Aromatic hydrocarbon Questions in English

(C) In the nitration of benzene,the mixture of concentrated $H_2SO_4$ and $HNO_3$ is used.
$H_2SO_4$ is a stronger acid than $HNO_3$,so it donates a proton $(H^+)$ to $HNO_3$.
$H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H_2NO_3^+$
Here,$H_2SO_4$ acts as an acid (proton donor) and $HNO_3$ acts as a base (proton acceptor).
The $H_2NO_3^+$ species then loses a water molecule to form the electrophile,the nitronium ion $(NO_2^+)$:
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$

(A) The acidity of a compound depends on the stability of its conjugate base.
When cyclopentadiene loses a proton $(H^+)$,it forms a cyclopentadienyl anion.
This anion has $6 \pi$ electrons and is cyclic and planar,making it aromatic according to $H$ückel's rule.
Since the conjugate base is aromatic,it is exceptionally stable,which makes cyclopentadiene the strongest acid among the given options.

(C) Sulphonation is an electrophilic aromatic substitution reaction. The rate of this reaction 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 susceptible to electrophilic attack.
$1$. $CH_3$ group in Toluene is an electron-donating group ($+I$ and hyperconjugation effect),which activates the benzene ring.
$2$. $Cl$ in Chlorobenzene is deactivating due to its strong $-I$ effect.
$3$. $NO_2$ in Nitrobenzene is a strongly deactivating group due to its $-I$ and $-M$ effects.
$4$. Benzene has no substituent.
Therefore,Toluene is the most reactive towards electrophilic substitution among the given options.

(C) To determine if a compound is aromatic,it must follow $H$ückel's rule:
$1$. The molecule must be cyclic.
$2$. The molecule must be planar.
$3$. The molecule must be fully conjugated.
$4$. The molecule must have $(4n + 2) \pi$ electrons,where $n = 0, 1, 2, ...$
Let's analyze each option:
- $A$: Cycloheptatrienyl cation has $6 \pi$ electrons $(n=1)$,is cyclic,planar,and conjugated. It is aromatic.
- $B$: Cyclopropenyl cation has $2 \pi$ electrons $(n=0)$,is cyclic,planar,and conjugated. It is aromatic.
- $C$: Cyclooctatetraene has $8 \pi$ electrons. It is not planar (it adopts a tub shape) and does not follow the $(4n + 2) \pi$ rule. It is non-aromatic.
- $D$: Cyclopentadienyl anion has $6 \pi$ electrons $(n=1)$,is cyclic,planar,and conjugated. It is aromatic.
Therefore,the compound that is not aromatic is cyclooctatetraene.

(B) compound is aromatic if it is cyclic,planar,fully conjugated,and follows $H$ückel's rule ($4n+2$ $\pi$ electrons).
$A$ (Cyclopentadienyl anion): It has $6$ $\pi$ electrons ($4$ from double bonds + $2$ from the lone pair on the $sp^3$ carbon,which becomes $sp^2$ due to conjugation). It is cyclic,planar,and aromatic.
$B$ (Tropylium cation): It has $6$ $\pi$ electrons. It is cyclic,planar,and aromatic.
$C$ (Cyclooctatetraene derivative): It is non-planar (tub-shaped) to avoid anti-aromaticity,hence it is non-aromatic.
$D$ (Cycloheptatrienyl anion): It has $8$ $\pi$ electrons ($4n$ system),making it anti-aromatic.
Therefore,only $A$ and $B$ are aromatic.

(B) Benzenoid compounds are organic compounds that contain at least one benzene ring.
$A$ (Phenanthrene),$C$ (Naphthalene),and $D$ (Aniline) all contain benzene rings in their structures.
$B$ (Cyclooctatetraene) is an eight-membered cyclic polyene $(C_8H_8)$ that does not contain a benzene ring.
Therefore,cyclooctatetraene is not a benzenoid compound.

(C) $[10]$ Annulene,although it follows the $(4n + 2) \pi$ electron rule,is non-aromatic due to its non-planar nature.
It is non-planar because of the steric repulsion between the two $C-H$ bonds present inside the ring.

(C) When benzene $(C_{6}H_{6})$ reacts with excess chlorine $(Cl_{2})$ in the presence of sunlight ($UV$ light),an addition reaction occurs.
This reaction results in the formation of benzene hexachloride $(C_{6}H_{6}Cl_{6})$,also known as gammaxene or lindane.
The product $X$ is $C_{6}H_{6}Cl_{6}$.
In this molecule,each of the $6$ carbon atoms is bonded to one hydrogen atom and one chlorine atom.
Therefore,the total number of hydrogen atoms in $X$ is $6$.

(D) $[6]-$Annulene (benzene) is aromatic as it is planar,cyclic,and has $6\pi$ electrons ($4n+2$ rule,$n=1$).
$[8]-$Annulene is non-aromatic because it is non-planar (tub-shaped) to avoid angle strain,despite having $8\pi$ electrons ($4n$ rule).
Cis$-10-$Annulene is non-aromatic because the internal hydrogen atoms at positions $1$ and $6$ experience steric hindrance,forcing the molecule out of planarity,which prevents continuous conjugation.
Thus,Assertion $A$ is incorrect because it labels cis$-10-$Annulene as aromatic.
Reason $R$ is correct because planarity is indeed a fundamental requirement for a molecule to be aromatic ($H$ückel's rule).
Therefore,$A$ is not correct but $R$ is correct.

(B) The reactivity towards electrophilic aromatic substitution (like nitration) depends on the electron density of the benzene ring.
$1$. $-NO_2$ (in $D$) is a strongly electron-withdrawing group (deactivating),making the ring least reactive.
$2$. $-Br$ (in $B$) is an electron-withdrawing group due to the inductive effect (deactivating),but less so than $-NO_2$.
$3$. $E$ (benzene) is the reference point.
$4$. $-CH_3$ groups are electron-donating (activating) due to hyperconjugation and inductive effects.
$5$. $p$-xylene $(A)$ has two $-CH_3$ groups,and mesitylene $(C)$ has three $-CH_3$ groups. More electron-donating groups increase the reactivity further.
Thus,the order of increasing reactivity is: $D < B < E < A < C$.

(C) Ozonolysis of a benzene ring with fixed double bonds involves the cleavage of the $C=C$ bonds to form carbonyl compounds.
For $1,2$-dimethylbenzene ($o$-xylene),the two resonance structures have different positions of the double bonds relative to the methyl groups.
In the first structure,the double bonds are between $C_1-C_2$,$C_3-C_4$,and $C_5-C_6$. Ozonolysis yields $CH_3-CO-CO-CH_3$ and $2 \ CHO-CHO$.
In the second structure,the double bonds are between $C_2-C_3$,$C_4-C_5$,and $C_6-C_1$. Ozonolysis yields $2 \ CH_3-CO-CHO$ and $CHO-CHO$.
Since the products are different,the pair of $o$-xylene structures gives different products upon ozonolysis.

(D)
Non-benzenoid aromatic compounds are those compounds which do not contain a benzene ring in their structure. They exhibit aromaticity due to the presence of alternate $\pi$-bonds in the cyclic system.
$o$-xylene,Phenanthrene,and Indole all contain at least one benzene ring fused or substituted in their structure.
Thiophene is a five-membered heterocyclic compound containing sulfur,which does not contain a benzene ring,yet it is aromatic.
Therefore,Thiophene is a non-benzenoid aromatic compound.

(C)
In electrophilic aromatic substitution reactions of chlorobenzene,the $ortho/para$-directing ability of chlorine is due to its $+R$-effect.
Chlorine has lone pairs of electrons that can be delocalized into the benzene ring through resonance.
The resonating structures of chlorobenzene are as follows:
Due to this effect,the electron density increases more at $ortho$ and $para$-positions than at $meta$-position,making these positions more susceptible to electrophilic attack.

(C) The reaction is a Friedel-Crafts alkylation. The alkyl halide $CH_3-CH(CH_3)-CH_2Br$ reacts with $AlCl_3$ to form a primary carbocation,which undergoes a $1,2-$hydride shift to form a more stable tertiary carbocation,$(CH_3)_3C^+$. This tertiary carbocation acts as the electrophile and attacks the toluene ring. Since the $-CH_3$ group is ortho/para directing and the tert-butyl group is bulky,the para-substituted product is the major product due to less steric hindrance. The major product is $1-$methyl$-4-(1,1-$dimethylethyl$)$benzene.

(C) The reaction of benzene with succinic anhydride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction. This produces $X$,which is $4$-oxo-$4$-phenylbutanoic acid (also known as $\beta$-benzoylpropionic acid).
In the second step,$X$ undergoes intramolecular cyclization (Friedel-Crafts acylation) in the presence of phosphoric acid $(H_3PO_4)$ and heat $(\Delta)$. This results in the formation of $Y$,which is $1$-tetralone ($3,4$-dihydro-$1(2H)$-naphthalenone).
Therefore,$X$ is $4$-oxo-$4$-phenylbutanoic acid and $Y$ is $1$-tetralone.

(D) When benzene reacts with an electrophile $E^{+}$,the electrophile attacks the $\pi$-electron system of the benzene ring to form a resonance-stabilized carbocation intermediate known as the $\sigma$-complex or arenium ion. In this intermediate,the carbon atom at the site of attack becomes $sp^3$ hybridized,bearing both the electrophile $E$ and the hydrogen atom $H$. The positive charge is delocalized over the remaining five carbon atoms of the ring,specifically at the ortho and para positions relative to the site of attack. Among the given options,the structure that correctly represents the delocalization of the positive charge in the arenium ion is option $(d)$.

(D) The correct answer is $ (D) $.
Friedel-Crafts acylation is an electrophilic aromatic substitution reaction. In this process,an acyl group $( RCO- )$ is introduced into an aromatic ring by reacting the aromatic compound with an acyl chloride or an acid anhydride in the presence of a Lewis acid catalyst,such as $ AlCl_3 $.
The reaction can be represented as:
$ C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl $

(C) compound is aromatic if it satisfies the following conditions:
$1$. It must be cyclic.
$2$. It must be planar.
$3$. It must have a fully conjugated system of $\pi$ electrons.
$4$. It must follow $H$ückel's rule,i.e.,it must contain $(4n+2) \pi$ electrons,where $n = 0, 1, 2, ...$.
Let us analyze the given options:
- Option $(A)$: The cyclopentadienyl cation has $4 \pi$ electrons. It is anti-aromatic.
- Option $(B)$: The cyclohexadienyl cation is non-planar and non-aromatic.
- Option $(C)$: The tropylium cation (cycloheptatrienyl cation) has $6 \pi$ electrons $(n=1)$. It is cyclic,planar,and fully conjugated. Thus,it is aromatic.
- Option $(D)$: The cyclooctatrienyl cation is non-planar and non-aromatic.
Therefore,the correct option is $(C)$.

(D) The aromatic compounds are those that satisfy the following conditions:
$(i)$ The compound must be cyclic and planar.
$(ii)$ There must be complete delocalization of $\pi$ electrons in the ring.
$(iii)$ It must follow $H$ückel's rule,i.e.,it must contain $(4n+2) \pi$ electrons,where $n$ is an integer $(n = 0, 1, 2, \dots)$.
Analysis of the given structures:
$(I)$ Fulvene: It is non-aromatic because the exocyclic double bond makes the structure non-planar or prevents continuous conjugation.
$(II)$ Azulene: It has $10 \pi$ electrons ($4n+2$ with $n=2$),is cyclic,and planar. It is aromatic.
$(III)$ Cycloheptatriene: It is non-aromatic because one carbon atom is $sp^3$ hybridized,which breaks the continuous conjugation.
$(IV)$ Thiophene: It has $6 \pi$ electrons ($4n+2$ with $n=1$) including the lone pair on sulfur participating in conjugation. It is aromatic.
Thus,compounds $(II)$ and $(IV)$ are aromatic.
Therefore,the correct option is $(d).$

(A) The correct answer is $A$.
This reaction is a Friedel-Crafts acylation.
In this reaction,benzene reacts with an acyl halide (such as $2$-nitrobenzoyl chloride) in the presence of a Lewis acid catalyst like $AlCl_3$ to yield the corresponding acylbenzene derivative.

(A) The reaction sequence is as follows:
$1$. Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ (a Lewis acid catalyst) to form toluene. This is a Friedel-Crafts alkylation reaction.
$2$. Toluene then undergoes nitration using a nitrating mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ to form $2$-nitrotoluene (ortho-isomer) and $4$-nitrotoluene (para-isomer).
Therefore,$X = CH_3Cl$,$Y = \text{Anhydrous } AlCl_3$,and $Z = HNO_3 + H_2SO_4$.

(A) The reaction sequence is as follows:
$1$. Heating coke $(C)$ with lime $(CaO)$ gives calcium carbide $(CaC_2)$,which is compound $X$.
$CaO + 3C \xrightarrow{\Delta} CaC_2 + CO$
$2$. Calcium carbide reacts with water to form acetylene $(C_2H_2)$,which is compound $Y$.
$CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2$
$3$. Passing acetylene over red-hot iron at $873 \ K$ leads to cyclic trimerization to produce benzene $(C_6H_6)$,which is compound $Z$.
$3C_2H_2 \xrightarrow{\text{Fe, } 873 \ K} C_6H_6$ (Benzene)
Thus,the correct option is $A$.

(C) The reactivity of aromatic compounds towards electrophilic aromatic substitution like Friedel-Crafts alkylation depends on the electron density of 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.
(iv) Anisole $(-OCH_3)$ has an $EDG$ which strongly activates the ring.
(ii) Benzene is the reference compound.
(iii) Methyl benzoate $(-COOCH_3)$ has an $EWG$ which deactivates the ring.
$(i)$ Nitrobenzene $(-NO_2)$ has a very strong $EWG$ which strongly deactivates the ring.
The order of reactivity is: $(iv) > (ii) > (iii) > (i)$.

(B) The conditions for a compound to be aromatic are:
$(i)$ The molecule should be planar.
$(ii)$ It should be cyclic with a continuous system of conjugated $\pi$ electrons.
$(iii)$ It should follow Huckel's rule,i.e.,it should have $(4n+2) \pi$ electrons,where $n = 0, 1, 2, \dots$
Let us analyze the given compounds:
$(i)$ Pyrrole: It has $6 \pi$ electrons ($4$ from double bonds + $2$ from the lone pair on $N$),which follows $(4n+2)$ rule $(n=1)$. It is aromatic.
$(ii)$ Pentalene: It has $8 \pi$ electrons. This follows the $4n$ rule $(n=2)$,making it anti-aromatic,not aromatic.
$(iii)$ Azulene: It has $10 \pi$ electrons,which follows $(4n+2)$ rule $(n=2)$. It is aromatic.
$(iv)$ Oxazole: It has $6 \pi$ electrons ($4$ from double bonds + $2$ from the lone pair on $O$),which follows $(4n+2)$ rule $(n=1)$. It is aromatic.
Thus,compound $(ii)$ is not aromatic.

(C) Assertion $A$: Benzene is more stable than hypothetical cyclohexatriene due to resonance energy. This statement is true.
Reason $R$: The delocalized $\pi$ electron cloud in benzene is spread over the entire ring and is held more strongly by the carbon nuclei compared to localized double bonds in cyclohexatriene,contributing to its stability. This statement is true and provides the correct explanation for the stability of benzene.
Therefore,both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.

(A) The reaction involves the electrophilic addition of benzene to methylenecyclohexane in the presence of an acid catalyst $(HF)$.
$1$. First,the $HF$ protonates the double bond of methylenecyclohexane to form a stable tertiary carbocation,the $1$-methylcyclohexyl cation.
$2$. This carbocation then acts as an electrophile and undergoes an electrophilic aromatic substitution $(EAS)$ reaction with benzene.
$3$. The final product formed is $1$-methyl-$1$-phenylcyclohexane.

(C) The reaction involves the oxidation of side chains on a naphthalene ring using alkaline $KMnO_4$ followed by acidic workup.
$1$. The ethyl group $(-CH_2CH_3)$ at the benzylic position has $\alpha$-hydrogens,so it is oxidized to a carboxylic acid group $(-COOH)$.
$2$. The vinyl group $(-CH=CH_2)$ is also oxidized to a carboxylic acid group $(-COOH)$ by strong oxidizing agents like alkaline $KMnO_4$.
$3$. The ester group $(-COOCH_3)$ is hydrolyzed under alkaline conditions to a carboxylate salt,which upon acidic workup $(H_3O^+)$ becomes a carboxylic acid group $(-COOH)$.
$4$. Therefore,all three side chains are converted into carboxylic acid groups,resulting in the product shown in option $C$.

(D) Statement $I$ is false: Tropolone is an aromatic compound,but it follows $H$ückel's rule ($4n+2$ $\pi$ electrons). It has $6$ $\pi$ electrons in the seven-membered ring,which makes it aromatic. It does not have $8$ $\pi$ electrons involved in the aromatic system.
Statement $II$ is false: The $\pi$ electrons of the $ > C = O $ group are not involved in the aromaticity of the ring. The aromaticity arises from the $6$ $\pi$ electrons present within the seven-membered ring system (as shown in the resonance structure where the oxygen atom carries a negative charge and the ring carbon carries a positive charge).
Therefore,both statements are false.

(B) $H$ückel's rule states that a planar,cyclic,fully conjugated system is aromatic if it contains $(4n + 2) \pi$ electrons,where $n = 0, 1, 2, \dots$
Let's analyze each species:
$(i)$ Naphthalene: $10 \pi$ electrons $(n=2)$,aromatic.
$(ii)$ Cyclopentadienyl anion: $6 \pi$ electrons $(n=1)$,aromatic.
$(iii)$ Cyclobutadiene: $4 \pi$ electrons ($n=1$ for anti-aromatic),not $H$ückel's rule.
$(iv)$ Cyclopropenyl anion: $4 \pi$ electrons,anti-aromatic.
$(v)$ Cyclopropenyl cation: $2 \pi$ electrons $(n=0)$,aromatic.
$(vi)$ Cyclooctatetraene: $8 \pi$ electrons,non-planar,non-aromatic.
$(vii)$ Anthracene: $14 \pi$ electrons $(n=3)$,aromatic.
The species that obey $H$ückel's rule are $(i), (ii), (v),$ and $(vii)$.
Thus,the total number is $4$.

(B) To determine if a compound is aromatic,it must be cyclic,planar,fully conjugated,and follow $H$ückel's rule ($4n+2$ $\pi$ electrons).
$(A)$ Decalin derivative: Not fully conjugated,non-aromatic.
$(B)$ Benzylcyclohexene: The benzene ring is aromatic,but the whole molecule is not considered a single aromatic system due to the $sp^3$ carbon bridge. However,in the context of identifying aromatic components,the benzene ring is aromatic.
$(C)$ Indene anion (indenyl anion): The cyclopentadienyl ring is conjugated with the benzene ring,and the lone pair on the $sp^3$ carbon becomes part of the $10$ $\pi$ electron system $(n=2)$,making it aromatic.
$(D)$ Fluorenyl anion: It has $14$ $\pi$ electrons $(n=3)$,is planar,and fully conjugated,making it aromatic.
Thus,compounds $(C)$ and $(D)$ are aromatic. The total number is $2$.

(B) Meta-directing groups are those that withdraw electron density from the benzene ring through the inductive effect or resonance,making the ring less reactive towards electrophilic substitution and directing the incoming electrophile to the meta position.
From the given list:
$1$. $-NO_2$ (Nitro group): Meta-directing
$2$. $-CN$ (Cyano group): Meta-directing
$3$. $-COR$ (Acyl group): Meta-directing
$4$. $-COOH$ (Carboxyl group): Meta-directing
The groups $-OCH_3$,$-CH_3$,$-NHCOCH_3$,$-OH$,and $-Cl$ are ortho/para-directing.
Thus,the total number of meta-directing groups is $4$.

(D) The reaction between benzene and benzoyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction.
$1$. The anhydrous $AlCl_3$ acts as a Lewis acid and reacts with benzoyl chloride to generate the electrophile,the benzoyl cation $(C_6H_5CO^+)$.
$2$. This electrophile then attacks the benzene ring.
$3$. The final product formed is benzophenone $(C_6H_5COC_6H_5)$.

(D) The reactivity towards electrophilic substitution depends on the electron density of the benzene ring.
$A$ is benzene (reference).
$B$ is toluene ($-CH_3$ group): It shows $+M$ (hyperconjugation) and $+I$ effects,which increase electron density,making it more reactive than benzene.
$C$ is chlorobenzene ($-Cl$ group): It shows $+M$ and $-I$ effects. The $-I$ effect dominates,making it less reactive than benzene.
$D$ is nitrobenzene ($-NO_2$ group): It shows $-M$ and $-I$ effects,which strongly decrease electron density,making it the least reactive.
Therefore,the order of reactivity is $B > A > C > D$.

(B) In aromatic electrophilic substitution reactions,groups that withdraw electron density from the benzene ring are called deactivating groups. These typically exhibit a $-M$ (negative mesomeric) or $-I$ (negative inductive) effect.
Let us analyze the given groups:
$1$. $-COOCH_3$: This group has a carbonyl group attached to the ring,which exerts a $-M$ effect. It is a deactivating group.
$2$. $-NHCOCH_3$: The nitrogen atom has a lone pair that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
$3$. $-NHCH_3$: The nitrogen atom has a lone pair that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
$4$. $-CN$: The cyano group has a triple bond between carbon and nitrogen,exerting a strong $-M$ effect. It is a deactivating group.
$5$. $-OCH_3$: The oxygen atom has lone pairs that can be donated to the ring via resonance ($+M$ effect). It is an activating group.
The deactivating groups are $-COOCH_3$ and $-CN$.
Therefore,the total number of deactivating groups is $2$.

(C) Meta-directing groups are those that withdraw electrons from the benzene ring through the inductive effect $(-I)$ and/or resonance effect $(-M)$.
In option $C$,all the groups $(-NO_2, -CHO, -SO_3H, -COR)$ are electron-withdrawing groups that exhibit a $-M$ effect,making them meta-directing.
In other options,groups like $-NH_2, -NHR, -OCH_3$ and $-NHCOCH_3$ are ortho/para-directing due to their $+M$ effect.

(B) In the nitration of benzene,concentrated $H_2SO_4$ and $HNO_3$ are used as reagents,which generate the electrophile $NO_2^+$ in the following steps:
$H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H_2O^+-NO_2$
$H_2O^+-NO_2 \rightleftharpoons H_2O + NO_2^+$
Statement $I$ is correct as it shows the dissociation of the protonated nitric acid into water and the nitronium ion $(NO_2^+)$.
Statement $II$ is incorrect because Lewis acids (like $AlCl_3$,$FeBr_3$) are used to promote electrophilic substitution by generating electrophiles,not Lewis bases.

(A) Electrophilic aromatic substitution is facilitated by electron-donating groups $(EDG)$ and inhibited by electron-withdrawing groups $(EWG)$.
$1$. The $-OCH_3$ group in compound $(III)$ is a strong electron-donating group due to the $+M$ effect,which significantly increases the electron density on the benzene ring.
$2$. The $-CH_3$ group in compound $(I)$ is an electron-donating group due to the $+H$ (hyperconjugation) and $+I$ effects.
$3$. Compound $(II)$ is benzene,which serves as the reference.
$4$. The $-CF_3$ group in compound $(IV)$ is a strong electron-withdrawing group due to the $-I$ effect,which decreases the electron density on the benzene ring.
Therefore,the reactivity order towards electrophilic substitution is: $(III) > (I) > (II) > (IV)$.

(A) 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.
$A$: This is $1,4$-dihydronaphthalene. It is non-aromatic because it is not fully conjugated.
$B$: This is styrene (vinylbenzene). It contains a benzene ring,which is aromatic.
$C$: This is $[10]$-annulene. It is non-aromatic due to steric hindrance between internal hydrogen atoms,which prevents the molecule from being planar.
$D$: This is $[14]$-annulene. It is aromatic as it follows $H$ückel's rule ($n=3$,$4(3)+2 = 14$ $\pi$ electrons) and is planar.
Thus,$B$ and $D$ are aromatic.

(A) Step $1$: Ethylbenzene reacts with alkaline $KMnO_4$ followed by heating to undergo oxidation of the alkyl side chain to form potassium benzoate $(A)$.
Step $2$: Potassium benzoate $(A)$ upon treatment with nitrating mixture $(conc. HNO_3 + conc. H_2SO_4)$ undergoes electrophilic aromatic substitution. Since the $-COOH$ group is meta-directing,the nitro group enters the meta position to form $m$-nitrobenzoic acid $(B)$.
Step $3$: The structure of $B$ ($m$-nitrobenzoic acid) contains:
- $3$ $\pi$-bonds in the benzene ring.
- $1$ $\pi$-bond in the $C=O$ group of the carboxylic acid.
- $1$ $\pi$-bond in the $N=O$ group of the nitro group.
Total $\pi$-bonds = $3 + 1 + 1 = 5$.

(A) To determine the aromaticity of the given compounds,we check for $H$ückel's rule ($4n+2$ $\pi$ electrons),planarity,and cyclic conjugation:
$1$. $1,4$-dihydronaphthalene: Non-aromatic (not fully conjugated).
$2$. Fulvalene: Non-aromatic (not fully conjugated).
$3$. Cyclopentadienyl cation: Aromatic ($2$ $\pi$ electrons,$n=0$,planar,cyclic,conjugated).
$4$. Cyclooctatetraene: Non-aromatic (tub-shaped,non-planar,$8$ $\pi$ electrons).
$5$. Pyridine: Aromatic ($6$ $\pi$ electrons,$n=1$,planar,cyclic,conjugated).
$6$. Cycloheptatriene: Non-aromatic ($sp^3$ carbon present).
Thus,there are $2$ aromatic compounds (Cyclopentadienyl cation and Pyridine).

(A) Reaction $(A)$: Benzene reacts with $t$-butyl bromide in the presence of $NaOC_2H_5$. Since $NaOC_2H_5$ is a strong base,it causes dehydrohalogenation of $t$-butyl bromide to form isobutylene. Thus,no Friedel-Crafts alkylation occurs.
Reaction $(B)$: Benzene reacts with isobutyl chloride in the presence of $AlCl_3$. The $AlCl_3$ induces a rearrangement of the isobutyl carbocation to a more stable $t$-butyl carbocation,which then attacks the benzene ring to form tert-butylbenzene.
Reaction $(C)$: Benzene reacts with isobutylene in the presence of $H_2SO_4$. The acid protonates the alkene to form a $t$-butyl carbocation,which undergoes electrophilic aromatic substitution to form tert-butylbenzene.
Reaction $(D)$: Benzene reacts with isobutyl alcohol in the presence of $BF_3 \cdot OEt_2$. The Lewis acid $BF_3$ facilitates the formation of a $t$-butyl carbocation via rearrangement,which then attacks the benzene ring to form tert-butylbenzene.
Therefore,reactions $(B)$,$(C)$,and $(D)$ yield tert-butylbenzene as the major product.

(A) To determine if a compound is aromatic,it must satisfy $H$ückel's rule: it must be planar,cyclic,fully conjugated,and contain $(4n + 2) \pi$ electrons,where $n = 0, 1, 2, ...$.
Let us analyze the given structures:
$1$. Cyclooctatetraene: Non-planar (tub-shaped),$8 \pi$ electrons. Not aromatic.
$2$. Cyclopropenyl anion: $4 \pi$ electrons. Anti-aromatic.
$3$. Cyclopropenyl cation: $2 \pi$ electrons $(n=0)$. Aromatic.
$4$. Cyclohexadiene: Not fully conjugated. Not aromatic.
$5$. Cycloheptatrienyl cation (tropylium ion): $6 \pi$ electrons $(n=1)$. Aromatic.
$6$. Cyclopentadienyl cation: $4 \pi$ electrons. Anti-aromatic.
$7$. Cyclopentadienyl anion: $6 \pi$ electrons $(n=1)$. Aromatic.
$8$. $1,2$-dihydronaphthalene: Not fully conjugated. Not aromatic.
$9$. Phenanthrene: $14 \pi$ electrons $(n=3)$. Aromatic.
The aromatic compounds are: cyclopropenyl cation,cycloheptatrienyl cation,cyclopentadienyl anion,and phenanthrene.
Total number of aromatic compounds = $4$.
Since $4$ is not in the options,let us re-evaluate the structures provided in the image.
Looking closely at the provided solution image,it identifies: cyclopropenyl cation,cycloheptatrienyl cation,cyclopentadienyl anion,and phenanthrene as aromatic. The total count is $4$. Given the options,there might be a discrepancy in the question's provided options or the image interpretation. Based on standard chemistry,the count is $4$.

(A) The stability of cyclic conjugated systems can be predicted by $H$ückel's rule.
$(A)$ $1,3$-cyclohexadiene is a stable non-aromatic molecule.
$(B)$ Cyclobutadiene is a $4n$ $\pi$-electron system $(n=1)$,making it antiaromatic. It is highly reactive and unstable at room temperature.
$(C)$ Cyclopentadienone is antiaromatic because it has $4$ $\pi$-electrons in the ring (the carbonyl oxygen withdraws electron density,but the ring itself is antiaromatic). It dimerizes rapidly at room temperature.
$(D)$ Cycloheptatrienone (Tropone) is stable due to the contribution of the aromatic tropylium cation resonance structure.
Therefore,molecules $(B)$ and $(C)$ are unstable at room temperature.

(D) Step $1$: Reaction $(A)$ is the acid-catalyzed trimerization of acetone to form $1,3,5-$trimethylbenzene (mesitylene).
Step $2$: Reaction $(B)$ is the cyclotrimerization of propyne over a heated iron tube to form $1,3,5-$trimethylbenzene.
Step $3$: Reaction $(C)$ involves the haloform reaction followed by decarboxylation,which yields benzene,not $1,3,5-$trimethylbenzene.
Step $4$: Reaction $(D)$ is the Clemmensen reduction of $1,3,5-$triformylbenzene,which reduces the aldehyde groups to methyl groups,yielding $1,3,5-$trimethylbenzene.
Therefore,reactions $(A), (B),$ and $(D)$ lead to the formation of $1,3,5-$trimethylbenzene.

(A) Step $1$: Benzene reacts with $CO / HCl$ in the presence of $AlCl_3 / CuCl$ (Gattermann-Koch reaction) to form benzaldehyde. Subsequent reaction with $Ac_2O / NaOAc$ (Perkin reaction) yields cinnamic acid as compound $X$ $(C_6H_5-CH=CH-COOH)$.
Step $2$: $X$ reacts with $Br_2 / Na_2CO_3$ to form the dibromo derivative,which upon heating with moist $KOH$ at $473 \ K$ undergoes dehydrohalogenation to form phenylacetylene $(C_6H_5-C \equiv CH)$ as compound $Y$. This corresponds to option $C$ in the first set of images.
Step $3$: $X$ undergoes hydrogenation with $H_2 / Pd-C$ to form $3-$phenylpropanoic acid. Treatment with $H_3PO_4$ causes intramolecular Friedel-Crafts acylation to form $1-$indanone $(Z)$. This corresponds to option $A$ in the second set of images.
Therefore,$Y$ is $C$ and $Z$ is $A$. The correct option is $A$.

(D) $1$. $P$: Cyclopropenyl chloride reacts with $AlCl_3$ to form the cyclopropenyl cation,which is aromatic ($2\pi$ electrons,planar,cyclic,conjugated).
$2$. $Q$: Cyclopentadiene reacts with $NaH$ to form the cyclopentadienyl anion,which is aromatic ($6\pi$ electrons,planar,cyclic,conjugated).
$3$. $R$: The reaction of hexane$-2,5-$dione with $(NH_4)_2CO_3$ (Paal-Knorr synthesis) yields $2,5-$dimethylpyrrole,which is aromatic ($6\pi$ electrons,planar,cyclic,conjugated).
$4$. $S$: Tropone reacts with $HCl$ to form the hydroxy-tropylium cation,which is aromatic ($6\pi$ electrons,planar,cyclic,conjugated).
Thus,all $P$,$Q$,$R$,and $S$ are aromatic compounds.

(C) The reaction of $p$-hydroxybenzenesulfonic acid with aqueous $Br_2$ ($3.0$ equivalents) involves electrophilic aromatic substitution.
First,the highly activating $-OH$ group directs bromination to the ortho positions relative to itself.
Since the para position is already occupied by the $-SO_3H$ group,the two ortho positions are brominated.
Subsequently,the $-SO_3H$ group is a good leaving group in the presence of excess bromine and water,leading to its displacement by a third bromine atom.
This results in the formation of $2,4,6$-tribromophenol,which corresponds to structure $R$.

(B) $1$. The first step is the Friedel-Crafts alkylation of benzene with propene in the presence of an acid catalyst $(H^+)$. This reaction proceeds via the formation of an isopropyl carbocation,which attacks the benzene ring to form cumene (isopropylbenzene) as intermediate $T$.
$2$. The second step is the autoxidation of cumene. In the presence of oxygen $(O_2)$ and a radical initiator,the benzylic hydrogen is abstracted to form a benzylic radical,which then reacts with $O_2$ to form cumene hydroperoxide as the major product $U$.
$3$. The structure of cumene hydroperoxide is $C_6H_5-C(CH_3)_2-O-OH$.

(B) The starting material is $oct-4-ene$ $(CH_3CH_2CH_2CH=CHCH_2CH_2CH_3)$.
Step $1$: Ozonolysis $(O_3, Zn/H_2O)$ cleaves the double bond to form two moles of butanal $(CH_3CH_2CH_2CHO)$.
Step $2$: Oxidation with $KMnO_4$ converts butanal into butanoic acid $(CH_3CH_2CH_2COOH)$.
Step $3$: Kolbe's electrolysis of the sodium salt of butanoic acid $(CH_3CH_2CH_2COONa)$ leads to the coupling of the propyl radical $(CH_3CH_2CH_2\cdot)$ to form $n-hexane$ $(CH_3CH_2CH_2CH_2CH_2CH_3)$.
Step $4$: Aromatization of $n-hexane$ using $Cr_2O_3$ at $770 \ K$ and $20 \ atm$ yields benzene $(C_6H_6)$.
In the final product,benzene,there are no $-CH_2-$ (methylene) groups as all carbon atoms are $sp^2$ hybridized and part of the aromatic ring.
Therefore,the number of $-CH_2-$ groups is $0$.

(D) $1$. $L$-Glucose on reduction with $HI$ and red $P$ gives $n$-hexane.
$2$. $n$-Hexane,when heated with $Cr_2O_3$ at $775 \ K$ and $10-20 \ atm$ pressure,undergoes aromatization to form benzene $(P)$.
$3$. Benzene $(P)$ reacts with excess $Cl_2$ in the presence of $UV$ light (sunlight) to undergo electrophilic addition,forming benzene hexachloride $(C_6H_6Cl_6)$,also known as $BHC$ or Lindane.
$4$. Therefore,the major product $Q$ is benzene hexachloride.