Aromatic hydrocarbon Questions in English

(B) The reaction proceeds via the formation of a carbocation intermediate. $n$-propyl chloride reacts with $AlCl_3$ to form a primary carbocation,which undergoes a $1,2$-hydride shift to form a more stable secondary isopropyl carbocation.
$CH_3-CH_2-CH_2-Cl + AlCl_3 \rightarrow CH_3-CH_2-CH_2^+ + AlCl_4^-$
$CH_3-CH_2-CH_2^+ \xrightarrow{\text{1,2-hydride shift}} CH_3-CH^+-CH_3$
This secondary carbocation then attacks the benzene ring to form isopropylbenzene (cumene) as the major product.

(C) The oxidation of an alkyl group attached to a benzene ring,which possesses at least one benzylic hydrogen,using $KMnO_4/H^+$ results in the formation of benzoic acid.
In the case of ethylbenzene $(C_6H_5CH_2CH_3)$,the side chain is oxidized to a carboxylic acid group,yielding benzoic acid $(C_6H_5COOH)$.

(B) The Friedel-Crafts reaction $(FCR)$ is a type of electrophilic aromatic substitution reaction where an alkyl or acyl group is introduced into an aromatic ring in the presence of a Lewis acid catalyst like anhydrous $AlCl_3$.
Option $(B)$ represents the Friedel-Crafts alkylation of benzene with methyl chloride $(CH_3Cl)$ in the presence of anhydrous $AlCl_3$ to form toluene,which is a classic example of a Friedel-Crafts reaction.

(D) The nitro group $(-NO_2)$ is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It withdraws electron density from the benzene ring,thereby reducing the electron density available for electrophilic attack.
Consequently,it deactivates the benzene ring towards electrophilic substitution reactions.

(D) Aromatic compounds must follow $H$ückel's rule,which states they must be cyclic,planar,fully conjugated,and contain $(4n + 2) \pi$ electrons.
$1,3,5$-Heptatriene is an open-chain polyene. Since it is not cyclic,it cannot be aromatic.

(C) The Friedel-Crafts reaction is an electrophilic aromatic substitution reaction.
Strongly deactivating groups,such as the nitro group $(-NO_2)$,withdraw electron density from the benzene ring,making it insufficiently nucleophilic to attack the electrophile.
Therefore,nitrobenzene does not undergo the Friedel-Crafts reaction.

(A) Coal tar is a thick,dark liquid obtained as a byproduct during the destructive distillation of coal. It is a complex mixture of various organic compounds,primarily serving as the main industrial source of aromatic compounds such as benzene,toluene,naphthalene,and anthracene.

(D) The Friedel-Crafts reaction requires a Lewis acid catalyst to generate the electrophile.
$FeCl_3$ and $FeBr_3$ are common Lewis acid catalysts used in electrophilic substitution reactions.
$FeBr_2$ is not a strong Lewis acid and is not typically used as a catalyst for Friedel-Crafts reactions.
$NaCl$ is an ionic salt and does not act as a Lewis acid catalyst.
However,among the given options,$NaCl$ is the most distinct non-catalyst,but in the context of Friedel-Crafts specifically,$FeBr_2$ is also not a standard catalyst. Given the standard curriculum,$NaCl$ is the correct answer as it is a simple salt.

(D) The nitration of toluene is an electrophilic aromatic substitution reaction.
Since the methyl group $(-CH_3)$ is an ortho/para-directing group,the nitration proceeds in steps.
With excess concentrated $HNO_3$ and concentrated $H_2SO_4$,the reaction continues until all available ortho and para positions are substituted.
The final product formed is $2,4,6$-trinitrotoluene,commonly known as $TNT$.

(C) In the laboratory,benzene is prepared by heating sodium benzoate $(C_6H_5COONa)$ with soda lime $(NaOH + CaO)$.
This reaction is known as decarboxylation.
The chemical equation is: $C_6H_5COONa + NaOH \xrightarrow{CaO, \Delta} C_6H_6 + Na_2CO_3$.

(A) The reaction $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$ is an example of Friedel-Crafts alkylation.
In this reaction,benzene reacts with an alkyl halide in the presence of a Lewis acid catalyst like $AlCl_3$ to introduce an alkyl group onto the benzene ring.

(B) Arenes are aromatic hydrocarbons characterized by the presence of a cyclic,planar structure with $(4n+2) \pi$-electrons,which leads to the delocalization of $\pi$-electrons. This delocalization provides extra stability to the molecule,known as resonance energy. While resonance and high stability are properties of arenes,the defining characteristic that allows for their unique reactivity (electrophilic substitution rather than addition) is the delocalization of $\pi$-electrons. Therefore,all options are technically properties,but delocalization is the fundamental electronic feature.

(B) The nitro group $(-NO_2)$ is a strongly deactivating and meta-directing group.
When nitrobenzene undergoes nitration using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$,the incoming nitro group is directed to the meta position.
Therefore,the product formed is $m-$dinitrobenzene.

(B) In the Friedel-Crafts reaction,anhydrous $AlCl_3$ acts as a Lewis acid.
$AlCl_3$ is electron-deficient because the aluminum atom has only $6$ electrons in its valence shell.
It accepts a lone pair of electrons from the alkyl or acyl halide to form a carbocation,which acts as an electrophile in the substitution reaction.

(B) In the Friedel-Crafts alkylation of benzene with propene,the reaction proceeds via the formation of a carbocation intermediate.
Propene reacts with the electrophile $(H^+)$ to form a carbocation.
The secondary carbocation,$CH_3CH^+CH_3$,is more stable than the primary carbocation,$CH_3CH_2CH_2^+$,due to hyperconjugation and inductive effects.
Therefore,the reaction proceeds through the formation of the isopropyl carbocation $(CH_3CH^+CH_3)$.

(A) Aromatic compounds must follow $H$ückel's rule ($4n + 2$ $\pi$ electrons) and must be planar,cyclic,and fully conjugated.
$Cyclohexane$ $(C_6H_{12})$ is a saturated cyclic alkane with $sp^3$ hybridized carbon atoms.
It lacks a conjugated $\pi$ system and does not satisfy $H$ückel's rule,therefore it is non-aromatic.
$Trinitrotoluene$,$Picric$ $acid$,and $Xylene$ are all derivatives of benzene and are aromatic.

(D) The resonance energy is defined as the difference between the calculated heat of hydrogenation (assuming three isolated double bonds) and the experimental heat of hydrogenation.
Calculated heat of hydrogenation for benzene = $3 \times 28.6 \, kcal \, mol^{-1} = 85.8 \, kcal \, mol^{-1}$.
Experimental heat of hydrogenation for benzene = $50 \, kcal \, mol^{-1}$.
Resonance energy = $85.8 - 50 = 35.8 \, kcal \, mol^{-1}$.

(A) In the presence of a Lewis acid like $FeCl_3$,the chlorination of benzene proceeds via an electrophilic aromatic substitution mechanism.
$FeCl_3$ acts as a catalyst and reacts with $Cl_2$ to generate the electrophile $Cl^{+}$ (chloronium ion).
The reaction is: $Cl_2 + FeCl_3 \rightarrow Cl^{+} + [FeCl_4]^{-}$.
Thus,the electrophile required is $Cl^{+}$.

(C) Electrophilic aromatic substitution is facilitated by electron-donating groups $(EDG)$ attached to the benzene ring.
These groups increase the electron density of the ring through the inductive effect $(+I)$ or resonance effect ($+R$ or $+M$).
Among the given options:
$1$. $-CN$,$-CHO$,and $-NO_2$ are electron-withdrawing groups $(EWG)$ that decrease the electron density of the ring.
$2$. $-CH_3$ is an electron-donating group due to the hyperconjugation effect and inductive effect $(+I)$,which increases the electron density of the benzene ring,thereby activating it towards electrophilic substitution.

(D) The reaction of benzene $(C_6H_6)$ with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
Since benzene is in excess,the reaction proceeds in two steps:
$1$. $C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2Cl + HCl$
$2$. $C_6H_5-CH_2Cl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + HCl$
The final product obtained is diphenylmethane $(C_6H_5-CH_2-C_6H_5)$.

(B) The reaction is an electrophilic aromatic substitution reaction. The reactant is $1,2$-diphenylethan-$1$-one. The carbonyl group $(-CO-)$ is a strongly electron-withdrawing group and is meta-directing. The phenyl ring attached directly to the carbonyl group is deactivated,while the other phenyl ring (attached to the $-CH_2-$ group) is relatively more activated towards electrophilic substitution. The $-CH_2-$ group is ortho/para-directing. Therefore,the electrophile $Br^+$ will attack the para-position of the phenyl ring attached to the $-CH_2-$ group to form the major product,which is $1$-phenyl-$2$-($4$-bromophenyl)ethan-$1$-one.

(D) The complete nitration of toluene involves the reaction of toluene with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$.
Since the methyl group $(-CH_3)$ is an ortho/para directing group,it activates the ring towards electrophilic substitution.
Upon complete nitration,the nitro groups $(-NO_2)$ occupy all available ortho and para positions,resulting in the formation of $2,4,6$-trinitrotoluene $(TNT)$.

(C) Friedel-Crafts reactions (alkylation and acylation) involve the electrophilic substitution of aromatic rings.
These reactions require a strong Lewis acid catalyst to generate the electrophile.
Anhydrous $AlCl_3$ is the most commonly used Lewis acid catalyst for these reactions.

(B) The reaction of benzene with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ is a classic example of Friedel-Crafts acylation.
In this reaction,the acetyl group $(CH_3CO-)$ replaces a hydrogen atom on the benzene ring to form acetophenone $(C_6H_5COCH_3)$.

(D) Groups that withdraw electrons from the benzene ring via inductive or resonance effects are $m-$directing.
$-SO_3H$,$-NO_2$,and $-CN$ are electron-withdrawing groups and are $m-$directing.
$-NH_2$ is an electron-donating group due to the lone pair on nitrogen,which increases electron density at the $o-$ and $p-$ positions through resonance.
Therefore,$-NH_2$ is an $o, p-$directing group,not a $m-$directing group.

(B) The reaction of benzene with a mixture of concentrated $HNO_3$ and $H_2SO_4$ (nitrating mixture) at $353 \, K$ is known as nitration.
In this reaction,the nitro group $(-NO_2)$ replaces a hydrogen atom on the benzene ring to form nitrobenzene.

(D) $1$. Nitration of aromatic compounds is an electrophilic aromatic substitution reaction where the rate-determining step involves the formation of the sigma complex (arenium ion).
$2$. In the case of benzene and hexadeuterobenzene $(C_6D_6)$,the $C-H$ or $C-D$ bond is not broken in the rate-determining step,so there is no primary kinetic isotope effect. Thus,the rate of nitration is the same.
$3$. Toluene contains a methyl group $(-CH_3)$,which is an electron-donating group by inductive and hyperconjugation effects. This increases the electron density on the benzene ring,making it more reactive towards electrophilic substitution compared to benzene.
$4$. Therefore,both statements $B$ and $D$ are technically correct in a general context. However,in standard chemistry examinations,if multiple correct statements are present,the most fundamental characteristic is often sought. Given the options,$B$ and $D$ are both correct. If this is a single-choice question,$D$ is the most fundamental definition,while $B$ is a specific comparative observation.

(A) Friedel-Crafts alkylation involves the reaction of an aromatic hydrocarbon (like benzene) with an alkyl halide in the presence of a Lewis acid catalyst such as $AlCl_3$.
In the given options,$C_6H_6 + CH_3Cl$ represents the reaction between benzene and methyl chloride,which is a classic example of Friedel-Crafts alkylation.
$C_6H_6 + CH_3COCl$ represents Friedel-Crafts acylation,not alkylation.
Therefore,the correct option is $A$.

(A) Benzene $(C_6H_6)$ was first isolated by Michael Faraday in $1825$ from the oily residue left after illuminating gas was compressed. Therefore,the correct answer is $A$.

(A) The nitration of benzene involves the reaction of benzene with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $323-333 \ K$.
In this reaction,the nitronium ion $(NO_2^+)$ acts as an electrophile.
This electrophile attacks the electron-rich benzene ring,leading to the replacement of a hydrogen atom.
Therefore,this process is classified as an electrophilic substitution reaction.

(A) The Friedel-Crafts alkylation of benzene involves the reaction of benzene $(C_6H_6)$ with an alkyl halide (such as $CH_3Cl$) in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$.
This reaction substitutes a hydrogen atom on the benzene ring with an alkyl group.
For the preparation of toluene $(C_6H_5CH_3)$,benzene reacts with methyl chloride $(CH_3Cl)$:
$C_6H_6 + CH_3Cl \xrightarrow{Anhydrous \ AlCl_3} C_6H_5CH_3 + HCl$
Therefore,the correct option is $A$.

(B) Benzene and its derivatives are electron-rich due to the presence of a delocalized $\pi$-electron cloud.
Because of this,they are susceptible to attack by electrophiles (electron-deficient species).
Therefore,the characteristic reaction of benzene and its derivatives is $Electrophilic \ substitution$ reaction,where an electrophile replaces a hydrogen atom on the aromatic ring to maintain aromaticity.

(B) The given reaction involves the alkylation of benzene with methyl chloride $(CH_3Cl)$ in the presence of an anhydrous Lewis acid catalyst,$AlCl_3$.
This specific type of electrophilic aromatic substitution is known as the Friedel-Crafts alkylation reaction.

(D) The oxidation of toluene with alkaline $KMnO_4$ followed by acidification results in the formation of benzoic acid.
The reaction is: $C_6H_5CH_3 + [O]$ $\xrightarrow{KMnO_4/OH^-} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$.

(B) In the presence of $U.V.$ light,benzene undergoes an addition reaction with bromine to form benzene hexabromide $(C_6H_6Br_6)$,also known as hexabromocyclohexane.
This is a free radical addition reaction.
The reaction is: $C_6H_6 + 3Br_2 \xrightarrow{U.V. \text{ light}} C_6H_6Br_6$.

(B) The reaction of benzene with isobutylene in the presence of $H_2SO_4$ is an electrophilic aromatic substitution reaction (Friedel-Crafts alkylation).
$1$. $H_2SO_4$ protonates isobutylene $(CH_3)_2C=CH_2$ to form a stable tert-butyl carbocation $(CH_3)_3C^+$.
$2$. This carbocation acts as an electrophile and attacks the benzene ring.
$3$. The final product formed is tert-butylbenzene.

(A) When benzene reacts with ozone $(O_3)$,it forms benzene triozonide as an intermediate product.
Upon further hydrolysis (reductive ozonolysis) of benzene triozonide,it breaks down into three molecules of glyoxal $(CHO-CHO)$.
Therefore,the final stable product obtained is glyoxal.

(C) The oxidation of alkylbenzenes with alkaline $KMnO_4$ followed by acidification leads to the formation of benzoic acid,regardless of the length of the alkyl side chain,provided that the benzylic carbon has at least one hydrogen atom.
Ethylbenzene $(C_6H_5CH_2CH_3)$ undergoes oxidation to form benzoic acid $(C_6H_5COOH)$.

(B) The $-CH_3$ group present in Toluene is an electron-donating group ($+I$ effect and hyperconjugation),which increases the electron density of the benzene ring.
This makes the ring more reactive towards electrophilic substitution compared to Benzene.
In contrast,$-COOH$ and $-NO_2$ groups are electron-withdrawing groups,which deactivate the benzene ring towards electrophilic nitration.
Therefore,Toluene is the most reactive among the given options.

(A) The reactivity towards electrophilic aromatic substitution $(ESR)$ depends on the electron density in the benzene ring. Substituents that increase electron density (electron-donating groups) activate the ring,while those that decrease it (electron-withdrawing groups) deactivate the ring.
$(I)$ Benzene: Reference compound.
$(II)$ Toluene $(-CH_3)$: $-CH_3$ is an electron-donating group by $+I$ and hyperconjugation effects,activating the ring.
$(III)$ Anisole $(-OCH_3)$: $-OCH_3$ is a strong electron-donating group by $+M$ effect,which strongly activates the ring.
$(IV)$ Trifluoromethylbenzene $(-CF_3)$: $-CF_3$ is a strong electron-withdrawing group by $-I$ effect,which strongly deactivates the ring.
Order of activation: $-CF_3$ (deactivating) < Benzene < $-CH_3$ (activating) < $-OCH_3$ (strongly activating).
Therefore,the increasing order of reactivity is $IV < I < II < III$.

(D) The reactivity of aromatic compounds towards electrophilic substitution depends on the electron density of the benzene ring.
An electron-donating group (like $-NH_2$ in Aniline) increases the electron density,making the ring more reactive.
An electron-withdrawing group (like $-NO_2$ in Nitrobenzene) decreases the electron density,making the ring less reactive.
Benzene has no substituent,so its reactivity is intermediate.
Thus,the order of reactivity is: Aniline $(I) >$ Benzene $(II) >$ Nitrobenzene $(III)$.

(D) The molecular formula of benzene is $C_6H_6$.
In the structure of benzene,there are $6$ $C-C$ sigma bonds and $6$ $C-H$ sigma bonds,making a total of $12$ sigma bonds.
Benzene contains $3$ alternating double bonds,where each double bond consists of one sigma bond and one pi bond.
Therefore,there are $3$ pi $(\pi)$ bonds in benzene.
Thus,the number of sigma bonds is $12$ and the number of pi bonds is $3$.

(B) The Friedel-Crafts reaction involves the electrophilic substitution of an aromatic ring.
Option $B$ represents the Friedel-Crafts acylation reaction,where benzene reacts with acetyl chloride in the presence of an anhydrous $AlCl_3$ catalyst to form acetophenone.

(C) The reaction of benzene with an alkyl halide (like ethyl chloride,$CH_3CH_2Cl$) in the presence of a Lewis acid catalyst such as anhydrous aluminum chloride $(AlCl_3)$ is known as the Friedel-Crafts alkylation reaction.
In this reaction,the ethyl group $(-CH_2CH_3)$ replaces a hydrogen atom on the benzene ring to form ethylbenzene.
The chemical equation is:
$C_6H_6 + CH_3CH_2Cl \xrightarrow{anhydrous \ AlCl_3} C_6H_5CH_2CH_3 + HCl$
Comparing this with the given options,the structure corresponding to ethylbenzene is shown in option $C$.

(A) In the nitration of benzene,the mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ is used.
$H_2SO_4$ is a stronger acid than $HNO_3$,so it protonates the $HNO_3$ molecule.
The reaction is: $HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^- \rightleftharpoons H_2O + NO_2^+ + HSO_4^-$.
Here,$H_2SO_4$ acts as an acid (proton donor) and $HNO_3$ acts as a base (proton acceptor) to generate the electrophile $NO_2^+$ (nitronium ion).

(B) Benzene $(C_6H_6)$ is a planar molecule with a hexagonal ring structure.
Each carbon atom in the benzene ring is $sp^2$ hybridized.
Due to $sp^2$ hybridization,the geometry around each carbon atom is trigonal planar,which results in bond angles of $120^o$ for both $C-C-C$ and $C-C-H$ bonds.

(D) Benzene is a highly stable molecule due to the phenomenon of resonance.
The $6 \pi$-electrons are delocalized over the entire ring,which provides extra stability to the molecule.
Addition reactions would involve the breaking of this delocalized $\pi$-electron system,which would result in the loss of resonance energy.
Therefore,benzene prefers substitution reactions over addition reactions to preserve its aromatic character and resonance stability.

(D) Sulfonation 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,making the ring more reactive towards electrophiles.
$Toluene$ $(C_6H_5CH_3)$ contains a methyl group $(-CH_3)$,which is an electron-donating group due to the $+I$ effect and hyperconjugation.
$Benzene$ has no substituent.
$Nitrobenzene$ $(-NO_2)$ and $Benzoic$ $acid$ $(-COOH)$ contain electron-withdrawing groups,which deactivate the ring.
Therefore,$Toluene$ undergoes sulfonation most easily.