Class 11 Chemistry MCQs | Chapter 13: Hydrocarbons – Part 3

Explanation: Kekulé’s original structure represented benzene as a six-membered ring with alternating single and double bonds. To account for the observed equivalence of the hydrogen atoms, he later proposed that the double and single bonds were rapidly interchanging or oscillating between two equivalent structures.

Explanation: The actual ${C}-{C}$ bond length in benzene is $1.39 { Å}$, which is exactly intermediate between a single bond ($1.54 { Å}$) and a double bond ($1.34 { Å}$). This evidence strongly supports the idea of resonance and delocalization, where the bonds have partial double bond character.

Explanation: In benzene, each of the six ${sp}^2$ hybridized carbon atoms has one unhybridized ${p}$ orbital perpendicular to the ring. These six ${p}$ orbitals overlap laterally around the entire ring, allowing the six $pi$ electrons to be completely delocalized across all six carbon atoms in the ring.

Explanation: The calculated heat of hydrogenation for a hypothetical cyclohexatriene (Kekulé structure) is much higher than the experimentally measured value for benzene. The difference in energy (approximately $150 { kJ/mol}$ or $36 { kcal/mol}$) is the Delocalization Energy (or Resonance Energy), which represents the extra stability gained by the molecule due to the $pi$-electron delocalization.

Explanation: Benzene has 6 ${C}-{C}$ single $sigma$ bonds forming the ring, and 6 ${C}-{H}$ single $sigma$ bonds (one for each ${C}$). This gives a total of $12$ $sigma$ bonds. There are six $pi$ electrons (one from each carbon) which form a total of 3 $pi$ bonds that are delocalized over the ring.

Explanation: For a compound to be planar and allow for full $pi$-electron delocalization, every atom in the ring must have a ${p}$ orbital available to participate in the cyclic cloud. This requires that every ring atom be ${sp}^2$ or ${sp}$ hybridized (or sometimes ${sp}^3$ hybridized if it contains a lone pair that can be promoted to a ${p}$ orbital), but it absolutely cannot contain only ${sp}^3$ hybridized carbon atoms, as this would break the continuous $pi$-system.

Explanation: The concept of resonance means the true structure of benzene is not oscillating between the Kekulé forms, but is a single, static state that is an average—or hybrid—of all equivalent contributing structures. This hybrid correctly accounts for the observation that all ${C}-{C}$ bond lengths are identical.

Explanation: Benzene has $6$ $pi$ electrons. If we substitute $n=1$ into Hückel’s rule: $4(1) + 2 = 6$. Therefore, the correct $n$ value is $n=1$.

Explanation: To form a planar ring with three single bonds (${C}-{C}$ and ${C}-{H}$) and one unhybridized ${p}$ orbital, each carbon atom in the benzene ring must be ${sp}^2$ hybridized.

Explanation: The primary source of benzene’s exceptional stability, known as aromaticity, is the delocalization of its $6$ $pi$ electrons across the entire ring. This delocalization results in a significant lowering of the overall energy of the molecule relative to the localized Kekulé structure, making it highly stable.

Explanation: Hückel’s rule states that a planar, cyclic, fully conjugated system is only aromatic if its number of $pi$ electrons is equal to $4n+2$, where $n$ is any non-negative integer ($0, 1, 2, 3, ldots$).

Explanation: Substituting $n=0$ into the $4n+2$ rule gives $4(0)+2 = 2$. Therefore, a compound with $2$ $pi$ electrons (like the cyclopropenyl cation) can be aromatic, provided it meets the other structural criteria (cyclic, planar, fully conjugated). Compounds with $4, 8,$ or $12$ $pi$ electrons are generally considered anti-aromatic or non-aromatic.

Explanation: The immense stability of aromatic compounds stems from the continuous, overlapping orbital cloud that allows the delocalization of the $pi$ electrons over the entire ring system, resulting in a large resonance energy that stabilizes the molecule.

Explanation: The presence of $4n$ $pi$ electrons in a cyclic, planar, fully conjugated system causes a high degree of electron repulsion in the higher energy orbitals, making the system highly unstable. These compounds are classified as anti-aromatic and are significantly less stable than their open-chain counterparts.

Explanation: While Cyclooctatetraene has $4n$ $pi$ electrons ($n=2$), which suggests anti-aromaticity, it avoids this highly unstable state by twisting out of planarity and adopting a non-planar, tub-like structure. Breaking planarity breaks the cyclic conjugation, thus classifying it as non-aromatic instead of anti-aromatic.

Explanation: Pyridine is a six-membered ring containing one nitrogen atom. The ring has three double bonds, contributing $6$ $pi$ electrons. The lone pair on the nitrogen atom is in an ${sp}^2$ orbital that points away from the ring and is not part of the $pi$-system. Thus, the $pi$-electron count is $6$.

Explanation: The cyclopentadienyl anion has two double bonds, contributing $4$ $pi$ electrons. The negative charge represents a lone pair of electrons (2 $pi$ electrons) in a ${p}$ orbital that participates in the cyclic conjugation. Total $pi$ electrons $= 4 + 2 = mathbf{6}$, satisfying the $4n+2$ rule for $n=1$.

Explanation: The Cyclopropenyl cation has one double bond, contributing $2$ $pi$ electrons. The positive charge means the third ${C}$ atom has an empty ${p}$ orbital, which completes the continuous cyclic conjugation. Since $4n+2 = 2$ for $n=0$, the system is aromatic with $2$ $pi$ electrons.

Explanation: A compound is non-aromatic if it is either non-cyclic, or if it is cyclic but fails the criteria for continuous cyclic conjugation (usually due to the presence of an ${sp}^3$ hybridized carbon atom that interrupts the $pi$-system), regardless of the $pi$-electron count.

Explanation: The $4n$ $pi$ electron systems are classified as anti-aromatic and are highly destabilized. Among the options, only $4$ $pi$ electrons ($4n$ for $n=1$) represents an anti-aromatic system. The other counts ($6, 10, 14$) represent highly stable aromatic systems ($4n+2$).

Explanation: Benzene’s stability due to aromaticity makes it resist addition reactions. It preferentially undergoes electrophilic substitution, where an electron-deficient species (the electrophile) attacks the $pi$-electron system, and a hydrogen atom is replaced by the electrophile.

Explanation: The nitrating mixture generates the powerful electrophile, the nitronium ion (${NO}_2^+$), through the reaction of concentrated nitric acid and concentrated sulfuric acid. This ion attacks the electron-rich benzene ring.

Explanation: Concentrated sulfuric acid acts as a stronger acid than nitric acid. It protonates ${HNO}_3$, which then loses water to produce the required electrophile, the nitronium ion (${NO}_2^+$).

Explanation: The electrophile attack breaks the aromaticity temporarily, forming a high-energy, but resonance-stabilized, carbocation intermediate known as the ${sigma}$ complex (or arenium ion).

Explanation: The nitro group (${-NO}_2$) is a powerful deactivating group but is also meta-directing. If the temperature is raised, the reaction rate increases, and a second nitration step occurs preferentially at the meta-position, leading to the formation of $m$-Dinitrobenzene.

Explanation: The final, fast step involves a base (often ${HSO}_4^-$) removing the proton (${H}^+$) from the ${sp}^3$ carbon of the ${sigma}$ complex. This allows the two electrons of the ${C}-{H}$ bond to reform the $pi$ system, thus re-establishing aromaticity.

Explanation: In this reaction, ${H}_2{SO}_4$ acts as a strong acid, protonating ${HNO}_3$. By accepting a proton, ${HNO}_3$ is acting as a base. It then loses water to form the nitronium ion.

Explanation: High yields of mononitration require two key factors: efficient generation of the electrophile (using concentrated acids) and a controlled, low temperature to suppress the second, slower nitration reaction.

Explanation: The methyl group (${-CH}_3$) is an activating group and is an $ortho$-/$para$-director. Therefore, the nitronium ion will be directed primarily to the $o$- and $p$-positions, forming a mixture of $o$-Nitrotoluene and $p$-Nitrotoluene as the major products.

Explanation: Electrophilic aromatic substitution reactions, including nitration, are generally exothermic reactions. Heat is evolved because the $sigma$-complex intermediate is stabilized by resonance, and the final re-establishment of aromaticity releases a large amount of energy, leading to a net release of heat.

Explanation: The actual electrophile in the sulfonation of benzene is the neutral, electron-deficient sulfur trioxide (${SO}_3$) molecule. This ${SO}_3$ is typically generated in two ways: either directly from the dissociation of fuming sulfuric acid (${H}_2{S}_2{O}_7$, or ${H}_2{SO}_4 cdot {SO}_3$), or through the reaction of concentrated sulfuric acid, where two molecules of ${H}_2{SO}_4$ react to yield ${H}_3{O}^+$, ${HSO}_4^-$, and ${SO}_3$.

Explanation: The sulfonation reaction is reversible, meaning the sulfonic acid group (${-SO}_3{H}$) can be easily removed. The reverse reaction, known as desulfonation, is favored by heating the product (benzenesulfonic acid) with hot water or dilute acid (steam). This process is often utilized in synthetic chemistry to temporarily block a position on an aromatic ring and then remove the group later.

Explanation: Like all electrophilic aromatic substitution reactions, the initial, slow step is the attack of the electrophile (${SO}_3$) on the electron-rich benzene ring. This breaks the aromaticity temporarily and forms a resonance-stabilized intermediate, a positive ion called the ${sigma}$ complex (or arenium ion). The positive charge is delocalized over the three $ortho$ and $para$ positions of the ring.

Explanation: To ensure a high yield of benzenesulfonic acid, the reaction is typically performed with fuming sulfuric acid (Oleum), which provides a high concentration of the ${SO}_3$ electrophile. The reaction is also carried out at an elevated temperature (usually around $100^circ {C}$ or more) to favor the thermodynamic product and increase the reaction rate.

Explanation: The electrophile in sulfonation is the neutral sulfur trioxide (${SO}_3$) molecule. When it attacks the benzene ring, the $pi$-electrons form a bond with the Sulfur, pushing a pair of electrons onto an Oxygen atom. This results in a positive charge delocalized on the ring carbons and a negative charge on the oxygen atom of the sulfonate group (${-SO}_3^-$), creating a dipolar ion or zwitterion intermediate.

Explanation: At the lower temperature ($80^circ {C}$), the reaction is under kinetic control, meaning the product that forms fastest (${o}$-toluenesulfonic acid, despite steric hindrance) is the major one. At the higher temperature ($160^circ {C}$), the reaction is reversible and under thermodynamic control, favoring the formation of the most stable product (${p}$-toluenesulfonic acid, which avoids steric hindrance).

Explanation: The sulfonic acid group (${-SO}_3{H}$) is a bulky group that can be placed on the ring, direct a second substitution reaction to a difficult position, and then be easily removed (desulfonation). This method is a crucial strategy for synthesizing specific $ortho$- or $para$-disubstituted compounds that would be otherwise inaccessible due to directing effects.

Explanation: The overall reaction is a substitution reaction because a hydrogen atom (and its electron) on the benzene ring is formally replaced by the sulfonic acid group (${-SO}_3{H}$). Specifically, it is an Electrophilic Aromatic Substitution (EAS) reaction.

Explanation: In the mechanism of Electrophilic Aromatic Substitution, the initial attack of the electrophile and the subsequent breaking of the stable aromatic $pi$-system to form the ${sigma}$ complex is the highest energy step. Therefore, the formation of the ${sigma}$ complex is the slow, rate-determining step for sulfonation and most other EAS reactions.

Explanation: The sulfonic acid group (${-SO}_3{H}$) is a versatile intermediate. Treatment with molten ${NaOH}$ at high temperatures converts the group into the phenoxide ion (${C}_6{H}_5{O}^-$), which upon acidification yields phenol (${C}_6{H}_5{OH}$). This demonstrates its value as a group that can be substituted by other functionalities.

Explanation: Benzene’s aromatic stability makes it unreactive toward a neutral halogen molecule (${Br}_2$ or ${Cl}_2$). The role of the Lewis acid catalyst (such as ${FeBr}_3$, ${AlCl}_3$, or ${FeCl}_3$) is to react with the halogen molecule to create a highly electron-deficient, polarized species—the strong electrophile (${X}^+$)—which is then powerful enough to attack the benzene ring.

Explanation: For the electrophilic aromatic substitution of chlorine, the standard reagents are chlorine gas (${Cl}_2$) and a Lewis acid catalyst, most commonly anhydrous aluminum chloride (${AlCl}_3$) or ferric chloride (${FeCl}_3$). The catalyst generates the powerful electrophile ${Cl}^+$ (or a highly polarized ${Cl}-{AlCl}_3$ species).

Explanation: The ${sigma}$ complex (arenium ion) is a carbocation intermediate where the positive charge is delocalized over three specific carbon atoms in the ring (the $ortho$ and $para$ positions relative to the point of attack). This charge delocalization across the $pi$-system makes the ${sigma}$ complex significantly stabilized by resonance, although it is still much less stable than the original aromatic ring.

Explanation: When UV light or high heat is used without a Lewis acid, the reaction proceeds via a free-radical mechanism. This mechanism causes the halogen to add across the double bonds, leading to the formation of hexachlorocyclohexane (${C}_6{H}_6{Cl}_6$)—a process that destroys the aromaticity—rather than the desired substitution product.

Explanation: Fluorine is the most reactive halogen. Its reaction with benzene is extremely fast and highly exothermic, making it difficult to control and often resulting in complex mixtures and hazardous conditions. Special methods, such as the Balz-Schiemann reaction, are used for aromatic fluorination instead of direct halogenation.

Explanation: In the fast final step, the base (the anion formed from the catalyst, e.g., ${FeBr}_4^-$) removes the hydrogen atom (${H}$) from the ${sp}^3$ hybridized carbon in the ${sigma}$ complex. The electrons from the ${C}-{H}$ bond then move back into the ring to reform the $pi$-system and re-establish aromaticity.

Explanation: While both reactions are mechanistically similar, elemental ${Bromine}$ (${Br}_2$) is a deep red-brown liquid that is easier to measure, weigh, and handle safely than the toxic ${Chlorine}$ (${Cl}_2$) gas. This makes bromination a more convenient reaction for many laboratory-scale syntheses.

Explanation: While the forward step is highly favorable, the reverse reaction (desubstitution or dehalogenation) would require breaking the newly formed, stable ${C}-{X}$ bond and reforming the original aromatic system, which has an extremely high activation energy under the reaction conditions. The stability of the aromatic ${C}-{X}$ bond makes the substitution effectively irreversible.

Explanation: Halogens (${Cl}$, ${Br}$, ${I}$) are unique: they are deactivating groups (due to electron-withdrawing inductive effects) but are also $ortho$/$para$-directors (due to resonance effects that stabilize the ${sigma}$ complex when substitution occurs at $o$/$p$ positions). Therefore, the second substitution will occur primarily at the $ortho$ and $para$ positions.

Explanation: The Lewis acid accepts a lone pair of electrons from the halogen, polarizing the ${Br}-{Br}$ bond and leading to the ionization. The overall result is the generation of the electron-deficient ${Br}^+$ electrophile and the formation of the complex anion ${FeBr}_4^-$ (${FeBr}_3 + {Br}_2 rightleftharpoons {Br}^+[{FeBr}_4^-]$).

Explanation: The most common and effective Lewis acid catalyst for both Friedel–Crafts alkylation and acylation is anhydrous aluminum chloride (${AlCl}_3$). The catalyst’s role is to coordinate with the halogen atom of the alkyl halide (${R}-{X}$), helping the ${R}$ group leave as a positive electrophile (${R}^+$) or a highly polarized complex. Other halides like ${FeCl}_3$ or ${BF}_3$ can also be used, but ${AlCl}_3$ is the standard.

Explanation: In Friedel–Crafts Alkylation, the reaction proceeds via the formation of a carbocation electrophile (${R}^+$), particularly when starting with secondary or tertiary alkyl halides. Carbocations are known to undergo rearrangements (hydride or alkyl shifts) to convert a less stable carbocation (e.g., primary) into a more stable one (e.g., secondary or tertiary) before reacting with the benzene ring. This leads to a mixture of unwanted alkylated products.

Explanation: The electrophile in acylation is the acylium ion (${RCO}^+$), which is highly stabilized by resonance (${R}-{C}equiv {O}^+ leftrightarrow {R}-{C}^+={O}$). Because this ion is resonance-stabilized, it does not rearrange, unlike the simple alkyl carbocation (${R}^+$) formed in alkylation. This ensures a clean synthesis of a single acylated product.

Explanation: The Clemmensen Reduction is a powerful method used specifically to reduce the carbonyl group (${C}={O}$) of a ketone (the acylation product) to a methylene group (${CH}_2$). This provides an indirect route to synthesize a pure, non-rearranged alkylbenzene (${C}_6{H}_5{CH}_2{R}$), overcoming the rearrangement limitation of direct Friedel–Crafts alkylation.

Explanation: Both Friedel–Crafts reactions are highly sensitive to the electron density of the aromatic ring. They fail if the ring is substituted with a strong deactivating group (like ${-NO}_2$, ${-CN}$, ${-SO}_3{H}$, or ${-COOH}$). These groups remove so much electron density from the ring that it is no longer nucleophilic enough to attack the electrophile, rendering the catalyst ineffective.

Explanation: The acyl group (${-COR}$) is a moderately deactivating group because the highly electronegative carbonyl oxygen pulls electron density out of the ring through induction and resonance. It is also a $meta$-director because it destabilizes positive charge intermediates at the $ortho$ and $para$ positions, directing further attack to the $meta$ position.

Explanation: The alkyl group (${-R}$) that is introduced is a mildly activating group (due to electron donation via hyperconjugation and induction). This means the monoalkylated product (e.g., toluene) is more reactive than the starting material (benzene), leading to the favored addition of a second and third alkyl group, resulting in an undesired mixture of polysubstituted products.

Explanation: The reaction initially forms the primary carbocation ${CH}_3{CH}_2{CH}_2^+$. This unstable primary carbocation immediately undergoes a 1,2-hydride shift (rearrangement) to form the more stable secondary carbocation, ${CH}_3{C}^+{HCH}_3$. This secondary carbocation then attacks the benzene ring to yield the rearranged product, Isopropylbenzene.

Explanation: The carbonyl oxygen of the final ketone product is a Lewis base and forms a highly stable complex with the Lewis acid catalyst (${AlCl}_3$). This complex ties up the catalyst and makes it inactive. Therefore, slightly more than one equivalent of ${AlCl}_3$ must be used to ensure enough remains active to catalyze the reaction itself, in addition to the amount complexed with the product.

Explanation: Friedel–Crafts Alkylation typically uses an alkyl halide (${RX}$) (e.g., ${CH}_3{Cl}$ or ${CH}_3{CH}_2{Br}$) as the source of the alkyl electrophile (${R}^+$). However, note that alkenes (${C}$) and alcohols (${A}$) can also be used as starting materials if a strong acid is employed to generate the carbocation. The most standard reagent is the alkyl halide.

Explanation: The directing influence of a substituent is fundamentally determined by its electronic effect (both inductive and resonance) on the stability of the carbocation intermediate, the ${sigma}$ complex. Groups that stabilize the positive charge intermediates when substitution occurs at the $ortho$ or $para$ positions are $o/p$-directors, while groups that destabilize those positions but leave the $meta$ position relatively unaffected are $m$-directors.

Explanation: The Carboxyl group (${-COOH}$) is a strong electron-withdrawing group primarily through resonance with the ring. This effect destabilizes the ${sigma}$ complex when the electrophile attacks the $ortho$ or $para$ positions, making those attacks highly unfavorable. As a result, the incoming electrophile is directed to the $meta$ position, which has comparatively higher electron density. All other options (${-OCH}_3$, ${-NH}_2$, ${-OH}$) are $o/p$-directing and activating.

Explanation: An activating group (e.g., ${-OH}$, ${-NH}_2$, ${-CH}_3$) donates electron density to the ring, either through resonance or induction. This increased electron density makes the entire ring a better nucleophile, thereby lowering the activation energy for the initial electrophilic attack and significantly increasing the reaction rate compared to benzene.

Explanation: Halogens exhibit conflicting electronic effects. Their strong inductive effect (electron-withdrawal through the $sigma$-bond) deactivates the ring by reducing overall electron density. However, their resonance effect (donation of a lone pair into the ring) stabilizes the ${sigma}$ complex specifically when the electrophile attacks the $ortho$ or $para$ positions. This results in the unusual combination of being deactivating yet $o/p$-directing.

Explanation: Groups with a lone pair directly attached to the ring, such as the Dialkylamino group (${-NR}_2$), are powerful electron-donating groups via resonance. They create a highly stabilized ${sigma}$ complex when the positive charge lands on the carbon bearing the substituent during $o/p$ attack. This makes them strongly activating and thus $o/p$-directing.

Explanation: The Nitro group (${-NO}_2$) is a powerful electron-withdrawing group via both induction and resonance, making it a strong deactivating group and a $meta$-director. When the ${NO}_2^+$ electrophile attacks the $ortho$ or $para$ positions, it places the positive charge immediately adjacent to the existing ${NO}_2$ group, causing extreme electronic repulsion and instability. Substitution at the $meta$ position avoids this instability, making the $meta$ product the favored, though low-yield, result.

Explanation: Resonance-donating groups (like ${-OH}$ or ${-NH}_2$) stabilize the ring by pushing electron density toward it. When drawing the resonance structures, the lone pair placement on the ring generates a negative charge specifically at the $ortho$ and $para$ positions. This localized increase in electron density is what makes these positions the preferred targets for the incoming positive electrophile.

Explanation: The Methyl group is classified as an activating and $o/p$-directing group. While it has a weak inductive effect, its primary source of electron donation and ${sigma}$ complex stabilization is hyperconjugation. This is the overlap between the $sigma$-bonds of the ${-CH}_3$ group and the adjacent $p$-orbital of the ring, which effectively helps to delocalize the positive charge created during the $o/p$ electrophilic attack.

Explanation: When two or more directing groups are present, the general rule is that the group with the strongest activating influence (the most electron-donating group) dictates the position of the incoming electrophile. Since ${-NH}_2$ is a strongly activating group and ${-Cl}$ is a deactivating group, the next substitution will occur at the $ortho$ or $para$ positions relative to the ${-NH}_2$ group.

Explanation: Almost all $meta$-directing groups (e.g., ${-NO}_2$, ${-COOH}$, ${-CN}$, ${-SO}_3{H}$) are characterized by having an atom with a double or triple bond to a more electronegative atom immediately adjacent to the ring carbon. This structural feature allows the group to withdraw electrons from the ring via resonance, resulting in deactivation and $meta$-direction.

Explanation: PAHs, such as naphthalene, anthracene, and benzopyrene, are typically formed when organic materials (like wood, coal, oil, or tobacco) are subjected to high temperatures in an environment with insufficient oxygen. This process, known as incomplete combustion or pyrolysis, results in the fusion of benzene rings. Common sources include vehicle exhaust, industrial emissions, and charcoal-grilled foods.

Explanation: Naphthalene is the smallest and simplest PAH. Its structure consists of two benzene rings sharing a common ${C}-{C}$ bond. While it is classified as a PAH, it is generally considered to be non-carcinogenic in its pure form, unlike many larger PAHs.

Explanation: PAHs themselves are often not the ultimate carcinogens. The toxicity arises when the body attempts to detoxify them using enzymes (like cytochrome P450). This process converts the PAH into highly reactive intermediates, particularly di-ol epoxides. These epoxides are powerful electrophiles that can covalently bind to and damage the nucleophilic sites on the cellular genetic material (${DNA}$), leading to mutations and cancer.

Explanation: Benzo[a]pyrene (often written as ${B}[{a}]{P}$) is a five-ring PAH and is notorious for being a Group 1 human carcinogen (according to the International Agency for Research on Cancer, IARC). It is a major component in the smoke from various sources and is historically known for causing scrotal cancer in chimney sweeps.

Explanation: Anthracene has three benzene rings fused in a linear manner. While it is a PAH, it is considered much less carcinogenic than its non-linear isomer, Phenanthrene, or the larger Benzo[a]pyrene. Its lower stability is reflected in its lower resonance energy (per ring) compared to benzene, and it undergoes addition reactions more readily than benzene, acting almost like two benzene rings connected by a central double bond.

Explanation: The highly reactive di-ol epoxide metabolites of PAHs act as electrophiles. They are strongly attracted to the electron-rich sites (nucleophiles) on ${DNA}$ bases (particularly Guanine). The covalent bonding of the PAH metabolite to the ${DNA}$ base is called adduction (or alkylation), which permanently distorts the ${DNA}$ helix and leads to errors during ${DNA}$ replication, initiating carcinogenesis.

Explanation: Naphthalene has $10$ carbon atoms in its structure, each contributing one ${p}$ orbital and one $pi$ electron. Therefore, the total number of $pi$ electrons is $10$. This value satisfies Hückel’s rule for aromaticity ($4n+2$, where $n=2$ gives $4(2)+2 = 10$), confirming its aromatic nature.

Explanation: PAHs are combustion by-products. Activities involving the burning of organic matter (A, C, D) are known sources of PAHs. Living in a forested rural area is generally associated with cleaner air and is the least likely scenario among the options to result in high exposure to PAHs, although wildfire smoke could be an exception.

Explanation: The term polynuclear (or polycyclic) in ${PAH}$ specifically refers to the fact that the compounds consist of two or more fused benzene rings, where adjacent rings share a common pair of carbon atoms.

Explanation: PAHs are nonpolar molecules, which makes them essentially insoluble in water but highly soluble in nonpolar solvents and lipids (fats). This property, known as lipophilicity or hydrophobicity, allows them to be effectively removed from aqueous media through processes like adsorption onto activated carbon or bioaccumulation in fatty tissues, making them environmental contaminants of concern.

Explanation: This is a challenging sequence demonstrating the versatility of the Benzenesulfonic acid intermediate.

1. Step 1 (${Benzene} rightarrow {A}$): Treatment of Benzene with hot concentrated ${H}_2{SO}_4$ results in Sulfonation, forming Benzenesulfonic acid (${C}_6{H}_5{SO}_3{H}$).

2. Step 2 (${A} rightarrow {X}$): Benzenesulfonic acid is fused with ${NaOH}$ at high temperature ($300^circ {C}$) and pressure. This is a crucial industrial method for synthesizing Phenol (the Dow process or its precursor). The sulfonic acid group is replaced by a hydroxyl group (${-OH}$) upon acidification (${H}_3{O}^+$). Therefore, the final product (X) is Phenol (${C}_6{H}_5{OH}$).

Explanation: This is an Electrophilic Aromatic Substitution (EAS) reaction.

1. The substituent on the ring, the Methyl group (${-CH}_3$), is a mildly activating and $ortho$/$para$-directing group.

2. The reagents (${Br}_2$ and ${FeBr}_3$) are the standard system for Aromatic Halogenation.

3. The Bromine electrophile (${Br}^+$) will attack the most activated positions, which are $ortho$ and $para$ to the methyl group. Even with excess reagent, the reaction is EAS, and the rate is limited by the activation of the Toluene ring, preventing complete tribromination unless a strong activator like ${-OH}$ or ${-NH}_2$ is present.

4. Therefore, the major product is a mixture of $o$-Bromotoluene and $p$-Bromotoluene. (If sunlight/heat were used without ${FeBr}_3$, side-chain bromination would occur).

Explanation: This reaction is a dehalogenation or catalytic hydrogenation of the aromatic halide.

1. Aromatic Halides (Aryl Halides) are generally unreactive towards typical dehalogenation reagents used for alkyl halides.

2. However, the halogen atom can be removed through Catalytic Hydrogenation (reduction) using ${H}_2$ gas in the presence of a powerful, non-selective catalyst like Nickel or Raney Nickel, usually at elevated temperature and pressure. This replaces the ${C}-{Cl}$ bond with a ${C}-{H}$ bond.

3. Option B is the Clemmensen reduction, used for reducing ${C}={O}$ to ${CH}_2$. Option C is sulfonation. Option D is incorrect as Chlorobenzene is highly inert to nucleophilic substitution under these mild conditions.

Explanation: Friedel–Crafts reactions fail when the aromatic ring is substituted with a strong deactivating group.

1. Benzoic Acid contains the Carboxyl group (${-COOH}$), which is a strong deactivating group (via resonance and induction). It pulls so much electron density from the ring that the ring is no longer nucleophilic enough to attack the acylium ion electrophile.

2. Anisole (activating), Toluene (activating), and Chlorobenzene (mildly deactivating, but still reactive) will all undergo the reaction successfully.

Explanation: This sequence tests both Friedel-Crafts rearrangement and radical halogenation regioselectivity.

1. Step 1 (${Benzene} rightarrow {Q}$): Friedel–Crafts Alkylation with 1-Chloropropane leads to carbocation rearrangement (primary ${C}^+$ to secondary ${C}^+$). The product (Q) is Isopropylbenzene (Cumene) (${C}_6{H}_5{CH}({CH}_3)_2$).

2. Step 2 (${Q} rightarrow {P}$): Halogenation using ${Cl}_2$ and UV light (or heat) promotes Free-Radical Halogenation, not Electrophilic Substitution (EAS). Free-radical halogenation occurs exclusively on the side chain.

3. The halogenation selectively occurs at the most substituted (${3}^circ$ or ${2}^circ$) benzylic position because the resulting Benzylic radical is highly resonance-stabilized. Isopropylbenzene has a tertiary ${C}-{H}$ bond at the benzylic position, making it the most reactive site.

4. Therefore, ${Cl}$ replaces the ${H}$ on the benzylic carbon, yielding $mathbf{alpha}$-Chloro-isopropylbenzene (${C}_6{H}_5{CCl}({CH}_3)_2$).

Explanation: This is a tricky conversion requiring the removal of the entire carbonyl group (${-CHO}$) and an associated carboxyl group.

1. First, oxidize Benzaldehyde to Benzoic Acid (${C}_6{H}_5{COOH}$) using ${KMnO}_4$ or ${Tollen}'{s}$ reagent (not shown, but implied as a standard intermediate step for this conversion type).

2. The product Benzoic Acid is then converted to its sodium salt (${C}_6{H}_5{COONa}$).

3. The sodium salt undergoes Decarboxylation when heated with Soda Lime (${NaOH}+{CaO}$). This removes the ${CO}_2$ portion, replacing the carboxyl group with a hydrogen atom, yielding Benzene.

4. Options A and B (Clemmensen and Wolff-Kishner) would only reduce the aldehyde to Toluene (${C}_6{H}_5{CH}_3$), not Benzene. Option C would reduce the aldehyde to Benzyl Alcohol.

Explanation: This question addresses the low reactivity of Aryl Halides towards nucleophilic substitution.

1. Aryl Halides (like Chlorobenzene) are inert towards nucleophilic substitution (${S}_{{N}}1$ or ${S}_{{N}}2$) under normal conditions because the ${C}-{X}$ bond is strong (due to ${sp}^2$ hybridization) and the lone pairs on the halogen participate in resonance with the ring.

2. To force the substitution of ${Cl}$ with ${OH}^-$ (to make Phenol), extremely harsh conditions are required, typically $300-350^circ {C}$ and $300 { atm}$ (the Dow process) for Chlorobenzene.

3. Benzyl Chloride (A) is highly reactive and undergoes ${S}_{{N}}1$ or ${S}_{{N}}2$ easily. Picryl Chloride (C) is highly activated by the three nitro groups and undergoes substitution readily under mild conditions.

Explanation: This sequence is the superior, two-step method for synthesizing a non-rearranged alkylbenzene.

1. Step 1 (Friedel–Crafts Acylation): ${CH}_3{CH}_2{COCl}$ forms the acylium ion (${CH}_3{CH}_2{CO}^+$), which is resonance-stabilized and does not rearrange. It attacks Benzene to form the ketone, Propanophenone (${C}_6{H}_5{COCH}_2{CH}_3$).

2. Step 2 (Clemmensen Reduction): The ketone is reduced by ${Zn}({Hg})/{HCl}$ (or the Wolff-Kishner method) to remove the carbonyl oxygen. The ${-COCH}_2{CH}_3$ group is reduced to ${-CH}_2{CH}_2{CH}_3$.

3. The final product (M) is straight-chain $n$-Propylbenzene. This avoids the rearrangement to Isopropylbenzene that occurs in direct alkylation.

Explanation: This is a standard Electrophilic Aromatic Substitution (EAS) reaction on a substituted benzene ring.

1. The substituent on the ring is the Carboxyl group (${-COOH}$).

2. The ${-COOH}$ group is a strong deactivating group (electron-withdrawing by resonance and induction).

3. All strong deactivating groups are $meta$-directors because attack at the $ortho$ and $para$ positions results in a highly unstable ${sigma}$ complex where the positive charge is placed directly adjacent to the ${-C}^+={OH}$ resonance structure, which is extremely unfavorable.

4. Therefore, the incoming ${NO}_2^+$ electrophile is directed to the $meta$ position, forming $m$-Nitrobenzoic acid.

Explanation: This question focuses on the complete reduction of the aromatic ring.

1. Using a potent, non-poisoned catalyst like Palladium on Carbon (${Pd}/{C}$) along with ${H}_2$ under pressure will lead to the complete saturation of the aromatic ring. The stable aromatic system is entirely reduced.

2. This process converts the benzene ring into a saturated cyclohexane ring.

3. The two methyl substituents (${CH}_3$) remain unchanged and retain their $1,2$ relative positions.

4. Therefore, $o$-Xylene is completely reduced to $1,2$-Dimethylcyclohexane.