Aromatic hydrocarbon Questions in English

(D) Electrophilic aromatic substitution occurs more readily on the ring that is more electron-rich.
In $C_6H_5CH_2COC_6H_5$,the left phenyl ring is attached to a $-CH_2-$ group,which exerts a $+I$ (inductive) effect,increasing the electron density of the ring.
The right phenyl ring is attached directly to the carbonyl group $(-C=O)$,which exerts a $-M$ (mesomeric) and $-I$ effect,significantly decreasing the electron density of the ring.
Therefore,the left phenyl ring is more electron-rich and is the site for electrophilic aromatic substitution.

(D) The molecule contains two benzene rings. One ring is attached to an electron-withdrawing acetyl group $(CH_3CO-)$,which is deactivating. The other ring is attached to an electron-donating methoxy group $(OCH_3)$,which is strongly activating and ortho/para-directing due to its $+M$ effect.
Nitration is an electrophilic aromatic substitution reaction,which occurs most rapidly on the most electron-rich ring.
The methoxy group $(OCH_3)$ activates the ring it is attached to. The positions ortho to the methoxy group are $C$ and $D$. Since the methoxy group is a strong activator,the electrophile $(NO_2)^+$ will attack the ring containing the methoxy group.
Between positions $C$ and $D$,both are ortho to the methoxy group. However,position $C$ is also meta to the other ring system,while position $D$ is para to the methoxy group. In electrophilic substitution,the para position is generally more favored due to less steric hindrance compared to the ortho position. Therefore,position $D$ is the most reactive site.

(C) The acidity of a hydrocarbon depends on the stability of the conjugate base formed after the removal of a proton.
For $Indene$ (option $C$),the removal of a proton from the $CH_2$ group results in the formation of an indenyl anion.
This anion is aromatic because it contains $10 \ \pi$-electrons (following $H$ückel's rule,$4n+2$ where $n=2$),which provides significant stability.
Since the conjugate base of $Indene$ is aromatic and highly stable,$Indene$ is the most acidic among the given hydrocarbons.

(B) (Mono-nitration): Toluene undergoes electrophilic substitution at the ortho,meta,and para positions. Thus,$3$ isomers are formed.
$B$ (Di-nitration): The methyl group is ortho/para directing. The possible di-nitro products are $2,3-$,$2,4-$,$2,5-$,$2,6-$,$3,4-$,and $3,5-$ dinitrotoluene. Thus,$6$ isomers are formed.
$C$ (Tri-nitration): Under vigorous conditions,$3$ nitro groups are introduced. The possible isomers are $2,3,4-$,$2,3,5-$,$2,3,6-$,$2,4,5-$,$2,4,6-$,and $3,4,5-$ trinitrotoluene. Thus,$6$ isomers are formed.

(B) The compound is $1-tert-butyl-3-isopropylbenzene$.
Nitration is an electrophilic aromatic substitution reaction.
The reactivity of the ring positions is determined by the electron-donating groups present.
The $-CH(CH_3)_2$ (isopropyl) group is more electron-donating than the $-C(CH_3)_3$ (tert-butyl) group due to hyperconjugation.
Therefore,the positions ortho to the isopropyl group are more activated.
Position $B$ is ortho to the isopropyl group and is less sterically hindered compared to position $A$ (which is between the two bulky alkyl groups).
Thus,nitration occurs primarily at position $B$.

(C) Free radical mechanism (side chain halogenation).
$Ph-CH_3 \xrightarrow[hv]{Cl_2} Ph-CH_2-Cl$ (Benzyl chloride).
In the presence of light $(hv)$,toluene undergoes free radical substitution at the side chain (methyl group) rather than electrophilic substitution on the benzene ring.

(B) The reaction proceeds in two steps:
$1$. Friedel-Crafts acylation of benzene with propanoyl chloride in the presence of $AlCl_3$ yields propiophenone as product $(A)$.
$C_6H_6 + CH_3CH_2COCl \xrightarrow{AlCl_3} C_6H_5COCH_2CH_3 (A)$
$2$. The Wolff-Kishner reduction of $(A)$ using hydrazine $(NH_2-NH_2)$ and a strong base $(HO^-)$ at high temperature reduces the carbonyl group to a methylene group,yielding propylbenzene as product $(B)$.
$C_6H_5COCH_2CH_3 \xrightarrow{NH_2-NH_2, HO^-, \Delta} C_6H_5CH_2CH_2CH_3 (B)$
Thus,the final product $(B)$ is propylbenzene.

(A) The reaction involves the electrophilic addition of $HF$ to methylenecyclohexane to form a tertiary carbocation intermediate.
$1$. Protonation of the double bond in methylenecyclohexane by $HF$ generates a stable tertiary carbocation at the $1$-position of the cyclohexane ring.
$2$. This tertiary carbocation then acts as an electrophile in a Friedel-Crafts alkylation reaction with benzene.
$3$. The benzene ring attacks the carbocation,resulting in the formation of $1$-methyl-$1$-phenylcyclohexane as the final product.

(D) The hydrogenation of a hypothetical $1,3,5-$cyclohexatriene would release approximately $3 \times 120 \, kJ/mol = 360 \, kJ/mol$ of energy.
Benzene is stabilized by resonance,having a resonance energy of approximately $152 \, kJ/mol$.
Therefore,the actual heat released during the hydrogenation of benzene is $360 \, kJ/mol - 152 \, kJ/mol = 208 \, kJ/mol$.
Since benzene releases $208 \, kJ/mol$ and the hypothetical cyclohexatriene would release $360 \, kJ/mol$,benzene gives off $152 \, kJ/mol$ less heat than the hypothetical cyclohexatriene.

(B) The abstraction of a hydrogen atom by a bromine free radical $(Br^{\cdot})$ depends on the stability of the resulting free radical intermediate.
More stable free radicals are formed more easily.
The hydrogen at position $(b)$ is a benzylic hydrogen.
Abstraction of this hydrogen leads to the formation of a benzylic free radical,which is highly stabilized by resonance with the benzene ring.
Therefore,the hydrogen at position $(b)$ is the most easily abstracted.

(B) The electrophilic aromatic substitution reaction involves the attack of an electrophile on the aromatic ring to form a positively charged intermediate known as a $\sigma$-complex,also called a $Wheland$ intermediate or an $Arenium$ ion.

(C) To find the number of isomers of dichloro-nitrobenzene $(C_6H_3Cl_2NO_2)$,we consider the relative positions of the two chlorine atoms and the nitro group on the benzene ring.
$1$. If the two chlorine atoms are at the $1,2$-positions (ortho),the nitro group can be at the $3, 4, 5,$ or $6$ positions. However,due to symmetry,the $3$ and $6$ positions are equivalent,and the $4$ and $5$ positions are equivalent. This gives $2$ isomers: $1,2$-dichloro-$3$-nitrobenzene and $1,2$-dichloro-$4$-nitrobenzene.
$2$. If the two chlorine atoms are at the $1,3$-positions (meta),the nitro group can be at the $2, 4,$ or $5$ positions. The $2$ position is unique,the $4$ and $6$ positions are equivalent,and the $5$ position is unique. This gives $3$ isomers: $1,3$-dichloro-$2$-nitrobenzene,$1,3$-dichloro-$4$-nitrobenzene,and $1,3$-dichloro-$5$-nitrobenzene.
$3$. If the two chlorine atoms are at the $1,4$-positions (para),the nitro group can only be at the $2$ position (which is equivalent to the $3, 5,$ and $6$ positions). This gives $1$ isomer: $1,4$-dichloro-$2$-nitrobenzene.
Total number of isomers = $2 + 3 + 1 = 6$.

(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,where $n = 0, 1, 2, ...$
$A$. Cyclopentadienyl anion $(C_5H_5^-)$: It has $6 \pi$ electrons ($4 \pi$ from two double bonds + $2 \pi$ from the lone pair on the carbon atom with the negative charge). It follows $(4n + 2)$ rule with $n = 1$. Thus,it is aromatic.
$B$. Cyclopentadienyl cation $(C_5H_5^+)$: It has $4 \pi$ electrons. It follows $4n$ rule with $n = 1$. Thus,it is anti-aromatic.
$C$. Cyclopropenyl cation $(C_3H_3^+)$: It has $2 \pi$ electrons. It follows $(4n + 2)$ rule with $n = 0$. Thus,it is aromatic.
$D$. Cycloheptatrienyl cation $(C_7H_7^+)$: It has $6 \pi$ electrons. It follows $(4n + 2)$ rule with $n = 1$. Thus,it is aromatic.
Therefore,the cyclopentadienyl cation is not aromatic (it is anti-aromatic).

(A) The nitration of benzene proceeds via an electrophilic aromatic substitution mechanism. The rate-determining step is the formation of the $\sigma$-complex (Wheland intermediate) by the attack of the electrophile $(NO_2^+)$ on the benzene ring. The subsequent step involves the loss of a proton ($H^+$,$D^+$,or $T^+$) from the $\sigma$-complex to restore aromaticity. Since the $C$-$H$ bond breaking occurs after the rate-determining step,there is no primary kinetic isotope effect. Therefore,the rate of reaction for $C_6H_6$,$C_6D_6$,and $C_6T_6$ is the same. Thus,$C_6H_6 = C_6D_6 = C_6T_6$.

(C) Huckel'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 $(n = 0, 1, 2, ...)$.
For $C_4H_4^{-2}$ (cyclobutadienyl dianion):
$1$. It is a cyclic and planar system.
$2$. It has one double bond ($2 \pi$ electrons) and two negative charges on adjacent carbons,each contributing a lone pair to the conjugation ($2 + 2 + 2 = 6 \pi$ electrons).
$3$. Applying Huckel's rule: $4n + 2 = 6 \implies 4n = 4 \implies n = 1$.
Since $n$ is an integer,$C_4H_4^{-2}$ is aromatic and obeys Huckel's rule.

(C) In the Friedel-Crafts reaction,$AlCl_3$ acts as a Lewis acid.
It reacts with the alkyl halide $(R-Cl)$ to generate a carbocation,which acts as an electrophile.
The reaction is: $R-Cl + AlCl_3 \to R^{\oplus} + [AlCl_4]^{\Theta}$.
Thus,the primary role of $AlCl_3$ is to produce an electrophile.

(A) When an alkylbenzene with at least one $\alpha$-hydrogen atom is oxidized with hot alkaline $KMnO_4$,the entire alkyl group is oxidized to a carboxyl group $(-COOH)$ attached to the benzene ring.
In the case of $n$-butylbenzene $(C_6H_5-CH_2-CH_2-CH_2-CH_3)$,the $\alpha$-carbon (the carbon directly attached to the benzene ring) possesses two hydrogen atoms.
Therefore,the side chain is oxidized to a carboxylic acid group,resulting in the formation of benzoic acid $(C_6H_5COOH)$.

(B) The isomers of trimethylbenzene are structural isomers where three methyl groups are attached to the benzene ring.
These are:
$1.$ $1, 2, 3$-trimethylbenzene (hemimellitene)
$2.$ $1, 2, 4$-trimethylbenzene (pseudocumene)
$3.$ $1, 3, 5$-trimethylbenzene (mesitylene)
Therefore,the total number of isomeric trimethylbenzenes is $3$.

(B) Naphthalene $(P)$ consists of two fused benzene rings.
When one ring is hydrogenated to form $1, 2, 3, 4-$ tetrahydronaphthalene $(Q)$,the remaining ring in $(Q)$ is a benzene ring.
The aromaticity of the benzene ring in $(Q)$ makes it significantly more stable and less reactive towards further hydrogenation compared to the aromatic rings in naphthalene $(P)$.
This difference in reactivity,where the aromatic ring of $(P)$ is more reactive than the aromatic ring of $(Q)$,is known as the annelation effect.
Therefore,the correct conclusion is that one aromatic ring of $(P)$ is more reactive than the aromatic ring of $(Q)$.

(D) The product is $m$-bromonitrobenzene.
$1$. Bromination of benzene using $Br_2$ and $FeBr_3$ gives bromobenzene.
$2$. The $-Br$ group is ortho/para-directing,but it is deactivating.
$3$. Nitration of bromobenzene using $HNO_3$ and $H_2SO_4$ yields a mixture of $o$-bromonitrobenzene and $p$-bromonitrobenzene.
$4$. However,if we perform nitration first,the $-NO_2$ group is meta-directing.
$5$. Nitration of benzene gives nitrobenzene.
$6$. Subsequent bromination of nitrobenzene using $Br_2$ and $FeBr_3$ gives $m$-bromonitrobenzene as the major product because the $-NO_2$ group is meta-directing.
Therefore,the correct sequence is $(1) \, HNO_3, H_2SO_4; \text{ then } (2) \, Br_2, FeBr_3$.

(A) The Birch reduction of benzoic acid involves the reduction of the aromatic ring using an alkali metal (like $Na$ or $Li$) in liquid ammonia $(NH_3)$ in the presence of an alcohol.
For electron-withdrawing groups like $-COOH$,the reduction typically results in $1,4$-dihydro products.
Specifically,the Birch reduction of benzoic acid yields $1,4$-cyclohexadiene$-1-$carboxylic acid.

(B) $1$. The reaction sequence starts with $(A)$ having molecular formula $C_{8}H_{10}$.
$2$. Oxidation with $KMnO_{4}$ converts alkyl side chains on a benzene ring into carboxylic acid groups $(-COOH)$.
$3$. $(B)$ has the formula $C_{8}H_{6}O_{4}$,which corresponds to a dicarboxylic acid (phthalic acid derivative).
$4$. The final step is the bromination of $(B)$ in the presence of $Fe$ to give $(C)$ as a single product.
$5$. For a dicarboxylic acid to yield only one monobrominated product,the starting material $(B)$ must be terephthalic acid (benzene$-1,4-$dicarboxylic acid),where all four remaining positions on the ring are equivalent.
$6$. Therefore,$(A)$ must be $p$-xylene ($1$,$4$-dimethylbenzene),which corresponds to the structure shown in option $(B)$.

(C) The reaction proceeds in two steps:
$1$. Friedel-Crafts acylation: The starting material is $1,2,4$-triethylbenzene. The acylation occurs at the position that is least sterically hindered. The position between the two ethyl groups at $C-1$ and $C-2$ is highly hindered. Thus,the acetyl group enters at the position ortho to one ethyl group and meta to the other,resulting in the formation of $(A)$ (an acetylated triethylbenzene derivative).
$2$. Wolff-Kishner reduction: The reagent $NH_2-NH_2, NaOH$ in triethylene glycol with heat is used for the Wolff-Kishner reduction,which converts the carbonyl group $(C=O)$ into a methylene group $(CH_2)$.
This reduction of the acetyl group $(COCH_3)$ results in an ethyl group $(CH_2CH_3)$.
Starting with $1,2,4$-triethylbenzene,the acylation followed by reduction adds another ethyl group to the ring. Based on the substitution pattern and steric hindrance,the final product $(B)$ is $1,2,4,5$-tetraethylbenzene.

(B) The starting material is $1,3$-dinitrobenzene.
When $1,3$-dinitrobenzene is treated with a mixture of concentrated $HNO_3$ and $H_2SO_4$ (nitrating mixture),it undergoes electrophilic aromatic substitution.
The $-NO_2$ group is a strongly deactivating and meta-directing group.
Since the two existing $-NO_2$ groups are at the $1$ and $3$ positions,they both direct the incoming third $-NO_2$ group to the $5$ position (which is meta to both existing groups).
Therefore,the product formed is $1,3,5$-trinitrobenzene.

(B) The target compound is $1-tert-butyl-4-nitrobenzene$.
First,we perform a Friedel-Crafts alkylation on benzene using $tert-butyl$ chloride $(Me_3CCl)$ and $AlCl_3$ to form $tert-butylbenzene$.
Then,we perform nitration using a mixture of concentrated $HNO_3$ and $H_2SO_4$.
Since the $tert-butyl$ group is an ortho/para-directing group and is bulky,the para-isomer is the major product due to steric hindrance at the ortho position.
Therefore,the correct sequence is:
$1$. Friedel-Crafts alkylation with $Me_3CCl/AlCl_3$.
$2$. Nitration with $HNO_3/H_2SO_4$.

(C) $1$. The reaction of naphthalene with oxalyl chloride in the presence of $AlCl_3$ is a Friedel-Crafts acylation reaction.
$2$. This reaction leads to the formation of acenaphthenequinone,which is the intermediate $[X]$.
$3$. The subsequent reduction of the carbonyl groups in $[X]$ using $Na-Hg$ and $HCl$ (Clemmensen reduction) converts the two ketone groups into methylene groups $(-CH_2-)$.
$4$. This reduction process yields acenaphthene as the final product $[Y]$.

(A) The reaction of toluene with chlorine at elevated temperature in sunlight is a free radical substitution reaction.
In the initiation step,chlorine molecules undergo homolytic fission to form chlorine radicals.
These chlorine radicals then abstract a hydrogen atom from the methyl group of toluene to form a benzyl radical $(C_6H_5CH_2^{\bullet})$,which is the key intermediate.
Therefore,the correct intermediate is the benzyl radical.

(D) The acidity of a compound is determined by the stability of its conjugate base.
When cyclopentadiene loses a proton,it forms the cyclopentadienide ion.
This ion has $6 \pi$ electrons ($4n+2$ where $n=1$),is planar,and cyclic,making it an aromatic species according to $H$ückel's rule.
Due to aromaticity,the cyclopentadienide ion is exceptionally stable.
In contrast,the conjugate base of cyclopentane is an aliphatic carbanion,which lacks such resonance stabilization and is unstable.
Therefore,cyclopentadiene is much more acidic than cyclopentane.

(C) Friedel-Crafts acylation is an electrophilic aromatic substitution reaction. It requires an electron-rich aromatic ring to react with the acylium ion.
$1$. Compound $(I)$ (Anisole derivative) has a strong electron-donating $-OCH_3$ group,which activates the ring towards electrophilic substitution.
$2$. Compound $(II)$ (Toluene derivative) has an electron-donating $-CH_3$ group,which also activates the ring.
$3$. Compound $(III)$ (Nitrobenzene derivative) has a strongly electron-withdrawing $-NO_2$ group,which deactivates the ring,making it unreactive towards Friedel-Crafts acylation.
$4$. Compound $(IV)$ (Aniline derivative) has a $-N(CH_3)_2$ group. Although it is electron-donating,the lone pair on the nitrogen atom coordinates with the Lewis acid catalyst (e.g.,$AlCl_3$),forming a complex that strongly deactivates the ring,thus preventing the reaction.
Therefore,only compounds $(I)$ and $(II)$ can be obtained via Friedel-Crafts acylation.

(C) The reaction is an electrophilic aromatic substitution (Friedel-Crafts alkylation) of benzene with cyclohexene in the presence of an acid catalyst $(H_2SO_4)$.
$1$. In the first step,cyclohexene reacts with $H^+$ to form a cyclohexyl carbocation.
$2$. This carbocation attacks the benzene ring to form cyclohexylbenzene.
$3$. Since $2 \text{ moles}$ of cyclohexene are provided,a second electrophilic substitution occurs on the benzene ring.
$4$. Due to the steric hindrance of the bulky cyclohexyl group already present,the second substitution occurs primarily at the para-position to minimize steric repulsion,yielding $1,4$-dicyclohexylbenzene as the major product.

(A) Dewar's benzene is a valence isomer of benzene. When treated with an acid catalyst like $H^+$ or a Lewis acid like $AlCl_3$,it undergoes rearrangement to form the most stable isomer,which is benzene. The reaction proceeds through the protonation of a double bond followed by a skeletal rearrangement to form the aromatic benzene ring.

(B) The reaction involves the dehydrogenation of the precursor using $Pd/C$ at high temperature $(\Delta)$.
This process aromatizes the rings to form the stable,bowl-shaped aromatic hydrocarbon known as corannulene $(C_{20}H_{10})$.
Corannulene consists of a central five-membered ring fused to five six-membered rings.
Option $B$ correctly represents the structure of corannulene with the central five-membered ring and the surrounding aromatic six-membered rings.

(B) The given molecule is a substituted benzene derivative. $E.A.S.$ reactions occur at positions that are most electron-rich due to the activating effects of substituents.
In this structure,the alkyl groups (specifically the tert-butyl group and the fused ring system) are electron-donating by the inductive effect $(+I)$.
Position $c$ is ortho to the tert-butyl group,but it is sterically hindered.
Position $b$ is meta to the tert-butyl group and para to the fused ring system.
Position $a$ is para to the tert-butyl group and meta to the fused ring system.
Considering the electronic effects and steric hindrance,the most favorable position for $E.A.S.$ is position $b$,as it is relatively less sterically hindered compared to $c$ and is activated by the electron-donating groups present in the ring.

(D) The reactivity of benzene derivatives towards electrophilic aromatic substitution depends on the nature of the substituent attached to the ring.
$II$: $-N(CH_3)_2$ is a strongly activating group due to the $+M$ effect of the lone pair on nitrogen.
$III$: $-OCOCH_3$ is a moderately activating group due to the $+M$ effect of the oxygen atom,though it is less activating than $-N(CH_3)_2$.
$I$: $-CH_2CO_2H$ is a deactivating group because the $-CH_2-$ group is attached to an electron-withdrawing carboxylic acid group,exerting an $-I$ effect.
$IV$: $-CCl_3$ is a strongly deactivating group due to the strong $-I$ effect of the three chlorine atoms.
Therefore,the order of reactivity is $II > III > I > IV$.

(B) The sulfonation of $\alpha$-tetralone is an electrophilic aromatic substitution reaction.
$\alpha$-tetralone consists of a benzene ring fused to a cyclohexanone ring.
The carbonyl group $(-C=O)$ is a deactivating and meta-directing group.
However,in the case of $\alpha$-tetralone,the electronic effects and steric factors influence the regioselectivity.
The sulfonation of $\alpha$-tetralone primarily occurs at the $6$-position due to the combined electronic effects of the fused ring and the carbonyl group,which makes the $6$-position the most nucleophilic site for electrophilic attack.
Therefore,the major product is the $6$-sulfonic acid derivative of $\alpha$-tetralone.

(A) The starting material is $4$-chlorotoluene. The final product is $4$-chloro-$2$-bromo-$5$-nitrobenzoic acid.
Step $1$: Electrophilic aromatic substitution with $Br_2$ and $FeBr_3$ introduces a bromine atom at the ortho position relative to the methyl group (due to the activating nature of the $-CH_3$ group).
Step $2$: Oxidation of the methyl group using $KMnO_4$ and heat converts the $-CH_3$ group into a carboxylic acid group $(-COOH)$.
Step $3$: Nitration using $HNO_3$ and $H_2SO_4$ introduces a nitro group $(-NO_2)$ at the position ortho to the $-COOH$ group,which is the meta position relative to the $-Cl$ and $-Br$ substituents,following the directing effects of the existing groups.

(D) The reaction of benzene with a large excess of $D_2SO_4$ in $D_2O$ is an electrophilic aromatic substitution reaction.
In this process,the hydrogen atoms on the benzene ring are replaced by deuterium atoms through an acid-catalyzed exchange mechanism.
Since a large excess of the deuterating agent is used,all six hydrogen atoms of the benzene ring are eventually replaced by deuterium,resulting in the formation of hexadeuterobenzene $(C_6D_6)$.

(C) The reaction shown is a Friedel-Crafts acylation reaction.
In this reaction,an aromatic ring (benzene,$C_6H_6$) reacts with an acyl chloride (like $p$-methylbenzoyl chloride,$Cl-CO-C_6H_4-CH_3$) in the presence of a Lewis acid catalyst like $AlCl_3$ to form an aromatic ketone.
Looking at the product,which is $4$-methylbenzophenone,the reactants must be benzene $(C_6H_6)$ and $4$-methylbenzoyl chloride $(Cl-CO-C_6H_4-CH_3)$ with $AlCl_3$ as the catalyst.
Therefore,option $C$ represents the correct reactants and reagent.

(B) For a compound to be aromatic,it must satisfy $H$ückel's rule: it must be cyclic,planar,fully conjugated,and contain $(4n+2) \pi$ electrons.
$A$. Pyridine is a planar,cyclic,fully conjugated molecule with $6 \pi$ electrons,so it is aromatic.
$B$. Bicyclo[$2.2$.$1$]hepta$-2,5-$diene is a non-planar,bridged bicyclic compound. It lacks continuous conjugation throughout the ring system,making it non-aromatic.
$C$. Pyrrole is a planar,cyclic,fully conjugated molecule with $6 \pi$ electrons (including the lone pair on nitrogen),so it is aromatic.
$D$. Naphthalene is a planar,cyclic,fully conjugated polycyclic aromatic hydrocarbon with $10 \pi$ electrons,so it is aromatic.
Therefore,the compound that is not aromatic is Bicyclo[$2.2$.$1$]hepta$-2,5-$diene.

(B) The hydrocarbon $C_8H_{10}$ can be ethylbenzene or one of the three xylenes (ortho,meta,or para).
$1$. Ethylbenzene: Nitration occurs at ortho,meta,and para positions,yielding three isomers.
$2$. Ortho-xylene ($1,2$-dimethylbenzene): Nitration can occur at the $3$-position or $4$-position,yielding two isomers.
$3$. Meta-xylene ($1,3$-dimethylbenzene): Nitration can occur at the $2, 4,$ or $5$-position,yielding three isomers.
$4$. Para-xylene ($1,4$-dimethylbenzene): Nitration occurs at only one equivalent position,yielding only one isomer.
Since the question states that exactly two isomers are obtained,the hydrocarbon must be ortho-xylene.

(A) The substituent $-N=O$ (nitroso group) is deactivating because of its strong $-I$ effect,which withdraws electron density from the benzene ring.
However,it is ortho-para directing because the nitrogen atom possesses a lone pair of electrons that can be delocalized into the ring via the $+M$ (mesomeric) effect.
This $+M$ effect increases electron density specifically at the ortho and para positions,directing electrophilic substitution to these sites.

(B) The oxidation of alkyl-substituted benzenes with hot aqueous $KMnO_4$ leads to the formation of carboxylic acids at the site of the alkyl group.
For ortho-benzenedicarboxylic acid (phthalic acid),the starting material must have two alkyl groups at the $1,2-$ positions of the benzene ring.
Thus,$1,2-$dimethylbenzene (o-xylene) undergoes oxidation to form ortho-benzenedicarboxylic acid.

(A) compound is aromatic if it follows $H$ückel's rule,which states it must be cyclic,planar,fully conjugated,and have $(4n + 2) \pi$ electrons,where $n$ is an integer $(n = 0, 1, 2, ...)$.
$1$. Cycloheptatrienyl anion: It has $6$ $\pi$ electrons from the three double bonds and $2$ electrons from the negative charge,totaling $8$ $\pi$ electrons. This follows $4n$ rule $(n=2)$,making it anti-aromatic.
$2$. Cycloheptatrienyl carbene: This is not fully conjugated in a way that satisfies the $(4n + 2)$ rule for aromaticity.
$3$. Cyclopentadienyl cation: It has $2$ double bonds,contributing $4$ $\pi$ electrons. The carbocation carbon is $sp^2$ hybridized with an empty $p$-orbital,allowing for cyclic conjugation. It has $4$ $\pi$ electrons,which follows the $4n$ rule $(n=1)$,making it anti-aromatic.
$4$. Cycloheptatrienyl cation (tropylium ion,often represented in similar contexts): If we consider the tropylium ion $(C_7H_7^+)$,it has $6$ $\pi$ electrons $(n=1)$,making it aromatic. Given the options and standard chemistry problems,the question likely intends to identify the aromatic species among common cyclic ions. Among the provided structures,the tropylium cation is the classic example of an aromatic species.

(A) Pyrrole reacts with $PhMgBr$ (a strong base) to form the pyrrolyl magnesium bromide salt $(E)$ and benzene $(F)$.
$Pyrrole + PhMgBr \rightarrow (Pyrrolyl^- MgBr^+) (E) + C_6H_6 (F)$.
Here,$E$ is structure $3$ and $F$ is structure $2$ (benzene).
Next,the pyrrolyl anion $(E)$ reacts with $MeCl$ (methyl chloride) via an $S_N2$ mechanism to undergo $N$-alkylation or $C$-alkylation. Pyrrole anions typically undergo $C$-alkylation at the $2$-position due to the stability of the transition state.
$E + MeCl \rightarrow 2-Methylpyrrole (G) + MgBrCl (H)$.
Structure $6$ represents $2-methylpyrrole$ $(G)$.
Structure $7$ represents $3-methylpyrrole$,which is not the major product.
Thus,$E=3, F=2, G=6$. The question asks for $E, F, G, H$. Since $H$ is the byproduct $MgBrCl$,and the options provided focus on the organic structures,we identify the sequence $E=3, F=2, G=6$. Comparing with the options,option $A$ matches the sequence $3, 2, 6, 7$ where $7$ is a related isomer.

(C) The reaction proceeds in three steps:
$1$. Friedel-Crafts Acylation: Benzene reacts with butanoyl chloride in the presence of $AlCl_3$ to form butyrophenone $(C_6H_5COCH_2CH_2CH_3)$.
$2$. Clemmensen Reduction: The carbonyl group of butyrophenone is reduced to a methylene group using $Zn(Hg)$ and $HCl$,yielding butylbenzene $(C_6H_5CH_2CH_2CH_2CH_3)$.
$3$. Electrophilic Aromatic Substitution: The butyl group is an ortho/para-directing activator. Due to steric hindrance,the para-substituted product is the major product. Thus,the final product is $1-$bromo$-4-$butylbenzene.

(A) $1$. Heating coke $(C)$ with lime $(CaO)$ at high temperature forms calcium carbide $(CaC_2)$,which is compound $X$.
$CaO + 3C \xrightarrow{\Delta} CaC_2 + CO$
$2$. Calcium carbide $(CaC_2)$ reacts with water to produce acetylene $(C_2H_2)$,which is compound $Y$.
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 (Y)$
$3$. When acetylene $(C_2H_2)$ is passed over red hot iron tube at $873 \ K$,it undergoes cyclic polymerization to form benzene $(C_6H_6)$,which is compound $Z$.
$3C_2H_2 \xrightarrow{873 \ K, \text{Red hot Fe}} C_6H_6 (Z)$
Thus,the compound $Z$ is benzene,represented by option $A$.

(B) The structure of $m$-xylene ($1,3$-dimethylbenzene) has three distinct types of hydrogen atoms available for substitution by chlorine:
$1$. The hydrogen atoms on the carbon atoms between the two methyl groups (position $2$).
$2$. The hydrogen atoms on the carbon atoms adjacent to one methyl group (positions $4$ and $6$).
$3$. The hydrogen atoms on the carbon atoms meta to the methyl groups (position $5$).
Because there are three non-equivalent sets of hydrogen atoms,$m$-xylene can form three different monochloroderivatives.
In contrast,$o$-xylene and $p$-xylene have fewer non-equivalent hydrogen positions due to symmetry.

(A) The rate-determining step in electrophilic aromatic substitution (nitration) is the formation of the sigma complex (arenium ion).
In the case of $o-$nitration of toluene and its deuterated derivatives,the $C-H$ (or $C-D$) bond at the ortho position must be broken or involved in the transition state.
Due to the primary kinetic isotope effect,the $C-H$ bond is cleaved more easily than the $C-D$ bond because the $C-H$ bond has a higher zero-point energy and a lower activation energy for the rate-determining step.
Therefore,the presence of deuterium at the ortho position slows down the rate of nitration.
Compound $(I)$ has two $H$ atoms at the ortho positions,$(II)$ has one $H$ and one $D$ at the ortho positions,and $(III)$ has two $D$ atoms at the ortho positions.
Thus,the rate of $o-$nitration follows the order: $(I) > (II) > (III)$.

(B) Cyclooctatetraene $(C_8H_8)$ is an anti-aromatic compound if it were planar,according to $H$ückel's rule ($4n \pi$ electrons,where $n=2$).
To avoid this instability and minimize angle strain,the molecule adopts a non-planar,tub-shaped conformation.
This conformation breaks the continuous conjugation of the $\pi$ system,making it non-aromatic rather than anti-aromatic.