Aromatic hydrocarbon Questions in English

(D) The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ (or $CuCl$) is known as the $Gattermann-Koch$ reaction.
This reaction is used to introduce a formyl group $(-CHO)$ into the benzene ring.
The overall reaction is: $C_6H_6 + CO + HCl \xrightarrow{anhydrous \ AlCl_3 / CuCl} C_6H_5CHO$.
Therefore,the product formed is benzaldehyde.

(C) The structure of ethylbenzene is $C_6H_5-CH_2-CH_3$.
Benzylic hydrogen atoms are the hydrogen atoms attached to the carbon atom that is directly bonded to the benzene ring.
In ethylbenzene,the carbon atom of the ethyl group attached to the benzene ring is the benzylic carbon.
This carbon is bonded to two hydrogen atoms.
Therefore,the number of benzylic hydrogen atoms is $2$.

(A) The reaction of benzene with hydrogen in the presence of a nickel catalyst at high temperature and high pressure is a catalytic hydrogenation reaction.
Benzene is an aromatic compound,and the hydrogenation of its three double bonds leads to the formation of a saturated six-membered ring.
The reaction is as follows:
$C_6H_6 + 3H_2 \xrightarrow{Ni, \Delta, P} C_6H_{12}$
Here,$C_6H_6$ is benzene and $C_6H_{12}$ is cyclohexane.
Therefore,the product $(A)$ is cyclohexane.

(D) The catalytic oxidation of benzene is an industrial process used to produce maleic anhydride.
When a mixture of benzene vapour and oxygen is passed over a $V_2O_5$ catalyst at $775\, K$,benzene undergoes controlled oxidation to form maleic anhydride.
The chemical reaction is: $2C_6H_6 + 9O_2 \xrightarrow{V_2O_5, 775\, K} 2C_4H_2O_3 + 4CO_2 + 4H_2O$.

(D) The given molecule is pyrrole.
In pyrrole,there are two double bonds,which contribute $2 \times 2 = 4$ $\pi$ electrons.
The nitrogen atom has a lone pair of electrons that participates in resonance to make the ring aromatic.
Thus,the lone pair on the nitrogen atom also contributes $2$ $\pi$ electrons.
Total number of $\pi$ electrons = $4$ (from double bonds) + $2$ (from lone pair) = $6$ $\pi$ electrons.

(B) To determine if a compound is aromatic,it must follow $H$ückel's rule: it must be cyclic,planar,fully conjugated,and have $(4n+2) \pi$ electrons.
$A$. Thiophene: Cyclic,planar,fully conjugated,and has $6 \pi$ electrons ($4$ from double bonds + $2$ from lone pair on $S$). It is aromatic.
$B$. $1,3$-Cyclohexadiene: It is cyclic and planar,but it is not fully conjugated (the $sp^3$ carbons interrupt the conjugation). It has $4 \pi$ electrons. It is not aromatic.
$C$. Tropylium cation: Cyclic,planar,fully conjugated,and has $6 \pi$ electrons. It is aromatic.
$D$. Furan: Cyclic,planar,fully conjugated,and has $6 \pi$ electrons ($4$ from double bonds + $2$ from lone pair on $O$). It is aromatic.
Therefore,$1,3$-Cyclohexadiene is not an aromatic compound.

(B) Sulphonation is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring. Alkyl groups are electron-donating groups ($+I$ effect and hyperconjugation) that increase the electron density of the ring,thereby activating it towards electrophilic substitution.
The activating power of alkyl groups follows the order: $CH_3- > CH_3CH_2- > (CH_3)_2CH- > (CH_3)_3C-$.
This is because the primary alkyl group (methyl) has the most effective hyperconjugation and the least steric hindrance compared to the bulky tertiary alkyl group (tert-butyl). Therefore,toluene $(CH_3-C_6H_5)$ will undergo sulphonation at the fastest rate.

(A) Sulphonation is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the aromatic ring.
Electron-donating groups (like alkyl groups) increase the electron density of the ring,making it more reactive towards electrophiles.
$m$-Xylene has two methyl groups at the $1$ and $3$ positions. These groups are meta-directing relative to each other and reinforce the activation of the ring at the $4$-position.
In $m$-xylene,the two methyl groups provide a greater inductive effect and hyperconjugation compared to toluene (one methyl group) and $o$-/$p$-xylene (where steric hindrance and electronic effects are less favorable for the specific electrophilic attack).
Therefore,$m$-xylene is the most reactive among the given options.

(D) Sulphonation is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring. Electron-donating groups increase the electron density and thus increase the rate of reaction,while electron-withdrawing groups decrease the electron density and decrease the rate of reaction.
$1$. $C_6H_5NO_2$: The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which strongly deactivates the benzene ring.
$2$. $C_6H_5CH_2NO_2$: The $-CH_2NO_2$ group is electron-withdrawing due to the $-I$ effect of the $-NO_2$ group,which deactivates the ring.
$3$. $C_6H_5CH_2CH_2NO_2$: The $-CH_2CH_2NO_2$ group is also electron-withdrawing due to the $-I$ effect of the $-NO_2$ group,but the effect is weaker than in the previous cases because the $-NO_2$ group is further from the ring.
$4$. $C_6H_6$ (Benzene): Benzene has no substituent attached to it. Compared to the other substituted benzenes where the substituents are electron-withdrawing,benzene has a higher electron density in the ring.
Therefore,$C_6H_6$ undergoes sulphonation at the fastest rate among the given options.

(C) The nitration of dimethylbenzene (xylene) depends on the symmetry of the molecule and the directing effects of the two methyl groups.
$1$. $1,2-$Dimethylbenzene ($o-$xylene): The two methyl groups are at positions $1$ and $2$. Nitration can occur at positions $3$ or $4$,leading to two different isomers.
$2$. $1,3-$Dimethylbenzene ($m-$xylene): The methyl groups are at positions $1$ and $3$. Nitration can occur at positions $2, 4,$ or $5$,leading to multiple isomers.
$3$. $1,4-$Dimethylbenzene ($p-$xylene): The methyl groups are at positions $1$ and $4$. Due to the high symmetry of the $p-$xylene molecule,all four available ring positions ($2, 3, 5,$ and $6$) are equivalent. Therefore,nitration at any of these positions results in the formation of only one product,$2-$nitro-$1,4-$dimethylbenzene.

(B) In $p$-methoxytoluene,there are two substituents on the benzene ring: a methyl group $(-CH_3)$ and a methoxy group $(-OCH_3)$.
Both groups are ortho/para directing.
The methoxy group $(-OCH_3)$ is a much stronger activating group than the methyl group $(-CH_3)$ due to the resonance effect ($+R$ effect) of the oxygen lone pair.
Therefore,the electrophilic substitution (nitration) will be directed by the $-OCH_3$ group.
The positions ortho to the $-OCH_3$ group are positions $2$ and $6$ relative to the methoxy group,which are also ortho to the methyl group.
Since both ortho positions are equivalent,nitration occurs at one of these positions to form $2$-nitro-$4$-methoxytoluene (or $3$-nitro-$4$-methylanisole).
Comparing the options,the structure in option $B$ represents the nitration ortho to the methoxy group.

(A) The number of isomers formed upon nitration depends on the number of non-equivalent positions available in the benzene ring for electrophilic substitution.
For $o$-dibromobenzene ($1,2$-dibromobenzene),there are $3$ non-equivalent positions available for nitration,yielding $3$ isomers.
For $m$-dibromobenzene ($1,3$-dibromobenzene),there are $4$ non-equivalent positions available,yielding $4$ isomers.
For $p$-dibromobenzene ($1,4$-dibromobenzene),there is only $1$ non-equivalent position available,yielding $1$ isomer.
Since the question states that $3$ isomers were obtained,the original compound must be $o$-dibromobenzene.

(B) When two hydrogen atoms in a benzene ring $(C_6H_6)$ are replaced by two substituents,three distinct isomers are formed based on the relative positions of the substituents.
These positions are known as:
$1.$ Ortho $(o-)$ position: $1,2-$ substitution.
$2.$ Meta $(m-)$ position: $1,3-$ substitution.
$3.$ Para $(p-)$ position: $1,4-$ substitution.
Therefore,there are $3$ possible disubstituted products of benzene.

(B) In an electrophilic aromatic substitution reaction,the reaction proceeds through the formation of a carbocation intermediate,also known as the arenium ion or sigma complex.
This intermediate is represented by species $(III)$ in the given reaction mechanism.
The formation of this intermediate involves the loss of aromaticity of the benzene ring,which makes it highly unstable and high in energy compared to the reactants and the final product.
Therefore,the arenium ion (species $(III)$) has the maximum enthalpy (potential energy) among the species shown in the reaction coordinate diagram.

(A) In electrophilic aromatic substitution,the rate-determining step is the formation of the sigma complex (arenium ion) by the attack of the electrophile on the aromatic ring.
Since the $C-H$ (or $C-D$ or $C-T$) bond is not broken in the rate-determining step,there is no primary kinetic isotope effect.
Therefore,the rates of nitration for benzene $(C_6H_6)$,hexadeuterobenzene $(C_6D_6)$,and hexatritiobenzene $(C_6T_6)$ are equal.
Thus,$k_{C_6H_6} = k_{C_6D_6} = k_{C_6T_6}$.

(A) In electrophilic aromatic substitution,the rate-determining step for sulphonation is the removal of the proton (or deuteron/triton) from the sigma complex intermediate.
Since the $C-H$ bond is weaker than the $C-D$ bond,and the $C-D$ bond is weaker than the $C-T$ bond due to the primary kinetic isotope effect,the reaction rate is fastest for benzene $(C_6H_6)$ and slowest for tritiated benzene $(C_6T_6)$.
Therefore,the correct order is $k_{C_6H_6} > k_{C_6D_6} > k_{C_6T_6}$.

(B) compound is aromatic if it follows $H$ückel's rule: it must be cyclic,planar,fully conjugated,and contain $(4n+2) \pi$ electrons,where $n$ is an integer $(0, 1, 2, ...)$.
$1$. Option $A$: This is the pyrrolium cation. It has $4 \pi$ electrons (from two double bonds). It is anti-aromatic.
$2$. Option $B$: This is an imidazolium-like cation. The nitrogen atom with the lone pair participates in the aromatic sextet. The ring has $6 \pi$ electrons ($4$ from two double bonds and $2$ from the lone pair on the nitrogen atom). It is planar and cyclic. Thus,it is aromatic.
$3$. Option $C$: This is a pyrylium-like cation. It is not fully conjugated in a way that satisfies the $(4n+2)$ rule for the entire ring system.
$4$. Option $D$: Styrene is a benzene derivative. While the benzene ring is aromatic,the question asks for the compound as a whole. However,in the context of these specific choices,the imidazolium-type cation $(B)$ is a classic example of a heterocyclic aromatic system.

(C) The rate of electrophilic aromatic substitution depends on the electron density of the benzene ring. Electron-donating groups (EDGs) increase the electron density and thus increase the rate of substitution.
In the given compounds:
- Compound $(IV)$ has two $-NH$ groups,which are strong electron-donating groups due to the resonance effect ($+R$ effect) of the lone pair on nitrogen.
- Compound $(II)$ has one $-NH$ group and one $-O$ group. The $-NH$ group is a stronger electron donor than the $-O$ group.
- Compound $(I)$ has two $-O$ groups,which are electron-donating but less effective than $-NH$ groups.
- Compound $(III)$ has only alkyl groups,which are weak electron donors via the inductive effect ($+I$ effect).
Therefore,the electron-donating ability follows the order: two $-NH$ groups $(IV)$ > one $-NH$ and one $-O$ group $(II)$ > two $-O$ groups $(I)$ > alkyl groups $(III)$.
Thus,the decreasing order of the rate of electrophilic aromatic substitution is $IV > II > I > III$.

(B) In the nitration of benzene,$H_2SO_4$ acts as a strong acid and $HNO_3$ acts as a base.
The reaction between them is:
$H_2SO_4 + HO-NO_2 \rightleftharpoons H_2SO_4^- + H_2O^+-NO_2$
The $H_2O^+-NO_2$ then loses water to form the nitronium ion $(NO_2^+)$,which is the electrophile.

(A) The reaction is a Friedel-Crafts acylation of a substituted benzene ring. The substrate is $1,1,3$-trimethylindane (or a similar derivative). The incoming electrophile is the acetyl cation,$CH_3CO^+$,generated from $CH_3COCl$ and $AlCl_3$. The substitution occurs at the position that is most activated and least sterically hindered. In the given structure,the position ortho to the alkyl group on the benzene ring is the most favorable site for electrophilic aromatic substitution due to the activating effect of the alkyl groups and the relative lack of steric hindrance compared to other positions. Therefore,the acetyl group attaches to the carbon atom that is ortho to the $tert$-butyl group and meta to the other alkyl group,which corresponds to the structure shown in option $A$.

(B) The starting material is $n$-hexane $(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3)$.
When $n$-hexane is heated with $Al_2O_3$ (aromatization/dehydrocyclization),it forms benzene.
Reaction: $C_6H_{14} \xrightarrow{Al_2O_3 / \Delta} C_6H_6$ (Benzene).
Thus,$(A)$ is benzene.
Next,the reaction of benzene with $CH_3Cl$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation.
Reaction: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5-CH_3$ (Toluene).
However,looking at the options provided,there seems to be a mismatch in the reaction sequence or the product structure. If the starting material was $n$-heptane,it would form toluene,and further alkylation would yield xylenes. Given the options,the question likely implies the formation of a dimethylbenzene derivative. Assuming the starting material is $n$-heptane,the aromatization yields toluene,and subsequent Friedel-Crafts alkylation with $CH_3Cl$ yields a mixture of $o$-,$m$-,and $p$-xylenes,with the $p$-isomer ($1$,$4$-dimethylbenzene) often being the major product due to steric factors.

(B) The reactivity of an aromatic ring towards electrophilic aromatic substitution $(EAS)$ depends on the nature of the substituents attached to it. Electron-donating groups $(EDG)$ increase reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$1$. Ring $A$ is attached to an $-N(CH_3)-$ group,which is a strongly activating group due to the strong $+M$ effect of the nitrogen lone pair.
$2$. Ring $B$ is attached to an alkyl chain,which acts as a weakly activating group due to the $+I$ effect and hyperconjugation.
$3$. Ring $C$ is attached to a carbonyl group $(-C=O)$,which is a strongly deactivating group due to the $-M$ and $-I$ effects.
$4$. Ring $D$ is attached to an oxygen atom of an ester group $(-O-C=O-)$,which acts as a moderately activating group due to the $+M$ effect of the oxygen lone pair,although it is also electron-withdrawing by the $-I$ effect.
Comparing these,the order of reactivity is $C$ (deactivated) $< B$ (weakly activated) $< D$ (moderately activated) $< A$ (strongly activated). Thus,the correct order is $C < B < D < A$.

(A) The rate of $EAS$ depends on the electron density of the benzene ring. Electron-donating groups increase the rate,while electron-withdrawing groups decrease it.
$1$. $-CH_3$ (in $C$) is an electron-donating group ($+I$ and hyperconjugation),which activates the ring.
$2$. Benzene $(D)$ has no substituent.
$3$. $-Cl$ (in $B$) is a deactivating group due to its strong $-I$ effect,although it is ortho/para directing.
$4$. $-NO_2$ (in $A$) is a strongly deactivating group due to its strong $-I$ and $-M$ effects.
Therefore,the order of reactivity is $C$ (activated) $> D$ (benzene) $> B$ (weakly deactivated) $> A$ (strongly deactivated).
The correct order is $C > D > B > A$.

(A) To determine the aromaticity of a compound,we use $H$ückel's rule,which states that a planar,cyclic,fully conjugated system with $(4n+2) \pi$ electrons is aromatic.
$A$. Pyrrole cation $(C_4H_6N^+)$: The nitrogen atom is $sp^3$ hybridized because it is bonded to four atoms (two carbons and two hydrogens). Since it is not $sp^2$ hybridized,the ring is not fully conjugated and thus it is nonaromatic.
$B$. Benzene $(C_6H_6)$: It is a planar,cyclic,fully conjugated system with $6 \pi$ electrons $(n=1)$,so it is aromatic.
$C$. Cyclopentadienyl cation $(C_5H_5^+)$: It is a planar,cyclic,fully conjugated system with $4 \pi$ electrons ($n=1$ in $4n$ rule),so it is antiaromatic.
$D$. Cyclopropenyl cation $(C_3H_3^+)$: It is a planar,cyclic,fully conjugated system with $2 \pi$ electrons $(n=0)$,so it is aromatic.
Therefore,the pyrrole cation is nonaromatic.

(A) $H$ückel's rule states that for a planar,cyclic,conjugated system to be aromatic,it must contain $(4n+2) \pi$ electrons,where $n$ is an integer $(n = 0, 1, 2, ...)$.
Thus,the correct representation is $(4n+2) \pi$ electrons.

(A) The reactivity toward electrophilic aromatic substitution is determined by the electron density on the benzene ring.
Groups that donate electrons to the ring increase its electron density,making it more reactive toward electrophiles.
$Anisole$ $(C_6H_5OCH_3)$ contains a methoxy group $(-OCH_3)$,which is a strong electron-donating group due to the resonance effect ($+R$ effect).
$Benzene$ has no substituent.
$Nitrobenzene$ $(C_6H_5NO_2)$ contains a nitro group $(-NO_2)$,which is a strong electron-withdrawing group ($-I$ and $-R$ effects).
$Chlorobenzene$ $(C_6H_5Cl)$ contains a chlorine atom,which is deactivating due to its strong inductive effect ($-I$ effect) despite its resonance donation ($+R$ effect).
Therefore,$Anisole$ is the most reactive among the given options.

(A) The reactivity of a benzene ring towards electrophilic aromatic substitution depends on the electron density of the ring. Groups that donate electron density to the ring (activating groups) increase its reactivity,while groups that withdraw electron density (deactivating groups) decrease its reactivity.
$1$. The $-OH$ group in phenol is a strong activating group due to its $+M$ (mesomeric) effect,which significantly increases the electron density on the benzene ring.
$2$. The $-CH_3$ group in toluene is a weakly activating group due to the $+I$ (inductive) effect and hyperconjugation.
$3$. The $-NO_2$ group in nitrobenzene is a strongly deactivating group due to its $-M$ and $-I$ effects.
$4$. The $-CHO$ group in benzaldehyde is a deactivating group due to its $-M$ and $-I$ effects.
Since the $-OH$ group is the strongest activating group among the given options,phenol is the most reactive towards an electrophile.

(B) compound is defined as homocyclic if its ring consists only of carbon atoms. It is aromatic if it follows $H$ückel's rule ($4n+2$ $\pi$ electrons) and is planar/cyclic. It is unsaturated if it contains double or triple bonds.
$A$. Cyclohexane $(C_6H_{12})$ is homocyclic and saturated (no double bonds).
$B$. Benzene $(C_6H_6)$ is homocyclic (only carbon atoms in the ring),aromatic (follows $H$ückel's rule with $6$ $\pi$ electrons),and unsaturated (contains three double bonds).
$C$. Tetrahydropyran is a heterocyclic compound because it contains an oxygen atom in the ring.
$D$. $1,3-$Butadiene is an open-chain (acyclic) unsaturated hydrocarbon,not a cyclic compound.
Therefore,the correct answer is Benzene.

(D) To determine if a compound is aromatic,it must satisfy $H$ückel's rule ($4n+2$ $\pi$ electrons),be planar,and have a continuous cyclic conjugation.
$1$. Cyclobutadiene: $4$ $\pi$ electrons (anti-aromatic).
$2$. Cyclopentadienyl anion: $6$ $\pi$ electrons (aromatic).
$3$. Azulene: $10$ $\pi$ electrons (aromatic).
$4$. Cyclooctatetraene: Non-planar,$8$ $\pi$ electrons (non-aromatic).
$5$. Cyclopropenone: $2$ $\pi$ electrons (aromatic).
$6$. Furan: $6$ $\pi$ electrons (aromatic).
$7$. Cyclobutene: Not fully conjugated (non-aromatic).
$8$. Fulvene derivative: $6$ $\pi$ electrons (aromatic).
The aromatic compounds are: Cyclopentadienyl anion,Azulene,Cyclopropenone,Furan,and the Fulvene derivative.
Total number of aromatic species = $5$.

(A) To be aromatic,a compound must follow $H$ückel's rule ($4n+2$ $\pi$ electrons),be planar,cyclic,and fully conjugated.
$(I)$ Cyclopentadiene: It is non-aromatic because it has an $sp^3$ hybridized carbon atom.
$(II)$ Cyclooctatetraene: It is non-aromatic (tub-shaped,non-planar) and has $8$ $\pi$ electrons (anti-aromatic if planar,but it is non-planar).
$(III)$ Borazine $(B_3N_3H_6)$: It is isoelectronic with benzene,planar,cyclic,and has $6$ $\pi$ electrons. It is aromatic.
$(IV)$ $1,4-$diboracyclohexa$-2,5-$diene: It has $4$ $\pi$ electrons and is not fully conjugated in a way that satisfies $H$ückel's rule for aromaticity.
Thus,only structure $(III)$ is aromatic.
The correct option is $A$.

(C) Friedel-Crafts alkylation of benzene with ethyl chloride gives ethylbenzene. This is correct.
$(b)$ Friedel-Crafts alkylation of benzene with isopropyl chloride gives isopropylbenzene (cumene). This is correct.
$(c)$ Electrophilic addition of propene to benzene in the presence of $H^+$ gives isopropylbenzene. This is correct.
$(d)$ Ozonolysis of $o$-xylene ($1,2$-dimethylbenzene) yields glyoxal,methylglyoxal,and dimethylglyoxal. This is correct.
Therefore,all the given reactions are correctly matched.

(B) The process of converting an alkane with $6$ or more carbon atoms into an aromatic hydrocarbon by heating it with a catalyst like $Cr_2O_3$,$V_2O_5$,or $Mo_2O_3$ supported over $Al_2O_3$ at high temperature and pressure is known as aromatization or reforming.
When $n-$heptane $(C_7H_{16})$ is passed over $(Al_2O_3 + Cr_2O_3)$ catalyst at $773\ K$,it undergoes cyclization and dehydrogenation to form toluene $(C_6H_5CH_3)$.

(C) For any compound to be aromatic,all carbon atoms must be $sp^2$ hybridized,the molecule must be planar,and it must follow $H$ückel's rule.
$H$ückel's Rule $\Rightarrow 4n + 2 = \text{number of delocalized } \pi\text{-electrons } (n = 0, 1, 2, 3, \dots)$
$(i)$ Naphthalene: $10 \pi\text{-electrons}$ $\Rightarrow 4n + 2 = 10$ $\Rightarrow n = 2$ (Aromatic)
$(ii)$ Anthracene: $14 \pi\text{-electrons}$ $\Rightarrow 4n + 2 = 14$ $\Rightarrow n = 3$ (Aromatic)
$(iii)$ Phenanthrene: $14 \pi\text{-electrons}$ $\Rightarrow 4n + 2 = 14$ $\Rightarrow n = 3$ (Aromatic)
All three compounds are planar,have $sp^2$ hybridized carbon atoms,and follow $H$ückel's rule.
Therefore,all are aromatic compounds.
Hence,the correct answer is option $C$.

(C) In structure $(i)$,which is $1,3$-dimethylbenzene,there are two methyl groups attached to the benzene ring. Each methyl group is attached to an $sp^2$ hybridized carbon of the ring,making the carbon of the methyl group an $\alpha$-carbon. Each methyl group has $3$ $\alpha$-hydrogens. Thus,total $\alpha$-hydrogens $= 3 + 3 = 6$.
In structure $(ii)$,which is $1,3$-diethylbenzene,there are two ethyl groups attached to the benzene ring. The carbon of the ethyl group directly attached to the benzene ring is the $\alpha$-carbon. Each $\alpha$-carbon in an ethyl group has $2$ $\alpha$-hydrogens. Thus,total $\alpha$-hydrogens $= 2 + 2 = 4$.
The sum of $\alpha$-hydrogens $= 6 + 4 = 10$.

(A) The reactivity towards electrophilic substitution $(ESR)$ depends on the electron density in the benzene ring.
Electron donating groups $(EDG)$ increase the electron density and thus increase reactivity,while electron withdrawing groups $(EWG)$ decrease the electron density and decrease reactivity.
- $(III)$ Anisole $(-OCH_3)$: The $-OCH_3$ group is a strong $EDG$ due to its $+M$ effect.
- $(I)$ Toluene $(-CH_3)$: The $-CH_3$ group is a weak $EDG$ due to $+I$ and hyperconjugation effects.
- $(II)$ Benzene: This is the reference compound with no substituents.
- $(IV)$ Trifluoromethylbenzene $(-CF_3)$: The $-CF_3$ group is a strong $EWG$ due to its $-I$ effect.
Therefore,the order of reactivity is: $(III) > (I) > (II) > (IV)$.

(B) The reactions are as follows:
$(i)$ Phenol reacts with zinc dust $(Zn)$ upon distillation to form benzene: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$. Thus,$A = Zn$.
(ii) Benzenesulfonic acid undergoes desulfonation with superheated steam $(H_2O)$ at $150-200^{\circ}C$ in the presence of $HCl$ to form benzene: $C_6H_5SO_3H + H_2O \rightarrow C_6H_6 + H_2SO_4$. Thus,$B = H_2O \ (\text{steam})$.
(iii) Chlorobenzene undergoes catalytic hydrogenation using $H_2$ gas in the presence of $Ni-Al$ alloy (Raney nickel type catalyst) to form benzene: $C_6H_5Cl + H_2 \xrightarrow{Ni-Al} C_6H_6 + HCl$. Thus,$C = H_2 \ (Ni-Al \ \text{alloy})$.
Therefore,the correct option is $B$.

(A) The reaction of benzene with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
First,benzene reacts with $CH_2Cl_2$ to form benzyl chloride $(C_6H_5CH_2Cl)$ as the intermediate product '$A$'.
Then,this intermediate '$A$' (benzyl chloride) reacts with another molecule of benzene in the presence of $AlCl_3$ to form diphenylmethane as the final product.
Therefore,the intermediate '$A$' is benzyl chloride.

(C) In nitrobenzene,the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
When an electrophile $E^{+}$ attacks the benzene ring,the positive charge in the $\sigma$-complex is delocalized over the ring.
For ortho and para attacks,one of the resonance structures places the positive charge directly on the carbon atom attached to the electron-withdrawing $-NO_2$ group,which is highly destabilizing.
For meta attack,the positive charge is never placed on the carbon atom attached to the $-NO_2$ group.
Therefore,the $\sigma$-complex formed by meta attack is the most stable (lowest energy) among the three,which explains why the $-NO_2$ group is meta-directing in electrophilic substitution reactions $(ESR)$.

(B) The reaction involves the protonation of the exocyclic double bond of $1$-methylenecyclohexa-$2,5$-diene.
This leads to the formation of a resonance-stabilized carbocation intermediate.
Subsequent loss of a proton $(H^+)$ from the $sp^3$ hybridized carbon results in the formation of the aromatic compound,toluene $(C_6H_5CH_3)$.

(B) The reaction involves the treatment of $1,4-\text{dibromocyclohexa-1,4-diene}$ with $2Na$.
Sodium $(Na)$ acts as a reducing agent,removing the two bromine atoms.
This process results in the formation of two radical centers at the $1$ and $4$ positions.
These radicals then undergo rearrangement to form a fully conjugated aromatic system,which is benzene $(C_6H_6)$.

(A) The reaction of benzene with $Cl_2$ in the presence of a Lewis acid like $FeCl_3$ is an electrophilic aromatic substitution reaction (chlorination),which yields chlorobenzene $(C_6H_5Cl)$ as product $X$.
The reaction of benzene with concentrated $H_2SO_4$ and $SO_3$ (fuming sulfuric acid or oleum) is an electrophilic aromatic substitution reaction (sulfonation),which yields benzenesulfonic acid $(C_6H_5SO_3H)$ as product $Y$.
Therefore,$X = C_6H_5Cl$ and $Y = C_6H_5SO_3H$.

(A) In the nitration of benzene,the reaction between concentrated $HNO_3$ and concentrated $H_2SO_4$ occurs as follows:
$HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^-$
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$
Here,$HNO_3$ accepts a proton $(H^+)$ from $H_2SO_4$ to form $H_2NO_3^+$. According to the Bronsted-Lowry theory,a substance that accepts a proton acts as a Bronsted base.

(B) For a compound to be aromatic,it must follow $H$ückel's rule,which states that the compound must be cyclic,planar,fully conjugated,and contain $(4n + 2) \pi$ electrons,where $n$ is an integer $(n = 0, 1, 2, ...)$.
$A$. Cyclopentadienyl cation: It has $4 \pi$ electrons. It is anti-aromatic.
$B$. Cycloheptatrienyl cation (Tropylium ion): It is cyclic,planar,fully conjugated,and has $6 \pi$ electrons $(n = 1)$. Thus,it is aromatic.
$C$. Pyrazine: It is a heterocyclic aromatic compound with $6 \pi$ electrons.
$D$. Cyclooctatetraene: It is non-planar (tub-shaped) and has $8 \pi$ electrons. It is non-aromatic.
Note: Both $B$ and $C$ are aromatic. However,in the context of typical multiple-choice questions of this type,the tropylium cation is a classic example of an aromatic ion.

(B) The reaction involves the oxidation of alkyl groups attached to a benzene ring using $KMnO_4$ in an acidic medium under heating.
Any alkyl group attached to the benzene ring,provided it has at least one benzylic hydrogen atom,is oxidized to a carboxylic acid group $(-COOH)$.
In the given reactant,$1$-ethyl-$3$-methylbenzene,both the ethyl group $(-CH_2CH_3)$ and the methyl group $(-CH_3)$ have benzylic hydrogen atoms.
Therefore,both groups are oxidized to $-COOH$ groups,resulting in benzene-$1,3$-dicarboxylic acid (isophthalic acid).

(A) The reactivity of benzene derivatives towards $ESR$ depends on the electron density of the ring.
Electron-donating groups $(EDG)$ increase electron density and reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$(1)$ Benzene: Reference compound.
$(2)$ Toluene: Contains a $-CH_3$ group,which is an $EDG$ due to $+I$ and hyperconjugation effects,increasing reactivity.
$(3)$ Chlorobenzene: Contains a $-Cl$ group,which is deactivating due to its strong $-I$ effect,although it is ortho/para directing.
$(4)$ Nitrobenzene: Contains a $-NO_2$ group,which is a strong $EWG$ due to $-I$ and $-M$ effects,significantly decreasing reactivity.
Therefore,the order of reactivity is: $Toluene (2) > Benzene (1) > Chlorobenzene (3) > Nitrobenzene (4)$.

(A) The reactivity of an aromatic ring towards an electrophilic substitution reaction depends on the electron density of the ring.
$1$. Phenol $(d)$: The $-OH$ group is a strongly activating group due to its strong $+M$ (mesomeric) effect,which significantly increases the electron density on the ring.
$2$. Toluene $(b)$: The $-CH_3$ group is an activating group due to $+I$ (inductive) effect and hyperconjugation,which increases the electron density on the ring,but less than the $-OH$ group.
$3$. Benzene $(a)$: This is the reference compound with no substituents.
$4$. Chlorobenzene $(c)$: The $-Cl$ group is a deactivating group due to its strong $-I$ effect,which outweighs its $+M$ effect,thereby decreasing the electron density on the ring.
Therefore,the order of decreasing reactivity is $d > b > a > c$.

(C) To determine aromaticity,a compound must be cyclic,planar,fully conjugated,and follow $H$ückel's rule ($4n+2$ $\pi$ electrons).
$(I)$ Tropone: It is aromatic because the oxygen atom pulls electrons,creating a $6$ $\pi$ electron system in the seven-membered ring.
$(II)$ Benzimidazole: It is a bicyclic aromatic compound with $10$ $\pi$ electrons.
$(III)$ Isothiazole: It is a five-membered heterocyclic aromatic compound with $6$ $\pi$ electrons.
$(IV)$ $2H$-Pyran: It is non-aromatic because the $sp^3$ hybridized carbon atom breaks the conjugation.
$(V)$ $1,4$-Naphthohydroquinone (or $1,4$-dihydroxynaphthalene): The naphthalene core is aromatic ($10$ $\pi$ electrons).
$(VI)$ Cyclooctatetraene: It is non-aromatic (tub-shaped,non-planar) with $8$ $\pi$ electrons.
Thus,compounds $(I, II, III, V)$ are aromatic.

(D) To identify aromatic compounds,we apply $H$ückel's rule: the compound must be cyclic,planar,fully conjugated,and contain $(4n+2) \pi$ electrons.
$1$. Cyclobutadiene: $4 \pi$ electrons (anti-aromatic).
$2$. Cyclopentadienyl anion: $6 \pi$ electrons (aromatic).
$3$. Azulene: $10 \pi$ electrons (aromatic).
$4$. Cyclooctatetraene: Non-planar,$8 \pi$ electrons (non-aromatic).
$5$. Cyclopropenone: $2 \pi$ electrons in the ring (aromatic).
$6$. Furan: $6 \pi$ electrons (aromatic).
$7$. Cyclobutadiene dication: $2 \pi$ electrons (aromatic).
$8$. Fulvalene derivative: Not fully conjugated across the system (non-aromatic).
The aromatic compounds are: Cyclopentadienyl anion,Azulene,Cyclopropenone,Furan,and Cyclobutadiene dication.
Total number of aromatic compounds = $5$.

(C) $1$. Wurtz reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with $Na$ gives $n$-hexane $(A)$: $2CH_3CH_2CH_2Br + 2Na \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3 + 2NaBr$.
$2$. Aromatization of $n$-hexane over $Cr_2O_3/Al_2O_3$ at $773 \ K$ gives benzene $(B)$: $C_6H_{14} \rightarrow C_6H_6 + 4H_2$.
$3$. Benzene $(B)$ reacts with $Cl_2$ in the presence of $UV$ light (addition reaction) to form benzene hexachloride $(C)$: $C_6H_6 + 3Cl_2 \xrightarrow{UV} C_6H_6Cl_6$.
$4$. Benzene $(B)$ undergoes Friedel-Crafts alkylation with $CH_3Cl/AlCl_3$ to form toluene $(D)$: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$.
$5$. Toluene $(D)$ reacts with $Cl_2/hv$ (free radical substitution) to form benzylidyne trichloride $(E)$: $C_6H_5CH_3 + 3Cl_2 \xrightarrow{hv} C_6H_5CCl_3 + 3HCl$.

(D) $ESR$ (Electrophilic Substitution Reaction) rate depends on the electron density of the benzene ring. Groups that donate electrons to the ring via resonance ($+R$ effect) or induction ($+I$ effect) increase the electron density,making the ring more reactive towards electrophiles.
$1$. $N,N-Dimethylaniline$ ($-N(CH_3)_2$ group): The nitrogen atom has a lone pair of electrons which it donates to the benzene ring via strong $+R$ effect. This significantly increases the electron density on the ring,making it the most reactive.
$2$. $Toluene$ ($-CH_3$ group): The methyl group donates electrons via hyperconjugation and $+I$ effect,which is weaker than the $+R$ effect of the $-N(CH_3)_2$ group.
$3$. $Chlorobenzene$ ($-Cl$ group): The chlorine atom has a $-I$ effect (electron-withdrawing) which dominates over its $+R$ effect,making the ring less reactive than benzene.
$4$. $Benzaldehyde$ ($-CHO$ group): The aldehyde group is a strong electron-withdrawing group ($-R$ and $-I$ effects),which significantly decreases the electron density on the ring,making it the least reactive.
Therefore,$N,N-Dimethylaniline$ gives $ESR$ at the fastest rate.