Aromatic hydrocarbon Questions in English

(C) $(i)$ Benzene reacts with $O_3$ to form benzene triozonide. This is a correct statement.
$(ii)$ Benzene is a planar molecule with $sp^2$ hybridized carbon atoms. This is an incorrect statement.
$(iii)$ Benzene undergoes Friedel-Crafts acylation with $CH_3COCl$ in the presence of anhydrous $AlCl_3$ to form acetophenone (phenylethanone),which is a monosubstituted product. This is a correct statement.
$(iv)$ Benzene reacts with $Cl_2$ under photochemical conditions ($UV$ light) via addition reaction to form $1,2,3,4,5,6$-hexachlorocyclohexane (gammexane),not hexachlorobenzene. This is an incorrect statement.
Therefore,statements $(i)$ and $(iii)$ are correct.

(B) When $Glucose$ is heated with $HI$,it undergoes reduction to form $n-hexane$.
The reaction is: $C_6H_{12}O_6 + HI \xrightarrow{\Delta} CH_3-(CH_2)_4-CH_3$ $(n-hexane)$.
Further,when $n-hexane$ is heated with $Mo_2O_3$ at $773 \ K$ and $10-20 \ atm$ pressure,it undergoes aromatization (dehydrocyclization) to form $Benzene$.
Therefore,the major product is $Benzene$.

(D) Alkaline $KMnO_4$ oxidizes any alkyl group directly attached to a benzene ring to a carboxylic acid group $(-COOH)$,provided the alkyl group has at least one $\alpha$-hydrogen atom.
In both ethylbenzene and propylbenzene,the $\alpha$-carbon has hydrogen atoms,so both are oxidized to benzoic acid.
Therefore,both $Y$ and $Z$ are benzoic acid.

(B) The cyclopentadienyl anion is a cyclic,planar species with $6 \pi \ e^-$ (where $n=1$ in the $4n+2$ $H$ückel rule).
Since it contains $6 \pi \ e^-$,it follows $H$ückel's rule and is aromatic.
Because it does not contain a benzene ring,it is classified as non-benzenoid.
Therefore,it is non-benzenoid and aromatic.

(D) Reactivity towards an electrophilic aromatic substitution reaction depends on the electron density of the benzene ring. Electron-donating groups $(EDG)$ increase the electron density,making the ring more reactive,while electron-withdrawing groups $(EWG)$ decrease it.
The substituents are:
$(iv) \text{ Phenol } (-OH): \text{ Strong EDG due to } +M \text{ effect.}$
$(ii) \text{ Toluene } (-CH_3): \text{ Weak EDG due to } +I \text{ and hyperconjugation.}$
$(i) \text{ Benzene: Reference compound.}$
$(iii) \text{ Chlorobenzene } (-Cl): \text{ EWG due to strong } -I \text{ effect (despite } +M \text{ effect, the net effect is deactivating).}$
Therefore,the order of decreasing reactivity is: $(iv) > (ii) > (i) > (iii)$.

(A) The stability of polycyclic aromatic hydrocarbons is determined by the resonance energy per ring.
Benzene $(A)$: $36 \ kcal/mol$ per ring.
Naphthalene $(B)$: $61 \ kcal/mol$ for $2$ rings,so $61/2 = 30.5 \ kcal/mol$ per ring.
Anthracene $(C)$: $84 \ kcal/mol$ for $3$ rings,so $84/3 = 28 \ kcal/mol$ per ring.
Phenanthrene $(D)$: $92 \ kcal/mol$ for $3$ rings,so $92/3 = 30.67 \ kcal/mol$ per ring.
Comparing the values: $28 (C) < 30.5 (B) < 30.67 (D) < 36 (A)$.
Therefore,the order of stability is $C < B < D < A$.

(A) To determine if a species is aromatic,it must follow $H$ückel's rule: it must be cyclic,planar,fully conjugated,and contain $(4n+2) \pi$ electrons,where $n$ is an integer $(0, 1, 2, ...)$.
$A$. Cyclopentadienyl cation $(C_5H_5^+)$: It has $4 \pi$ electrons (anti-aromatic).
$B$. Cyclopentadienyl anion $(C_5H_5^-)$: It has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$C$. Cycloheptatrienyl cation $(C_7H_7^+)$: It has $6 \pi$ electrons $(n=1)$,so it is aromatic.
$D$. Naphthalene $(C_{10}H_8)$: It has $10 \pi$ electrons $(n=2)$,so it is aromatic.
Therefore,the cyclopentadienyl cation is not aromatic.

(A) To determine the number of $\pi$-electrons in each species:
$1$. Phenanthrene: It consists of three fused benzene rings. It has $14$ $\pi$-electrons.
$2$. Naphthalene: It consists of two fused benzene rings. It has $10$ $\pi$-electrons.
$3$. Cyclopentadienyl anion: It has two double bonds ($4$ $\pi$-electrons) and one lone pair on the carbon atom involved in resonance,contributing $2$ $\pi$-electrons,totaling $6$ $\pi$-electrons.
$4$. Cycloheptatrienyl cation: It has three double bonds ($6$ $\pi$-electrons) and a vacant $p$-orbital on the positively charged carbon,totaling $6$ $\pi$-electrons.
Comparing these,Phenanthrene has the maximum number of $\pi$-electrons $(14)$.
Therefore,the correct option is $A$.

(B) To identify aromatic compounds,we apply $H$ückel's rule,which states that a compound is aromatic if it is cyclic,planar,fully conjugated,and contains $(4n + 2) \pi$ electrons (where $n = 0, 1, 2, \dots$).
$1$. Aniline: It is a benzene derivative,which is cyclic,planar,fully conjugated,and has $6 \pi$ electrons. It is aromatic.
$2$. Naphthalene: It is a polycyclic aromatic hydrocarbon with $10 \pi$ electrons. It is aromatic.
$3$. Cyclopentadienyl cation: It has $4 \pi$ electrons,which follows the $4n$ rule. It is anti-aromatic.
$4$. $2,5-$Dihydrofuran: It is not fully conjugated due to the $sp^3$ hybridized carbon atom. It is non-aromatic.
$5$. Pyridine: It is a heterocyclic compound with $6 \pi$ electrons. It is aromatic.
Thus,the aromatic compounds are Aniline,Naphthalene,and Pyridine. The total number of aromatic compounds is $3$.

(B) Key Idea: Conditions for a species to be aromatic are:
$(I)$ The structure must follow $H$ückel's rule,i.e.,it must have $(4n+2) \pi$-electrons,where $n$ is an integer $(0, 1, 2, 3, \dots)$.
$(II)$ The structure must be a planar cyclic ring.
$(A)$ Cyclopentadienyl anion: It has $6 \pi$-electrons $(n=1)$ and is planar. Thus,it is aromatic.
$(B)$ $1,2-$dihydronaphthalene: It is not fully conjugated and has $8 \pi$-electrons. Thus,it is not aromatic.
$(C)$ Tropylium cation: It has $6 \pi$-electrons $(n=1)$ and is planar. Thus,it is aromatic.
$(D)$ Phenanthrene: It has $14 \pi$-electrons $(n=3)$ and is planar. Thus,it is aromatic.
$(E)$ Cyclopentadienyl cation: It has $4 \pi$-electrons ($n=1$ for $4n$ rule). Thus,it is anti-aromatic (not aromatic).
$(F)$ Cyclopropenyl anion: It has $4 \pi$-electrons ($n=0$ for $4n$ rule). Thus,it is anti-aromatic (not aromatic).
Therefore,species $(B)$,$(E)$,and $(F)$ are not aromatic. Hence,option $(b)$ is the correct answer.

(C) Tropolone is a seven-membered ring compound containing a carbonyl group and a hydroxyl group.
It does not contain a benzene ring,so it is non-benzenoid.
It exhibits aromatic character due to the resonance structure where the seven-membered ring becomes a tropylium cation $(C_7H_7^+)$,which follows $H$ückel's rule ($4n+2$ $\pi$ electrons,where $n=1$).
Thus,tropolone is non-benzenoid and aromatic.

(B) Linear alkanes containing $6$ or more carbon atoms undergo cyclization and aromatization when heated at $773 \ K$ under $10-20 \ atm$ pressure in the presence of catalysts like $Cr_2O_3$,$V_2O_5$,or $Mo_2O_3$ supported over alumina. This process is known as aromatization or reforming. For example,$n$-hexane gives benzene under these conditions.

(A) When acetylene $(C_2H_2)$ is passed through a red hot iron tube,it undergoes cyclic polymerization to form benzene $(C_6H_6)$,which is compound $X$.
$3C_2H_2 \xrightarrow{\text{red hot iron tube}} C_6H_6 (X)$
Option $A$ represents the reduction of phenol using zinc dust,which is a standard laboratory method to produce benzene:
$C_6H_5OH + Zn \xrightarrow{\text{distillation}} C_6H_6 + ZnO$
Thus,reaction $A$ yields $X$ as the major product.

(D) $1$. The reaction of $C_2 H_2$ (acetylene) with a red hot iron tube at $500^{\circ}C$ is a cyclic polymerization reaction that produces $C_6 H_6$ (benzene). Therefore,$A = C_6 H_6$.
$2$. The reaction of benzene $(A)$ with a mixture of concentrated $HNO_3$ and concentrated $H_2 SO_4$ at $70^{\circ}C$ is a nitration reaction,which produces $C_6 H_5 NO_2$ (nitrobenzene). Therefore,$B = C_6 H_5 NO_2$.
$3$. The reduction of nitrobenzene $(B)$ with $LiAlH_4$ can lead to the formation of azobenzene $(C_6 H_5-N=N-C_6 H_5)$ under specific conditions.
$4$. Thus,$A = C_6 H_6$ and $B = C_6 H_5 NO_2$.

(C) $1$. The starting material is $2$-methylhexane. Upon heating with $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$,it undergoes aromatization to form $A$ (Toluene).
$2$. Toluene $(A)$ reacts with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ to undergo oxidation of the methyl group to a carboxylic acid group,forming $B$ (Benzoic acid).
$3$. Benzoic acid $(B)$ reacts with concentrated $HNO_3$ and $H_2SO_4$ (nitration). Since the $-COOH$ group is meta-directing,the $-NO_2$ group attaches at the meta position,forming $C$ ($m$-nitrobenzoic acid).
$4$. Compound $C$ has substituents $D = -COOH$ and $E = -NO_2$.

(D) $1$. The starting material is $n$-butyl bromide $(CH_3CH_2CH_2CH_2Br)$.
$2$. Reaction with $Na$ in dry ether is the Wurtz reaction,which couples the alkyl group to form $n$-octane $(CH_3(CH_2)_6CH_3)$,so $X$ is $n$-octane.
$3$. Aromatization of $n$-octane using $\text{Cr}_2\text{O}_3$ at $773 \text{ K}$ and $10-20 \text{ atm}$ leads to the formation of ethylbenzene $(Y)$.
$4$. Friedel-Crafts acylation of ethylbenzene with $\text{CH}_3\text{COCl}$ in the presence of anhydrous $\text{AlCl}_3$ gives $p$-methylacetophenone (or $o$-methylacetophenone,but $p$- is major) as $Z$. The structure of $p$-methylacetophenone is $CH_3-C_6H_4-COCH_3$.
$5$. In $p$-methylacetophenone $(CH_3-C_6H_4-COCH_3)$:
- $sp^3$ carbons: The methyl group on the ring $(1)$,the methyl group on the carbonyl $(1)$. Total $a = 2$.
- $sp^2$ carbons: The $6$ carbons of the benzene ring $(6)$ and the carbonyl carbon $(1)$. Total $b = 7$.
$6$. The value of $(a+b) = 2 + 7 = 9$.

(D) Aromatic compounds,such as $C_6H_6$ (benzene),have a high carbon-to-hydrogen ratio.
Due to incomplete combustion,they produce a sooty flame.

(D) In benzene,all the $C-C$ bond lengths are equal to $1.39 \ Å$ due to resonance. Therefore,the statement that benzene has two different bond lengths of $1.54 \ Å$ and $1.34 \ Å$ is incorrect.

(A) $1$. $C_2H_4 + Br_2/CCl_4 \rightarrow BrCH_2-CH_2Br$ ($A$ is $1,2$-dibromoethane).
$2$. $BrCH_2-CH_2Br \xrightarrow[(ii) NaNH_2]{(i) alc. KOH} HC \equiv CH$ ($B$ is ethyne).
$3$. $3HC \equiv CH \xrightarrow{\text{cyclic polymerization}} C_6H_6$ ($C$ is benzene).
$4$. $C_6H_6 + 3Cl_2 \xrightarrow{\text{dry } AlCl_3, \text{dark, cold}} C_6H_6Cl_6$ ($D$ is benzene hexachloride,also known as gammaxene or lindane).
$5$. The molecular formula of $D$ is $C_6H_6Cl_6$. The empirical formula is the simplest ratio of atoms,which is $CH Cl$.

(C) The given reaction is the Friedel-Crafts alkylation of benzene with ethyl chloride $(C_2H_5Cl)$ in the presence of anhydrous $AlCl_3$ to form ethylbenzene $(X)$.
The chemical formula of ethylbenzene $(X)$ is $C_8H_{10}$.
Atomic mass of Carbon $(C)$ $= 12.01 \ g/mol$.
Atomic mass of Hydrogen $(H)$ $= 1.008 \ g/mol$.
Molar mass of $C_8H_{10} = (8 \times 12.01) + (10 \times 1.008) = 96.08 + 10.08 = 106.16 \ g/mol$.
Percentage of carbon in $C_8H_{10} = \frac{\text{Mass of Carbon}}{\text{Total Molar Mass}} \times 100$.
Percentage of carbon $= \frac{96.08}{106.16} \times 100 \approx 90.505 \% \approx 90.6 \%$.

(C) The reaction of benzene $(C_6H_6)$ with ozone $(O_3)$ is an ozonolysis reaction.
Benzene reacts with three molecules of ozone to form benzene triozonide $(X)$.
Benzene triozonide on reductive hydrolysis with $Zn / H_2O$ yields three molecules of glyoxal $(Y)$ $(CHO-CHO)$.
Therefore,$X$ is triozonide and $Y$ is glyoxal.

(C) An electrophilic substitution reaction involves the replacement of an atom or group on a molecule by an electrophile.
Option $A$ is a nucleophilic addition reaction.
Option $B$ is a nucleophilic substitution reaction $(S_N1)$.
Option $C$ is a Friedel-Crafts acylation reaction,which is a classic example of electrophilic aromatic substitution where the electrophile is the acylium ion $(CH_3CO^+)$.
Option $D$ is an elimination reaction (dehalogenation).

(A) The reaction of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction,which produces toluene. Thus,$X = CH_3Cl$.
The reaction of benzene with $CH_3COCl$ (acetyl chloride) in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction,which produces acetophenone. Thus,$Y = CH_3COCl$.
Therefore,$X$ and $Y$ are $CH_3Cl$ and $CH_3COCl$ respectively.

(B) $1$. The reaction of $C_2H_4Br_2$ ($1$,$2$-dibromoethane) with $Alc. KOH$ followed by $NaNH_2$ leads to dehydrohalogenation to form ethyne $(X = HC \equiv CH)$.
$2$. Passing ethyne through a red-hot iron tube at $873 \ K$ results in cyclic polymerization to form benzene $(Y = C_6H_6)$.
$3$. Benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ via Friedel-Crafts acylation to produce acetophenone $(Z = C_6H_5COCH_3)$.

(C) $1$. The reaction of $2CH_3CH_2CH_2Br$ with $Na/Ether$ is a Wurtz reaction,which produces $n$-hexane $(X = CH_3CH_2CH_2CH_2CH_2CH_3)$.
$2$. The aromatization of $n$-hexane in the presence of $Mo_2O_3$ at $773K$ and $10-20 \ atm$ yields benzene $(Y = C_6H_6)$.
$3$. The reaction of benzene $(Y)$ with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation,which produces toluene $(Z = C_6H_5CH_3)$.

(A) The reactions are identified as follows:
$A$. $\text{Benzene} + Cl_2 \xrightarrow{AlCl_3} \text{Chlorobenzene}$ (Electrophilic substitution),so $A-III$.
$B$. $\text{Benzene} + CH_3Cl \xrightarrow{\text{anhyd. } AlCl_3} \text{Toluene}$ (Friedel-Crafts alkylation),so $B-IV$.
$C$. $\text{Benzene} + R-CO-Cl \xrightarrow{\text{anhyd. } AlCl_3} \text{Alkyl phenyl ketone}$ (Friedel-Crafts acylation),so $C-II$.
$D$. $\text{Toluene} \xrightarrow{(i) KMnO_4/KOH, (ii) H_3O^+} \text{Benzoic acid}$ (Oxidation of side chain),so $D-I$.
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(C) The reaction shown is a Friedel-Crafts acylation of benzene $(A)$ to form acetophenone,which is then further substituted to form $B$. The electrophile in the Friedel-Crafts acylation reaction is the acylium ion,$CH_3-C\equiv O^{\oplus}$ (or $CH_3CO^{\oplus}$),which is generated by the reaction of acetyl chloride $(CH_3COCl)$ with the Lewis acid catalyst $AlCl_3$.

(A) In reaction $(A)$,Friedel-Crafts alkylation of benzene with $n$-propyl chloride in the presence of $AlCl_3$ leads to the formation of a carbocation intermediate. The primary propyl carbocation rearranges to a more stable secondary isopropyl carbocation,which then attacks the benzene ring to form isopropylbenzene as the major product $P$.
In reaction $(B)$:
$(i)$ Friedel-Crafts alkylation of benzene with $CH_3Cl$ in the presence of $AlCl_3$ gives toluene.
$(ii)$ Free radical halogenation of toluene with $Br_2$ in the presence of $hv$ (light) gives benzyl bromide $(C_6H_5CH_2Br)$.
$(iii)$ Nucleophilic substitution of benzyl bromide with aqueous $KOH$ gives benzyl alcohol $(C_6H_5CH_2OH)$ as the major product $Q$.

(C) Step $1$: Dehydrohalogenation of $C_6H_5-CHBr-CH_2Br$ with $alc. KOH$ followed by $NaNH_2$ leads to the formation of phenylacetylene $(C_6H_5-C \equiv CH)$.
Step $2$: The cyclic trimerization of phenylacetylene occurs when passed through a red hot iron tube at $873 \ K$.
Step $3$: The trimerization of $C_6H_5-C \equiv CH$ yields $1,3,5-triphenylbenzene$ as the major product due to steric hindrance,which favors the meta-substituted product.

(D) When benzene $(C_6H_6)$ reacts with excess chlorine $(Cl_2)$ in the presence of ultraviolet light,an addition reaction occurs.
This results in the formation of benzene hexachloride $(C_6H_6Cl_6)$,also known as $BHC$ or Gammaxene.
The reaction is: $C_6H_6 + 3Cl_2 \xrightarrow{uv/hv} C_6H_6Cl_6$.

(B) This reaction is known as Friedel-Crafts alkylation. In the presence of anhydrous $AlCl_3$,$n-$propyl chloride acts as an alkylating agent. It is an electrophilic aromatic substitution reaction.
Step $1$: Generation of electrophile:
$CH_3-CH_2-CH_2-Cl + AlCl_3 \rightarrow CH_3-CH_2-CH_2^+ + AlCl_4^-$
Step $2$: Carbocation rearrangement:
The primary carbocation $(CH_3-CH_2-CH_2^+)$ formed in step $1$ rearranges to form a more stable secondary carbocation $(CH_3-CH^+-CH_3)$ via a $1,2-$hydride shift.
Step $3$: Electrophilic attack:
This secondary carbocation attacks the benzene ring to form the intermediate.
Step $4$: Deprotonation:
The final step is the abstraction of a proton from the intermediate to restore aromaticity,yielding isopropyl benzene (cumene) as the major product.

(A) The given reactions are analyzed as follows:
$1.$ Friedel-Crafts alkylation of benzene with $t$-butyl chloride in the presence of $AlCl_3$ yields $t$-butyl benzene.
$2.$ Benzene reacts with $2$-methylpropene (isobutylene) in the presence of $BF_3/HF$ to form $t$-butyl benzene via a carbocation intermediate.
$3.$ Benzene reacts with $t$-butyl alcohol in the presence of $H_2SO_4$ to form $t$-butyl benzene through the formation of a $t$-butyl carbocation.
$4.$ Benzene reacts with butanoyl chloride via Friedel-Crafts acylation to form $1$-phenylbutan-$1$-one. Subsequent Clemmensen reduction $(Zn/Hg, HCl)$ yields $n$-butylbenzene,not $t$-butyl benzene.
Thus,reactions $1, 2,$ and $3$ produce $t$-butyl benzene.

(B) The reaction of benzene with acetyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction,which yields acetophenone $(C_6H_5COCH_3)$ as the major product.
In the structure of acetophenone:
$1$. The benzene ring contains $3$ $\pi$-bonds.
$2$. The carbonyl group $(C=O)$ contains $1$ $\pi$-bond.
Total number of $\pi$-bonds = $3 + 1 = 4$.
$3$. The oxygen atom in the carbonyl group has $2$ lone pairs of electrons.
Total number of lone pairs = $2$.
Thus,the number of $\pi$-bonds and lone pairs are $4$ and $2$ respectively.

(A) The reaction of benzene with $n$-propyl chloride $(CH_3CH_2CH_2Cl)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
In this reaction,the primary carbocation $(CH_3CH_2CH_2^+)$ formed initially is unstable and undergoes a $1,2-$hydride shift to form a more stable secondary carbocation $(CH_3CH^+CH_3)$.
This secondary carbocation then attacks the benzene ring to form isopropylbenzene (cumene) as the major product $P$.

(A) In the process of nitration,the nitronium ion $(NO_2^+)$ is generated by the reaction between sulfuric acid $(H_2SO_4)$ and nitric acid $(HNO_3)$.
$H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H_2NO_3^+$
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$
According to the Brønsted-Lowry acid-base theory,$H_2SO_4$ acts as a proton donor (acid),while $HNO_3$ acts as a proton acceptor (base) because it accepts a proton from $H_2SO_4$ to form $H_2NO_3^+$,which subsequently dehydrates to form the nitronium ion.

(B) The given reaction is a Friedel-Crafts alkylation reaction. In this reaction,benzene reacts with methyl chloride $(CH_3Cl)$ in the presence of anhydrous aluminum chloride $(Anhy. AlCl_3)$ as a Lewis acid catalyst to form toluene $(C_6H_5CH_3)$ as the product $(Z)$.
The reaction is:
$C_6H_6 + CH_3Cl \xrightarrow{Anhy. AlCl_3} C_6H_5CH_3 + HCl$
Thus,the product $(Z)$ is toluene.

(D) Friedel-Crafts Acylation: This reaction involves the treatment of benzene with an acylating agent such as acetyl chloride $(CH_3COCl)$ or acetic anhydride in the presence of a Lewis acid catalyst like anhydrous aluminum chloride $(AlCl_3)$.
The reaction is as follows:
$C_6H_6 + CH_3COCl \xrightarrow{anhydrous \ AlCl_3} C_6H_5COCH_3 + HCl$
Thus,benzene reacts with acetyl chloride $(CH_3COCl)$ to form acetophenone.

(A) Acetophenone can be prepared from benzene by its electrophilic substitution (Friedel-Crafts acylation) reaction as follows:
$C_6H_6 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} C_6H_5COCH_3 + HCl$
In this reaction,the hydrogen atom of the benzene ring is replaced by an acetyl group $(-COCH_3)$,which is a characteristic electrophilic substitution reaction.