Aromatic hydrocarbon Questions in English

(B) Benzoic acid $(C_6H_5COOH)$ reacts with soda lime $(NaOH + CaO)$ to produce benzene via decarboxylation: $C_6H_5COOH + NaOH \xrightarrow{CaO} C_6H_6 + Na_2CO_3$. Thus,$X$ is soda lime.
Phenol $(C_6H_5OH)$ reacts with zinc dust to produce benzene via reduction: $C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$. Thus,$Y$ is zinc dust.
Therefore,$X$ and $Y$ are respectively soda lime and zinc dust.

(B) When $n$-heptane $(CH_3(CH_2)_5CH_3)$ is heated to $773 \ K$ under high pressure in the presence of $V_2O_5$,$Cr_2O_3$,or $Mo_2O_3$ supported over $Al_2O_3$,it undergoes cyclization and dehydrogenation to form toluene $(C_6H_5CH_3)$.
This process is known as aromatization or reforming.

(C) The correct option is $(C)$.
Friedel-Crafts acylation involves the reaction of an aromatic compound (like benzene) with an acyl halide (like $CH_3COCl$) in the presence of a Lewis acid catalyst (like anhydrous $AlCl_3$) to form an aromatic ketone.
The reaction is: $C_6H_6 + CH_3COCl \xrightarrow{\text{anhydrous } AlCl_3} C_6H_5COCH_3 + HCl$.

(A) In the nitration of benzene,the mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ is used.
Here,$H_2SO_4$ acts as a stronger acid and protonates $HNO_3$ to generate the electrophile $NO_2^+$.
The reaction is as follows:
$HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^-$
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$
Since $HNO_3$ accepts a proton from $H_2SO_4$,it acts as a base in this reaction.

(B) Ring monobromination is an electrophilic aromatic substitution reaction.
The reactivity of the benzene ring depends on the nature of the substituent group.
Activating groups (electron-donating groups like $-CH_3$) increase the electron density on the ring and increase reactivity,while deactivating groups (electron-withdrawing groups like $-COOH$ and $-NO_2$) decrease the electron density and decrease reactivity.
Benzene $(III)$ is more reactive than rings with deactivating groups but less reactive than rings with activating groups.
Among the deactivating groups,$-NO_2$ is more strongly deactivating than $-COOH$.
Therefore,the correct decreasing order of reactivity is: $C_6H_5CH_3$ $(I)$ > $C_6H_6$ $(III)$ > $C_6H_5COOH$ $(II)$ > $C_6H_5NO_2$ $(IV)$.

(C) The correct answer is $(C)$.
Benzene is obtained by the cyclic polymerization of three molecules of acetylene $(C_2H_2)$ when passed through a red-hot iron or copper tube at $500\ ^oC$.
The reaction is as follows:
$3CH \equiv CH \xrightarrow{\text{Red hot } Cu/Fe, 500\ ^oC} C_6H_6$

(A) In a benzene ring,the relative positions of substituents are defined as follows:
$1, 2$ or $1, 6$ positions are known as $ortho$.
$1, 3$ or $1, 5$ positions are known as $meta$.
$1, 4$ position is known as $para$.
Therefore,the $1, 3$ position is called $meta$.

(B) The biological oxidation of benzene in the body of a dog leads to the formation of $Cinnamic \ acid$ $(C_6H_5CH=CHCOOH)$.

(C) The reaction shown is the aromatization of cyclohexane to benzene. This process involves dehydrogenation and cyclization. The catalyst commonly used for this industrial process is a mixture of chromium oxide $(Cr_{2}O_{3})$ supported on alumina $(Al_{2}O_{3})$. Therefore,the correct option is $(C)$.

(B) The reaction of benzene with methyl chloride $(CH_3Cl)$ in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$ is known as the Friedel-Crafts alkylation reaction.
The chemical equation is: $C_6H_6 + CH_3Cl \xrightarrow{\text{Anhydrous } AlCl_3} C_6H_5CH_3 + HCl$.
Therefore,anhydrous $AlCl_3$ is used as the catalyst.

(C) $B.H.C.$ stands for $Benzene$ $hexachloride$,which has the chemical formula $C_6H_6Cl_6$.
In the structure of $B.H.C.$,all carbon atoms are $sp^3$ hybridized.
Since there are no double or triple bonds present in the molecule,the number of $\pi$-bonds is $0$.

(B) The reaction of benzene $(C_6H_6)$ with chlorine $(Cl_2)$ in the presence of $UV$ light is an addition reaction.
$C_6H_6 + 3Cl_2 \xrightarrow{UV \ \text{light}} C_6H_6Cl_6$.
The product formed is $C_6H_6Cl_6$,which is commonly known as $BHC$ (Benzene Hexachloride) or Gammaxene.

(C) The reaction of benzene with chlorine in the presence of ultraviolet light or bright sunlight is an addition reaction.
In this reaction,three molecules of chlorine add to the benzene ring to form benzene hexachloride $(C_6H_6Cl_6)$,also known as $BHC$ or gammaxene.
The reaction is:
$C_6H_6 + 3Cl_2 \xrightarrow{\text{Sunlight}} C_6H_6Cl_6$
Therefore,the correct option is $(C)$.

(C) When chlorine is passed through benzene in the presence of sunlight,an addition reaction occurs to form benzene hexachloride $(C_6H_6Cl_6)$.
This product is commonly known as Gammexane,$BHC$,or Lindane.
It is widely used as an insecticide.

(D) In the Friedel-Crafts alkylation of benzene with $n$-propyl bromide in the presence of $AlCl_3$,the primary carbocation $(CH_3-CH_2-CH_2^+)$ formed initially undergoes a $1, 2$-hydride shift to form a more stable secondary carbocation $(CH_3-CH^+-CH_3)$.
This secondary carbocation then attacks the benzene ring to form isopropyl benzene (cumene) as the major product.

(A) When benzene $(C_6H_6)$ reacts with chlorine $(Cl_2)$ in the presence of sunlight (ultraviolet light),an addition reaction occurs.
This reaction results in the formation of benzene hexachloride $(C_6H_6Cl_6)$,which is commonly known as $B.H.C.$ or gammaxene.
The chemical equation is: $C_6H_6 + 3Cl_2 \xrightarrow{\text{sunlight}} C_6H_6Cl_6$ $(B.H.C.)$.

(D) $TNT$ stands for $2,4,6$-trinitrotoluene.
Its structure consists of a benzene ring with a methyl group $(-CH_3)$ at position $1$ and three nitro groups $(-NO_2)$ at positions $2, 4,$ and $6$.
This corresponds to the structure shown in image $353-$s157.

(C) The correct answer is $(C)$.
Acetophenone is prepared from benzene by the Friedel-Crafts acylation reaction.
In this reaction,benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ to form acetophenone $(C_6H_5COCH_3)$ and hydrogen chloride $(HCl)$.
This is an electrophilic aromatic substitution reaction.

(C) The catalytic oxidation of $o$-xylene in the presence of vanadium pentoxide $(V_2O_5)$ at high temperatures leads to the oxidation of both methyl groups attached to the benzene ring into carboxylic acid groups.
This reaction produces phthalic acid (benzene$-1,2-$dicarboxylic acid).
The reaction is represented as:
$o$-xylene $\xrightarrow{V_2O_5, \text{Oxidation}}$ Phthalic acid.

(A) The aerial oxidation of $Naphthalene$ in the presence of a catalyst like $V_2O_5$ at high temperature leads to the formation of phthalic anhydride,which upon hydrolysis yields phthalic acid.
$C_{10}H_8 + 4.5 O_2 \xrightarrow{V_2O_5, \Delta} C_8H_4O_3 (\text{Phthalic anhydride}) + 2CO_2 + 2H_2O$
$C_8H_4O_3 + H_2O \rightarrow C_6H_4(COOH)_2 (\text{Phthalic acid})$
Therefore,the correct option is $A$.

(D) Toluene can be oxidized to benzoic acid using strong oxidizing agents such as alkaline $KMnO_4$ or acidic $K_2Cr_2O_7$.
Both $KMnO_4$ and $K_2Cr_2O_7$ are effective oxidizing agents for the side-chain oxidation of alkylbenzenes to carboxylic acids.
Therefore,the correct option is $(d)$.

(B) The hydrogenation of $C_6H_5CHOHCOOH$ (mandelic acid) over a $Rh-Al_2O_3$ catalyst is a selective reaction that reduces the aromatic benzene ring to a cyclohexane ring while leaving the carboxylic acid $(-COOH)$ and hydroxyl $(-OH)$ groups intact.
The reaction is as follows:
$C_6H_5CHOHCOOH + 3H_2 \xrightarrow{Rh-Al_2O_3} C_6H_{11}CHOHCOOH$
Thus,the product formed is $C_6H_{11}CHOHCOOH$ (cyclohexylmandelic acid).

(B) In the nitration of benzene,the reaction between conc. $HNO_3$ and conc. $H_2SO_4$ occurs as follows:
$HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^-$
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$
Here,$H_2SO_4$ is a stronger acid than $HNO_3$,so it protonates $HNO_3$. Thus,$HNO_3$ acts as a base in the presence of conc. $H_2SO_4$ to generate the electrophile $NO_2^+$.

(C) The nitration of benzene involves the reaction of benzene with a nitrating mixture $(conc. \ HNO_3 + conc. \ H_2SO_4)$.
The rate-determining step of this electrophilic aromatic substitution reaction is the formation of the electrophile,the nitronium ion $(NO_2^+)$,from the nitrating mixture.

(B) . The $-NO_2$ group is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It decreases the electron density at the $o-$ (ortho) and $p-$ (para) positions of the benzene ring.
As a result,the incoming electrophile is directed to the $m-$ (meta) position,where the electron density is relatively higher compared to the $o-$ and $p-$ positions.
Therefore,it is a $m-$directing group.

(C) When benzene is treated with fuming nitric acid $(HNO_3)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$,it undergoes electrophilic aromatic substitution to form $1,3,5$-trinitrobenzene,also known as sym-trinitrobenzene.
The reaction is as follows:
$C_6H_6 + 3HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_3(NO_2)_3 + 3H_2O$
Therefore,the correct option is $(C)$.

(C) The nitration of benzene is an electrophilic aromatic substitution reaction.
Each $-NO_2$ group introduced into the benzene ring is strongly electron-withdrawing,which deactivates the ring towards further electrophilic substitution.
After the introduction of three $-NO_2$ groups,the ring becomes so deactivated that further nitration under standard conditions is extremely difficult.
Therefore,the maximum number of $-NO_2$ groups that can be introduced by direct nitration is $3$,resulting in $1,3,5$-trinitrobenzene.

(B) The reaction of nitrobenzene with concentrated $H_2SO_4$ at high temperature (sulfonation) yields $3$-nitrobenzenesulfonic acid as the major product.
This is because the $-NO_2$ group is a strong electron-withdrawing group and is meta-directing for electrophilic aromatic substitution reactions.
Therefore,the electrophile $SO_3$ attacks the meta-position relative to the $-NO_2$ group.

(A) The reactivity towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
$1$. The $-NH_2$ group in aniline $(I)$ is a strong electron-donating group due to its $+M$ effect,which significantly increases the electron density on the ring,making it highly reactive.
$2$. Benzene $(II)$ has no substituents,serving as the reference point.
$3$. The $-NO_2$ group in nitrobenzene $(III)$ is a strong electron-withdrawing group due to its $-M$ and $-I$ effects,which significantly decreases the electron density on the ring,making it the least reactive.
Therefore,the correct order of reactivity is $I > II > III$.

(C) The order of reactivity towards electrophilic aromatic substitution is governed by the presence of electron-donating or electron-withdrawing groups attached to the benzene ring.
$Toluene$ contains a methyl group $(-CH_3)$,which is an electron-donating group due to hyperconjugation and the inductive effect,making the ring more reactive towards electrophiles.
$Benzene$ is the standard reference.
$Nitrobenzene$ $(C_6H_5NO_2)$ has a strongly electron-withdrawing nitro group $(-NO_2)$,which deactivates the ring.
$Chlorobenzene$ has a chloro group $(-Cl)$,which is deactivating due to its strong $-I$ effect.
Thus,$Toluene$ is the most reactive towards nitration.

(B) Styrene is primarily produced from ethylbenzene.
Ethylbenzene is synthesized via the Friedel-Crafts alkylation of benzene with ethylene $(C_2H_4)$ in the presence of an acid catalyst.
Subsequently,styrene is obtained by the catalytic dehydrogenation of ethylbenzene:
$C_6H_5CH_2CH_3 \rightarrow C_6H_5CH=CH_2 + H_2$.
Thus,ethylene (ethene) is a key starting material for the synthesis of styrene.

(B) Styrene $(C_6H_5CH=CH_2)$ is a colorless,oily liquid at room temperature.

(D) Naphthalene $(C_{10}H_8)$ consists of two fused benzene rings.
In its Kekule structure,there are $5$ double bonds.
Each double bond contains one $\pi$ bond.
Therefore,the total number of $\pi$ bonds in naphthalene is $5$.

(C) The chemical formula of toluene is $C_6H_5CH_3$.
In the benzene ring,there are $6$ carbon atoms and $5$ hydrogen atoms attached to the ring.
The methyl group $(-CH_3)$ has $1$ carbon and $3$ hydrogen atoms.
Total $\sigma$ bonds:
- $6$ $C-C$ $\sigma$ bonds (including ring and methyl attachment).
- $5$ $C-H$ $\sigma$ bonds in the ring.
- $3$ $C-H$ $\sigma$ bonds in the methyl group.
Total $\sigma$ bonds = $6 + 5 + 3 = 15$.
Total $\pi$ bonds:
- There are $3$ double bonds in the benzene ring,each containing $1$ $\pi$ bond.
Total $\pi$ bonds = $3$.
Thus,toluene has $15\sigma$ and $3\pi$ bonds.

(A) In a benzene molecule,each carbon atom undergoes $sp^2$ hybridization.
Due to this $sp^2$ hybridization,the bond angle between the carbon atoms is $120^o$.

(A) In $Toluene$,the $C-H$ bond of the methyl group is the weakest.
This is because the resulting benzyl radical (after homolytic cleavage) or benzyl carbocation (after heterolytic cleavage) is resonance-stabilized by the benzene ring.

(D) $1$. Activating groups (like $-OH, -NH_2, -CH_3$) increase electron density on the ring and are $o-, p-$ directing.
$2$. Halogens are deactivating due to the $-I$ effect but are $o-, p-$ directing due to the $+M$ effect.
$3$. $m-$directing groups (like $-NO_2, -CHO, -COOH$) are strongly deactivating,which is more than the deactivating effect of halogens.
$4$. Regarding $t-$butylbenzene vs toluene: Toluene has a methyl group $(-CH_3)$ which is a stronger activator than the $t-$butyl group $(-C(CH_3)_3)$ due to steric hindrance and the nature of the inductive effect in the transition state. Therefore,toluene undergoes nitration more easily than $t-$butylbenzene. Thus,statement $D$ is incorrect.

(A) The reactivity towards electrophilic substitution depends on the electron density of the benzene ring.
In Aniline $(I)$,the $-NH_2$ group is a strong electron-donating group due to its $+M$ effect,which increases electron density at the ortho and para positions,making it highly reactive.
In Benzene $(II)$,there is no substituent to alter the electron density.
In Nitrobenzene $(III)$,the $-NO_2$ group is a strong electron-withdrawing group due to both $-I$ and $-M$ effects,which significantly decreases the electron density of the ring,making it the least reactive.
Therefore,the order of reactivity is $I > II > III$.

(C) When benzene reacts with fuming nitric acid $(HNO_3)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$,it undergoes electrophilic aromatic substitution to form $1,3,5$-trinitrobenzene,also known as symmetric trinitrobenzene. The reaction is: $C_6H_6 + 3HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_3(NO_2)_3 + 3H_2O$.

(A) The nitration of nitrobenzene using fuming $HNO_3$ and concentrated $H_2SO_4$ at high temperatures $(373 \ K)$ leads to the formation of $m$-dinitrobenzene.
$m$-Dinitrobenzene is a solid compound at room temperature.
Therefore,the correct option is $A$.

(C) The reactivity of aromatic compounds towards electrophilic substitution (like nitration) depends on the electron density of the ring.
Groups that donate electrons to the ring (activating groups) increase reactivity,while electron-withdrawing groups decrease it.
$1$. $Toluene$ $(CH_3-C_6H_5)$ has a methyl group,which is an electron-donating group ($+I$ and hyperconjugation effect),making the ring more reactive than benzene.
$2$. $Benzene$ $(C_6H_6)$ is the reference.
$3$. $Chlorobenzene$ $(C_6H_5Cl)$ has a chlorine atom,which is deactivating due to its strong $-I$ effect,despite being ortho/para directing.
$4$. $Nitrobenzene$ $(C_6H_5NO_2)$ has a nitro group,which is a strongly deactivating group due to its $-I$ and $-M$ effects.
Therefore,$Toluene$ is the most reactive towards nitration.

(A) In the nitration of benzene,the nitrating mixture consists of concentrated $HNO_3$ and concentrated $H_2SO_4$.
$H_2SO_4$ acts as a strong acid and protonates $HNO_3$.
The reaction is: $HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^- \rightleftharpoons NO_2^+ + H_2O + HSO_4^-$.
Here,$HNO_3$ accepts a proton from $H_2SO_4$,therefore $HNO_3$ acts as a base.

(C) $1$. Nitration of aromatic compounds is an electrophilic aromatic substitution reaction where the rate-determining step involves the formation of the sigma complex.
$2$. In the case of benzene and hexadeuterobenzene $(C_6D_6)$,the rate-determining step does not involve the breaking of the $C-H$ or $C-D$ bond.
$3$. Therefore,there is no primary kinetic isotope effect,and the rates of nitration for benzene and hexadeuterobenzene are essentially the same.
$4$. Toluene has a methyl group which is electron-donating ($+I$ and hyperconjugation),making the ring more reactive towards electrophilic substitution than benzene.
$5$. Thus,statement $C$ is incorrect because the rates are equal,not different.

(A) Benzene $(C_6H_6)$ is a highly symmetrical molecule where all six carbon atoms and all six hydrogen atoms are equivalent.
Due to this high degree of symmetry,replacing any one of the six hydrogen atoms with a substituent results in the same product.
Therefore,only $1$ monosubstituted product is possible for benzene.

(D) The reaction of benzene with chloroform $(CHCl_3)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
In this reaction,three molecules of benzene react with one molecule of chloroform to form triphenylmethane.
The chemical equation is: $3C_6H_6 + CHCl_3 \xrightarrow{AlCl_3} (C_6H_5)_3CH + 3HCl$.

(D) The reaction of benzene $(C_6H_6)$ with ethene $(CH_2=CH_2)$ in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$ is an example of Friedel-Crafts alkylation.
In this reaction,the ethene molecule acts as an alkylating agent,adding an ethyl group $(-C_2H_5)$ to the benzene ring.
The product formed is ethylbenzene ($C_6H_5-CH_2CH_3$ or $C_6H_5C_2H_5$).

(B) Styrene $(C_6H_5CH=CH_2)$ is a clear,colorless,oily liquid at room temperature.

(D) For compound $P$ ($4$-hydroxybenzoic acid): The $-OH$ group is strongly activating and ortho/para-directing,while the $-COOH$ group is deactivating and meta-directing. The $-OH$ group dominates,directing the nitro group to the ortho position relative to itself (position $3$ relative to $-COOH$).
For compound $Q$ ($1$-methoxy$-4-$methylbenzene): The $-OCH_3$ group is strongly activating and ortho/para-directing,while the $-CH_3$ group is weakly activating and ortho/para-directing. The $-OCH_3$ group dominates,directing the nitro group to its ortho position (position $2$ relative to $-OCH_3$).
For compound $S$ (phenyl benzoate): The benzoate group $(-O-CO-C_6H_5)$ is deactivating and meta-directing. However,the ring attached to the oxygen atom is activated by the oxygen lone pair (ortho/para-directing). Nitration occurs on the ring attached to the oxygen,specifically at the meta position relative to the ester linkage due to the electronic influence of the ester group on the phenoxy ring. The correct product is $3$-nitrophenyl benzoate.

(C) When benzene reacts with chlorine in the presence of ultraviolet $(UV)$ light,it undergoes an addition reaction rather than electrophilic substitution.
This reaction involves the addition of three molecules of $Cl_2$ to the benzene ring,resulting in the formation of benzene hexachloride $(C_6H_6Cl_6)$,also known as gammaxene or lindane.
The reaction is: $C_6H_6 + 3Cl_2 \xrightarrow{UV} C_6H_6Cl_6$.