Aromatic hydrocarbon Questions in English

(A) The nitration of benzene is a classic example of an electrophilic aromatic substitution reaction.
In this process,$HNO_3$ reacts with $H_2SO_4$ to generate the nitronium ion $(NO_2^+)$,which acts as a strong electrophile.
The benzene ring,being electron-rich,attacks the electrophile $(NO_2^+)$ to form a carbocation intermediate (arenium ion).
Finally,the loss of a proton restores the aromaticity of the benzene ring,resulting in the formation of nitrobenzene.

(B) The electrophilic aromatic substitution (halogenation) of benzene requires the presence of a Lewis acid catalyst,which acts as a halogen carrier.
$1.$ $A$ halogen carrier like $FeCl_3$ or $AlCl_3$ is necessary to polarize the halogen molecule $(X_2)$ to generate the electrophile $(X^+)$.
$2.$ The reaction is typically carried out in the dark to prevent free-radical substitution,which would otherwise occur at the side chain (benzylic position) in the presence of light.
$3.$ Therefore,the reaction conditions for electrophilic aromatic halogenation are the presence of a halogen carrier and the absence of light (dark conditions).

(A) The given reaction $C_6H_6 + CH_3Cl \xrightarrow{anhydrous \ AlCl_3} C_6H_5CH_3 + HCl$ involves the alkylation of benzene in the presence of a Lewis acid catalyst $(AlCl_3)$.
This is a classic example of a Friedel-Crafts alkylation reaction.

(B) The reaction of benzene with chlorine in the presence of an iron catalyst (or $FeCl_3$) is an electrophilic aromatic substitution reaction.
In this reaction,the iron catalyst reacts with chlorine to generate the electrophile $Cl^+$.
This electrophile then attacks the benzene ring to form chlorobenzene $(C_6H_5Cl)$ by replacing a hydrogen atom.
The reaction is represented as:
$C_6H_6 + Cl_2 \xrightarrow{Fe} C_6H_5Cl + HCl$
Therefore,the correct option is $(B)$.

(C) Michael Faraday first isolated and identified benzene in $1825$ from the oily residue derived from the production of illuminating gas,giving it the name bicarburet of hydrogen.

(C) The structure of benzene has been of interest since its discovery.
German chemist August Kekule von Stradonitz proposed the cyclic structure of benzene in $1866$,which consists of six carbon atoms with alternating single and double bonds.

(D) The centric structure of benzene was proposed by $Armstrong$ and $Baeyer$.

(A) In benzene $(C_6H_6)$,the structure is a regular hexagon due to the resonance and delocalization of $\pi$-electrons.
All $C-C$ bonds have an equal bond length of $1.39 \mathring{A}$.
All six carbon atoms are $sp^2$ hybridized and are equivalent in terms of their chemical environment.
Therefore,all six carbon atoms are of the same type.

(C) The reaction of sodium benzoate $(C_6H_5COONa)$ with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
When sodium benzoate is heated with sodalime,it undergoes decarboxylation to produce benzene $(C_6H_6)$ and sodium carbonate $(Na_2CO_3)$.
The chemical equation is:
$C_6H_5COONa + NaOH \xrightarrow{CaO} C_6H_6 + Na_2CO_3$
Therefore,the correct option is $(C)$.

(B) The nitration of benzene with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $50-60 \ ^oC$ yields nitrobenzene.
Further nitration of nitrobenzene at higher temperatures (around $100 \ ^oC$) leads to the formation of $m-$dinitrobenzene because the $-NO_2$ group is a deactivating and meta-directing group.
Thus,the final product is $m-$dinitrobenzene.

(D) The nitration of toluene with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ leads to the formation of $2, 4, 6-$trinitrotoluene $(TNT)$ as the final product.
The reaction is as follows:
$C_6H_5CH_3 + 3HNO_3 \xrightarrow{Conc. H_2SO_4} C_6H_2(NO_2)_3CH_3 + 3H_2O$
Thus,the correct option is $(d)$.

(B) The process of sulphonation of benzene is a reversible reaction.
In this reaction,benzene reacts with concentrated $H_2SO_4$ (or fuming $H_2SO_4$,i.e.,oleum) to form benzenesulphonic acid.
The reaction is as follows:
$C_6H_6 + H_2SO_4 \rightleftharpoons C_6H_5SO_3H + H_2O$
Since the reaction is reversible,the addition of superheated steam can reverse the process to regenerate benzene from benzenesulphonic acid.

(C) $m-xylene$ is most easily sulphonated because it has the most reactive positions available for electrophilic substitution.
In $m-xylene$,the two methyl groups are at $1,3-$ positions,which activates the $2, 4,$ and $6$ positions towards electrophilic attack.
Specifically,the $4$ position is highly activated and less sterically hindered compared to the positions in $o-xylene$ and $p-xylene$.

(D) The oxidation of alkyl benzenes (like toluene) with strong oxidizing agents such as alkaline $KMnO_4$ results in the oxidation of the alkyl side chain to a carboxylic acid group.
Even in the presence of dilute $HNO_3$,the strong oxidizing action of alkaline $KMnO_4$ converts the methyl group $(-CH_3)$ of toluene into a carboxyl group $(-COOH)$.
Thus,toluene is oxidized to benzoic acid $(C_6H_5COOH)$.

(C) When benzene vapour is mixed with air and passed over $V_2O_5$ catalyst at $775 \ K$,it undergoes catalytic oxidation to form maleic anhydride.
The reaction is as follows:
$C_6H_6 + \frac{9}{2} O_2 \xrightarrow{V_2O_5, 775 \ K} \text{Maleic anhydride} + 2CO_2 + 2H_2O$
This process involves the oxidation of the benzene ring followed by dehydration to yield maleic anhydride.

(B) Benzene and its derivatives possess extra stability due to resonance,which makes them less prone to addition reactions as that would disrupt the aromatic system.
The delocalized $\pi$-electron cloud above and below the plane of the ring is electron-rich,making it susceptible to attack by electrophiles.
Therefore,the characteristic reactions of benzene and its derivatives are $Electrophilic \ substitution \ reactions$.

(A) Most $o, p-$directing groups are Activating Groups.
If electrophilic aromatic substitution of a monosubstituted benzene is faster than that of benzene under identical conditions,the substituent in the monosubstituted benzene is called an Activating Group.

(A) Electrophilic aromatic substitution is facilitated by electron-donating groups on the benzene ring,which increase the electron density of the ring.
$1$. The $-CH_3$ group in Toluene is an electron-donating group due to the $+I$ effect and hyperconjugation.
$2$. Benzene has no substituents.
$3$. The $-COOH$ group in Benzoic acid and the $-NO_2$ group in Nitrobenzene are electron-withdrawing groups,which decrease the electron density of the ring.
Therefore,Toluene is the most reactive towards electrophilic nitration among the given options.

(C) Sulphonation is an electrophilic aromatic substitution reaction.
The rate of reaction depends on the electron density of the benzene ring.
Electron-donating groups (like $-CH_3$ in Toluene) increase the electron density of the ring,making it more reactive towards electrophilic attack.
Electron-withdrawing groups (like $-NO_2$ in Nitrobenzene) decrease the electron density,making it less reactive.
Therefore,Toluene is the most reactive among the given compounds towards electrophilic sulphonation.

(C) In a benzene molecule,all $C-C$ bond lengths are equal due to resonance.
Benzene does not readily undergo addition reactions because it is highly stable due to aromaticity.
There is a cyclic delocalisation of $\pi$ electrons in the benzene ring.
Monosubstitution of benzene gives only a single isomeric product because all six positions in the benzene ring are equivalent.

(A) In Friedel-Crafts alkylation,an alkyl group is introduced into the benzene ring by reacting benzene with an alkyl halide in the presence of a Lewis acid catalyst like $AlCl_3$.
For the methylation of benzene,the reactants are benzene $(C_6H_6)$ and methyl chloride $(CH_3Cl)$.
The reaction is: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$.

(A) The nitration of benzene is an example of an electrophilic aromatic substitution reaction.
In this process,the nitronium ion $(NO_2^+)$ acts as a strong electrophile.
Benzene,being electron-rich,attacks the electrophile to form a sigma complex (arenium ion).
Finally,the loss of a proton $(H^+)$ restores the aromaticity of the benzene ring.
Since the hydrogen atom is replaced by a nitro group,it is classified as an electrophilic displacement (substitution) reaction.

(D) Benzene is a stable aromatic compound due to the delocalization of $\pi$-electrons.
Because of this stability,it primarily undergoes electrophilic substitution reactions rather than addition reactions,as addition would destroy the aromaticity.
However,under specific conditions,benzene can undergo addition reactions (e.g.,hydrogenation to cyclohexane).
Furthermore,alkyl-substituted benzenes can undergo oxidation of the side chain using strong oxidizing agents like alkaline $KMnO_4$.
Since benzene and its derivatives exhibit all these types of reactions under appropriate conditions,the correct answer is $All \ of \ these$.

(B) The correct answer is $(B)$.
Benzene $(C_6H_6)$ is obtained by the cyclic trimerisation of ethyne $(C_2H_2)$.
When ethyne gas is passed through a red-hot iron tube at $873 \ K$,it undergoes cyclic polymerisation to form benzene.

(A) Thiophene is significantly more reactive toward electrophilic aromatic substitution than benzene due to the electron-donating effect of the sulfur atom.
When a mixture of benzene and thiophene is shaken with concentrated sulfuric acid $(H_2SO_4)$ at room temperature,thiophene undergoes sulfonation to form thiophene$-2-$sulfonic acid,which is water-soluble.
Benzene does not undergo sulfonation under these mild conditions.
The resulting mixture forms two layers: the organic layer containing benzene and the aqueous layer containing the thiophene-sulfonic acid.
These layers can be separated using a separating funnel,effectively isolating the benzene.

(B) hydrocarbon is an organic compound consisting entirely of hydrogen and carbon.
$(B)$ Benzene $(C_6H_6)$ consists only of carbon and hydrogen atoms,making it a hydrocarbon.
Urea $(NH_2CONH_2)$,Ammonium cyanate $(NH_4OCN)$,and Phenol $(C_6H_5OH)$ contain other elements like nitrogen or oxygen.

(C) Aromatic compounds have a high carbon-to-hydrogen ratio compared to aliphatic hydrocarbons.
Due to this high percentage of carbon,there is insufficient oxygen available for complete combustion,leading to the formation of unburnt carbon particles,which appear as soot.
Therefore,they burn with a sooty flame.

(C) Nitration of aromatic compounds is an electrophilic aromatic substitution reaction where the rate-determining step is the formation of the sigma complex (arenium ion).
Since the $C-H$ or $C-D$ bond cleavage occurs after the rate-determining step,there is no primary kinetic isotope effect observed.
Therefore,the rate of nitration of benzene $(C_6H_6)$ and hexadeuterobenzene $(C_6D_6)$ is essentially the same.
Statement $C$ is false because the rates are equal,not greater.

(C) $2,4,6-$Trinitrotoluene,commonly known as $TNT$,is a chemical compound widely used as an explosive.

(D) When benzene reacts with chlorine in the presence of sunlight ($UV$ light),an addition reaction occurs instead of substitution.
The delocalized $\pi$-electron system of the benzene ring is broken,and one chlorine atom adds to each of the six carbon atoms.
The product formed is $1,2,3,4,5,6-$hexachlorocyclohexane,commonly known as benzene hexachloride $(BHC)$ or gammexane,with the formula $C_6H_6Cl_6$.

(D) The reaction of benzene with $Cl_2$ in the presence of a Lewis acid catalyst like $AlCl_3$ is an electrophilic aromatic substitution reaction.
In this reaction,a hydrogen atom on the benzene ring is replaced by a chlorine atom to form chlorobenzene and $HCl$.
Therefore,it is specifically an example of electrophilic substitution (or halogenation).
Among the given options,'Substitution' is the most accurate classification of the reaction mechanism.

(B) Anthracene is a solid polycyclic aromatic hydrocarbon $(PAH)$ with the formula $C_{14}H_{10}$,which consists of three linearly fused benzene rings.
Naphthalene consists of two fused benzene rings.
Phenanthroline is a heterocyclic compound containing nitrogen atoms.
Triphenyl methane consists of three phenyl groups attached to a central carbon atom,not fused rings.
Therefore,the correct answer is Anthracene.

(A) The reaction of benzene with ozone $(O_3)$ leads to the formation of benzene triozonide as an intermediate product.
If the process is stopped after the addition of ozone and no hydrolysis (reduction) step is performed,the final product isolated is benzene triozonide.
Hydrolysis or reductive cleavage of benzene triozonide using $Zn/H_2O$ is required to obtain glyoxal $(CHO-CHO)$.

(C) The catalytic dehydrogenation and cyclization of $n-heptane$ in the presence of $Cr_2O_3/Al_2O_3$ at $750 \ K$ results in the formation of toluene.
This process is known as aromatization or reforming.
The reaction is: $n-C_7H_{16} \xrightarrow[750 \ K]{Cr_2O_3/Al_2O_3} \text{Toluene} + 4H_2$.

(B) $1$. The reaction of benzene $(C_6H_6)$ with nitrating mixture $(HNO_3 + H_2SO_4)$ is an electrophilic aromatic substitution reaction that yields nitrobenzene $(X)$.
$2$. Nitrobenzene $(X)$ contains a $-NO_2$ group,which is a strong electron-withdrawing group and a meta-directing group.
$3$. When nitrobenzene reacts with chlorine $(Cl_2)$ in the presence of a Lewis acid catalyst $(FeCl_3)$,the incoming electrophile $(Cl^+)$ is directed to the meta-position.
$4$. Therefore,the product $Y$ is $3-$nitrochlorobenzene (also known as $m-$nitrochlorobenzene).

(C) Knocking is the premature ignition of the fuel-air mixture in an engine,which causes a metallic sound. The knocking property is inversely related to the octane rating of the fuel.
The order of knocking tendency is: $\text{Straight chain paraffins} > \text{Branched chain paraffins} > \text{Olefins} > \text{Cycloalkanes} > \text{Aromatic hydrocarbons}$.
Since aromatic hydrocarbons have the highest octane rating,they exhibit the lowest knocking property.

(C) Benzene is prepared in the laboratory by the decarboxylation of sodium benzoate with soda lime $(NaOH + CaO)$.
The reaction is: $C_6H_5COONa + NaOH \xrightarrow{CaO, \Delta} C_6H_6 + Na_2CO_3$.
Thus,the correct option is $C$.

(A) The chlorination of benzene is an electrophilic aromatic substitution reaction.
In the presence of a Lewis acid catalyst like $FeCl_3$ or $AlCl_3$,the chlorine molecule $(Cl_2)$ reacts to generate the electrophile,which is the chloronium ion $(Cl^{+})$.
Therefore,the reactive species is $Cl^{+}$.

(A) The presence of delocalised $\pi$ electrons is a characteristic feature of aromatic compounds.
Benzene $(C_6H_6)$ is an aromatic compound that contains a cyclic system of $6$ delocalised $\pi$ electrons.

(B) In benzene,all the six carbon atoms are present in the same plane. All the carbon atoms are $sp^2$ hybridized. Hence,it is a planar molecule.

(A) In the chlorination of benzene,$FeCl_3$ acts as a Lewis acid catalyst.
It reacts with $Cl_2$ to generate the electrophile $Cl^{+}$ (chloronium ion).
The reaction is: $Cl_2 + FeCl_3 \rightarrow Cl^{+} + [FeCl_4]^{-}$.
Therefore,the correct option is $A$.

(D) The $-CH_3$ group in toluene is an electron-donating group due to the inductive effect and hyperconjugation.
It activates the benzene ring towards electrophilic substitution reactions and is ortho-para directing.
Therefore,nitration of toluene with concentrated $HNO_3$ and concentrated $H_2SO_4$ yields a mixture of $o-$nitrotoluene and $p-$nitrotoluene.
The reaction is:
$2C_6H_5CH_3 + 2HNO_3 \xrightarrow{Conc. H_2SO_4} C_6H_4(CH_3)(NO_2) (o-) + C_6H_4(CH_3)(NO_2) (p-) + 2H_2O$.

(B) $1$. The molecular formula $C_8H_{10}$ corresponds to the degree of unsaturation $4$,indicating an aromatic ring with two alkyl substituents.
$2$. $p-$Xylene ($1,4-$dimethylbenzene) has a high degree of symmetry. Due to the equivalence of positions,it yields only $1$ mononitro derivative.
$3$. Upon further nitration,$p-$Xylene can yield $3$ dinitro isomers depending on the sequence of substitution and steric factors,making it the correct choice.

(B) The catalytic hydrogenation of benzene $(C_6H_6)$ involves the addition of $3$ moles of hydrogen $(H_2)$ in the presence of a catalyst like nickel $(Ni)$,palladium $(Pd)$,or platinum $(Pt)$ at high temperature and pressure.
This reaction results in the saturation of the aromatic ring to form cyclohexane $(C_6H_{12})$.
The chemical equation is: $C_6H_6 + 3H_2 \xrightarrow{Ni} C_6H_{12}$ (Cyclohexane).

(A) Benzene was first discovered by the English scientist Michael Faraday in $1825$ in illuminating gas.
In $1834$,German chemist Eilhardt Mitscherlich heated benzoic acid with lime and produced benzene.
In $1845$,German chemist $A$.$W$. von Hofmann isolated benzene from coal tar.
Thus,coal tar is a primary industrial source of benzene.

(C) Aromatic hydrocarbons are obtained by fractional distillation of coal tar.
The first fraction (crude light oil) mainly contains benzene,toluene,and xylene.
The second fraction (middle oil) contains $C_{10}H_8$ (naphthalene).
The next fraction (heavy oil/green oil) contains mainly anthracene and phenanthrene.

(B) Lindane,also known as $BHC$ (Benzene Hexachloride) or gammaxene,is obtained by the additive chlorination of benzene in the presence of ultraviolet light (sunlight).
The reaction is as follows:
$C_6H_6 + 3Cl_2 \xrightarrow{h\nu} C_6H_6Cl_6$ (Lindane/$BHC$)
Therefore,the correct option is $(b)$.

(A) The fractional distillation of coal tar yields various fractions based on their boiling points. $Benzene$ is obtained from the $Light \ oil$ fraction,which distills over at a temperature range of $350 \ K$ to $440 \ K$. It is called light oil because its density is lower than that of water.