Aromatic hydrocarbon Questions in English

(B) The mixture of conc. $H_2SO_4$ and conc. $HNO_3$ is called nitrating mixture.
It is used in the nitration of aryl compounds,for example,the reaction of benzene with this mixture produces nitrobenzene:
$\text{Benzene} + HNO_3 \xrightarrow{\text{conc. } H_2SO_4} \text{Nitrobenzene} + H_2O$

(B) In benzene $(C_6H_6)$,each carbon atom is bonded to two other carbon atoms and one hydrogen atom via $\sigma$-bonds,and it also participates in a delocalized $\pi$-system.
Since each carbon atom forms $3 \sigma$-bonds and has no lone pairs,the steric number is $3$.
Therefore,the hybridization of each carbon atom in benzene is $sp^2$.

(C) Naphthalene consists of two fused benzene rings.
It has $5$ $\pi$ bonds in its structure.
Since each $\pi$ bond contains $2$ electrons,the total number of $\pi$ electrons is $5 \times 2 = 10$.

(D) The chemical formula of $o$-xylene is $C_8H_{10}$.
Structure: $A$ benzene ring with two methyl groups at the ortho $(1,2)$ positions.
Total $\sigma$ bonds = (bonds in benzene ring) + (bonds in two methyl groups) + (bonds between ring and methyl groups).
- $6$ $C-C$ bonds in the ring (including $3$ single and $3$ double,where each double bond has $1 \sigma$ and $1 \pi$ bond).
- $4$ $C-H$ bonds on the ring.
- $6$ $C-H$ bonds in the two methyl groups ($3$ per group).
- $2$ $C-C$ bonds connecting the methyl groups to the ring.
Total $\sigma$ bonds = $6$ (ring $C-C$) + $4$ (ring $C-H$) + $6$ (methyl $C-H$) + $2$ (methyl-ring $C-C$) = $18$.
Thus,the correct option is $D$.

(D) The chemical formula of benzene is $C_6H_6$.
In the benzene ring,there are $6$ $C-C$ $\sigma$ bonds and $6$ $C-H$ $\sigma$ bonds.
Total $\sigma$ bonds $= 6 + 6 = 12$.

(D) . In benzene,the structure consists of a hexagonal ring where each of the $6$ carbon atoms is bonded to two other carbon atoms and one hydrogen atom.
Each carbon atom forms three $\sigma$-bonds and one $\pi$-bond.
Since each carbon atom is involved in three $\sigma$-bonds,the hybridization of each carbon atom is $sp^2$.

(D) The chemical formula of toluene is $C_6H_5CH_3$.
In the benzene ring,there are $6$ $C-C$ bonds (including $3$ double bonds) and $5$ $C-H$ bonds.
In the methyl group $(-CH_3)$,there are $1$ $C-C$ bond and $3$ $C-H$ bonds.
Total $\sigma$ bonds = $6$ (ring $C-C$) + $5$ (ring $C-H$) + $1$ ($C-C$ bond to methyl) + $3$ ($C-H$ bonds in methyl) = $15$ $\sigma$ bonds.
Total $\pi$ bonds = $3$ (from the $3$ double bonds in the benzene ring) = $3$ $\pi$ bonds.
Therefore,toluene has $15$ $\sigma$ and $3$ $\pi$ bonds.

(C) Naphthalene $(C_{10}H_8)$ consists of two fused benzene rings.
In the structure of naphthalene,there are $5$ double bonds.
Each double bond contains one $\pi$-bond.
Therefore,there are $5$ $\pi$-bonds present in a naphthalene molecule.

(B) The structure of toluene $(C_6H_5CH_3)$ consists of a benzene ring attached to a methyl group $(-CH_3)$.
In the benzene ring $(C_6H_6)$,there are $6$ $C-C$ $\sigma$ bonds and $6$ $C-H$ $\sigma$ bonds. However,in toluene,one $H$ atom of the benzene ring is replaced by a $-CH_3$ group.
Thus,the benzene ring part has $6$ $C-C$ $\sigma$ bonds and $5$ $C-H$ $\sigma$ bonds.
The methyl group $(-CH_3)$ has $3$ $C-H$ $\sigma$ bonds and $1$ $C-C$ $\sigma$ bond (connecting the methyl group to the ring).
Total $\sigma$ bonds = ($6$ $C-C$ ring bonds) + ($5$ $C-H$ ring bonds) + ($3$ $C-H$ methyl bonds) + ($1$ $C-C$ bond between ring and methyl) = $6 + 5 + 3 + 1 = 15$ $\sigma$ bonds.
Therefore,the correct option is $(B)$.

(B) The chemical formula of benzene is $C_6H_6$.
In the structure of benzene,there are $6$ $C-C$ bonds (consisting of $3$ single bonds and $3$ double bonds) and $6$ $C-H$ bonds.
Each single bond is a $\sigma$ bond,and each double bond consists of $1$ $\sigma$ bond and $1$ $\pi$ bond.
Total $\sigma$ bonds = $6$ ($C-C$ $\sigma$ bonds) + $6$ ($C-H$ $\sigma$ bonds) = $12$ $\sigma$ bonds.
Total $\pi$ bonds = $3$ $\pi$ bonds (from the $3$ $C=C$ double bonds).
Therefore,benzene contains $12$ $\sigma$ bonds and $3$ $\pi$ bonds.

(D) The aromatic character of $benzene$ is explained by multiple theoretical frameworks:
$1$. The $Aromatic$ $sextet$ $theory$ explains the stability of the $6\pi$ electron system.
$2$. $Resonance$ $theory$ accounts for the delocalization of electrons and the equivalence of $C-C$ bond lengths.
$3$. $Molecular$ $orbital$ $theory$ provides a quantum mechanical description of the cyclic delocalization of $\pi$ electrons in the $p$-orbitals.
Therefore,all these theories contribute to our understanding of the aromatic properties of $benzene$.

(B) For a molecule to be aromatic,it must satisfy $H$ückel's rule ($4n + 2$ $\pi$ electrons),be planar,and have a fully conjugated cyclic system.
$1$. Decalin is non-planar and non-aromatic.
$2$. Benzene $(C_6H_6)$ is planar,cyclic,fully conjugated,and has $6$ $\pi$ electrons $(4(1) + 2 = 6)$,thus it is aromatic.
$3$. Cyclohexene is not fully conjugated and is non-aromatic.
$4$. Furan is aromatic,but among the given options,Benzene is the most standard example of a simple aromatic hydrocarbon.

(D) Benzene is relatively unreactive towards electrophilic addition reactions because the $6\pi$ electrons are delocalized over the entire ring due to resonance,which provides extra stability to the molecule.

(C) The ring structure of benzene $(C_6H_6)$ was proposed by August Kekule in $1865$.
He suggested that benzene consists of a hexagonal ring of carbon atoms with alternating single and double bonds.

(B) The reaction between concentrated $H_2SO_4$ and $HNO_3$ is as follows:
$HNO_3 + 2H_2SO_4 \rightleftarrows NO_2^+ + H_3O^+ + 2HSO_4^-$
Here,$NO_2^+$ is the nitronium ion,which acts as the electrophile or the active nitrating species in the nitration reaction.

(D) To determine aromaticity,we use $H$ückel's rule: a compound is aromatic if it is cyclic,planar,fully conjugated,and contains $(4n + 2) \pi$ electrons.
$A$. Benzene $(C_6H_6)$ has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$B$. Cyclooctatetraenyl dianion $(C_8H_8^{2-})$ has $10 \pi$ electrons $(n=2)$,so it is aromatic.
$C$. Tropylium cation $(C_7H_7^+)$ has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$D$. Cyclopentadienyl cation $(C_5H_5^+)$ has $4 \pi$ electrons ($n=1$ in $4n$ rule),making it anti-aromatic. Thus,it is not aromatic.

(A) The cyclopentadienyl anion $(C_5H_5^-)$ has a cyclic structure with $sp^2$ hybridized carbon atoms,making it planar. It contains $6 \pi$-electrons ($4$ from the two double bonds and $2$ from the lone pair on the negatively charged carbon),which follows $H$ückel's rule ($4n + 2 = 6$,where $n = 1$). Therefore,it is aromatic.

(A) The $C-C$ bond length in benzene is $1.39 \ \mathring{A}$.
This value is intermediate between the single bond length of $C-C$ $(1.54 \ \mathring{A})$ and the double bond length of $C=C$ $(1.34 \ \mathring{A})$.
This is due to the delocalization of $\pi$-electrons in the benzene ring,known as resonance.

(B) The Friedel-Crafts reaction involves the electrophilic substitution of an aromatic ring.
In Friedel-Crafts acylation,an acyl halide (such as $RCOCl$) reacts with an aromatic compound in the presence of a Lewis acid catalyst,typically $AlCl_3$.
Therefore,$RCOCl$ acts as the reagent for the acylation process.

(A) According to $H$ückel's rule,for a compound to be aromatic,it must contain $(4n + 2)$ $\pi$-electrons,where $n$ is an integer $(n = 0, 1, 2, 3, ...)$.
Since $n$ must be an integer,$0.5$ is not a valid value for $n$.

(C) The reaction of toluene with chlorine in the presence of light $(h\nu)$ is a classic example of side-chain chlorination.
This reaction proceeds via a free radical mechanism.
$1$. Initiation: $Cl_2 \xrightarrow{h\nu} 2Cl^\bullet$
$2$. Propagation: The chlorine radical abstracts a hydrogen atom from the methyl group of toluene to form a benzyl radical,which then reacts with $Cl_2$ to form benzyl chloride $(C_6H_5CH_2Cl)$.

(C) Nitrobenzene is meta-directing for an electrophilic substitution reaction because it contains an electron-withdrawing group (i.e.,$-NO_2$).
Due to the $-I$ and $-M$ effects of the $-NO_2$ group,the electron density at the ortho- and para- positions of the benzene ring is significantly reduced compared to the meta- position.
Consequently,the incoming electrophile preferentially attacks the meta- position,where the electron density is relatively higher.

(C) Aromatic compounds or arenes undergo substitution reactions,in which the aromatic hydrogen is replaced with an electrophile,hence their reactions proceed via electrophilic substitution.
Arenes contain double bonds just like alkenes but they do not undergo electrophilic addition because these would result in the loss of their ring aromaticity.

(D) In the Friedel-Crafts reaction,$AlCl_3$ acts as a Lewis acid catalyst.
It reacts with the alkyl halide or acyl halide to generate a carbocation or an acylium ion,which acts as an electrophile.
This electrophile then attacks the benzene ring to initiate the substitution reaction.
For example: $CH_3-CH_2-CH_2-Cl + AlCl_3 \rightarrow CH_3-CH_2-CH_2^+ + AlCl_4^-$.

(A) The electrophile in the nitration of benzene is the nitronium ion,$NO_2^+$.
It is generated by the reaction between concentrated $HNO_3$ and concentrated $H_2SO_4$ as follows:
$HNO_3 + H_2SO_4 \to NO_2^+ + HSO_4^- + H_2O$
Thus,the correct option is $A$.

(D) The reactivity of benzene towards electrophilic substitution depends on the electron density of the ring.
Groups that donate electrons to the ring (activating groups) increase the rate of electrophilic substitution,while groups that withdraw electrons (deactivating groups) decrease it.
$Nitrobenzene$,$Benzoic \text{ } acid$,and $Benzaldehyde$ contain electron-withdrawing groups ($-NO_2$,$-COOH$,and $-CHO$ respectively),which deactivate the benzene ring.
$Phenol$ contains an $-OH$ group,which is a strong electron-donating group due to resonance,thereby increasing the electron density of the ring and making it more reactive towards electrophilic substitution than benzene.
Therefore,the correct option is $(D)$.

(D) The given molecule is a bicyclic system where one ring is attached to an $-NH-$ group (which is an activating group due to the lone pair on nitrogen) and the other ring is attached to a $-C(=O)-$ group (which is a deactivating group).
Electrophilic aromatic substitution,such as bromination using $Br_2/Fe$,occurs preferentially on the more electron-rich ring.
The ring containing the $-NH-$ group is activated,making it more susceptible to electrophilic attack.
The $-NH-$ group is ortho/para directing. Due to steric hindrance from the methyl group at the ortho position,the bromination occurs at the para position relative to the $-NH-$ group.
Thus,the bromine atom will be substituted at the para position of the activated ring.

(B) The reactivity of a benzene ring towards electrophilic aromatic substitution depends on the electron density of the ring.
$1.$ $Toluene$ $(IV)$ has a methyl group $(-CH_3)$,which is an electron-donating group ($+I$ effect and hyperconjugation),increasing electron density.
$2.$ $Benzene$ $(II)$ is the reference compound.
$3.$ $Chlorobenzene$ $(I)$ has a chlorine atom,which is deactivating due to its strong $-I$ effect,although it is ortho/para directing due to resonance ($+M$ effect).
$4.$ $Anilinium$ $chloride$ $(III)$ contains the $-NH_3^+$ group,which is a strong electron-withdrawing group ($-I$ effect),significantly reducing electron density.
Therefore,the order of reactivity is $IV > II > I > III$.

(C) disubstituted benzene compound has two substituents attached to the benzene ring. Depending on the relative positions of these two substituents,there are three possible structural isomers:
$1$. Ortho ($1,2$-position)
$2$. Meta ($1,3$-position)
$3$. Para ($1,4$-position)
Therefore,the total number of isomers is $3$.

(B) The molecular formula for trimethyl benzene is $C_9H_{12}$.
There are $3$ possible structural isomers for trimethyl benzene:
$1.$ $1,2,3$-Trimethylbenzene (Hemimellitene)
$2.$ $1,2,4$-Trimethylbenzene (Pseudocumene)
$3.$ $1,3,5$-Trimethylbenzene (Mesitylene)
Therefore,the total number of possible isomeric trimethyl benzene is $3$.

(A) The reaction involves electrophilic aromatic substitution on a substituted benzene ring ${C_6H_5Y}$.
Groups that are electron-withdrawing by resonance or induction (deactivating groups) are meta-directing.
Among the given options,$-COOH$ is an electron-withdrawing group and acts as a meta-directing group.
Conversely,$-NH_2$,$-OH$,and $-Cl$ are electron-donating groups (by resonance) and act as ortho/para-directing groups.
Therefore,the correct option is $A$.

(A) According to $H$ückel's rule for aromaticity,a molecule must be planar,cyclic,and possess a delocalized $(4n + 2) \, \pi$ electron system,where $n$ is an integer $(n = 0, 1, 2, 3, \dots)$.
$1.$ The cyclopropenyl cation has $2 \, \pi$ electrons $(n = 0)$,making it aromatic.
$2.$ Cyclobutadiene has $4 \, \pi$ electrons ($4n$ system),making it anti-aromatic.
$3.$ The cyclopentadienyl cation has $4 \, \pi$ electrons ($4n$ system),making it anti-aromatic.
$4.$ The cyclopropenyl anion has $4 \, \pi$ electrons ($4n$ system),making it anti-aromatic.
Therefore,the cyclopropenyl cation is the aromatic compound.

(A) The cyclic hydrocarbon $A$ described is benzene $(C_6H_6)$.
In benzene,all carbon atoms are $sp^2$ hybridized and lie in the same plane.
The carbon-carbon bond length in benzene is $1.39 \ \mathring{A}$,which satisfies the condition $1.34 \ \mathring{A} < 1.39 \ \mathring{A} < 1.54 \ \mathring{A}$.
For $sp^2$ hybridized carbon atoms in a planar ring,the $C-C-C$ bond angle is $120^{\circ}$.

(A) The reaction of benzene $(C_6H_6)$ with ethene $(H_2C = CH_2)$ in the presence of an anhydrous $AlCl_3$ catalyst (often with a trace of $HCl$) is an example of Friedel-Crafts alkylation.
This reaction proceeds via the formation of an ethyl carbocation intermediate,which then attacks the benzene ring to form ethylbenzene $(C_6H_5CH_2CH_3)$.
The reaction is: $C_6H_6 + H_2C = CH_2 \xrightarrow{AlCl_3, HCl} C_6H_5CH_2CH_3$.

(A) When acetylene $(C_2H_2)$ is passed through a red-hot iron tube at $873 \ K$,it undergoes cyclic polymerization to form benzene $(C_6H_6)$.
The reaction is as follows:
$3C_2H_2 \xrightarrow{\text{red hot tube}} C_6H_6$
Thus,the correct option is $A$.

(D) The fractional distillation of coal tar yields several fractions based on their boiling point ranges:
$1$. Light oil (up to $200 \ ^\circ C$)
$2$. Middle oil $(200-250 \ ^\circ C)$
$3$. Heavy oil $(250-300 \ ^\circ C)$
$4$. Anthracene oil $(300-350 \ ^\circ C)$
$5$. Pitch (residue)
Mobil oil is a lubricant derived from petroleum,not from coal tar distillation.

(D) Xylenes are dimethyl derivatives of benzene. There are three isomers of xylene: $o$-xylene,$m$-xylene,and $p$-xylene.
When these isomers are oxidized with acidic $KMnO_4$,the methyl groups are oxidized to carboxylic acid groups $(-COOH)$.
$o$-Xylene gives phthalic acid (benzene$-1,2-$dicarboxylic acid).
$m$-Xylene gives isophthalic acid (benzene$-1,3-$dicarboxylic acid).
$p$-Xylene gives terephthalic acid (benzene$-1,4-$dicarboxylic acid).
Since all three isomers are xylenes,they yield their respective dicarboxylic acids upon oxidation.

(C) In the Friedel-Crafts reaction,anhydrous $AlCl_3$ acts as a Lewis acid.
It reacts with the alkyl halide or acyl halide to generate a carbocation or an acylium ion,which acts as an electrophile.
This electrophile then attacks the benzene ring to initiate the substitution reaction.
For example: $CH_3CH_2CH_2Cl + AlCl_3 \rightarrow CH_3CH_2CH_2^+ + AlCl_4^-$.
The generated $CH_3CH_2CH_2^+$ is the electrophile.

(D) The reaction of benzene with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ is known as the Friedel-Crafts acylation reaction.
In this reaction,the acetyl group $(-COCH_3)$ replaces a hydrogen atom on the benzene ring to form acetophenone $(C_6H_5COCH_3)$.
The chemical equation is: $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$.
Therefore,the correct option is $(d)$.

(C) In the Friedel-Crafts alkylation of benzene,the alkyl group introduced is electron-donating,which activates the benzene ring towards further electrophilic substitution. This leads to the formation of polyalkylated products.
In contrast,the acyl group introduced in acylation is electron-withdrawing,which deactivates the ring and prevents further substitution,ensuring the formation of a mono-acylated product.
Therefore,acylation is preferred,followed by the reduction of the carbonyl group to a methylene group to obtain the desired alkyl benzene.

(C) Benzene typically undergoes electrophilic aromatic substitution reactions (e.g.,nitration,sulfonation,halogenation).
Benzene can undergo addition reactions under specific conditions,such as hydrogenation to cyclohexane or addition of chlorine in the presence of $UV$ light.
Benzene can undergo oxidation reactions,such as catalytic oxidation to maleic anhydride or oxidation to phenol.
Benzene does not undergo elimination reactions because it lacks a leaving group attached to an $sp^3$ hybridized carbon that could be removed to form a double or triple bond within the ring system without destroying its aromatic stability.
Therefore,the correct answer is $C$.

(B) $o-$Xylene exists in two resonance structures.
Ozonolysis of $o-$xylene involves the cleavage of double bonds in the benzene ring.
Based on the two resonance structures,the products formed are methylglyoxal,$1,2-$dimethylglyoxal (also known as diacetyl),and glyoxal.
Ethyl glyoxal is not among the products formed during this reaction.
Therefore,the correct option is $B$.

(C) The chemical formula of benzene is $C_6H_6$.
In the benzene ring,there are $6$ $C-C$ $\sigma$ bonds and $6$ $C-H$ $\sigma$ bonds,totaling $12$ $\sigma$ bonds.
Additionally,there are $3$ $C-C$ $\pi$ bonds due to the alternating double bonds in the ring.
Therefore,a molecule of benzene contains $12$ $\sigma$ bonds and $3$ $\pi$ bonds.
Thus,the correct option is $(C)$.

(B) In a benzene molecule $(C_6H_6)$,there are $12$ sigma $(\sigma)$ bonds ($6$ $C-C$ and $6$ $C-H$ bonds) and $3$ pi $(\pi)$ bonds.
Ratio = $\frac{\sigma \text{ bonds}}{\pi \text{ bonds}} = \frac{12}{3} = 4$.

(A) In a benzene molecule $(C_6H_6)$,each carbon atom is $sp^2$ hybridized.
Due to $sp^2$ hybridization,the geometry around each carbon atom is trigonal planar.
Therefore,the bond angle between the carbon atoms in the benzene ring is $120^o$.

(C) When benzene reacts with excess $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$ or $I_2$,it undergoes electrophilic aromatic substitution.
In the presence of excess chlorine,all hydrogen atoms of the benzene ring are replaced by chlorine atoms,resulting in the formation of $C_6Cl_6$,which is known as hexachlorobenzene.
Note: If the reaction were performed in the presence of $UV$ light (free radical addition),the product would be benzene hexachloride $(C_6H_6Cl_6)$. Since the question specifies $I_2$ (a Lewis acid catalyst),the product is hexachlorobenzene.

(B) Gammexane,also known as Benzene Hexachloride $(BHC)$ or Lindane,is produced by the additive chlorination of benzene.
When benzene reacts with $3$ moles of $Cl_2$ in the presence of bright sunlight (ultraviolet light),an addition reaction occurs to form $1,2,3,4,5,6$-hexachlorocyclohexane.
The reaction is: $C_6H_6 + 3Cl_2 \xrightarrow{\text{sunlight}} C_6H_6Cl_6$ (Gammexane).

(D) Benzene $(C_6H_6)$ has a planar hexagonal structure with delocalized $\pi$-electrons.
According to molecular orbital theory,the six $p$-orbitals overlap to form a single delocalized $\pi$-molecular orbital system covering the entire ring,not three separate ones.
Therefore,the statement that it has three delocalized $\pi$-molecular orbitals is incorrect.

(A) The Friedel-Crafts reaction involves the electrophilic substitution of an aromatic ring,typically using a Lewis acid catalyst like $AlCl_3$.
Option $A$ represents Friedel-Crafts alkylation,where benzene reacts with an alkyl halide $(C_2H_5Cl)$ in the presence of $AlCl_3$ to form ethylbenzene.
Option $B$ is a nucleophilic substitution reaction.
Option $C$ is chemically incorrect as it shows an impossible product for the given reactants.
Option $D$ is the formation of a Grignard reagent.

(B) When benzene reacts with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $350 \ K$,the electrophile $NO_2^+$ (nitronium ion) is generated.
This electrophile attacks the benzene ring to form nitrobenzene.
This process is known as nitration.
The reaction is: $C_6H_6 + HNO_3 \xrightarrow{Conc. H_2SO_4, 350 \ K} C_6H_5NO_2 + H_2O$.