Aromatic hydrocarbon Questions in English

(A) The benzyl radical $(C_6H_5CH_2^{\bullet})$ consists of a benzene ring attached to a $CH_2^{\bullet}$ group.
The benzene ring has $6$ carbon atoms,each contributing one $p$-orbital to the delocalized $\pi$-system,containing $6 \pi$-electrons.
The exocyclic carbon atom $(CH_2^{\bullet})$ is $sp^2$ hybridized and possesses one $p$-orbital containing one unpaired electron.
Therefore,the total number of $p$-orbitals involved in conjugation is $6 + 1 = 7$.
The total number of electrons in these $p$-orbitals is $6$ (from the ring) $+ 1$ (unpaired electron) $= 7$.
However,the aromaticity is determined by the $6 \pi$-electrons in the ring,and the radical is stabilized by resonance with the ring. The question asks for the characteristics that make it aromatic; it has $7$ $p$-orbitals in total (forming a conjugated system) and $7$ electrons in these $p$-orbitals (including the radical electron).

(B) The deactivating strength of a substituent depends on its electron-withdrawing power through inductive $(-I)$ and mesomeric $(-M)$ effects.
Among the given groups,the $-NO_2$ group exerts the strongest electron-withdrawing effect due to both a strong $-I$ effect and a strong $-M$ effect,making it the most deactivating group towards electrophilic aromatic substitution.

(C) The presence of an electron-releasing group like $-CH_3$,$-OH$,etc.,increases the electron density at the $o/p$ positions and thus makes the benzene ring more reactive towards electrophiles.
On the other hand,electron-withdrawing groups like $-COOH$,$-NO_2$,etc.,reduce the electron density of the benzene ring,thereby decreasing its reactivity towards electrophilic substitution.
Comparing the given compounds:
$1$. $-NO_2$ (in nitrobenzene) is a strong electron-withdrawing group.
$2$. $-COOH$ (in benzoic acid) is an electron-withdrawing group.
$3$. $-H$ (in benzene) is the reference.
$4$. $-CH_3$ (in toluene) is an electron-releasing group due to the $+I$ effect and hyperconjugation.
Therefore,the order of reactivity towards electrophilic nitration is: $\text{nitrobenzene} < \text{benzoic acid} < \text{benzene} < \text{toluene}$.
Thus,toluene is the most reactive.

(C) The reactivity towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
$1$. Phenol $(C_6H_5OH)$ has an $-OH$ group,which is a strong activating group due to its strong $+M$ (mesomeric) effect,making it the most reactive.
$2$. Toluene $(C_6H_5CH_3)$ has a $-CH_3$ group,which is an activating group due to $+I$ (inductive) and hyperconjugation effects,making it more reactive than benzene.
$3$. Benzene $(C_6H_6)$ is the reference compound.
$4$. Chlorobenzene $(C_6H_5Cl)$ has a $-Cl$ group,which is deactivating due to its strong $-I$ effect,although it is ortho/para-directing due to the $+M$ effect. Overall,it is less reactive than benzene.
Thus,the correct order of decreasing reactivity is: $(iv) > (ii) > (i) > (iii)$.

(D) In pyrrole,the lone pair of electrons on the nitrogen atom is involved in the aromatic sextet. Due to the resonance effect,the electron density is delocalized over the ring. When an electrophile attacks,the intermediate carbocation (sigma complex) is more stable when the electrophile attacks at the $C_{2}$ or $C_{5}$ positions because the positive charge can be delocalized over more atoms,including the nitrogen atom. Therefore,the electron density is maximum at the $C_{2}$ and $C_{5}$ positions.

(C) The reaction involves the electrophilic addition of a proton $(H^+)$ from $HF$ to cyclohexene to form a cyclohexyl carbocation.
This carbocation then acts as an electrophile and undergoes Electrophilic Substitution Reaction $(ESR)$ with benzene to form cyclohexylbenzene.
Thus,the product $P$ is cyclohexylbenzene.

(D) The enthalpy of hydrogenation is inversely proportional to the stability of the alkene/diene system.
$(I)$ is an aromatic compound (mesitylene),which is highly stable due to resonance energy.
$(II)$ is a conjugated diene system with some cross-conjugation,making it less stable than the aromatic system but more stable than the isolated exocyclic double bonds.
$(III)$ contains three isolated exocyclic double bonds,which are the least stable among the three.
Stability order: $I > II > III$.
Since enthalpy of hydrogenation is inversely proportional to stability,the order of enthalpy of hydrogenation is $III > II > I$.

(C) The $-CH_3$ group is an $o, p$-directing group. In $m$-xylene ($1,3$-dimethylbenzene),the positions available for electrophilic substitution are $2, 4, 5,$ and $6$.
$1$. The position between the two $-CH_3$ groups (position $2$) is sterically hindered,so no significant substitution occurs there.
$2$. The $4$ and $6$ positions are equivalent and are ortho to one $-CH_3$ group and para to the other,making them the most favorable sites for substitution.
$3$. The $5$ position is meta to both $-CH_3$ groups and is less activated.
Therefore,the major products formed are $4$-bromo-$1,3$-dimethylbenzene and $6$-bromo-$1,3$-dimethylbenzene (which are equivalent due to symmetry) or substitution at the $4$ and $6$ positions relative to the methyl groups.

(D) Molecules that do not satisfy the Huckel rule or $(4n+2) \pi$-electron rule are classified as non-aromatic or anti-aromatic depending on their cyclic conjugation.
$1$. Benzene,Naphthalene,and Thiophene are aromatic because they are cyclic,planar,fully conjugated,and follow the $(4n+2) \pi$-electron rule.
$2$. Cyclopentadiene (option $d$) has an $sp^3$ hybridized carbon atom in the ring,which breaks the continuous conjugation. Therefore,it is non-aromatic.

(D) The reaction of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is known as Friedel-Crafts alkylation.
In this reaction,the methyl group $(-CH_3)$ replaces a hydrogen atom on the benzene ring to form toluene $(C_6H_5CH_3)$.
The chemical equation is: $C_6H_6 + CH_3Cl \xrightarrow{Anhy. AlCl_3} C_6H_5CH_3 + HCl$.

(A) When benzene is oxidized by air in the presence of $V_2O_5$ at $773 \ K$,it undergoes catalytic oxidation to form maleic acid,which subsequently loses a water molecule to form maleic anhydride.
The reaction is as follows:
$C_6H_6 + \frac{9}{2} O_2$ $\xrightarrow{V_2O_5, 773 \ K} \text{Maleic acid}$ $\xrightarrow{\Delta, -H_2O} \text{Maleic anhydride}$

(B) According to the Bronsted-Lowry theory,an acid is a proton donor and a base is a proton acceptor.
In the nitrating mixture,conc. $H_2SO_4$ acts as a strong acid and donates a proton to conc. $HNO_3$.
The reaction is: $HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^-$.
Here,$HNO_3$ accepts a proton from $H_2SO_4$,therefore $HNO_3$ acts as a base.
Subsequently,$H_2NO_3^+$ decomposes to form the electrophile $NO_2^+$: $H_2NO_3^+ \rightarrow NO_2^+ + H_2O$.

(D) Aromatic compounds must follow $H$ückel's rule ($4n+2$ $\pi$ electrons),be planar,and cyclic.
$Pyrrole$,$Thiophene$,and $Furan$ are all five-membered heterocyclic aromatic compounds with $6$ $\pi$ electrons $(n=1)$.
$Pyran$ $(C_5H_6O)$ contains an $sp^3$ hybridized carbon atom in the ring,which breaks the continuous conjugation required for aromaticity.
Therefore,$Pyran$ is non-aromatic.

(B) compound is anti-aromatic if it is cyclic,planar,fully conjugated,and contains $4n$ $\pi$-electrons (where $n = 1, 2, ...$).
$A$. Pyrrole cation: It has $6$ $\pi$-electrons ($4$ from double bonds and $2$ from the lone pair on nitrogen,but here the nitrogen is protonated,so it has $4$ $\pi$-electrons). However,it is not planar due to $sp^3$ hybridization at the nitrogen atom.
$B$. Cyclobutadiene: It is a cyclic,planar,fully conjugated system with $4$ $\pi$-electrons ($4n$ where $n=1$). Thus,it is anti-aromatic.
$C$. Cyclopentadiene: It is not fully conjugated because of the $sp^3$ hybridized carbon atom. Thus,it is non-aromatic.
$D$. Cyclooctatetraene: It is non-planar (tub-shaped) to avoid anti-aromaticity. Thus,it is non-aromatic.
Therefore,the correct answer is $B$.

(D) The reaction of benzene $(C_6H_6)$ with ethene $(CH_2=CH_2)$ in the presence of $HCl$ and $AlCl_3$ is a Friedel-Crafts alkylation reaction.
First,$CH_2=CH_2$ reacts with $HCl$ in the presence of $AlCl_3$ to form an ethyl carbocation $(CH_3CH_2^+)$ or an activated complex that acts as an electrophile.
This electrophile then attacks the benzene ring to form ethylbenzene $(C_6H_5CH_2CH_3)$.
Therefore,the correct product is ethylbenzene.

(C) Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction. It occurs readily in benzene rings that are activated by electron-donating groups (EDGs).
$(i)$ $p$-Nitrotoluene: The $-NO_2$ group is a strong electron-withdrawing group $(EWG)$,which deactivates the ring.
$(ii)$ Ethylbenzene: The $-CH_2CH_3$ group is an electron-donating group (alkyl group),which activates the ring.
$(iii)$ Benzoic acid: The $-COOH$ group is a strong electron-withdrawing group $(EWG)$,which deactivates the ring.
$(iv)$ Phenol: The $-OH$ group is a strong electron-donating group $(EDG)$,which activates the ring.
Therefore,compounds $(ii)$ and $(iv)$ will undergo Friedel-Crafts alkylation.

(C) The reaction involves the electrophilic aromatic substitution of a polycyclic aromatic hydrocarbon (specifically,a derivative of $9,10-$dihydroanthracene or similar structure) with $Br_2$ in the presence of a Lewis acid catalyst,$AlCl_3$.
$1$. The $AlCl_3$ acts as a Lewis acid to generate the electrophile $Br^+$ from $Br_2$.
$2$. The electrophile $Br^+$ attacks the most electron-rich position on the aromatic ring.
$3$. In the given structure,the aromatic rings are activated towards electrophilic substitution. The position para to the alkyl group (or the most sterically accessible and electronically favored position) is typically preferred.
$4$. Comparing the options,the substitution occurs at the position that minimizes steric hindrance while maintaining the stability of the intermediate carbocation (sigma complex).
$5$. Option $C$ represents the substitution at the position para to the bridgehead,which is electronically favored due to the resonance stabilization provided by the fused ring system.

(B) Electrophilic aromatic substitution reactivity depends on the electron density of the benzene ring.
Groups that donate electrons via inductive $(+I)$ or hyperconjugation effects activate the ring,while electron-withdrawing groups deactivate it.
$II$ (Toluene) has a $-CH_3$ group,which activates the ring via $+I$ and hyperconjugation.
$III$ (Ethylbenzene) has a $-CH_2CH_3$ group,which also activates the ring via $+I$ and hyperconjugation,but the $-CH_3$ group in toluene is a slightly stronger activator due to less steric hindrance and slightly better hyperconjugative stabilization compared to the ethyl group in some contexts,or they are very similar; however,standard reactivity order for alkylbenzenes is $Toluene > Ethylbenzene$.
$IV$ (Benzene) has no substituents.
$I$ (Phenyl benzoate) has a $-COOC_6H_5$ group,which is a strong electron-withdrawing group ($-I$ and $-M$ effects),significantly deactivating the ring.
Thus,the order of reactivity is $II > III > IV > I$.

(B) Friedel-Crafts alkylation has several limitations:
$1$. Aniline forms a complex with the Lewis acid $(AlCl_3)$,which deactivates the ring,making it impossible to undergo Friedel-Crafts alkylation.
$2$. Polyalkylation is a major drawback because the alkyl group introduced is electron-donating,which activates the ring towards further substitution.
$3$. Carbocation rearrangement occurs when using primary alkyl halides in the presence of a strong Lewis acid like $AlCl_3$.
$4$. Nitrobenzene is a strongly deactivating group and is often used as a solvent in Friedel-Crafts reactions because it does not react with the Lewis acid catalyst.
Therefore,the statement that 'Polyalkylation is not possible' is incorrect.

(D) Let's analyze each reaction:
$(A)$ $3Me-C \equiv CH \xrightarrow[Fe \ tube]{Red \ hot} 1,3,5-trimethylbenzene$ (aromatic).
$(B)$ $n-Hexane \xrightarrow[Cr_2O_3/\Delta]{Al_2O_3} \text{Benzene}$ (aromatic).
$(C)$ Cyclopentadiene $\xrightarrow{CHCl_3/KOH} \text{Chlorobenzene}$ (aromatic). This is a ring expansion reaction.
$(D)$ Hexachlorocyclohexane $\xrightarrow{Alc. KOH} 1,2,4-trichlorobenzene$ (aromatic). However,the reaction with $Alc. KOH$ on hexachlorocyclohexane $(BHC)$ typically undergoes dehydrohalogenation to form $1,2,4-trichlorobenzene$,which is aromatic. Wait,let's re-evaluate. Actually,$1,2,3,4,5,6-hexachlorocyclohexane$ $(BHC)$ on treatment with $Alc. KOH$ undergoes elimination to form $1,2,4-trichlorobenzene$. All options produce aromatic products. Let's re-examine the question. Perhaps the question implies a specific condition or a different starting material. Actually,the reaction of cyclopentadiene with $CHCl_3/KOH$ is a carbene addition followed by rearrangement to form chlorobenzene,which is aromatic. All options listed are standard reactions that yield aromatic products. However,if we consider the most likely intended answer in a multiple-choice context,sometimes $(D)$ is considered to produce a mixture or not proceed cleanly to a single aromatic product compared to the others. Given the standard curriculum,all these are known to produce aromatic compounds.

(D) Let us analyze each option:
$A$. $-CCl_3$ is a strongly deactivating group due to the strong $-I$ effect of three chlorine atoms.
$B$. $-CH_3-COO^-$ (acetate ion) is an activating group because the oxygen atom with a negative charge can donate electron density into the ring via the $+M$ effect.
$C$. $-CH_3$ (methyl group) is an activating group because it donates electron density to the benzene ring via the $+I$ effect and hyperconjugation.
$D$. $-OH_2^+$ is a strongly deactivating group. While it has an oxygen atom,it carries a positive charge,making it a strong electron-withdrawing group via both $-I$ and $-M$ effects. Thus,it deactivates the ring.
Therefore,option $D$ is the correct statement.

(B) The reaction is a Friedel-Crafts acylation of $1$-methyl-$4$-isopropylbenzene (p-cymene).
In $p$-cymene,there are two alkyl groups: a methyl group $(-CH_3)$ and an isopropyl group $(-CH(CH_3)_2)$.
Both groups are ortho/para directing.
The methyl group has $3 \alpha$-hydrogens,while the isopropyl group has only $1 \alpha$-hydrogen.
The methyl group is a stronger activating group than the isopropyl group due to hyperconjugation.
Therefore,the incoming acyl group $(-COCH_3)$ will be directed by the more activating methyl group.
The position ortho to the methyl group is also ortho to the isopropyl group,but the steric hindrance from the bulky isopropyl group makes the position ortho to the methyl group (and meta to the isopropyl group) more favorable.
Thus,the major product is $2$-acetyl-$1$-methyl-$4$-isopropylbenzene,which corresponds to option $B$.

(C) compound is aromatic if it follows $H$ückel's rule ($4n+2$ $\pi$ electrons),is planar,and cyclic.
$A$: Cyclopentadienone has $4$ $\pi$ electrons in the ring (anti-aromatic).
$B$: Cycloheptatriene has an $sp^3$ hybridized carbon atom,so it is not planar and not aromatic.
$C$: $2-$aminopyridin$-4-$one is a tautomer of $2-$amino$-4-$hydroxypyridine. The ring is planar,fully conjugated,and contains $6$ $\pi$ electrons ($4$ from double bonds + $2$ from the lone pair on the nitrogen atom in the ring),satisfying $H$ückel's rule $(n=1)$. Thus,it is aromatic.
$D$: The structure shown is not aromatic due to the presence of $sp^3$ hybridized nitrogen and lack of continuous conjugation.

(C) When acetone $(CH_3COCH_3)$ is distilled with concentrated $H_2SO_4$,three molecules of acetone undergo self-aldol condensation followed by dehydration to form mesitylene ($1, 3, 5-$trimethylbenzene).
$3 CH_3COCH_3 \xrightarrow{\text{Conc. } H_2SO_4, \Delta} C_6H_3(CH_3)_3 + 3 H_2O$.

(A) Sodium benzoate $(C_6H_5COONa)$ undergoes decarboxylation when heated with sodalime $(NaOH + CaO)$.
The reaction is as follows:
$C_6H_5COONa + NaOH \xrightarrow{CaO, \Delta} C_6H_6 + Na_2CO_3$.
Thus,the product formed is benzene.

(B) $1$. Let us analyze the structures of the given compounds:
$2$. $Mesitylene$ is $1,3,5-trimethylbenzene$. Its structure consists of a benzene ring with three methyl groups at positions $1, 3,$ and $5$.
$3$. In $Mesitylene$ $(C_9H_{12})$:
- There are $3$ primary carbon atoms (the $3$ methyl groups attached to the ring).
- There are $3$ secondary carbon atoms (the $3$ $CH$ groups in the benzene ring).
- There are $3$ tertiary carbon atoms (the $3$ $C$ atoms in the ring attached to the methyl groups).
$4$. Since the number of primary,secondary,and tertiary carbon atoms is equal ($3$ each),$Mesitylene$ is the correct answer.

(C) The given species is the cyclopentadienyl anion $(C_5H_5^-)$.
It contains two double bonds,which contribute $2 \times 2 = 4$ $\pi$ electrons.
The negative charge on the carbon atom is part of the conjugated system,contributing an additional $2$ $\pi$ electrons.
Total $\pi$ electrons = $4 + 2 = 6$ $\pi$ electrons.
This species follows $H$ückel's rule ($4n + 2$ $\pi$ electrons,where $n=1$) and is aromatic.

(C) Naphthalene consists of two fused benzene rings.
Each benzene ring contains $3$ double bonds,contributing $6$ $\pi$ electrons.
However,the central bond is shared between the two rings.
In the structure of naphthalene,there are $5$ double bonds in total.
Since each double bond contributes $2$ $\pi$ electrons,the total number of $\pi$ electrons is $5 \times 2 = 10$.

(C) Benzene $(C_6H_6)$ is a colourless liquid with a characteristic aromatic smell and is highly inflammable.
Although benzene contains double bonds,it does not undergo addition reactions easily like typical alkenes due to its resonance stability (aromaticity).
Therefore,it does not show the typical properties of unsaturation (such as rapid decolourisation of bromine water or Baeyer's reagent).
Thus,'Unsaturation' is not considered a characteristic property of benzene in the context of its chemical reactivity.

(D) Benzene $(C_6H_6)$ is a highly stable aromatic compound due to the delocalization of $\pi$-electrons,a phenomenon known as resonance.
Resonance energy provides extra stability to the benzene ring.
Addition reactions would involve breaking the aromatic sextet and destroying the resonance stability,which is energetically unfavorable.
Substitution reactions,on the other hand,allow the benzene ring to retain its aromatic character and resonance stability,making them the preferred pathway for benzene.

$(A)$ Electrophilic aromatic substitution is facilitated by electron-donating groups on the benzene ring.
In the given compounds, the alkyl groups attached to the phenyl ring $(\phi)$ exhibit the $+I$ (inductive) effect and hyperconjugation.
As the number of $\alpha$-hydrogens increases, the electron-donating ability via hyperconjugation increases, thereby increasing the electron density on the ring and the reactivity towards electrophiles.
Number of $\alpha$-hydrogens:
$I. \phi-CH_3$: $3 \ \alpha-H$
$II. \phi-CH_2-CH_3$: $2 \ \alpha-H$
$III. \phi-CH(CH_3)_2$: $1 \ \alpha-H$
$IV. \phi-C(CH_3)_3$: $0 \ \alpha-H$
Thus, the order of reactivity is $I > II > III > IV$.

(D) The conversion of benzene to chlorobenzene is an electrophilic aromatic substitution reaction.
Benzene reacts with chlorine $(Cl_2)$ in the presence of a Lewis acid catalyst such as anhydrous aluminum chloride $(AlCl_3)$ or ferric chloride $(FeCl_3)$ at low temperatures to form chlorobenzene.
This reaction is known as electrophilic chlorination.
Option $D$ provides the correct reagents and conditions for this transformation.

(B) The reactivity of benzene derivatives toward electrophilic aromatic substitution (like monobromination) depends on the electron density of the ring.
$1$. $C_6H_5CH_3$ (Toluene): The $-CH_3$ group is an electron-donating group $(EDG)$ via hyperconjugation and inductive effect,which activates the ring.
$2$. $C_6H_6$ (Benzene): This is the reference compound with no substituents.
$3$. $C_6H_5COOH$ (Benzoic acid): The $-COOH$ group is an electron-withdrawing group $(EWG)$ via resonance and inductive effect,which deactivates the ring.
$4$. $C_6H_5NO_2$ (Nitrobenzene): The $-NO_2$ group is a strong electron-withdrawing group $(EWG)$ via resonance and inductive effect,which strongly deactivates the ring.
Thus,the order of decreasing reactivity is $I > III > II > IV$.

(B) The reaction of benzene with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ is known as nitration.
At $50-60^\circ C$,benzene undergoes mononitration to form nitrobenzene.
However,at higher temperatures like $90^\circ C$ or above,the reaction proceeds further to form $m-$dinitrobenzene because the nitro group $(-NO_2)$ is a deactivating and meta-directing group.

(C) Direct Friedel-Crafts alkylation of benzene often leads to polyalkylation because the alkyl group introduced is electron-donating,which activates the benzene ring towards further electrophilic substitution.
To avoid this and obtain the monoalkylated product,Friedel-Crafts acylation is preferred.
In acylation,an acyl group $(RCO-)$ is introduced,which is electron-withdrawing and deactivates the ring,preventing further substitution.
The resulting ketone can then be reduced to the corresponding alkyl group using Clemmensen reduction or Wolff-Kishner reduction.

(B) The reaction of benzene with $n-$propyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
During this reaction,the $n-$propyl carbocation $(CH_3CH_2CH_2^+)$ formed initially undergoes a $1,2-$hydride shift to form a more stable secondary isopropyl carbocation $(CH_3CH^+CH_3)$.
This secondary carbocation then attacks the benzene ring to form isopropylbenzene (cumene) as the major product.
Therefore,the correct option is $B$.

(B) $1$. In sulphonation of benzene,the effective electrophile is sulphur trioxide $(SO_3)$.
$2$. In Friedel-Crafts acetylation,the effective electrophile is the acylium ion $(CH_3-C \equiv O^{+})$.
$3$. In formylation (Gattermann-Koch reaction),the effective electrophile is the formyl cation $(HC \equiv O^{+})$,which is generated in situ from $CO$ and $HCl$ in the presence of a Lewis acid like $AlCl_3$.

(B) The molecular formula $C_7H_7Cl$ corresponds to a benzene ring substituted with a methyl group and a chlorine atom.
These are isomers of chlorotoluene.
There are three possible positions for the chlorine atom relative to the methyl group on the benzene ring:
$1$. Ortho-chlorotoluene $(o-Cl-C_6H_4-CH_3)$
$2$. Meta-chlorotoluene $(m-Cl-C_6H_4-CH_3)$
$3$. Para-chlorotoluene $(p-Cl-C_6H_4-CH_3)$
Additionally,there is one isomer where the chlorine is on the side chain: Benzyl chloride $(C_6H_5-CH_2Cl)$.
Therefore,the total number of benzene derivatives (isomers) is $3 + 1 = 4$.

(D) The molecular formula $C_7H_8O$ corresponds to a degree of unsaturation of $4$ $(7 - 8/2 + 1 = 4)$,which indicates a benzene ring.
Possible isomers are:
$1$. $o$-Cresol (ortho-methylphenol)
$2$. $m$-Cresol (meta-methylphenol)
$3$. $p$-Cresol (para-methylphenol)
$4$. Benzyl alcohol $(C_6H_5CH_2OH)$
$5$. Anisole (methoxybenzene,$C_6H_5OCH_3$)
Thus,there are $5$ possible benzene derivatives.

(B) The reaction shown is a Friedel-Crafts alkylation of toluene with methyl chloride $(CH_3Cl)$ in the presence of an anhydrous $AlCl_3$ catalyst.
Since the methyl group $(-CH_3)$ on the benzene ring is an ortho/para-directing group,the electrophilic substitution occurs at the ortho and para positions.
Due to steric hindrance at the ortho position,the para-isomer ($p-$xylene) is formed as the major product.

(B) The reaction $C_6H_6 + A \xrightarrow{AlCl_3} C_6H_5CONH_2$ represents a Friedel-Crafts acylation reaction.
In this reaction,benzene $(C_6H_6)$ reacts with an acylating agent in the presence of a Lewis acid catalyst $(AlCl_3)$ to form an amide derivative.
The reagent $A$ must be carbamoyl chloride $(ClCONH_2)$,which acts as the source of the $-CONH_2$ group.
The reaction proceeds as: $C_6H_6 + ClCONH_2 \xrightarrow{AlCl_3} C_6H_5CONH_2 + HCl$.

(C) The reaction of benzene with acetic anhydride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction.
In this reaction,the acetyl group $(-COCH_3)$ from acetic anhydride replaces a hydrogen atom on the benzene ring to form acetophenone $(C_6H_5COCH_3)$.
The reaction is represented as: $C_6H_6 + (CH_3CO)_2O \xrightarrow{Anhydrous \ AlCl_3} C_6H_5COCH_3 + CH_3COOH$.

(B) $1$. The reaction of benzene $(C_6H_6)$ with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction. This produces acetophenone $(C_6H_5COCH_3)$ as product $A$.
$2$. The reaction of acetophenone $(C_6H_5COCH_3)$ with $Zn-Hg/HCl$ is a Clemmensen reduction,which reduces the carbonyl group $(>C=O)$ to a methylene group $(-CH_2-)$.
$3$. The reduction of acetophenone yields ethyl benzene $(C_6H_5CH_2CH_3)$ as the final product $B$.

(B) In nitration using a mixture of conc. $HNO_3$ and conc. $H_2SO_4$,the electrophile generated is the nitronium ion,$NO_2^+$.
In sulphonation using fuming $H_2SO_4$ (oleum),the electrophile is sulphur trioxide,$SO_3$.
Therefore,the species responsible for nitration and sulphonation are $NO_2^+$ and $SO_3$ respectively.

(C) Electrophilic substitution reactions involve the replacement of a hydrogen atom on an aromatic ring by an electrophile.
Option $A$ is nitration,which is an electrophilic substitution.
Option $B$ is Friedel-Crafts alkylation,which is an electrophilic substitution.
Option $C$ is the addition reaction of chlorine to benzene in the presence of $UV$ light to form benzene hexachloride $(C_6H_6Cl_6)$,which is an electrophilic addition reaction,not substitution.
Option $D$ is the Gattermann-Koch reaction,which is an electrophilic substitution.

(A) The reaction is a Friedel-Crafts alkylation of benzene with $n$-butyl chloride in the presence of a Lewis acid catalyst,$AlCl_3$.
$1$. The reaction proceeds through the formation of a carbocation intermediate.
$2$. Initially,$n$-butyl chloride reacts with $AlCl_3$ to form a primary carbocation,$CH_3CH_2CH_2CH_2^+$.
$3$. Primary carbocations are unstable and undergo rearrangement to form more stable carbocations. The primary carbocation undergoes a $1,2$-hydride shift to form a secondary carbocation,$CH_3CH_2CH^+CH_3$.
$4$. This secondary carbocation then undergoes electrophilic aromatic substitution with benzene to form $sec$-butylbenzene as the major product.

(B) The reaction of $SO_3$ with an aromatic ring is an electrophilic aromatic substitution (sulfonation).
$SO_3$ acts as an electrophile.
The rate of electrophilic aromatic substitution depends on the electron density of the benzene ring.
Groups that donate electrons to the ring (activating groups) increase the electron density and make the ring more reactive towards electrophiles.
$CH_3$ (methyl group) in Toluene is an electron-donating group due to the inductive effect and hyperconjugation,which activates the ring.
Benzene has no substituents.
Nitrobenzene $(-NO_2)$ and Chlorobenzene $(-Cl)$ have electron-withdrawing groups that deactivate the ring.
Therefore,Toluene is the most reactive among the given options.

(C) The oxidation of alkyl benzenes (like toluene) to benzoic acid is a standard reaction in organic chemistry.
When toluene $(C_6H_5CH_3)$ is treated with a strong oxidizing agent like alkaline potassium permanganate $(KMnO_4)$,the methyl group attached to the benzene ring is oxidized to a carboxyl group $(-COOH)$.
This reaction produces potassium benzoate,which upon acidification yields benzoic acid $(C_6H_5COOH)$.

(A) The strength of an $o,p$-directing group is determined by its ability to donate electrons to the benzene ring via the resonance effect ($+R$ or $+M$ effect).
Among the given options,the $-OH$ group has a strong $+R$ effect due to the lone pairs on the oxygen atom,which are effectively donated to the ring.
Halogens like $-Cl$ and $-Br$ exhibit a $-I$ effect that outweighs their $+R$ effect,making them deactivating groups,although they remain $o,p$-directing.
The phenyl group $(-C_6H_5)$ is a weaker electron-donating group compared to $-OH$.
Therefore,$-OH$ is the strongest $o,p$-directing group among the choices.