electrophilic substitution. Thus,it yields only one monosubstituted product.

(D) The chlorine atom $(Cl)$ exerts an inductive effect ($-I$ effect) and a resonance effect ($+R$ effect).
Although the $-I$ effect is stronger,the $+R$ effect increases the electron density at the ortho and para positions.
Therefore,the chlorine atom is an ortho-para directing group.

(C) Nucleophilic aromatic substitution $(S_NAr)$ involves the replacement of a leaving group on an aromatic ring by a nucleophile.
$A$. Reaction of benzene with $Cl_2$ in sunlight is a free-radical substitution reaction.
$B$. Hydrolysis of benzyl bromide $(C_6H_5CH_2Br)$ is an aliphatic nucleophilic substitution ($S_N1$ or $S_N2$) reaction,not aromatic.
$C$. The reaction of $NaOH$ with dinitrofluorobenzene is a classic example of nucleophilic aromatic substitution,where the $OH^-$ nucleophile attacks the aromatic ring activated by electron-withdrawing nitro groups,displacing the fluoride ion.
Therefore,only $(c)$ represents a nucleophilic aromatic substitution reaction.

(D) The correct answer is $D$ $(CH_2 = CHCl)$.
In vinyl chloride $(CH_2 = CHCl)$,the lone pair of electrons on the chlorine atom is in conjugation with the $\pi$-bond of the double bond.
This leads to partial double bond character in the $C-Cl$ bond due to resonance.
As a result,the $C-Cl$ bond is stronger and shorter than in alkyl halides,making it very difficult to break in a nucleophilic substitution reaction.
Therefore,it is the least reactive among the given options.

(D) Electrophilic aromatic substitution reactivity depends on the electron density of the ring. Electron-donating groups $(EDG)$ increase reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$(A)$ Halogens are deactivating but ortho/para directing. Reactivity decreases as electronegativity decreases: $F > Cl > Br$.
$(B)$ Phenol ($-OH$ is strongly activating) $>$ $n$-propyl benzene (alkyl group is weakly activating) $>$ benzoic acid ($-COOH$ is strongly deactivating).
$(C)$ $p$-chlorotoluene has an activating $-CH_3$ and deactivating $-Cl$. $p$-nitrotoluene has a deactivating $-NO_2$. $2$-chloro-$4$-nitrotoluene has two deactivating groups. The order is correct.
$(D)$ Benzoic acid is the least reactive due to the $-COOH$ group. The correct order should be phenol $> n$-propyl benzene $>$ benzoic acid. Thus,option $D$ is not arranged in decreasing order.

(D) The reaction of toluene with chlorine in the presence of a Lewis acid like ferric chloride $(FeCl_3)$ is an electrophilic aromatic substitution reaction (chlorination).
The methyl group $(-CH_3)$ attached to the benzene ring is an ortho/para-directing group due to its electron-donating inductive effect and hyperconjugation.
Therefore,the chlorine electrophile $(Cl^+)$ attacks the ortho and para positions of the toluene ring,resulting in the formation of $o-$chlorotoluene and $p-$chlorotoluene as the major products.
The reaction is: $C_6H_5CH_3 + Cl_2 \xrightarrow{FeCl_3} C_6H_4(CH_3)Cl + HCl$.

(A) Nitrobenzene is least reactive towards bromine because the presence of the $-NO_2$ group decreases electron density at $o$ and $p$ positions due to its strong electron-withdrawing nature ($-I$ and $-M$ effects).
Consequently,the attack of the electrophile $(Br^+)$ on the benzene nucleus is difficult because the positive charge at $o$ and $p$ positions repels the incoming electrophile.

(B) Nitration is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring.
$Nitrobenzene$ contains a $-NO_2$ group,which is a strong electron-withdrawing group $(-EWG)$.
This group decreases the electron density of the benzene ring,making it deactivated towards electrophilic attack.
Therefore,$Nitrobenzene$ undergoes nitration with difficulty compared to $Benzene$,$Toluene$,and $Phenol$.

(D) $2, 4, 6-$ trinitrotoluene $(TNT)$ is a well-known chemical compound used as an explosive.

(B) The nitration of nitrobenzene involves the reaction of nitrobenzene with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$.
The $-NO_2$ group present in nitrobenzene is a deactivating and meta-directing group.
Therefore,the incoming nitro group enters the meta-position,resulting in the formation of $1, 3-$dinitrobenzene (also known as $m-$dinitrobenzene).

(C) $n$-propyl benzene is best prepared using the Wurtz-Fittig reaction.
In this reaction,an alkyl halide $(CH_3CH_2CH_2Br)$ reacts with an aryl halide $(C_6H_5Br)$ in the presence of sodium metal and dry ether.
The reaction is: $CH_3CH_2CH_2Br + C_6H_5Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_2CH_2CH_3 + 2NaBr$.
Friedel-Craft's alkylation is unsuitable because the $n$-propyl carbocation intermediate undergoes rearrangement to a more stable isopropyl carbocation,leading to isopropyl benzene as the major product.

(B) The correct answer is $(B)$.
Groups that withdraw electrons through the inductive effect ($-I$ effect) deactivate the benzene ring towards electrophilic aromatic substitution.
Halogens like $-Cl$ possess a strong $-I$ effect,which deactivates the ring. However,they also possess a $+M$ (mesomeric) effect due to the presence of lone pairs on the halogen atom,which allows them to donate electron density to the ring via resonance.
This resonance effect is most effective at the $o-$ (ortho) and $p-$ (para) positions,thereby directing the incoming electrophile to these positions.
Therefore,$-Cl$ is a deactivating group that is $o-, p-$ directing.

(C) The $o, p-$ directing groups are those that increase the electron density at the ortho and para positions of the benzene ring.
Most $o, p-$ directing groups are activating (e.g.,$-NH_2, -OH, -R$),meaning they increase the rate of electrophilic aromatic substitution.
However,halogens $(-X)$ are unique because they are $o, p-$ directing due to the resonance effect ($+M$ effect) but are deactivating due to their strong inductive effect ($-I$ effect),which withdraws electron density from the ring.
Therefore,the only $o, p-$ directing group that is deactivating in nature is the halogen group $(-X)$.

(B) Raschig's process is used for the industrial production of chlorobenzene by the oxychlorination of benzene.
The reaction involves the treatment of benzene with $HCl$ and $O_2$ in the presence of a catalyst.
The catalyst used in this process is $CuCl_2$ (Copper chloride).

(A) The group $-CCl_3$ is a strongly electron-withdrawing group due to the inductive effect of three chlorine atoms.
It is a deactivating group and is meta-directing in electrophilic aromatic substitution reactions.
Therefore,when benzotrichloride reacts with $Br_2$ in the presence of $Fe$ (a Lewis acid catalyst),the bromine atom is substituted at the meta-position to form $3-$bromobenzotrichloride.

(A) Chlorobenzene is less reactive than benzyl chloride towards nucleophilic substitution reactions.
In chlorobenzene,the lone pairs of electrons on the $Cl$ atom are involved in resonance with the $\pi$-electrons of the benzene ring. This gives the $C-Cl$ bond a partial double bond character,making it stronger and harder to break.
In contrast,benzyl chloride undergoes nucleophilic substitution more easily because the resulting carbocation is stabilized by resonance with the benzene ring.

(C) When chlorobenzene is fused with solid $NaOH$ at high temperature $(623 \ K)$ and high pressure $(300 \ atm)$,it undergoes nucleophilic substitution to form sodium phenoxide.
Sodium phenoxide on subsequent acidification (hydrolysis) yields phenol.
The reaction is as follows:
$C_6H_5Cl + 2NaOH \xrightarrow{623 \ K, 300 \ atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + H^+ \rightarrow C_6H_5OH + Na^+$
Thus,the final product is phenol.

(D) The synthesis of $DDT$ $(Dichloro-diphenyl-trichloro-ethane)$ involves the condensation reaction of two moles of chlorobenzene with one mole of chloral hydrate ($CCl_3CHO \cdot H_2O$ or $Trichloroethanal$).
This reaction occurs in the presence of concentrated $H_2SO_4$,which acts as a dehydrating agent.
The overall reaction is: $2C_6H_5Cl + CCl_3CHO \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$.

(C) $(i)$ Ethyl bromide reacts with aqueous $NaOH$ via nucleophilic substitution to form ethyl alcohol:
$C_2H_5Br + NaOH_{(aq)} \to C_2H_5OH + NaBr$
$(ii)$ Chlorobenzene is resistant to nucleophilic substitution under normal conditions due to resonance and $sp^2$ hybridized carbon. However,under drastic conditions $(623 \ K, 300 \ atm)$,it reacts with aqueous $NaOH$ to form sodium phenoxide,which upon acidification gives phenol:
$C_6H_5Cl + 2NaOH_{(aq)} \xrightarrow{623 \ K, 300 \ atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \to C_6H_5OH + NaCl$

(D) The reactivity of the halogen atom towards nucleophilic substitution depends on the strength of the $C-X$ bond.
In $chloroethane$ $(CH_3CH_2Cl)$,the $C-Cl$ bond is a pure single bond,making it the most reactive.
In $vinyl$ chloride $(CH_2=CHCl)$ and chlorobenzene $(C_6H_5Cl)$,the $C-Cl$ bond acquires partial double-bond character due to resonance,which makes the bond shorter and stronger,thus reducing reactivity.
Between $vinyl$ chloride and chlorobenzene,the $C-Cl$ bond in chlorobenzene is more stable due to the delocalization of electrons over the entire benzene ring.
Therefore,the ascending order of reactivity is: $Chlorobenzene < Vinyl \text{ } chloride < Chloroethane$.

(D) Aryl halides are less reactive than alkyl halides towards nucleophilic substitution due to two main factors:
$1$. Resonance stabilization: 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 shorter and stronger,making it difficult to break.
$2$. $sp^{2}$ hybridization: The carbon atom attached to the halogen is $sp^{2}$ hybridized,which is more electronegative than the $sp^{3}$ hybridized carbon in alkyl halides,leading to a shorter and stronger bond.

(B) In $CH_2 = CH - Cl$,the lone pair of electrons on the chlorine atom is in conjugation with the $C = C$ double bond.
This leads to the delocalization of electrons,resulting in resonance structures.
Due to this resonance,the $C - Cl$ bond acquires partial double bond character,which makes it stronger and shorter than a pure single bond.
Consequently,the $C - Cl$ bond is difficult to break,leading to lower reactivity towards nucleophilic substitution reactions.

(B) The reaction between $2$ moles of chlorobenzene and $1$ mole of chloral $(CCl_3CHO)$ in the presence of concentrated $H_2SO_4$ leads to the formation of $p,p$-dichlorodiphenyltrichloroethane,commonly known as $DDT$.
The reaction is: $2C_6H_5Cl + CCl_3CHO \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$.

(B) $C_6H_6Cl_6 + 3KOH \rightarrow C_6H_3Cl_3 + 3KCl + 3H_2O$
When $C_6H_6Cl_6$ (Benzene hexachloride) is treated with alcoholic $KOH$,it undergoes dehydrohalogenation to yield $1,2,4$-trichlorobenzene $(C_6H_3Cl_3)$.

(C) The 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.
The general equation is: $Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.
For example: $C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaBr$.

(A) The reaction of alkyl halides with aromatic compounds in the presence of anhydrous $AlCl_3$ is known as the Friedel-Crafts alkylation reaction.
This is a type of electrophilic aromatic substitution.
The general reaction is: $C_6H_6 + CH_3Cl \xrightarrow{\text{anhydrous } AlCl_3} C_6H_5CH_3 + HCl$.

(C) Chlorobenzene reacts with chloral $(CCl_3CHO)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$ to produce $1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane$,which is commonly known as $DDT$.
$DDT$ is widely used as an effective insecticide.

(D) The presence of electron-withdrawing groups like $-NO_2$ at the ortho and para positions relative to the halogen atom significantly activates the benzene ring toward nucleophilic aromatic substitution.
This occurs because the $-NO_2$ group stabilizes the carbanion intermediate (Meisenheimer complex) formed during the reaction by withdrawing electron density through both inductive and resonance effects.
In $2,4-$dinitrochlorobenzene,the two $-NO_2$ groups are located at the ortho and para positions,which effectively delocalize the negative charge of the intermediate,thereby lowering the activation energy for the substitution reaction.

(D) $p-$dichlorobenzene has a symmetrical structure.
It fits well in its crystal lattice due to its symmetry.
This leads to stronger intermolecular forces of attraction compared to the $o-$ and $m-$ isomers.
Therefore,it possesses the highest melting point.

(D) $(i)$ Ethyl bromide $(C_2H_5Br)$ undergoes nucleophilic substitution with aqueous $NaOH$ to form ethyl alcohol $(C_2H_5OH)$.
$(ii)$ Chlorobenzene $(C_6H_5Cl)$ is highly resistant to nucleophilic substitution under normal conditions due to the partial double bond character of the $C-Cl$ bond. Therefore,it does not react with aqueous $NaOH$ at room temperature.

(B) The reaction is an electrophilic aromatic substitution (bromination) of $o$-toluic acid.
$o$-Toluic acid contains two groups attached to the benzene ring: a methyl group $(-CH_3)$ and a carboxylic acid group $(-COOH)$.
$1$. The $-CH_3$ group is ortho/para-directing.
$2$. The $-COOH$ group is meta-directing.
In $o$-toluic acid,the positions ortho to the $-CH_3$ group are either occupied by the $-COOH$ group or are sterically hindered. The position para to the $-CH_3$ group is the most reactive site for electrophilic substitution.
Therefore,the bromine atom enters the position para to the $-CH_3$ group,which corresponds to the meta position with respect to the $-COOH$ group.
The product is $5$-bromo-$2$-methylbenzoic acid.

(D) The groups $-COOH$,$-CN$,and $-COCH_3$ are electron-withdrawing groups that deactivate the benzene ring and are meta-directing due to the presence of electron-withdrawing atoms or multiple bonds attached directly to the ring.
In contrast,the $-NHCOCH_3$ group is an ortho/para directing group.
This is because the nitrogen atom has a lone pair of electrons that can be delocalized into the benzene ring through resonance,which increases the electron density at the ortho and para positions.

(C) The $-NO_2$ group present in nitrobenzene is a strong electron-withdrawing group and is meta-directing in nature.
Therefore,during the electrophilic aromatic substitution reaction (nitration),the incoming electrophile $(-NO_2^+)$ attacks the meta-position of the benzene ring.
Thus,the nitration of nitrobenzene yields $m-$dinitrobenzene as the major product.

(D) Nitrobenzene is not an explosive.
Nitro compounds like $Trinitrotoluene$ $(TNT)$,$Trinitrobenzene$ $(TNB)$,and $Picric$ $acid$ $(2,4,6-trinitrophenol)$ are highly explosive due to the presence of multiple nitro groups attached to the aromatic ring,which provide oxygen for rapid combustion.
$Nitrobenzene$ $(C_6H_5NO_2)$ is a stable compound and is not used as an explosive.

(C) The reaction of $p-$nitrobromobenzene with $NaNH_2$ proceeds via an elimination-addition mechanism.
In the first step,the strong base $NH_2^-$ abstracts a proton from the ortho position to the bromine atom,followed by the elimination of the bromide ion $(Br^-)$ to form a highly reactive intermediate known as a benzyne (specifically,a substituted benzyne or aryne intermediate).
Subsequently,the nucleophile $NH_3$ attacks the benzyne intermediate to form $p-$nitroaniline.
Therefore,the intermediate formed is a benzyne.

(B) The $-NO_2$ group is a strong electron-withdrawing group and is meta-directing. When nitrobenzene undergoes further nitration with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$,the incoming nitro group is directed to the meta-position,resulting in the formation of $m-$dinitrobenzene. The reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(NO_2)_2 + H_2O$.

(A) The reaction of nitrobenzene with solid $KOH$ under heating conditions is a nucleophilic aromatic substitution reaction.
In this reaction,the $OH^{-}$ ion acts as a nucleophile and attacks the benzene ring,replacing a hydrogen atom (or hydride ion equivalent) to form ortho- and para-nitrophenol.
Since the attacking species is a nucleophile $(OH^{-})$,it is classified as a nucleophilic substitution reaction.
Therefore,the correct option is $A$.

(D) The reaction of chlorobenzene with ammonia in the presence of $Cu_2O$ at $570 \ K$ is an ammonolysis reaction used for the preparation of aromatic amines.
The chemical equation is:
$2C_6H_5Cl + 2NH_3 \xrightarrow{Cu_2O, 570 \ K} 2C_6H_5NH_2 + Cu_2Cl_2 + H_2O$
Thus,the product formed is aniline.

(A) In aryl halides $(ArX)$,the halogen atom is attached to the benzene ring through a $C-X$ bond that has partial double bond character due to resonance.
This makes the $C-X$ bond shorter and stronger,making nucleophilic substitution by $CN^-$ ions extremely difficult under normal conditions.
Therefore,the reaction $ArX + KCN$ does not proceed to form $ArCN$.
Other reactions provided are standard methods:
$ArN_2^+Cl^- + CuCN \to ArCN + N_2 + CuCl$ (Sandmeyer reaction).
$ArCONH_2 + P_2O_5 \to ArCN + H_2O$ (Dehydration).
$ArCONH_2 + SOCl_2 \to ArCN + SO_2 + 2HCl$ (Dehydration).

(C) Nitrobenzene is a strongly deactivating group due to the $-NO_2$ group,which is electron-withdrawing by both inductive and resonance effects.
It directs incoming electrophiles to the meta-position in electrophilic aromatic substitution reactions.
However,the question asks for $o$- and $p$-substituted products.
Nitrobenzene does not undergo standard electrophilic substitution to give $o$- and $p$-products easily.
Among the given options,none of these reactions typically yield $o$- and $p$-products as major products for nitrobenzene.
However,if the question implies a specific context like nucleophilic substitution (which is also difficult) or if there is an error in the premise,it is important to note that nitrobenzene is meta-directing.
Given the standard chemistry curriculum,nitrobenzene undergoes nitration (Option $C$) to give $m$-dinitrobenzene.
Since the question asks for $o$- and $p$-products,it is technically incorrect as nitrobenzene is meta-directing.

(D) Nitrobenzene contains a strongly electron-withdrawing $-NO_2$ group,which deactivates the benzene ring towards electrophilic substitution.
Friedel-Crafts reactions (alkylation and acylation) do not occur with nitrobenzene because the $-NO_2$ group is a strong deactivating group and the catalyst $AlCl_3$ forms a complex with the nitrogen atom of the $-NO_2$ group,further deactivating the ring.

(D) Nucleophilic aromatic substitution reactions typically occur in electron-deficient aromatic rings,such as nitrobenzene $(C_6H_5NO_2)$.
When nitrobenzene reacts with solid potassium hydroxide $(KOH)$,it undergoes a nucleophilic substitution reaction to form ortho- or para-nitrophenol derivatives.
Option $D$ represents this type of reaction,where the $OH^-$ ion acts as a nucleophile attacking the electron-deficient ring.

(D) Aryl halides are less reactive towards nucleophilic substitution reactions compared to alkyl halides due to the following reasons:
$1$. Resonance effect: The lone pair of electrons on the halogen atom participates in conjugation with the benzene ring,resulting in partial double bond character in the $C-X$ bond,which makes it stronger and shorter.
$2$. Difference in hybridization: In aryl halides,the carbon atom attached to the halogen is $sp^2$ hybridized,which is more electronegative and holds the electron pair more tightly than the $sp^3$ hybridized carbon in alkyl halides.
$3$. Instability of phenyl cation: The phenyl cation formed by the self-ionization of aryl halides is not stabilized by resonance.

(B) When chlorobenzene reacts with chloral in the presence of concentrated $H_2SO_4$,it undergoes a condensation reaction to form $DDT$ (dichlorodiphenyltrichloroethane).
The reaction is as follows:
$2C_6H_5Cl + CCl_3CHO \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$

(B) The reaction of $p$-chlorotoluene with $KNH_2$ in liquid $NH_3$ proceeds via the formation of a benzyne intermediate.
$1$. $KNH_2$ acts as a strong base and removes an ortho-hydrogen from $p$-chlorotoluene to form a benzyne intermediate.
$2$. The nucleophilic attack by $NH_2^-$ can occur at either the carbon atom originally attached to the chlorine or the adjacent carbon.
$3$. Due to the inductive effect of the methyl group,the attack results in a mixture of $p$-toluidine and $m$-toluidine.
$4$. In this specific reaction,$m$-toluidine is often formed as a significant product due to the regioselectivity of the nucleophilic attack on the benzyne intermediate.

(C) The $-Cl$ group is an ortho/para directing group in electrophilic aromatic substitution reactions.
Since the reaction is an electrophilic bromination,the bromine atom will be substituted at both the ortho and para positions relative to the chlorine atom.
Therefore,both $o-bromochlorobenzene$ and $p-bromochlorobenzene$ are formed as products.

(D) The chlorine atom in $CH_2=CH-Cl$ (vinyl chloride) is least reactive towards nucleophilic substitution.
This is due to the resonance effect,where the lone pair of electrons on the chlorine atom participates in conjugation with the $C=C$ double bond.
This results in a partial double-bond character between the carbon and chlorine atoms,making the $C-Cl$ bond shorter and stronger,thus preventing the nucleophilic attack.

(D) The reaction of toluene with chlorine in the presence of iron (a Lewis acid catalyst) and in the absence of light is an electrophilic aromatic substitution reaction (specifically,chlorination of the benzene ring).
In this reaction,the methyl group $(-CH_3)$ on the benzene ring is an ortho/para-directing group.
Therefore,the chlorine atom will substitute the hydrogen atoms at the ortho and para positions of the benzene ring.
This results in the formation of a mixture of $o$-chlorotoluene and $p$-chlorotoluene.