Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Phenols
From the Reimer-Tiemann mechanism to the kinetic vs thermodynamic battles of the Fries Rearrangement. Master the hyper-activated aromatic ring.
Phenols ($C_6H_5OH$) are a unique beast. The intense $+M$ (resonance) effect of the hydroxyl oxygen floods the aromatic ring with electron density, making it furiously reactive toward electrophiles. In strongly basic media, it becomes the phenoxide ion, pushing this reactivity to extreme limits. To conquer JEE Advanced, you must perfectly understand acidity anomalies, name reactions, and sigmatropic rearrangements.
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Step 1: Expected Inductive Effect
The nitro group exerts a powerful $-I$ (inductive) and $-M$ (resonance) effect. Because the inductive effect is heavily distance-dependent, one would expect the ortho isomer (where the $-NO_2$ is closer to the $-OH$) to exert a stronger pull, stabilizing the phenoxide anion more and making it more acidic.
Step 2: The Structural Trap (Ortho Isomer)
In o-nitrophenol, the acidic $-OH$ group is perfectly adjacent to the highly electronegative oxygen atoms of the $-NO_2$ group. This allows the formation of a strong Intramolecular Hydrogen Bond (chelation). This internal bond physically traps the acidic proton, resisting its release into the solution.
Step 3: The Para Isomer
In p-nitrophenol, the $-OH$ and $-NO_2$ groups are on opposite ends of the ring. Intramolecular H-bonding is impossible. They can only form intermolecular H-bonds with water/solvent, which does not prevent dissociation. The unhindered release of the proton, combined with massive $-M$ stabilization of the resulting p-nitrophenoxide anion, makes it the stronger acid.
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Step 1: Role of Water (High Polarity)
Water is highly polar and capable of hydrogen bonding. It forces phenol to heavily ionize into the phenoxide ion ($-O^-$). The phenoxide ion is an aggressively powerful activating group (much stronger than neutral $-OH$), flooding the ring with electron density. Furthermore, water strongly polarizes the $Br-Br$ bond, creating a highly concentrated, ferocious $Br^+$ electrophile. The combination of a hyper-activated ring and a strong electrophile leads to instant, uncontrollable tri-substitution.
Step 2: Role of $CS_2$ (Low Polarity)
Carbon disulfide ($CS_2$) is a non-polar solvent. Phenol does not ionize in $CS_2$, remaining as the neutral $-OH$ molecule, which is only moderately activating. Additionally, the non-polar solvent fails to strongly polarize the $Br_2$ molecule, yielding a very weak electrophile. This combination restricts the reaction to a slow, controlled mono-substitution at the least sterically hindered para position.
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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 a dichloromethyl anion intermediate attached to the ring, which quickly hydrolyzes to an aldehyde group.
Step 3: Ortho-Selectivity (The Cation Bridge)
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 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, lowering its activation energy immensely compared to the para attack.
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Step 1: Sodium Phenoxide (Kinetic / Chelation Control)
The Sodium ion ($Na^+$) has a small ionic radius and a high charge density. As the weak electrophile ($CO_2$) approaches, the $Na^+$ acts as a rigid anchor, strongly coordinating to both the phenoxide oxygen and the oxygen of the incoming $CO_2$. This tight, cyclic 6-membered transition state physically forces the $CO_2$ to attack the adjacent ortho position.
Step 2: Potassium Phenoxide (Loss of Chelation)
The Potassium ion ($K^+$) has a much larger ionic radius. Its charge density is too diffuse to effectively hold the phenoxide oxygen and the $CO_2$ oxygen together in a tight transition state. Without this chelating assistance, the $CO_2$ molecule faces severe steric hindrance at the crowded ortho position.
Step 3: High Temperature (Thermodynamic Control)
Without chelation, and supplied with high thermal energy ($> 200^{\circ}\text{C}$), the reaction becomes highly reversible. The system favors the most thermodynamically stable product. The para position is completely unhindered sterically, making p-Hydroxybenzoic acid the deepest energy well.
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Step 1: The Rearrangement Mechanism
The Lewis acid ($AlCl_3$) coordinates to the ester oxygen, causing the acyl group ($-COCH_3$) to detach as a highly reactive acylium ion ($CH_3-C \equiv O^+$). This electrophile then attacks the now-activated phenol ring.
Step 2: Kinetic Control (Low Temp, $20^{\circ}\text{C}$)
At low temperatures, the system lacks the energy to reverse the reaction. The product that forms fastest dominates. Due to severe steric hindrance at the ortho position (between the incoming acyl group and the $AlCl_3$-coordinated oxygen), attack at the unhindered para position has a lower activation energy.
Thus, Product A is p-Hydroxyacetophenone.
Step 3: Thermodynamic Control (High Temp, $160^{\circ}\text{C}$)
At high temperatures, the reaction becomes reversible. The para-isomer reverts back to the intermediate state. The acyl group eventually migrates to the ortho position. Why? Because in the ortho position, the resulting carbonyl oxygen can heavily coordinate/chelate with the aluminum atom (and eventually form a strong intramolecular hydrogen bond with the free $-OH$ after workup). This chelation makes the ortho-isomer vastly more thermodynamically stable.
Thus, Product B is o-Hydroxyacetophenone.
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Step 1: The Transition State
Heating triggers a concerted, cyclic shift of six electrons. The $\pi$-bond of the allyl group attacks the ortho-carbon of the benzene ring, the $C-O$ ether bond breaks to form a carbonyl ($C=O$), and the aromatic $\pi$-bond shifts.
Step 2: Tracking the Atoms
Let the allyl group attached to oxygen be $-O-CH_2^{(\alpha)}-CH^{(\beta)}=CH_2^{(\gamma)}$ (where $\gamma$ is the $^{14}C$ label).
In the cyclic 6-membered transition state, the terminal $\gamma$-carbon is the atom that physically reaches out and forms the new bond with the ortho-carbon of the benzene ring.
Step 3: The Inversion
Because the terminal carbon attaches to the ring, the entire allyl group effectively flips end-over-end during the migration. The new sequence attached to the ring is $-CH_2^{(\gamma)}-CH^{(\beta)}=CH_2^{(\alpha)}$.
Following tautomerization to restore aromaticity, the final product has the $^{14}C$ label directly bonded to the aromatic ring.
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Step 1: First [3,3]-Sigmatropic Shift
The molecule undergoes the standard Claisen rearrangement. The allyl group migrates to the blocked ortho position, undergoing its first inversion. This forms an intermediate cyclohexadienone.
Because there is no hydrogen at this ortho position (it has a methyl and the new allyl group), it cannot tautomerize back to a phenol.
Step 2: Second [3,3]-Sigmatropic Shift
To restore aromaticity, the molecule undergoes a second consecutive [3,3]-sigmatropic shift. The allyl group migrates from the crowded ortho position to the open para position.
Step 3: Second Inversion and Tautomerization
During this second migration, the allyl group flips end-over-end again. Since it has now inverted twice, it returns to its original orientation. The para position possesses a hydrogen atom, allowing standard enolization/tautomerization to restore the aromatic phenol ring.
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Step 1: Tautomerization and Addition
2-Naphthol exists in a tiny equilibrium with its keto-tautomer (a tetralone derivative). The bisulfite ion ($HSO_3^-$) is an incredibly powerful nucleophile. It attacks the double bond of this keto-tautomer, adding across it (Michael-type addition).
Step 2: The Crucial Intermediate
This addition completely destroys the aromaticity of that specific ring, yielding a highly stable, saturated tetralone-bisulfite adduct. The carbon bearing the oxygen is now a standard, saturated aliphatic ketone.
Step 3: Nucleophilic Acyl Substitution
Because it is now a simple ketone, Ammonia ($NH_3$) can easily attack the carbonyl carbon. The standard formation of an imine occurs, replacing the oxygen with nitrogen.
Step 4: Elimination and Rearomatization
The molecule expels the bisulfite ion and tautomerizes, driven by the massive thermodynamic payoff of restoring the full aromatic naphthalene system. The result is 2-Naphthylamine. (Note: The entire process is perfectly reversible, allowing amines to be converted to phenols!).
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Step 1: Nitrosation
$NaNO_2$ and $H_2SO_4$ generate nitrous acid ($HNO_2$), which produces the nitrosonium electrophile ($NO^+$). Phenol undergoes electrophilic aromatic substitution at the para-position to form p-nitrosophenol.
Step 2: Tautomerization and Condensation
p-Nitrosophenol tautomerizes into the quinone monoxime. In the presence of concentrated acid, this condenses with a second, unreacted molecule of Phenol. This condensation yields an indophenol derivative (a protonated oxonium sulfate salt) which exhibits the initial deep green/blue/red colors depending on hydration/protonation.
Step 3: Alkalization (The Deep Blue Salt)
Upon adding excess $NaOH$, the acidic phenolic proton of the indophenol derivative is removed. The resulting highly conjugated anion is the Sodium salt of Indophenol. The extreme extended conjugation spanning two benzene rings and the central nitrogen atom produces an intense, deep blue color.
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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: Activation by Acid
The carbonyl oxygen is protonated by the strong acid ($H_2SO_4$). This pulls electron density out of the ring, creating a carbocation character that delocalizes to the para-carbon (C4).
Step 2: The Blockade
The C4 position is a quaternary carbon, completely blocked by two methyl groups. It cannot simply lose a proton to restore aromaticity.
Step 3: The 1,2-Alkyl Shift
To resolve this highly unstable, non-aromatic carbocation, one of the methyl groups at C4 takes its bonding electrons and migrates to the adjacent C3 position (a 1,2-methyl shift). The positive charge shifts to C4.
Step 4: Rearomatization
The C3 position now has a hydrogen atom. The molecule rapidly loses this proton ($H^+$), which cascades electrons down to form a new $\pi$-bond, satisfying the criteria for a benzene ring (aromaticity). The carbonyl oxygen becomes a phenol $-OH$.
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Step 1: Structure in Acidic Medium (Colorless)
In acidic or neutral conditions, phenolphthalein exists as a lactone (a cyclic ester). The central carbon connecting the three benzene rings is $sp^3$ hybridized. Because an $sp^3$ carbon has no empty or overlapping p-orbitals, it acts as an insulator, breaking the conjugation between the three rings. Lacking extended conjugation, the molecule cannot absorb visible light and is colorless.
Step 2: Alkalization (The Pink Form)
When strong base ($NaOH$) is added, two things happen:
1. The acidic phenolic protons are stripped away, forming phenoxide anions.
2. The hydroxide ion attacks and breaks open the cyclic lactone ring.
Step 3: Establishment of the Chromophore
The opening of the lactone ring causes the central insulating carbon to eliminate water and convert into an $sp^2$ hybridized carbon. Suddenly, the p-orbitals of all three benzene rings are linked together through this central carbon. This creates a massive, continuous, highly conjugated anionic system. This extended $\pi$-network absorbs green light, transmitting the intense pink/magenta color.
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Step 1: The Oxidation Process
Chromic acid is a harsh oxidant. It removes the phenolic proton and an electron to form a phenoxy radical, which undergoes further oxidation and addition of oxygen to the para position.
Step 2: Destruction of Aromaticity
To accommodate the two oxygen atoms at opposite ends of the ring, the molecule sacrifices its aromaticity. The resulting structure contains a 6-membered ring with two ketone groups ($C=O$) directly opposite each other, and two isolated double bonds within the ring.
Step 3: The Conjugated Product
The product is a cross-conjugated diketone. Although non-aromatic, it is still highly stable due to the extensive conjugation between the two carbonyl pi-bonds and the two carbon-carbon pi-bonds.
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Step 1: Formation of Phenoxide
In alkaline medium, phenol exists as the highly reactive phenoxide ion ($-O^-$).
Step 2: Electrophilic Attack by Persulfate
The persulfate ion (${S_2O_8^{2-}}$) acts as the electrophile. It is an extremely bulky, massive molecule. Because the phenoxide oxygen itself is relatively large and surrounded by a hydration shell, the ortho positions are sterically blocked from the incoming persulfate titan.
Step 3: Para Substitution
The persulfate ion exclusively attacks the completely unhindered para position, forming an intermediate sulfate ester.
Step 4: Hydrolysis
Subsequent boiling with dilute acid hydrolyzes the sulfate ester group, replacing it with a hydroxyl group ($-OH$).
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Step 1: The Thermodynamic Rule
Reaction: $ArOH + NaHCO_3 \rightleftharpoons ArO^- Na^+ + H_2CO_3 \rightarrow H_2O + CO_2 \uparrow$.
For this equilibrium to proceed forward, the acid ($ArOH$) must have a lower $pK_a$ (be stronger) than carbonic acid ($pK_{a1} \approx 6.36$).
Step 2: Analyze Phenol
Phenol is a very weak acid ($pK_a \approx 10$). It is weaker than carbonic acid, so it cannot displace it. Phenol will not dissolve in $NaHCO_3$.
Step 3: Analyze 2,4-Dinitrophenol
The two strongly withdrawing nitro groups drastically lower the $pK_a$ to $\approx 4.1$. Because $4.1 < 6.36$, it is stronger than carbonic acid. It will dissolve with effervescence.
Step 4: Analyze Picric Acid
With three nitro groups, 2,4,6-trinitrophenol (Picric acid) is an exceptionally strong organic acid ($pK_a \approx 0.3$). It is vastly stronger than carbonic acid and will dissolve violently with effervescence.
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Function 1: Enhancing Nucleophilicity
Neutral phenol is a relatively poor nucleophile for attacking acyl chlorides. The aqueous $NaOH$ instantly deprotonates the phenol, converting it entirely into the phenoxide ion ($-O^-$). The phenoxide ion is a phenomenally powerful, aggressive nucleophile that attacks the benzoyl chloride with high kinetic speed.
Function 2: Neutralizing the Byproduct (Le Chatelier's Principle)
The nucleophilic acyl substitution reaction generates Hydrogen Chloride ($HCl$) gas as a byproduct. If allowed to accumulate, the strong acid would reprotonate the phenoxide back to phenol, halting the reaction and pushing the equilibrium backward. The $NaOH$ instantly neutralizes the $HCl$ to form water and $NaCl$, perpetually pulling the equilibrium forward toward a $100\%$ yield of Phenyl benzoate.
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Step 1: The Acyl Chloride Byproduct
If acetyl chloride is used, the byproduct of the esterification is $HCl$ gas. $HCl$ is a highly corrosive, toxic, fuming strong acid. In an industrial setting, managing massive quantities of $HCl$ gas is dangerous and requires expensive corrosion-resistant equipment.
Step 2: The Anhydride Byproduct
When acetic anhydride is used, the byproduct is Acetic acid ($CH_3COOH$). This is a weak, non-corrosive, easily manageable liquid that can actually be recycled or sold as a valuable secondary product.
Step 3: Side Reactions
Aspirin contains both an ester group and a carboxylic acid group. The strong $HCl$ byproduct from the chloride reaction could potentially catalyze unwanted side reactions or hydrolysis of the newly formed ester if moisture is present. The milder acetic acid byproduct avoids this.
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Step 1: Generation of the Electrophile
The $HCl$ gas and $ZnCl_2$ Lewis acid activate the nitrile, forming an electrophilic iminium cation intermediate: $[CH_3-C^+=NH]$.
Step 2: Electrophilic Attack (The Requirement)
This iminium cation is a very bulky, weak electrophile. Ordinary Phenol (with only one $-OH$ activating group) simply does not possess enough electron density in its $\pi$-ring to successfully attack this weak electrophile. The reaction fails.
Step 3: Polyhydric Activation
Resorcinol and Phloroglucinol have two or three $-OH$ groups arranged in a meta relationship. Their activating $+M$ resonance effects are synergistic, channeling immense overlapping electron density to the shared ortho/para positions. This hyper-activated ring easily attacks the weak iminium cation.
Step 4: Hydrolysis
The initial product is a ketimine hydrochloride salt attached to the ring. Boiling with water violently hydrolyzes the $C=N$ bond, expelling ammonia and yielding a ketone.
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Step 1: The Thermodynamic Barrier
A standard isolated alkene double bond has no resonance energy to protect it. However, the three double bonds in the phenol ring are part of a continuous conjugated system that grants the molecule $\approx 150 \text{ kJ/mol}$ of aromatic resonance stabilization energy. To force hydrogen to add across these bonds, this massive stabilization energy penalty must be overcome.
Step 2: Forcing the Reaction
To overcome the barrier, extreme kinetic forcing conditions are utilized: high temperatures ($160^{\circ}\text{C}$) to provide activation energy, high pressure to drive the collision frequency, and a finely divided transition metal catalyst ($Ni$) to weaken the $H-H$ bonds.
Step 3: The Product
All three $\pi$-bonds are fully reduced by $3 \text{ moles}$ of $H_2$. The planar aromatic ring puckers into a saturated, $sp^3$ hybridized non-planar chair conformation. The hydroxyl group remains intact.
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Step 1: Formation of the Monomers
Formaldehyde acts as an electrophile ($C^{\delta+}=O^{\delta-}$). The activated phenol ring attacks it at the ortho and para positions, yielding o-hydroxybenzyl alcohol and p-hydroxybenzyl alcohol.
Step 2: Linear Polymerization (Novolac)
Under mild heating, these monomers undergo dehydration condensation. The benzyl alcohol group ($-CH_2OH$) of one molecule reacts with the ortho/para hydrogen of another molecule, expelling water and creating a methylene bridge ($-CH_2-$) between the rings. Initial linear chains form, called Novolac.
Step 3: 3D Cross-Linking
Under high heat and further formaldehyde, the polymer chains begin cross-linking extensively at all available ortho and para positions. Every phenol ring becomes locked to multiple other rings via rigid methylene bridges, forming an impenetrable, chaotic 3-dimensional network.
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Step 1: Activation of HCN
Hydrogen cyanide ($H-C \equiv N$) reacts with $HCl$ gas in the presence of the Lewis acid catalyst ($AlCl_3$). The $HCl$ protonates the nitrogen, and the chloride (or $AlCl_4^-$) coordinates, effectively forming a highly reactive complex: Formimidoyl chloride ($HN=CH-Cl$).
Step 2: Generation of the Electrophile
The $AlCl_3$ pulls the chloride ion away from formimidoyl chloride, generating a powerful resonating electrophilic cation: The Formiminium cation ($[H-C^+=NH \leftrightarrow H-C=N^+H]$).
Step 3: Attack and Hydrolysis
The highly nucleophilic phenol ring attacks this electrophile predominantly at the para position (due to less steric hindrance). The resulting aldimine hydrochloride salt is then easily hydrolyzed by boiling water to replace the $=NH$ group with an oxygen atom ($=O$), yielding the final aldehyde.
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Step 1: The Standard Vector Sum
If the groups did not interact, the dipole of p-nitrophenol would simply be the linear addition of the phenol dipole vector and the nitrobenzene dipole vector (since they point in roughly the same linear axis across the ring).
Step 2: The "Push-Pull" Resonance Effect
The hydroxyl group ($-OH$) is a powerful electron donor via resonance ($+M$). The nitro group ($-NO_2$) is a powerful electron acceptor via resonance ($-M$). Because they are situated exactly para to each other, they engage in a continuous, synergistic "push-pull" electron cascade across the entire $\pi$-system of the benzene ring.
Step 3: The Resulting Structure
The lone pair from oxygen is pushed all the way through the ring and directly into the oxygen atoms of the nitro group. This creates a heavily weighted resonance structure with a full positive charge on the phenolic oxygen and a full negative charge on the nitro oxygen. This massive charge separation across the entire length of the molecule exponentially increases the dipole moment ($\mu = q \times d$).
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Step 1: The Reagent
The definitive test for the phenolic functional group is the addition of a few drops of neutral aqueous Ferric Chloride ($FeCl_3$).
Step 2: The Observation
- Phenol will instantly produce a deep violet/purple coloration.
- Benzyl alcohol ($Ph-CH_2OH$), being an aliphatic alcohol, will show absolutely no color change (remains yellowish/orange from the iron salt).
Step 3: The Chemistry
The mildly acidic phenol ionizes enough to provide phenoxide ligands. Six of these ligands coordinate with the central $Fe^{3+}$ ion, replacing its water molecules, to form a highly conjugated, hexacoordinated complex anion. The intense color is due to Ligand-to-Metal Charge Transfer (LMCT) within the complex.
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Step 1: The Reagents and Conditions
Phenol vapor is passed over hot Zinc dust. The $C-O$ bond in phenol is exceptionally strong due to partial double-bond character from resonance. Breaking it requires extreme thermal energy and a powerful oxophilic (oxygen-loving) metal.
Step 2: The Redox Process
At high temperatures, Zinc metal ($Zn^0$) forces a single-electron transfer mechanism, breaking the tough aromatic $C-O$ bond. The Zinc actively rips the oxygen atom away from the ring to form the highly stable lattice of solid Zinc Oxide ($ZnO$). The remaining phenyl radical instantly abstracts a hydrogen atom to restore the stable aromatic core.
Step 3: The Product
Stripped of its oxygen, the phenol is completely reduced down to the parent aromatic hydrocarbon, Benzene ($C_6H_6$), which boils off as a volatile liquid.
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Step 1: Formation of A (Friedel-Crafts Alkylation)
The acid protonates propene ($CH_3-CH=CH_2$) to form the stable secondary isopropyl carbocation. Benzene attacks this electrophile to yield Isopropylbenzene (commonly known as Cumene). So, A is Cumene.
Step 2: Formation of B (Auto-oxidation)
Cumene possesses a highly reactive tertiary benzylic hydrogen. Oxygen ($O_2$) undergoes a free-radical insertion directly into this specific $C-H$ bond, yielding Cumene Hydroperoxide ($Ph-C(CH_3)_2-O-O-H$). So, B is Cumene Hydroperoxide.
Step 3: The Acidic Rearrangement
The dilute acid protonates the terminal peroxide oxygen, which then departs as water, leaving a highly unstable, electron-deficient oxygen cation ($Ph-C(CH_3)_2-O^+$).
To relieve this, the adjacent Phenyl group migrates (a 1,2-aryl shift) from the carbon to the oxygen, forming a more stable carbocation on the carbon.
Step 4: Hydrolysis (Products C and D)
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). The remaining 3-carbon aliphatic fragment snaps into a stable $C=O$ double bond, yielding Acetone (Compound D).
Mastering the Chemistry of Phenols
Congratulations on conquering these 25 ultra-challenging problems on Phenols! The key to mastering this chapter for JEE Advanced is understanding the delicate balance between activating resonance and directing effects. Whether it's the chelation-controlled ortho-preference in Reimer-Tiemann, the double-inversion chaos of the Claisen rearrangement, or the explosive acidity of Picric Acid, the underlying theme is always thermodynamic stability. Keep honing your mechanistic intuition, and visit Chemca.in for more elite organic chemistry masterclasses!
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