Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Benzene & Aromaticity
From Electrophilic Aromatic Substitution (EAS) and the kinetic isotope effects, to the nuances of the Benzyne mechanism and steric inhibition of resonance. Conquer the aromatic ring.
Benzene represents a unique pinnacle of stability in organic chemistry due to its delocalized $\pi$-electron cloud. While alkenes undergo rapid addition reactions, benzene vehemently resists addition to preserve its aromaticity, opting instead for substitution. To conquer JEE Advanced, you must master the directing effects of substituents, the anomalies of Friedel-Crafts, and the fierce mechanisms of ${S_NAr}$ and Benzyne.
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Step 1: Analyze Neutral COT
Cyclooctatetraene has $8$ $\pi$-electrons. If it were planar, it would satisfy the $4n$ rule ($n=2$), making it highly unstable and anti-aromatic. To escape this massive thermodynamic penalty, the molecule adopts a non-planar "tub" shape, breaking conjugation and behaving as four isolated double bonds (a non-aromatic polyene).
Step 2: Electron Transfer from Potassium
Potassium is a strong reducing agent. It transfers two electrons to the lowest unoccupied molecular orbitals (LUMOs) of COT, forming the cyclooctatetraenide dianion (${C_8H_8^{2-}}$).
Step 3: The Aromatic Transformation
The dianion now possesses $8 + 2 = 10$ $\pi$-electrons. This perfectly satisfies Huckel's rule for aromaticity ($4n+2$, where $n=2$). Because becoming aromatic releases a massive amount of resonance stabilization energy, the molecule spontaneously flattens out into a perfect planar octagon to allow full $p$-orbital overlap.
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Step 1: Generation of the Electrophile
1-chloropropane (${CH_3-CH_2-CH_2-Cl}$) reacts with the Lewis acid ${AlCl_3}$ to form a complex, eventually polarizing the $C-Cl$ bond to generate a primary ($1^{\circ}$) $n$-propyl carbocation (${CH_3-CH_2-C^+H_2}$).
Step 2: The 1,2-Hydride Shift
Primary carbocations are highly unstable. The adjacent carbon (C2) possesses a hydrogen atom. A 1,2-hydride shift occurs rapidly, where the hydrogen and its bonding electron pair migrate to the terminal carbon. This transforms the $1^{\circ}$ carbocation into a vastly more stable secondary ($2^{\circ}$) isopropyl carbocation (${CH_3-C^+H-CH_3}$).
Step 3: Electrophilic Attack
The benzene ring attacks the more stable secondary carbocation. Following the restoration of aromaticity (loss of a proton), the resulting molecule has an isopropyl group attached to the ring.
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Step 1: General EAS Mechanism
EAS involves two main steps:
1. Attack of the electrophile to form the Wheland intermediate (sigma complex).
2. Loss of the proton ($H^+$ or $D^+$) to restore aromaticity.
Step 2: Analysis of Nitration
In nitration, the formation of the sigma complex (Step 1) is slow and highly endothermic, making it the RDS. The subsequent loss of the proton (Step 2) is extremely fast. Because the $C-H$ bond is NOT breaking during the RDS, substituting $D$ for $H$ has no effect on the overall rate ($k_H / k_D \approx 1$).
Step 3: Analysis of Sulfonation
Sulfonation is unique because it is highly reversible. The intermediate sigma complex faces a relatively high energy barrier to lose the proton. The rates of Step 1 and Step 2 are comparable, and under specific conditions, the breaking of the $C-H$ bond becomes the rate-determining step. Because a $C-D$ bond is stronger and has a lower zero-point energy than a $C-H$ bond, breaking it requires more activation energy, significantly slowing down the reaction for $C_6D_6$.
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Step 1: The Methyl Group (EDG) in Toluene
A methyl group is electron-donating via hyperconjugation. When the initial electron is transferred to the ring, a radical anion forms. To minimize severe electrostatic repulsion, the high electron density (the carbanion site that will be protonated) avoids the carbon bearing the EDG. Consequently, protonation occurs at the ortho and meta positions relative to the EDG. The EDG remains attached to one of the unreduced, $sp^2$ hybridized carbons on the resulting isolated double bond.
Step 2: The Carboxyl Group (EWG) in Benzoic Acid
The $-COOH$ group is strongly electron-withdrawing. It highly stabilizes the incoming negative charge of the radical anion via resonance. Therefore, the highest electron density localizes directly on the ipso carbon (the one bearing the EWG) and the para carbon. Protonation occurs precisely at these highly stabilized sites.
Step 3: Structural Outcome
Because protonation occurs at the ipso and para positions, those carbons become $sp^3$ hybridized. The EWG ends up attached to a saturated $sp^3$ carbon, flanked by two double bonds.
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Step 1: Analyze N,N-dimethylaniline
Without substituents at the ortho positions, the massive $-N(CH_3)_2$ group can easily lie coplanar with the benzene ring. The nitrogen lone pair perfectly overlaps with the ring's $\pi$-system, extensively delocalizing and becoming unavailable for protonation (low basicity).
Step 2: The Steric Clash in the 2,6-Derivative
In 2,6-dimethyl-N,N-dimethylaniline, two bulky methyl groups occupy the ortho positions flanking the $-N(CH_3)_2$ group. The extreme physical crowding (steric hindrance) makes it physically impossible for the $-N(CH_3)_2$ group to remain in the plane of the ring.
Step 3: The Loss of Resonance
To relieve this massive steric strain, the $C-N$ bond rotates, twisting the nitrogen's $p$-orbital out of alignment with the ring's $\pi$-system. This completely destroys orbital overlap, halting resonance entirely. The lone pair is now localized strictly on the nitrogen atom, making it highly available for protonation.
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Step 1: The Elimination Step
The exceptionally strong amide base (${NH_2^-}$) abstracts an ortho-proton from C2. The resulting carbanion rapidly expels the chloride ion from C1. This forces the formation of a highly strained, highly reactive formal triple bond between C1 and C2 in the aromatic ring. This intermediate is Benzyne.
Step 2: Symmetry of the Intermediate
The newly formed "triple bond" exists between the $^{14}C$-labeled C1 and the unlabeled C2. Crucially, the benzyne intermediate is highly symmetrical with respect to nucleophilic attack at this strained bond.
Step 3: The Addition Step
Ammonia (the solvent/nucleophile) attacks the benzyne intermediate to relieve the extreme ring strain. Because there are no other substituents to direct the attack sterically or electronically, the nucleophile has an exactly equal $50\%$ mathematical probability of attacking C1 or attacking C2.
Step 4: Product Distribution
- Attack at C1: Aniline is formed with the $-NH_2$ group directly on the $^{14}C$ atom (Direct Substitution).
- Attack at C2: Aniline is formed with the $-NH_2$ group attached adjacent to the $^{14}C$ atom (Cine Substitution).
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Step 1: The SNAr Mechanism
${S_NAr}$ involves two distinct steps:
1. Addition: The nucleophile attacks the electron-deficient aromatic carbon, breaking aromaticity and pushing the $\pi$-electrons onto the strongly electron-withdrawing nitro groups to form a negatively charged Meisenheimer Complex (a stable carbanion).
2. Elimination: The leaving group is expelled, restoring aromaticity.
Step 2: Identify the Rate-Determining Step (RDS)
Breaking the massive thermodynamic stability of the aromatic ring costs an enormous amount of energy. Therefore, the first step (Addition) is the slow, Rate-Determining Step. The second step (leaving group departure) is incredibly fast as it restores aromaticity.
Step 3: Evaluate Substituent Effects
Because the halogen does not leave during the RDS, carbon-halogen bond strength ($C-I$ vs $C-F$) is virtually irrelevant to the overall rate! Instead, the rate depends entirely on how quickly the nucleophile attacks. Fluorine is the most electronegative atom, exerting an extreme inductive electron-withdrawing effect ($-I$). This massively increases the electrophilicity of the attached carbon and stabilizes the developing negative charge in the Meisenheimer transition state.
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Step 1: Kinetic Control (Low Temperature, $80^{\circ}\text{C}$)
Electrophilic attack at the $\alpha$-position (C1) yields an intermediate carbocation stabilized by resonance structures that leave the second benzene ring fully intact (maintaining full aromaticity). Attack at the $\beta$-position (C2) yields fewer such structures. Thus, the activation energy for $\alpha$-attack is significantly lower, and the $\alpha$-isomer forms much faster. At lower temperatures, the reaction lacks the thermal energy to reverse. The product that forms fastest dominates (Kinetic Control).
Step 2: Thermodynamic Instability of the Alpha Isomer
While the $\alpha$-isomer forms faster, the massive sulfonic acid group ($-SO_3H$) at the C1 position suffers severe steric repulsion from the hydrogen atom at the adjacent C8 position (peri-interaction). This makes the final $\alpha$-isomer thermodynamically unstable (high internal potential energy).
Step 3: Thermodynamic Control (High Temperature, $160^{\circ}\text{C}$)
At higher temperatures, sufficient thermal energy is available to break the $C-S$ bond. The reaction becomes rapidly reversible. The $\alpha$-isomer continuously desulfonates back to naphthalene. Eventually, the system funnels into the deeper energy well: the $\beta$-isomer. The $\beta$-position has no peri-hydrogen interference, making naphthalene-2-sulfonic acid vastly more thermodynamically stable.
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Step 1: Generation of the Electrophile
The strong base (${OH^-}$) removes the acidic proton from chloroform to yield the trichloromethyl carbanion (${CCl_3^-}$). This unstable anion rapidly expels a chloride ion (alpha-elimination) to generate Dichlorocarbene (${:CCl_2}$). Dichlorocarbene is an uncharged, electron-deficient electrophile with an empty p-orbital.
Step 2: Electrophilic Attack
In basic medium, phenol exists as the highly nucleophilic phenoxide ion. The electron-rich ring attacks the empty orbital of the dichlorocarbene, forming an intermediate that quickly hydrolyzes to an aldehyde group.
Step 3: Ortho-Selectivity
While both ortho and para positions are activated, the ortho product forms almost exclusively. During the transition state, the developing negative charge on the carbene oxygen heavily coordinates with the positively charged Sodium ion (${Na^+}$), which simultaneously coordinates with the phenoxide oxygen atom. This 6-membered cyclic chelation state highly stabilizes the ortho transition state.
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Step 1: The Instability of Formyl Chloride
Unlike heavier acyl chlorides (like acetyl chloride), formyl chloride (${H-COCl}$) is completely unstable at room temperature. It violently and spontaneously decomposes into Carbon Monoxide (${CO}$) and Hydrogen Chloride (${HCl}$) gas. Therefore, it cannot be bought or stored, making direct Friedel-Crafts formylation impossible.
Step 2: The Gattermann-Koch Solution
To bypass this, chemists inject a high-pressure mixture of Carbon Monoxide (${CO}$) and Hydrogen Chloride (${HCl}$) gases directly into a vessel containing benzene, anhydrous ${AlCl_3}$, and a trace of cuprous chloride (${CuCl}$) catalyst.
Step 3: In Situ Generation
Under these specific catalytic conditions, the ${CO}$ and ${HCl}$ react in situ to transiently form the formyl cation electrophile ($[H-C \equiv O^+]$), which instantly attacks the benzene ring before it can decompose, successfully yielding Benzaldehyde.
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Step 1: Instability of Alkyl Diazonium Ions
In an aliphatic diazonium ion (${CH_3-CH_2-N^+ \equiv N}$), the $C-N$ bond is a pure sigma bond. The extreme leaving group ability of $N_2$ combined with the relatively stable carbocation formed leads to instant, spontaneous cleavage even below freezing.
Step 2: Orbital Resonance in Aryl Diazonium Ions
In benzenediazonium chloride (${C_6H_5-N^+ \equiv N Cl^-}$), the diazonium group is attached to an $sp^2$ hybridized carbon. More importantly, the empty $\pi^*$ anti-bonding orbitals of the diazonium group overlap extensively with the filled $\pi$-orbitals of the benzene ring.
Step 3: Bond Strengthening
This delocalization of electrons from the ring into the nitrogen group grants significant double-bond character to the Carbon-Nitrogen bond. Breaking a partial double bond is energetically far more difficult than breaking a single bond, rendering the aryl diazonium salt kinetically stable at temperatures between $0^{\circ}\text{C}$ and $5^{\circ}\text{C}$.
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Step 1: Extreme Activation
In aqueous solution (Bromine water), the phenolic $-OH$ group heavily activates the ring via its $+M$ effect. The ortho and para positions possess immense electron density, making them highly susceptible to electrophilic attack.
Step 2: Electrophilic Attack on Salicylic Acid (Ipso Attack)
Salicylic acid already has a $-COOH$ group occupying one of the ortho positions. When the highly reactive $Br^+$ electrophile attacks this specific ortho position (an "ipso" attack), a highly crowded intermediate Wheland complex forms.
Step 3: Elimination (Halodecarboxylation)
Normally, EAS restores aromaticity by losing a proton ($H^+$). However, at the ipso position bearing the $-COOH$ group, the intermediate can restore aromaticity by expelling the massive carboxyl group entirely as Carbon Dioxide (${CO_2}$) gas. This process is known as ipso-substitution or halodecarboxylation.
Step 4: Completion
Once the $-COOH$ group departs, the remaining ortho and para positions are rapidly brominated as usual, leading directly to the fully substituted product.
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Step 1: The Nature of the Reagents
Aniline (${C_6H_5-NH_2}$) contains a nitrogen atom with a highly basic lone pair of electrons. The required catalyst for the reaction, anhydrous ${AlCl_3}$, is an incredibly powerful Lewis Acid (electron pair acceptor).
Step 2: The Lewis Acid-Base Trap
Instead of reacting with the alkyl halide to generate the necessary carbocation, the ${AlCl_3}$ instantly and violently coordinates with the basic lone pair on the aniline nitrogen. This forms an unbreakable Lewis acid-base salt complex: $C_6H_5-NH_2^+ - AlCl_3^-$.
Step 3: Deactivation of the Ring
This complexation causes two fatal problems:
1. The catalyst is completely destroyed and consumed, preventing carbocation formation.
2. The nitrogen atom now bears a full positive formal charge. It transforms instantly from a powerful activating group (EDG) into a massively deactivating electron-withdrawing group (EWG). It violently pulls electron density out of the ring, rendering it too poor to attack any electrophiles.
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Step 1: Coupling with Phenol (Requires pH 9-10)
The neutral phenol molecule is not nucleophilic enough to attack the weak diazonium electrophile. By making the solution mildly basic, phenol is deprotonated to form the phenoxide ion (${Ph-O^-}$). The full negative charge on the oxygen pushes a massive amount of electron density into the ring via resonance, activating it heavily enough for the reaction to occur. If the solution is acidic, it remains as neutral phenol and fails to react.
Step 2: Coupling with Aniline (Requires pH 4-5)
Aniline is highly activated as a neutral molecule (${Ph-NH_2}$) and can couple perfectly. However, if the solution is strongly acidic (pH 1-2), the basic aniline is protonated to form the anilinium ion (${Ph-NH_3^+}$). The positive charge deactivates the ring entirely, stopping the reaction. Why not use a basic medium? In a strongly basic medium, the diazonium ion itself reacts with ${OH^-}$ to form a diazohydroxide (${Ph-N=N-OH}$), destroying the electrophile! Therefore, a delicate, mildly acidic balance is required to keep aniline unprotonated while keeping the diazonium ion intact.
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Step 1: The Benzyne Carbon Vulnerability
The carbon atom directly attached to the benzene ring is called the benzylic carbon. Due to resonance stabilization from the adjacent aromatic $\pi$-system, bonds formed at this position (specifically $C-H$ bonds) are unusually reactive toward radical abstraction and oxidation.
Step 2: The Requirement for a Benzylic Hydrogen
The mechanism of permanganate oxidation explicitly requires the presence of at least one benzylic hydrogen atom to initiate the oxidation sequence. Toluene has three, ethylbenzene has two, and isopropylbenzene has one. Once the oxidation initiates at this benzylic hydrogen, the extreme oxidative conditions ruthlessly cleave all subsequent Carbon-Carbon bonds in the side chain, burning it all the way down to a stable carboxyl group ($-COOH$) directly attached to the ring.
Step 3: The Tert-Butyl Blockade
In tert-butylbenzene, the benzylic carbon is bonded to three methyl groups and zero hydrogens. It is a quaternary carbon. Because it lacks the mandatory benzylic hydrogen, the oxidant has no point of attack to initiate the reaction. The molecule survives completely intact.
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Step 1: Reaction A (Lewis Acid, Dark)
Anhydrous ${FeCl_3}$ is a Lewis acid. It polarizes the ${Cl_2}$ molecule, generating a powerful chloronium electrophile ($Cl^+$). This triggers standard Electrophilic Aromatic Substitution (EAS) on the benzene ring. Because the methyl group is an activating, ortho/para director via hyperconjugation, the products are a mixture of ortho-chlorotoluene and para-chlorotoluene.
Step 2: Reaction B (UV Light, Heat)
UV light homolytically cleaves the ${Cl_2}$ molecule into highly reactive chlorine free radicals (${Cl^{\bullet}}$). Radicals do not attack the stable aromatic $\pi$-system. Instead, they abstract a hydrogen atom from the alkyl side chain. The benzylic hydrogen is abstracted because it forms a resonance-stabilized benzylic radical. If excess chlorine is used, sequential free-radical substitution occurs until all benzylic hydrogens are replaced, yielding (Trichloromethyl)benzene (Benzotrichloride).
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Step 1: Standard Electronic Effects
Normally, the $+I$ and hyperconjugative effects of a methyl group pump electron density into the ring, destabilizing the resulting carboxylate anion and making the molecule a weaker acid (as seen in the para-isomer).
Step 2: The Physical Reality of the Ortho Position
When the methyl group is at the ortho position, its physical bulk creates massive steric hindrance with the adjacent $-COOH$ group. To relieve this spatial crowding, the $-COOH$ group is forced to rotate out of the plane of the benzene ring.
Step 3: Loss of Conjugation (The Ortho Effect)
Because the carboxylate group is now twisted out of plane, its $p$-orbitals can no longer overlap with the $\pi$-system of the benzene ring. This completely severs the resonance connection between the ring and the carboxylate group.
Why does this increase acidity? The benzene ring is slightly electron-withdrawing via induction ($sp^2$ carbon), but normally it also donates electrons via resonance (+R) into the carbonyl group, destabilizing the anion. By twisting out of plane, the destabilizing resonance is eliminated, leaving only the stabilizing inductive withdrawal. This makes ALL ortho-substituted benzoic acids strictly stronger than benzoic acid itself.
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Step 1: The Electron-Deficient Ring
The nitrogen atom in pyridine is highly electronegative. It pulls electron density away from the carbon atoms of the ring via inductive and resonance effects. This makes the entire ring heavily electron-deficient (deactivated), which repels electrophiles but heavily invites attack from strong nucleophiles like the amide ion (${NH_2^-}$).
Step 2: Attack at C2 vs C3 vs C4
The nucleophile attacks the ring, forming an anionic intermediate. We must evaluate the resonance structures of this intermediate to see where the negative charge lands.
- If attack occurs at C3: The negative charge is localized only on carbon atoms.
- If attack occurs at C2 (or C4): One of the major resonance structures places the negative formal charge directly on the highly electronegative Nitrogen atom.
Step 3: Thermodynamic Outcome
Placing a negative charge on a highly electronegative atom provides massive thermodynamic stabilization. Therefore, the activation energy for attack at C2 (and C4) is much lower than at C3. C2 is statistically and kinetically favored over C4 due to proximity to the nitrogen pole.
Step 4: Hydride Loss
To restore aromaticity, the intermediate expels a highly basic Hydride ion ($H^-$), which instantly abstracts a proton from the newly attached $-NH_2$ group (forming $H_2$ gas and trapping the molecule as a stable salt until aqueous workup).
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Step 1: The Weak Electrophile
Carbon dioxide is a very weak electrophile. It can only attack exceptionally activated rings, like the phenoxide ion.
Step 2: The Role of the Sodium Cation
In the reaction mixture, the Sodium cation (${Na^+}$) remains closely associated with the negatively charged oxygen of the phenoxide. When the ${CO_2}$ molecule approaches, the highly charged Sodium cation simultaneously coordinates with the oxygen atoms of the incoming ${CO_2}$ molecule.
Step 3: The Cyclic Transition State
This dual coordination creates a rigid, 6-membered cyclic transition state complex. The Sodium ion physically holds the ${CO_2}$ electrophile directly over the ortho position of the benzene ring, forcing the nucleophilic attack to occur precisely there. This proximity effect massively lowers the activation energy for ortho-attack, completely overriding normal steric hindrance preferences.
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Step 1: The Sandmeyer Mechanism
In the Sandmeyer reaction, the catalyst is Cuprous Chloride (${Cu(I)Cl}$). It is a true molecular catalyst. The $Cu(I)$ transfers a single electron to the diazonium group, causing $N_2$ to depart and generating a highly reactive phenyl radical. The resulting $Cu(II)$ species then rapidly transfers a chlorine radical back to the phenyl radical, yielding chlorobenzene and regenerating $Cu(I)$.
Step 2: The Gattermann Mechanism
In the Gattermann reaction, finely divided Copper powder ($Cu(0)$) is used. For the reaction to work, the $Cu(0)$ must first react with the acid to generate a trace amount of $Cu(I)$ in situ, which then executes the same radical mechanism.
Step 3: The Yield Discrepancy
Because the Sandmeyer reaction uses a pre-formed, highly soluble molecular $Cu(I)$ complex, the electron transfer is extremely efficient and fast. The Gattermann reaction relies on a heterogeneous solid-surface reaction (copper dust) to generate the active catalyst, which is kinetically much slower and leads to competing side reactions (like phenol formation from water) that degrade the yield.
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Step 1: Analysis of Alkylation
When the first ethyl group is attached to the benzene ring, it acts as an Electron Donating Group (EDG) via induction and hyperconjugation. This makes the resulting ethylbenzene more reactive toward electrophiles than the original unreacted benzene. Consequently, the newly formed product competes aggressively with the starting material for the remaining electrophiles, leading to uncontrolled, cascading polyalkylation.
Step 2: Analysis of Acylation
When an acetyl group ($-COCH_3$) is attached to the benzene ring, it is a powerful Electron Withdrawing Group (EWG) via resonance ($-M$ effect). It aggressively pulls electron density out of the ring, heavily deactivating it.
Step 3: Synthetic Control
Because the resulting acetophenone is drastically less reactive than the original benzene, it effectively shuts down and refuses to undergo a second Friedel-Crafts reaction. The reaction stops cleanly after exactly one substitution.
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Step 1: Activation of the Acid
Treatment with ${SOCl_2}$ converts the terminal carboxylic acid into an acyl chloride (4-phenylbutanoyl chloride). This is necessary because carboxylic acids are poor substrates for Friedel-Crafts.
Step 2: Generation of Electrophile
The addition of anhydrous ${AlCl_3}$ abstracts the chloride ion, generating an acylium cation ($R-C \equiv O^+$) at the end of the flexible 3-carbon aliphatic chain tethered to the benzene ring.
Step 3: Intramolecular Cyclization
The tethered aliphatic chain bends back, allowing the acylium cation to come into perfect proximity with the ortho position of the benzene ring. The electron-rich ring attacks the cation, closing the loop. Because the tether is 3 carbons long, plus the carbonyl carbon, plus the two carbons of the benzene ring involved, it forms a perfectly stable, strain-free 6-membered ring fused to the benzene core.
Step 4: Final Structure
Loss of a proton restores aromaticity. The resulting molecule is composed of a benzene ring fused to a cyclohexanone ring.
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Step 1: The Deactivating Nature (Inductive Effect)
Halogens are highly electronegative. Through the sigma ($\sigma$) bond, they exert a powerful inductive electron-withdrawing effect ($-I$). This aggressive pull removes electron density from the entire $\pi$-system, making the ring less nucleophilic overall and drastically slowing down the rate of the reaction (Deactivation).
Step 2: The Directing Nature (Resonance Effect)
Once the electrophile commits to attacking, we must look at the stability of the resulting intermediate carbocation (Wheland intermediate). If attack occurs at the meta position, the positive charge circulates around the ring but never lands on the carbon bearing the halogen.
However, if attack occurs at the ortho or para position, one of the major resonance structures places the positive formal charge directly on the carbon bearing the halogen.
Step 3: The Lone Pair Rescue
At this exact moment, the halogen utilizes its lone pair of electrons to form a pi-bond with the positively charged carbon, donating electrons via resonance ($+M$ effect). Every atom in the resulting structure attains a complete octet. This massive thermodynamic stabilization is only available during ortho/para attack.
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Step 1: The Ring Rupture
Under extreme thermal and catalytic stress, the robust benzene ring cracks open. Oxygen attacks the double bonds, leading to a catastrophic cascade of oxidation.
Step 2: Carbon Loss
Two of the six carbon atoms are completely oxidized into Carbon Dioxide (${CO_2}$) gas and blown away. The remaining four-carbon fragment stabilizes itself with terminal carboxylic acid groups, forming an unsaturated dicarboxylic acid: Maleic Acid (cis-butenedioic acid).
Step 3: Dehydration to Anhydride
At $500^{\circ}\text{C}$, a dicarboxylic acid holding a cis-geometry cannot survive in its hydrated state. The two adjacent $-COOH$ groups instantly eliminate a molecule of water (${H_2O}$) in a thermal dehydration reaction, folding inward to form a highly stable, 5-membered cyclic anhydride ring.
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Step 1: Formation of A (Friedel-Crafts Alkylation)
Phosphoric acid protonates propene (${CH_3-CH=CH_2}$) to form the stable secondary isopropyl carbocation. Benzene attacks this to yield Isopropylbenzene (Cumene), which is compound A.
Step 2: Formation of B (Auto-oxidation)
Cumene possesses a highly reactive tertiary benzylic hydrogen. Oxygen undergoes a free-radical insertion into this specific $C-H$ bond, yielding Cumene Hydroperoxide (${Ph-C(CH_3)_2-O-O-H}$), which is compound B.
Step 3: The Acidic Rearrangement
The dilute acid protonates the terminal peroxide oxygen, which then departs as water, leaving a highly unstable oxygen cation (${Ph-C(CH_3)_2-O^+}$).
To relieve this, the adjacent Phenyl group migrates (1,2-aryl shift) from the carbon to the electron-deficient oxygen, forming a more stable carbocation on the carbon.
Step 4: Hydrolysis and Cleavage
Water attacks the new carbocation. Following a proton transfer, the unstable hemiketal collapses. The $O-C$ bond breaks, expelling the phenyl group attached to the oxygen as Phenol (Compound C), while the remaining 3-carbon fragment snaps into Acetone (Compound D).
Mastering the Aromatic Ring
Congratulations on conquering these 25 ultra-challenging problems on Benzene and Aromatic Chemistry! You have successfully navigated the stringent quantum mechanics of Huckel's rule, the chaotic carbocation rearrangements of Friedel-Crafts, the anomalous reactivity of halogens, and the fierce isotopic labeling proofs of the Benzyne mechanism. Remember, in JEE Advanced, the benzene ring is never just a simple reactant—it is a deeply conjugated thermodynamic fortress whose reactivity relies entirely on precisely positioned activating and deactivating electronic effects. Keep honing your mechanistic intuition, and visit Chemca.in for more elite masterclasses!
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