Aromatic hydrocarbon Questions in English

(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.

(A) In the presence of a Lewis acid (like $AlCl_3$),benzene undergoes an electrophilic substitution reaction with an alkyl halide. This reaction is known as Friedel-Crafts alkylation.
The reaction is as follows:
$C_6H_6 + C_2H_5Cl \xrightarrow{AlCl_3} C_6H_5-C_2H_5 + HCl$
The product formed is ethylbenzene,which has the molecular formula $C_8H_{10}$.

(D) Styrene $(C_6H_5CH=CH_2)$ is prepared from benzene $(C_6H_6)$ in two steps:
$1$. Ethylbenzene is prepared by the alkylation of benzene with ethene in the presence of an $AlCl_3$ catalyst: $C_6H_6 + CH_2=CH_2 \xrightarrow{AlCl_3} C_6H_5CH_2CH_3$.
$2$. Ethylbenzene is then dehydrogenated at high temperature $(700^{\circ}C)$ using a catalyst (typically iron oxide or $Al_2O_3$) to produce styrene: $C_6H_5CH_2CH_3 \xrightarrow{700^{\circ}C, \text{catalyst}} C_6H_5CH=CH_2 + H_2$.

(C) $o$-Xylene exists in two resonance structures. Ozonolysis of these structures breaks the double bonds to form dicarbonyl compounds.
Structure $I$ is $CH_3-CO-CO-CH_3$ (butane$-2,3-$dione).
Structure $II$ is $CH_3-CO-CHO$ ($2$-oxopropanal).
Structure $III$ is $CHO-CHO$ (ethanedial).
From the resonance structures of $o$-xylene,ozonolysis yields:
$1$ mole of $I$ (butane$-2,3-$dione),
$2$ moles of $II$ ($2$-oxopropanal),
$3$ moles of $III$ (ethanedial).
Therefore,the ratio of the products is $I:II:III = 1:2:3$.

(C) $1$. The starting material is $1,3,5$-trimethylbenzene (mesitylene).
$2$. Electrophilic aromatic substitution with $Br_2$ in $CCl_4$ occurs at the position ortho to the methyl groups. Since all positions are equivalent,it forms $2$-bromo-$1,3,5$-trimethylbenzene $(A)$.
$3$. Treatment of $A$ with $Mg$ in dry ether forms the Grignard reagent,$2,4,6$-trimethylphenylmagnesium bromide.
$4$. Reaction of the Grignard reagent with dry ice $(CO_2)$ followed by acidic workup $(H_3O^+)$ yields the corresponding carboxylic acid,$2,4,6$-trimethylbenzoic acid $(B)$.

(B) For a compound to be aromatic,it must satisfy $H$ückel's rule:
$(I)$ The molecule must be cyclic.
$(II)$ The molecule must be planar.
$(III)$ The molecule must be fully conjugated (all atoms in the ring must have an unhybridized $p$-orbital).
$(IV)$ The molecule must have $(4n+2) \pi$ electrons,where $n = 0, 1, 2, ...$.
Analysis of the given compounds:
$(A)$ Cyclopropenyl cation: Cyclic,planar,conjugated,and has $2 \pi$ electrons $(n=0)$. It is aromatic.
$(B)$ Cyclooctatetraene: It has $8 \pi$ electrons ($4n$ system,$n=2$). Furthermore,it adopts a non-planar 'tub' shape to avoid anti-aromaticity. Thus,it is non-aromatic.
$(C)$ Cyclopentadienyl anion: Cyclic,planar,conjugated,and has $6 \pi$ electrons $(n=1)$. It is aromatic.
$(D)$ $6-$(dimethylamino)fulvene: This is a substituted fulvene derivative that exhibits aromatic character due to the contribution of a dipolar resonance structure where the five-membered ring becomes a cyclopentadienyl anion ($6 \pi$ electrons).
Therefore,cyclooctatetraene is not aromatic.

(C) To determine the aromaticity,we use the Huckel rule: a cyclic,planar,fully conjugated system with $(4n+2)$ $\pi$-electrons is aromatic,where $n$ is an integer $(0, 1, 2, \dots)$.
$(i)$ Cyclopropenyl chloride reacts with $SbCl_5$ to form the cyclopropenyl cation,which has $2$ $\pi$-electrons $(n=0)$. This is aromatic.
$(ii)$ Cyclopentadiene reacts with sodium metal to form the cyclopentadienyl anion,which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.
$(iii)$ $7$-Bromocyclohepta-$1,3,5$-triene reacts with water to form the tropylium cation,which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.
$(iv)$ Cyclopentadienylamine reacts with $HNO_2$ to form a $4$ $\pi$-electron system (cyclopentadienyl cation),which is anti-aromatic (not aromatic).
Therefore,there are $3$ aromatic species generated.

(D) The Friedel-Crafts reaction is an electrophilic aromatic substitution reaction.
It requires an electron-rich aromatic ring to react efficiently with the electrophile $(CH_3^+)$.
Nitrobenzene and acetophenone contain electron-withdrawing groups ($-NO_2$ and $-COCH_3$ respectively),which deactivate the benzene ring towards electrophilic substitution.
Benzene is moderately reactive.
Toluene contains a methyl group $(-CH_3)$,which is an electron-donating group due to the $+I$ effect and hyperconjugation.
This increases the electron density of the benzene ring,making it the most reactive among the given options towards electrophilic substitution.

(C) The methyl group $(-CH_3)$ in toluene is an electron-donating group by inductive effect and hyperconjugation.
It is an ortho- and para-directing group.
Therefore,the nitration of toluene with a mixture of concentrated $HNO_3$ and $H_2SO_4$ at $25^{\circ} C$ yields a mixture of $o-$nitrotoluene and $p-$nitrotoluene as the major products,with a very small amount of $m-$nitrotoluene.
Thus,the reaction produces predominantly $o-$nitrotoluene and $p-$nitrotoluene.

(A) The reaction of hexadeuteriobenzene $(C_6D_6)$ with $Br_2$ in the presence of $Fe$ is an electrophilic aromatic substitution reaction.
Since $1 \ mole$ of $Br_2$ is used,one deuterium atom is replaced by a bromine atom to form bromopentadeuteriobenzene $(C_6D_5Br)$.
The addition of $H_2O$ is a standard workup procedure to remove inorganic salts or quench the reaction.
Thus,the major product is bromopentadeuteriobenzene.

(B) The reaction of $1,4-$dimethylbenzene ($p-$xylene) with anhydrous $AlCl_{3}$ and $HCl$ is an example of an isomerization reaction.
Under these acidic conditions,the methyl groups undergo rearrangement to form the more thermodynamically stable isomer.
$1,3-$dimethylbenzene ($m-$xylene) is more stable than $1,4-$dimethylbenzene ($p-$xylene) due to the reduced steric hindrance and electronic factors in the meta-substituted product.
Therefore,the reaction produces $1,3-$dimethylbenzene.

(D) Nuclear iodination of aromatic compounds is a reversible electrophilic substitution reaction.
$I_{2}$ is a weak electrophile and the reaction produces $HI$ as a byproduct,which is a strong reducing agent and can reduce the iodinated product back to the starting material.
To drive the reaction forward,an oxidizing agent like $HNO_{3}$ is used to oxidize $HI$ to $I_{2}$,thereby preventing the reverse reaction.
$2HI + 2HNO_{3} \rightarrow I_{2} + 2NO_{2} + 2H_{2}O$.
Thus,$I_{2} / HNO_{3}$ is the best reagent for this purpose.

(B) The synthesis of $1$-butyl-$4$-nitrobenzene involves the following steps:
$1$. Friedel-Crafts acylation of benzene with butanoyl chloride $(CH_3CH_2CH_2COCl)$ in the presence of $AlCl_3$ gives $1$-phenylbutan-$1$-one.
$2$. Clemmensen reduction of the ketone using $Zn/Hg$ and $HCl$ with heat reduces the carbonyl group to a methylene group,yielding butylbenzene.
$3$. Nitration of butylbenzene using a mixture of concentrated $HNO_3$ and $H_2SO_4$ introduces the nitro group at the para position because the alkyl group is an ortho/para-directing group.

(B) The reaction of benzene with $Me_{3}CCOCl$ (pivaloyl chloride) in the presence of anhydrous $AlCl_{3}$ is a Friedel-Crafts acylation reaction.
However,because the acyl group is bulky ($tert$-butyl group),the intermediate acylium ion is highly hindered.
Under these conditions,the reaction often proceeds with decarbonylation,leading to the formation of $tert$-butylbenzene $(Ph-CMe_{3})$ as the major product.
The reaction proceeds as follows:
$C_{6}H_{6} + Me_{3}CCOCl \xrightarrow{Anhy. AlCl_{3}} C_{6}H_{5}COCMe_{3} + HCl$
$C_{6}H_{5}COCMe_{3} \xrightarrow{-CO} C_{6}H_{5}CMe_{3}$

(C) Nitration is an electrophilic aromatic substitution reaction. The rate of reaction depends on the electron density of the benzene ring. Higher electron density leads to faster nitration.
$(I)$ Toluene has one methyl group $(-CH_3)$ which is electron-donating by the inductive effect $(+I)$ and hyperconjugation.
$(II)$ $m$-Xylene has two methyl groups. The electron density is increased by two methyl groups.
$(III)$ $p$-Xylene has two methyl groups. Similar to $m$-xylene,it has two electron-donating groups.
Comparing $m$-xylene and $p$-xylene: In $p$-xylene,the two methyl groups are at the para positions,which provides more effective stabilization of the intermediate arenium ion compared to $m$-xylene due to the resonance and hyperconjugation effects being more pronounced at the para position relative to the site of electrophilic attack.
Thus,the order of reactivity is $p$-xylene $(III)$ > $m$-xylene $(II)$ > toluene $(I)$.

(D) The $-NO_{2}$ group is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
$1$. It reduces the electron density of the benzene ring,thereby deactivating it towards electrophilic aromatic substitution (Statement $B$ is correct).
$2$. It increases the susceptibility of the ring towards nucleophilic attack by stabilizing the intermediate carbanion (Meisenheimer complex) through its $-M$ effect,thereby activating the ring towards nucleophilic aromatic substitution (Statement $C$ is correct).
Therefore,statements $B$ and $C$ are correct.

(A) The reaction sequence typically involves the oxidation of an alkyl side chain on a benzene ring. When an alkyl group (like an ethyl or propyl group) is attached to a benzene ring,treatment with strong oxidizing agents such as alkaline $KMnO_4$ followed by acidification results in the oxidation of the benzylic carbon to a carboxylic acid group $(-COOH)$. The remaining part of the side chain is cleaved. Therefore,the major product $P$ is a benzoic acid derivative.

(B) Friedel-Crafts alkylation of benzene with $n$-propyl chloride gives $iso$-propylbenzene as the major product because the $n$-propyl carbocation undergoes rearrangement to form a more stable $iso$-propyl carbocation: $CH_3CH_2CH_2^+ \rightarrow CH_3CH^+CH_3$.
Thus,Statement $B$ is correct.
Furthermore,alkylation of benzene increases the electron density on the ring,making it more reactive towards further electrophilic substitution,hence multiple substitution is a common byproduct. Thus,Statement $C$ is also correct.

(C) species is aromatic if it is cyclic,planar,fully conjugated,and follows $H$ückel's rule $(4n+2) \pi$ electrons.
$1$. $p$-Benzoquinone: Non-aromatic (contains $sp^3$ hybridized carbonyl carbons).
$2$. Cycloheptatrienyl cation: Aromatic ($6\pi$ electrons,cyclic,planar,fully conjugated).
$3$. Phenanthrene: Aromatic ($14\pi$ electrons,cyclic,planar,fully conjugated).
$4$. $1,4-$Cyclohexadiene: Non-aromatic (contains $sp^3$ hybridized carbons).
$5$. Cyclobutadiene dication: Aromatic ($2\pi$ electrons,cyclic,planar,fully conjugated).
$6$. Cyclohexadienyl anion: Non-aromatic (contains $sp^3$ hybridized carbon).
Thus,the aromatic species are: Cycloheptatrienyl cation,Phenanthrene,and Cyclobutadiene dication.
Total count = $3$.

(B) $1$. In the first reaction,benzene reacts with $6Cl_2$ in the presence of anhydrous $AlCl_3$ (a Lewis acid) under dark and cold conditions. This is an electrophilic substitution reaction where all hydrogen atoms of benzene are replaced by chlorine atoms,resulting in hexachlorobenzene $(C_6Cl_6)$. Thus,$X$ contains $6$ chlorine atoms.
$2$. In the second reaction,benzene reacts with $3Cl_2$ in the presence of $UV$ light at $500 \text{ K}$. This is a free radical addition reaction. Benzene undergoes addition to form benzene hexachloride $(C_6H_6Cl_6)$,which contains $6$ chlorine atoms. Thus,$Y$ contains $6$ chlorine atoms.
$3$. Therefore,the number of chlorine atoms in $X$ and $Y$ are $6$ and $6$,respectively.