Aromatic hydrocarbon Questions in English

(A) $Ethyl$ benzene cannot be prepared by the $Wurtz$ reaction.
The $Wurtz$ reaction involves the coupling of two alkyl halides to form a higher symmetric alkane.
It is not suitable for the synthesis of alkyl-substituted aromatic compounds like $Ethyl$ benzene.
$Wurtz-Fittig$ reaction,$Friedel-Crafts$ alkylation,and $Clemmensen$ reduction are all valid methods for preparing $Ethyl$ benzene.

(A) The nitro group $(-NO_2)$ is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It withdraws electron density from the benzene ring,specifically from the ortho $(o)$ and para $(p)$ positions through resonance.
As a result,the electron density at the $o$ and $p$ positions decreases significantly compared to the meta $(m)$ positions.
Since the electrophile is electron-deficient,it prefers to attack the position with relatively higher electron density,which is the meta position.
Therefore,the nitro group is meta-directing.

(B) Benzene reacts with chlorine in the presence of sunlight via a free radical addition reaction to give benzene hexachloride $(C_{6}H_{6}Cl_{6})$,also known as gammexane or lindane.
$C_{6}H_{6} + 3Cl_{2} \xrightarrow{\text{Sunlight}} C_{6}H_{6}Cl_{6}$

(D) The electrophile involved in the sulphonation of benzene is $SO_{3}$.
During the reaction,concentrated sulfuric acid acts as both a reagent and a catalyst to generate the electrophile.
The reaction is: $2 H_{2}SO_{4} \longrightarrow SO_{3} + H_{3}O^{+} + HSO_{4}^{-}$

(B) Benzene does not contain $3$ distinct single bonds and $3$ distinct double bonds. Due to resonance,all carbon-carbon bonds in benzene are equivalent,having a bond order of $1.5$.

(B) The reaction of benzene with excess chlorine in the presence of anhydrous $AlCl_3$ (a Lewis acid) in the dark and at cold conditions leads to the electrophilic aromatic substitution of all six hydrogen atoms of the benzene ring by chlorine atoms. This process is known as exhaustive chlorination. The reaction is as follows:
$C_6H_6 + 6Cl_2 \xrightarrow{\text{Anhydrous } AlCl_3, \text{dark, cold}} C_6Cl_6 + 6HCl$
Here,$C_6Cl_6$ is hexachlorobenzene. Therefore,the correct option is $B$.

(B) The nitration of benzene involves the generation of the nitronium ion $(NO_2^{+})$ as the active electrophile.
Step $1$: Sulfuric acid $(H_2SO_4)$ acts as a proton donor and nitric acid $(HNO_3)$ acts as a base. The reaction is: $HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^{+} + HSO_4^{-}$.
Step $2$: The protonated nitric acid $(H_2NO_3^{+})$ loses a water molecule to form the nitronium ion: $H_2NO_3^{+} \rightarrow NO_2^{+} + H_2O$.
This nitronium ion $(NO_2^{+})$ then attacks the benzene ring to form nitrobenzene.

(D) In graphite,each carbon atom is $sp^{2}$ hybridised and linked to three other carbon atoms in a hexagonal planar structure.
In diamond,each carbon atom is $sp^{3}$ hybridised and linked to four other carbon atoms in a tetrahedral geometry via strong $C-C$ $\sigma$-bonds.
Therefore,the hybridisation states for graphite and diamond are $sp^{2}$ and $sp^{3}$ respectively.

(A) The bond dissociation energy of a $C-H$ bond depends on the stability of the radical formed after the homolytic cleavage of the bond.
In toluene $(C_6H_5CH_3)$,the removal of a hydrogen atom from the methyl group results in the formation of a benzyl radical $(C_6H_5CH_2^{\bullet})$.
This benzyl radical is highly stabilized by resonance with the benzene ring,as shown in the provided resonance structures.
Since the resulting radical is very stable,the energy required to break the $C-H$ bond in the methyl group of toluene is the lowest among the given options.

(D) Phenyl magnesium bromide is a Grignard reagent. The carbon atom attached to the metal $Mg$ is strongly basic and abstracts $D^{+}$ from heavy water $(D_2O)$. The reaction of a Grignard reagent with $D_2O$ provides a convenient method for introducing a deuterium atom into a molecule at a specific location.
The reaction is:
$C_6H_5MgBr + D_2O \rightarrow C_6H_5D + Mg(OD)Br$
Thus,the product formed is deuterobenzene.

(D) The starting material is glucose $(CHO(CHOH)_4CH_2OH)$.
$1$. Reaction with $HI$ and $\Delta$ (reduction) converts glucose into $n$-hexane $(X = CH_3(CH_2)_4CH_3)$.
$2$. Reaction with $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$ is an aromatization reaction (dehydrocyclization),which converts $n$-hexane into benzene $(Y = C_6H_6)$.
$3$. Reaction of benzene with $Cl_2$ in the presence of $UV$ light is a free-radical addition reaction,which yields hexachlorocyclohexane ($Z = C_6H_6Cl_6$,also known as $BHC$ or Gammaxene).

(B) The reaction involves the oxidation of alkyl groups attached to a benzene ring using alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$.
Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes any alkyl side chain attached to the benzene ring to a carboxylic acid group $(-COOH)$,provided that the benzylic carbon has at least one hydrogen atom.
In the given reactant,$1$-ethyl-$4$-methylbenzene,both the ethyl group $(-CH_2CH_3)$ and the methyl group $(-CH_3)$ are attached to the benzene ring at the $1$ and $4$ positions.
Both of these alkyl groups have benzylic hydrogens,so both will be oxidized to carboxylic acid groups.
Therefore,the product $X$ is benzene-$1,4$-dicarboxylic acid,also known as terephthalic acid.

(A) The reaction sequence is as follows:
$1$. Oxidation of toluene with $KMnO_4/KOH$ followed by acidic workup $(H_3O^+, \Delta)$ yields benzoic acid $(C_6H_5COOH)$.
$2$. The $-COOH$ group is a strongly deactivating group and is meta-directing for electrophilic aromatic substitution.
$3$. Electrophilic aromatic bromination of benzoic acid using $Br_2/FeBr_3$ introduces the bromine atom at the meta-position,resulting in $3$-bromobenzoic acid.

(B) The methoxy group $(-OCH_3)$ is an activating group and is ortho/para-directing due to the resonance effect.
Nitration of anisole (methoxybenzene) using a mixture of concentrated $HNO_3$ and $H_2SO_4$ yields a mixture of ortho- and para-nitroanisole.
Due to steric hindrance at the ortho position,the para-isomer is formed as the major product.
Therefore,the major product is $p$-nitromethoxybenzene.

(C) Cumene (isopropylbenzene) has the chemical formula $C_6H_5CH(CH_3)_2$.
$1$. The benzene ring consists of $6$ carbons,all of which are $sp^2$-hybridised.
$2$. The isopropyl group attached to the ring consists of $3$ carbons: one $CH$ group and two $CH_3$ groups.
$3$. All $3$ carbons in the isopropyl group are $sp^3$-hybridised.
Therefore,the number of $sp^2$-hybridised carbons is $6$ and the number of $sp^3$-hybridised carbons is $3$.

(C) The structure of isopropyl benzene is $C_6H_5CH(CH_3)_2$.
It is commonly known as cumene.
Hence,option $(C)$ is correct.

(C) Cumene is chemically known as (propan$-2-$yl)benzene or isopropylbenzene. Its chemical formula is $C_9H_{12}$. The structure consists of a benzene ring attached to an isopropyl group $(-CH(CH_3)_2)$. Option $C$ correctly represents this structure.

(C) Cumene,also known as isopropylbenzene,is an organic compound consisting of a benzene ring substituted with an isopropyl group. Its chemical formula is $C_6H_5CH(CH_3)_2$. Option $C$ correctly represents this structure,where a benzene ring is attached to a $CH(CH_3)_2$ group.

(B) The given compound is $1,2,3,4$-tetrahydronaphthalene (tetralin).
By analyzing the structure,we can count the hydrogen atoms attached to each carbon atom:
- The benzene ring part has $4$ hydrogen atoms.
- The saturated cyclohexane ring part has $8$ hydrogen atoms.
Total number of $H$-atoms $= 4 + 8 = 12$.

(A) Meta-directing groups are electron-withdrawing groups (EWGs) that decrease the electron density on the benzene ring through $-M$ or $-I$ effects.
$(a) -CN$ is meta-directing.
$(b) -COR$ is meta-directing.
$(c) -NHCOR$ is ortho/para-directing due to the $+M$ effect of the nitrogen atom.
$(d) -SO_3H$ is meta-directing.
$(e) -OCH_3$ is ortho/para-directing due to the $+M$ effect of the oxygen atom.
Therefore,the meta-directing groups are $a, b,$ and $d$.

(D) An aromatic species must be planar,cyclic,fully conjugated,and possess $(4n+2) \pi$ electrons,where $n$ is an integer.
$A$. Cyclopentadienyl anion: It is cyclic,planar,fully conjugated,and has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$B$. Tropylium cation: It is cyclic,planar,fully conjugated,and has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$C$. Cyclopropenyl cation: It is cyclic,planar,fully conjugated,and has $2 \pi$ electrons $(n=0)$,so it is aromatic.
$D$. Cycloheptatriene: It has a $sp^3$-hybridized carbon atom in the ring,which breaks the continuous conjugation of the $\pi$ system. Therefore,it is not aromatic.

(B) Aromatic compounds must be cyclic,planar,fully conjugated,and contain $(4n + 2) \pi$ electrons ($H$ückel's rule).
$1$. Pyridine: Cyclic,planar,fully conjugated,and has $6 \pi$ electrons $(n=1)$. It is aromatic.
$2$. Cyclopentadienyl cation: Cyclic,planar,but has only $4 \pi$ electrons ($n=1$ for $4n$ rule). It is antiaromatic.
$3$. Anthracene: Cyclic,planar,fully conjugated,and has $14 \pi$ electrons $(n=3)$. It is aromatic.
$4$. Furan: Cyclic,planar,fully conjugated (lone pair on oxygen participates),and has $6 \pi$ electrons $(n=1)$. It is aromatic.
Therefore,the cyclopentadienyl cation is not aromatic.

(C) Aromatic compounds consist of a conjugated planar ring system accompanied by delocalized $\pi$-electron clouds. Compounds that follow $H$ückel's rule,i.e.,$(4n+2) \pi$ electrons,are aromatic.
$(i)$ Furan: Has $6 \pi$ electrons ($4$ from double bonds + $2$ from lone pair on $O$),follows $(4n+2) \pi$ rule $(n=1)$,hence aromatic.
$(ii)$ Cyclobutadiene: Has $4 \pi$ electrons,follows $4n \pi$ rule,hence anti-aromatic.
$(iii)$ Cyclopropenyl anion: Has $4 \pi$ electrons,follows $4n \pi$ rule,hence anti-aromatic.
$(iv)$ Pyridine: Has $6 \pi$ electrons,follows $(4n+2) \pi$ rule $(n=1)$,hence aromatic.
$(v)$ Tropylium cation: Has $6 \pi$ electrons,follows $(4n+2) \pi$ rule $(n=1)$,hence aromatic.
Therefore,$(i)$,$(iv)$ and $(v)$ are aromatic.

(B) non-benzenoid aromatic compound is an aromatic system that does not contain a benzene ring.
$Furan$ is a heterocyclic aromatic compound (aromatic,planar,$6\pi$ electrons) and is a classic example of a non-benzenoid aromatic compound.
$Tetrahydrofuran$ is non-aromatic.
$Cyclooctatetraene$ is non-aromatic (tub-shaped,non-planar).
$Heptafulvene$ is non-aromatic due to the lack of continuous conjugation and planarity in the system.

(B) compound is non-aromatic if it is not cyclic,not planar,or does not have a continuous system of $p$-orbitals for cyclic conjugation.
In option $B$,the molecule is $1,3$-cyclopentadiene. It contains an $sp^3$ hybridized carbon atom,which breaks the continuous conjugation of the $p$-orbitals in the ring.
Therefore,it is non-aromatic.
The other options are aromatic as they follow $Huckel's$ $(4n+2)$ $\pi$-electron rule.

(C) compound is aromatic if it follows $H$ückel's rule: it must be planar,cyclic,fully conjugated,and contain $(4n + 2) \pi$ electrons.
$A$. Cyclopentadienyl anion has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$B$. Benzene has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$C$. Cyclopentadienyl cation has $4 \pi$ electrons $(n=1)$,which follows the $4n$ rule,making it anti-aromatic.
$D$. Cycloheptatrienyl cation has $6 \pi$ electrons $(n=1)$,so it is aromatic.
Therefore,the cyclopentadienyl cation is not aromatic.

(D) compound is aromatic if it follows $H$ückel's rule ($4n+2$ $\pi$ electrons),is planar,and cyclic with continuous conjugation.
$a$: Cyclopentadienyl cation has $4$ $\pi$ electrons ($4n$ system),so it is anti-aromatic.
$b$: Cyclopentadienyl anion has $6$ $\pi$ electrons ($4n+2$ where $n=1$),so it is aromatic.
$c$: Cycloheptatrienyl anion has $8$ $\pi$ electrons ($4n$ system),so it is anti-aromatic.
$d$: Tropone (cycloheptatrienone) has $6$ $\pi$ electrons in the ring (due to polarization of $C=O$ bond),so it is aromatic.
$e$: Thiophene has $6$ $\pi$ electrons ($4$ from double bonds + $2$ from lone pair on $S$),so it is aromatic.
Thus,structures $b, d, e$ are aromatic. However,based on the provided options,the most appropriate choice is $b, d$ if we consider the specific set provided in the options.

(D) For a compound to be aromatic,it must satisfy $H$ückel's rule and structural requirements:
$(1)$ The molecule must be cyclic and planar to allow for continuous delocalization of $\pi$-electrons.
$(2)$ It must possess $(4n+2) \pi$-electrons,where $n$ is an integer $(n = 0, 1, 2, \dots)$.
$(3)$ It must have a fully conjugated $\pi$-electron system.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(C) The reactivity of aromatic compounds towards electrophilic substitution depends on the electron-donating ability of the substituents attached to the benzene ring.
Groups that donate electrons via resonance (like $-OH$,$-OCH_3$,$-NHCOCH_3$) activate the ring.
The $-OH$ group is a strong activating group due to the $+M$ effect,which significantly increases the electron density on the benzene ring.
In $2$-methylphenol,the $-OH$ group is a strong activator,and the $-CH_3$ group is a weak activator (via hyperconjugation and $+I$ effect).
Comparing the given options:
$A$: $1,2$-dimethylbenzene (two weak activators).
$B$: $1$-methoxy-$2$-(hydroxymethyl)benzene (one strong activator $-OCH_3$ and one deactivating/weakly activating $-CH_2OH$).
$C$: $2$-methylphenol (one strong activator $-OH$ and one weak activator $-CH_3$).
$D$: $N$-($2$-methylphenyl)acetamide (one moderately activating $-NHCOCH_3$ and one weak activator $-CH_3$).
Among these,the $-OH$ group is the most powerful activating group,making $2$-methylphenol $(C)$ the most reactive towards electrophilic reagents.

(B) The molecular formula $C_6H_4Cl_2$ represents a dichlorobenzene derivative.
For a benzene ring,there are three possible positions for the two chlorine atoms relative to each other:
$1$. Ortho-dichlorobenzene ($1,2$-dichlorobenzene)
$2$. Meta-dichlorobenzene ($1,3$-dichlorobenzene)
$3$. Para-dichlorobenzene ($1,4$-dichlorobenzene)
These are the three distinct aromatic benzenoid isomers for $C_6H_4Cl_2$.

(B) For the compound having formula $C_7 H_8 O$,there are $5$ aromatic isomers:
$(1)$ Benzyl alcohol $(C_6H_5CH_2OH)$
$(2)$ $o$-methyl phenol ($2$-methylphenol)
$(3)$ $m$-methyl phenol ($3$-methylphenol)
$(4)$ $p$-methyl phenol ($4$-methylphenol)
$(5)$ Anisole (Methoxybenzene,$C_6H_5OCH_3$)
Hence,the correct option is $(B)$.

(D) The preparation of hexachlorobenzene $(C_6Cl_6)$ from benzene $(C_6H_6)$ involves the electrophilic substitution of all six hydrogen atoms of the benzene ring with chlorine atoms. This is achieved by reacting benzene with excess chlorine in the presence of a Lewis acid catalyst like anhydrous $FeCl_3$ or $AlCl_3$ in the dark.
Note: The reaction shown in the image provided is the free radical addition of chlorine to benzene to form benzene hexachloride ($C_6H_6Cl_6$,also known as gammaxene or lindane),which occurs under $UV$ light. Hexachlorobenzene $(C_6Cl_6)$ is a substitution product,not an addition product. Given the options provided,none correctly describe the substitution process for $C_6Cl_6$,but option $D$ describes the addition reaction shown in the image.

(B) The starting material is $p$-nitrotoluene. The $-CH_3$ group is ortho/para directing,while the $-NO_2$ group is meta directing. In $p$-nitrotoluene,the para position is already occupied by the $-NO_2$ group. Therefore,the incoming electrophile $(NO_2^+)$ from the nitration reaction will be directed to the ortho position with respect to the $-CH_3$ group. This position is also meta to the existing $-NO_2$ group,which further supports the substitution at this site. Thus,the product $A$ is $2,4$-dinitrotoluene (or $1$-methyl-$2,4$-dinitrobenzene).

(D) $1$. $n$-propyl bromide $(CH_3CH_2CH_2Br)$ reacts with $Na$ in dry ether (Wurtz reaction) to form $n$-hexane $(X)$: $2CH_3CH_2CH_2Br + 2Na \rightarrow CH_3(CH_2)_4CH_3 + 2NaBr$.
$2$. $n$-hexane $(X)$ undergoes catalytic aromatization with $V_2O_5$ at $773 \ K$ and $20 \ atm$ to form benzene $(Y)$: $C_6H_{14} \rightarrow C_6H_6 + 4H_2$.
$3$. Benzene $(Y)$ reacts with $Cl_2$ under $UV$ light at $500 \ K$ (free radical substitution/addition conditions). Given the context of industrial synthesis,this typically leads to chlorobenzene or hexachlorocyclohexane. However,standard textbook sequences for this path lead to chlorobenzene $(C_6H_5Cl)$ $(Z)$.
$4$. The molecular formula of chlorobenzene is $C_6H_5Cl$. The empirical formula is also $C_6H_5Cl$.
$5$. Empirical formula weight = $(6 \times 12) + (5 \times 1) + 35.5 = 72 + 5 + 35.5 = 112.5$.
$6$. Re-evaluating the sequence: If $Z$ is $C_6H_6Cl_6$ $(BHC)$,empirical formula is $CHCl$. Weight = $12 + 1 + 35.5 = 48.5$. This matches option $D$.

(A) The first step is the aromatization of $n$-heptane to toluene. $n$-Alkanes having six or more carbon atoms,when heated to $773 \ K$ at $10-20 \ atm$ pressure in the presence of oxides of vanadium $(V_2O_5)$,molybdenum $(Mo_2O_3)$,or chromium $(Cr_2O_3)$ supported over alumina,undergo dehydrogenation and cyclization to form benzene and its homologs. Thus,$X$ is $Cr_2O_3, 773 \ K, 10-20 \ atm$.
The second step is the free radical bromination of the side chain of toluene. Treatment of toluene with $Br_2$ in the presence of heat (or light) leads to the substitution of a hydrogen atom on the methyl group,resulting in the formation of benzyl bromide $(C_6H_5CH_2Br)$ as $Y$.

(B) $1$. The reaction of $n$-heptane $(CH_3(CH_2)_5CH_3)$ with $V_2O_5$ or $Cr_2O_3$ or $Mo_2O_3$ at $773 \ K$ and $10-20 \ atm$ pressure leads to aromatization,producing toluene $(C_7H_8)$.
$2$. The subsequent reaction of toluene with $Br_2$ in the presence of heat $(\Delta)$ or light undergoes free radical substitution at the benzylic position to form benzyl bromide $(C_6H_5CH_2Br)$.

(D) Key Idea: Functional groups that are electron-withdrawing due to resonance (often containing a double bond to a more electronegative atom) are meta-directing. Groups that donate electrons via resonance (due to the presence of lone pairs on the atom directly attached to the benzene ring) are ortho/para-directing.
$(A)$ $-COOH$: The carbon atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(B)$ $-NO_2$: The nitrogen atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(C)$ $-CHO$: The carbon atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(D)$ $-OCH_3$: The oxygen atom attached to the benzene ring has lone pairs of electrons,which it donates to the ring via resonance ($+M$ effect). Therefore,it is ortho/para-directing,not meta-directing.
Thus,option $(D)$ is the correct answer.

(C) $1$. Vinylbenzene $(C_6H_5-CH=CH_2)$ on oxidation with alkaline $KMnO_4$ followed by heating undergoes oxidative cleavage of the side chain to form Potassium benzoate $(C_6H_5COOK)$,which on acidification gives Benzoic acid $(C_6H_5COOH)$. Thus,$X$ is Benzoic acid.
$2$. Benzoic acid $(C_6H_5COOH)$ reacts with Sodalime $(NaOH+CaO)$ upon heating (decarboxylation) to form Benzene $(C_6H_6)$. Thus,$Y$ is Benzene.
$3$. Benzene can be prepared by the aromatisation (dehydrocyclisation) of $n-$hexane $(C_6H_{14})$ in the presence of catalysts like $Cr_2O_3/Al_2O_3$ at high temperature and pressure.

(D) The reaction of benzoic acid $(C_6H_5COOH)$ with $Br_2$ in the presence of $Fe$ (dark) is an electrophilic aromatic substitution. The $-COOH$ group is a deactivating and meta-directing group,so the major product $X$ is $m$-bromobenzoic acid.
The reaction of toluene $(C_6H_5CH_3)$ with $Br_2$ in the presence of $Fe$ (dark) is also an electrophilic aromatic substitution. The $-CH_3$ group is an activating and ortho/para-directing group. Due to steric hindrance at the ortho position,the major product $Y$ is $p$-bromotoluene.
Therefore,the major products are $m$-bromobenzoic acid and $p$-bromotoluene.

(A) Groups that donate electrons to the aromatic ring via the $+M$ (mesomeric) or $+H$ (hyperconjugative) effect are known as ortho/para directing groups or activating groups.
These groups increase the electron density specifically at the ortho and para positions,facilitating the attack of the electrophile at these sites.
Among the given groups,$-OH$,$-OCH_3$,and $-NHCOCH_3$ are electron-donating groups and are ortho-para directing.
Conversely,$-CN$,$-CO_2H$,and $-CHO$ are electron-withdrawing groups and are meta-directing.
Therefore,the correct set is $I, IV, V$.

(D) The starting material is $2-$methylhexane.
Upon reaction with $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$,it undergoes cyclization and aromatization to form toluene (compound $X$).
Reaction of toluene with $KMnO_4$ in alkaline medium followed by acidification $(H_3O^+)$ oxidizes the methyl group to a carboxylic acid group,forming benzoic acid (compound $Y$).
Nitration of benzoic acid using concentrated $HNO_3$ and $H_2SO_4$ at $< 60^{\circ}C$ introduces a nitro group at the meta-position due to the deactivating and meta-directing nature of the $-COOH$ group,forming $m-$nitrobenzoic acid (compound $Z$).
Comparing the structure of $Z$ with the given substituents,$P$ is $-COOH$ and $Q$ is $-NO_2$.

(A) When benzene reacts with excess chlorine in the presence of ultraviolet $(UV)$ light at $500 \ K$,addition reaction occurs.
The delocalized $\pi$-electron system of the benzene ring is broken,and one chlorine atom adds to each of the six carbon atoms.
This results in the formation of $1,2,3,4,5,6-\text{hexachlorocyclohexane}$,commonly known as benzene hexachloride $(BHC)$ or gammaxene.
Therefore,$(P) = \text{Benzene hexachloride}$.

(C) The reactivity of benzene derivatives towards electrophilic substitution depends on the nature of the substituent attached to the ring.
Electron-donating groups (activating groups) increase the electron density in the ring,thereby increasing reactivity.
Electron-withdrawing groups (deactivating groups) decrease the electron density in the ring,thereby decreasing reactivity.
- The $-OH$ group exerts a strong $+M$ (mesomeric) effect,which strongly activates the ring.
- The $-CH_3$ group exerts $+I$ (inductive) and $+H$ (hyperconjugation) effects,which moderately activate the ring.
- Benzene $(A)$ has no substituent.
- The $-NO_2$ group exerts a strong $-M$ and $-I$ effect,which strongly deactivates the ring.
Thus,the order of reactivity is: Phenol $(C)$ > Toluene $(B)$ > Benzene $(A)$ > Nitrobenzene $(D)$.

(A) $(I)$ In a nitrating mixture $(HNO_3 + H_2SO_4)$,$HNO_3$ acts as a base because it accepts a proton from $H_2SO_4$ to form the nitronium ion $(NO_2^+)$. Thus,statement $I$ is incorrect.
$(II)$ The $\sigma$-complex (arenium ion) is the intermediate formed during the electrophilic aromatic substitution of benzene when an electrophile $(E^+)$ attacks the benzene ring. In this complex,one carbon atom becomes $sp^3$-hybridized. Thus,statement $II$ is correct.
$(III)$ During the Friedel-Crafts alkylation of benzene with $n$-propyl chloride,the primary carbocation $(CH_3CH_2CH_2^+)$ rearranges to a more stable secondary carbocation $(CH_3CH^+CH_3)$. Consequently,the major product formed is isopropyl benzene. Thus,statement $III$ is correct.
Therefore,statements $II$ and $III$ are correct,making option $(A)$ the correct answer.

(B) Benzene reacts with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $333 \ K$ $(X)$ to undergo nitration,forming nitrobenzene.
Benzene reacts with fuming sulphuric acid $(H_2SO_4 + SO_3)$ $(Y)$ to undergo sulphonation,forming benzene sulphonic acid.

(D) The reaction sequence is as follows:
$1$. Hexane undergoes aromatization in the presence of $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$ to form benzene $(X)$.
$C_6H_{14} \xrightarrow{Cr_2O_3, 773 \ K, 10-20 \ atm} C_6H_6 + 4H_2$
$2$. Benzene $(X)$ then undergoes Friedel-Crafts acylation with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ to form acetophenone $(Y)$.
$C_6H_6 + CH_3COCl \xrightarrow{Anhydrous \ AlCl_3} C_6H_5COCH_3 + HCl$
Thus,$Y$ is acetophenone,which is $C_6H_5COCH_3$.

(D) The reaction sequence is as follows:
$1$. Hexane $(C_6H_{14})$ undergoes aromatization in the presence of $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$ pressure to form Benzene $(C_6H_6)$,which is $X$.
$\underset{\text{Hexane}}{C_6H_{14}}$ $\xrightarrow[\substack{773 \ K \\ 10-20 \ atm}]{Cr_2O_3} \underset{\text{Benzene}}{C_6H_6} (X)$
$2$. Benzene $(X)$ then undergoes Friedel-Crafts acylation with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ to form Acetophenone $(C_6H_5COCH_3)$,which is $Y$.
$\underset{\text{Benzene}}{C_6H_6}$ $\xrightarrow[\text{Anhydrous } AlCl_3]{CH_3COCl} \underset{\text{Acetophenone}}{C_6H_5COCH_3} (Y)$
This is an electrophilic substitution reaction.

(B) $1$. Ethylene $(CH_2=CH_2)$ reacts with $Br_2/CCl_4$ to form $A$,which is $1,2-dibromoethane$ $(CH_2Br-CH_2Br)$.
$2$. $1,2-dibromoethane$ reacts with $(i) Alc. KOH$ and $(ii) NaNH_2$ to undergo dehydrohalogenation,forming $B$,which is acetylene $(HC \equiv CH)$.
$3$. Acetylene undergoes cyclic polymerization in a hot metal tube to form $C$,which is benzene $(C_6H_6)$.
$4$. Benzene reacts with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $60^{\circ} C$ (nitration) to form $D$,which is nitrobenzene $(C_6H_5NO_2)$.

(B) In $C_{60}$, all carbon atoms are $sp^2$-hybridised.
It contains $12$ rings of five carbons each and $20$ rings of six carbons each.
It is an aromatic compound due to the delocalization of $\pi$-electrons in a spherical cage structure.
It is a pure form of carbon (allotrope).
The $C-C$ bond lengths are $143.5 \ pm$ and $138.3 \ pm$.
It is prepared by heating graphite in an electric arc under an inert atmosphere (e.g., $He$ gas), not in the presence of oxygen.
Thus, statements $(i)$, $(ii)$, $(iv)$, and $(v)$ are correct.