Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Haloarenes
From the unyielding strength of Aryl-Halogen bonds to the chaotic symmetry of Benzyne intermediates. Master Nucleophilic Aromatic Substitution (${S_NAr}$) and Electrophilic Substitution.
Unlike haloalkanes, haloarenes (aryl halides) vigorously resist standard nucleophilic substitution. To conquer JEE Advanced, you must perfectly understand the dual nature of halogens (deactivating yet ortho/para directing), the kinetic requirements of the ${S_NAr}$ Meisenheimer complex, and the extreme conditions that trigger the highly strained Benzyne mechanism.
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Step 1: Failure of ${S_N2}$ (Resonance & Sterics)
In chlorobenzene, the Chlorine atom possesses three lone pairs. These lone pairs are directly adjacent to the benzene $\pi$-system and delocalize into the ring via resonance ($+M$ effect). This imparts partial double-bond character to the Carbon-Chlorine bond. A partial double bond is vastly stronger and shorter than a pure single bond, making it nearly impossible for a nucleophile to break. Furthermore, standard backside attack ($180^{\circ}$ inversion) is physically blocked by the bulky, planar electron cloud of the benzene ring.
Step 2: Failure of ${S_N1}$ (Electronegativity & Instability)
An ${S_N1}$ mechanism requires the spontaneous departure of the Chloride ion to form a carbocation. In chlorobenzene, the carbon is $sp^2$ hybridized ($33\%$ s-character), meaning it is highly electronegative and holds the $C-Cl$ bond electrons very tightly. If the bond were to break, it would form a phenyl cation. Placing a positive charge on an highly electronegative $sp^2$ orbital (which cannot be stabilized by resonance since the empty orbital is orthogonal to the $\pi$-system) is thermodynamically disastrous.
View Solution
Step 1: The ${S_NAr}$ Mechanism
The nucleophile (${OH^-}$) attacks the carbon bearing the chlorine, breaking aromaticity and pushing the $\pi$-electrons onto adjacent carbons. This forms a resonance-stabilized carbanion (Meisenheimer complex).
Step 2: Analyzing the Para-Isomer (1-chloro-4-nitrobenzene)
When attack occurs at C1, the negative charge delocalizes to the ortho (C2, C6) and para (C4) positions. Because the powerful Electron Withdrawing Group ($-NO_2$) is located exactly at the para position (C4), it can accept the negative charge directly into its own oxygen atoms via extended resonance. This massive thermodynamic stabilization drastically lowers the activation energy.
Step 3: Analyzing the Meta-Isomer (1-chloro-3-nitrobenzene)
When attack occurs at C1, the negative charge again delocalizes strictly to the ortho and para positions (C2, C4, C6). The $-NO_2$ group is sitting at the meta position (C3). The negative formal charge never lands on C3! Therefore, the $-NO_2$ group can only stabilize the complex weakly via its inductive effect ($-I$), but provides absolutely zero resonance stabilization ($-M$).
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Step 1: Identify the Rate-Determining Step
Breaking the massive thermodynamic stability of the aromatic ring costs an enormous amount of energy. Therefore, the first step (nucleophilic Addition to form the Meisenheimer complex) is the slow, Rate-Determining Step. The second step (leaving group Elimination) is incredibly fast as it instantly restores aromaticity.
Step 2: Evaluate the Role of the Halogen
Because the halogen does not leave during the RDS, the $C-X$ bond strength ($C-I$ vs $C-F$) is virtually irrelevant to the overall reaction rate! Instead, the rate depends entirely on how quickly the nucleophile attacks.
Step 3: The Fluorine Advantage
Fluorine is the most electronegative atom. It exerts an extreme inductive electron-withdrawing effect ($-I$). This massively increases the electrophilicity ($\delta+$) of the attached carbon, drawing the nucleophile in faster. Furthermore, it highly stabilizes the developing negative charge in the Meisenheimer transition state.
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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 structurally symmetrical with respect to nucleophilic attack at this strained $\pi$-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 $50\%$ 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).
View Solution
Step 1: Formation of Benzyne
The base can abstract a proton from either C2 (between Br and OCH3) or C4. The proton at C2 is the most acidic because it is flanked by two highly electronegative, -I withdrawing groups. Abstraction at C2 followed by loss of $Br^-$ forms a benzyne triple bond specifically between C2 and C3.
Step 2: Nucleophilic Attack (Addition)
The nucleophile (${NH_2^-}$) can attack the benzyne intermediate at either C2 or C3.
- Path A (Attack at C3): The nucleophile attaches to C3. The electrons from the triple bond are pushed onto C2, forming a carbanion at C2. This carbanion is directly adjacent to the strongly electronegative oxygen of the methoxy group ($-I$ effect), making it highly stabilized.
- Path B (Attack at C2): The nucleophile attaches to C2. The electrons are pushed onto C3, forming a carbanion at C3. While closer to the oxygen, a carbanion directly on the carbon bearing the methoxy group suffers from intense electron-pair repulsion from the oxygen lone pairs, destabilizing it.
Step 3: Conclusion
Because the carbanion at C2 is vastly more stable than at C3, Path A dominates heavily.
View Solution
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 initial electrophilic attack compared to benzene (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 resonance structure attains a complete octet. This massive thermodynamic stabilization is only available during ortho/para attack.
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Step 1: Formation of the Organosodium Species
Sodium metal reacts primarily via single-electron transfer to form carbanions (or highly polar organosodium complexes). Alkyl halides (aliphatic) are vastly more reactive towards Sodium than aryl halides (due to the partial double-bond character of the aryl $C-Cl$ bond preventing easy cleavage).
Step 2: The Kinetic Difference
Because the alkyl halide is significantly more reactive, Sodium preferentially attacks the chloromethane first, rapidly generating Methylsodium ($CH_3^-Na^+$) as the dominant reactive intermediate in the solution.
Step 3: The Coupling Step
The highly nucleophilic Methyl carbanion now acts as an attacker. While it could attack another chloromethane ($S_N2$ to form ethane), its high concentration allows it to successfully attack the unreacted, less-reactive chlorobenzene via an Addition-Elimination (or benzyne) pathway to yield Toluene.
Step 4: Suppression of Biphenyl
Because chlorobenzene reacts very sluggishly with Sodium, very little Phenylsodium ($C_6H_5^-Na^+$) is ever generated in the solution. Without Phenylsodium, the self-coupling reaction to form Biphenyl ($C_6H_5-C_6H_5$) is drastically suppressed.
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Step 1: Reagents of Ullmann Coupling
The Ullmann reaction specifically uses finely divided Copper powder ($Cu$) heated in a sealed tube to couple aryl halides.
Step 2: Halide Specificity
The reaction works efficiently ONLY with Iodobenzene (Aryl iodides).
Reaction: $2C_6H_5-I + 2Cu \xrightarrow{\Delta} C_6H_5-C_6H_5 + Cu_2I_2$.
Step 3: Why Chlorides Fail
Copper is a much milder reducing agent than Sodium. The $C-Cl$ bond in chlorobenzene has significant partial double-bond character due to resonance ($+M$ effect) and is incredibly strong. Copper simply lacks the thermodynamic reducing power to break the aryl $C-Cl$ (or $C-Br$) bond. The $C-I$ bond is exceptionally long and weak, allowing Copper to easily insert and facilitate the radical/organocuprate coupling mechanism.
View Solution
Step 1: The Activation Energy Barrier
Chlorobenzene has no Electron Withdrawing Groups (like $-NO_2$) to stabilize the incoming negative charge of the nucleophile. Therefore, forming the Meisenheimer complex (Addition step) has a phenomenally high activation energy. To provide sufficient thermal kinetic energy to overcome this barrier, extreme temperatures ($623\text{ K}$) are required. To keep water liquid at this temperature, a crushing pressure of $300\text{ atm}$ is strictly necessary.
Step 2: The Alkaline Trap
Under these extreme conditions, the ${OH^-}$ eventually substitutes the $Cl^-$, forming phenol ($C_6H_5OH$). However, the reaction medium is highly concentrated aqueous $NaOH$ (a strong base).
Step 3: Formation of the Intermediate
Phenol is a weak acid. As soon as it is formed, it immediately reacts with the excess $NaOH$ in an acid-base neutralization to form a highly stable, water-soluble salt. This pulls the equilibrium entirely to the right.
Step 4: The Final Intermediate
The species present in the reactor before acidification is Sodium Phenoxide ($C_6H_5O^- Na^+$). Acidification with $HCl$ is then required to reprotonate it and isolate pure phenol.
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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 solid $Cu(0)$ 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 allows competing side reactions (like phenol formation from water) to degrade the yield.
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Step 1: Diazotization
Aniline ($C_6H_5NH_2$) is treated with a cold ($0-5^{\circ}\text{C}$) mixture of Sodium Nitrite ($NaNO_2$) and Fluoroboric acid ($HBF_4$). Alternatively, standard diazotization with $HCl$ is performed first, followed by the addition of $HBF_4$.
Reaction: $C_6H_5NH_2 + HNO_2 + HBF_4 \rightarrow C_6H_5N_2^+ BF_4^- + 2H_2O$
Step 2: The Stable Intermediate
The product is Benzenediazonium fluoroborate ($C_6H_5N_2^+ BF_4^-$). Unlike standard diazonium chlorides which are dangerously explosive if dried, this specific fluoroborate salt is remarkably stable and insoluble. It precipitates out and can be safely filtered, washed, and dried.
Step 3: Thermal Decomposition
The dry salt is gently heated. It smoothly decomposes, expelling Nitrogen gas and Boron trifluoride gas, leaving behind the pure fluorobenzene.
Reaction: $C_6H_5N_2^+ BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 \uparrow + BF_3 \uparrow$
2) $Ph-N_2^+ BF_4^- \xrightarrow{\Delta} Ph-F + N_2 + BF_3$.
The stable intermediate is Benzenediazonium fluoroborate.
View Solution
Step 1: The Inert Reactants
The ${Cl_2}$ molecule is completely non-polar. The electron cloud of the benzene ring cannot induce a strong enough dipole in ${Cl_2}$ to force it to react. A true electrophile must be generated.
Step 2: Lewis Acid Coordination
Anhydrous ${FeCl_3}$ is a powerful Lewis acid; the Iron atom has an empty d-orbital and is electron-deficient. One of the Chlorine atoms in the ${Cl_2}$ molecule uses a lone pair of electrons to coordinate directly into the empty orbital of Iron: $Cl-Cl \cdots FeCl_3$.
Step 3: Heterolytic Cleavage
The Iron atom pulls electron density violently towards itself, forming the stable complex anion $[FeCl_4]^-$. This aggressive pull heterolytically cleaves the $Cl-Cl$ bond, stripping both bonding electrons away from the terminal Chlorine atom.
Step 4: The Active Electrophile
This leaves behind a "naked" Chloronium ion ($Cl^+$). This incredibly electron-deficient cation is the true, ferocious electrophile that attacks the $\pi$-cloud of the benzene ring to initiate the substitution.
View Solution
Step 1: Reaction A (UV Light, Heat)
UV light homolytically cleaves the ${Cl_2}$ molecule into highly reactive chlorine free radicals (${Cl^{\bullet}}$). Radicals do not attack the highly 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. Sequential free-radical substitution occurs until all benzylic hydrogens are replaced, yielding Benzotrichloride (${C_6H_5CCl_3}$).
Step 2: Reaction B (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 and +I effects, the products are a mixture of ortho-chlorotoluene and para-chlorotoluene.
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Step 1: Generation of the Electrophile
Concentrated $H_2SO_4$ protonates the carbonyl oxygen of chloral ($CCl_3-CH=O \rightarrow CCl_3-CH=O^+H$). This draws electron density away from the carbonyl carbon, creating a highly electrophilic carbocation intermediate ($CCl_3-C^+H-OH$).
Step 2: First Electrophilic Attack
This massive carbocation attacks a molecule of chlorobenzene. Because Chlorine is an ortho/para director, and the para position is far less sterically hindered, the attack occurs almost exclusively at the para position. Following deprotonation, an alcohol intermediate forms.
Step 3: Second Electrophilic Attack
The strong acid protonates the newly formed $-OH$ group, which leaves as water, generating a second carbocation. This carbocation attacks a second molecule of chlorobenzene, again at the para position, linking two chlorobenzene rings to the central carbon.
Step 4: Final Structure
The final molecule has two p-chlorophenyl rings and one trichloromethyl group attached to a central $CH$ group.
View Solution
Step 1: The Hybridization Effect (Charge Separation)
In cyclohexyl chloride, the carbon is $sp^3$ hybridized ($25\%$ s-character). In chlorobenzene, the carbon is $sp^2$ hybridized ($33\%$ s-character). Higher s-character means the carbon nucleus is closer and more effectively pulls electron density. Therefore, an $sp^2$ carbon is significantly more electronegative than an $sp^3$ carbon. Because the electronegativity difference between $Cl$ and the $sp^2$ carbon is smaller, the bond is less polar (smaller charge separation, $q$).
Step 2: The Resonance Effect (Bond Length)
In chlorobenzene, the lone pairs of chlorine delocalize into the benzene ring via resonance ($+M$ effect). This introduces partial double-bond character to the $C-Cl$ bond. Double bonds are physically shorter than single bonds. Therefore, the bond length ($d$) in chlorobenzene is shorter than in cyclohexyl chloride.
Step 3: Conclusion
Since both $q$ (charge separation) and $d$ (bond length) are smaller in chlorobenzene due to hybridization and resonance, the overall dipole moment ($\mu = q \times d$) is significantly reduced.
View Solution
Step 1: Analyze Boiling Points
Boiling points depend on dipole-dipole interactions. While the ortho isomer has a higher dipole moment than the meta, and the para isomer has a dipole moment of zero, their overall London dispersion forces (due to identical mass and similar size) are dominant. Hence, their boiling points are nearly identical.
Step 2: Analyze Melting Points (Crystal Packing)
Melting a solid requires breaking the rigid crystal lattice. The strength of a crystal lattice is dictated by how closely and neatly the molecules can stack together.
Step 3: The Symmetry of the Para Isomer
The para-isomer (1,4-dichlorobenzene) is highly symmetrical. Its linear, balanced shape allows it to fit perfectly and tightly into a highly ordered crystal lattice. This tight packing maximizes intermolecular van der Waals forces in the solid state, requiring significantly more thermal energy to break apart, leading to a much higher melting point.
The ortho and meta isomers are asymmetrical and bulky on one side, preventing them from packing efficiently, resulting in weaker lattices and lower melting points.
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Step 1: The Chlorobenzene Failure
As discussed, removing $Cl^-$ from chlorobenzene forms a highly unstable phenyl cation on an $sp^2$ hybridized carbon. It cannot be stabilized by resonance.
Step 2: The Benzyl Chloride Ionization
In benzyl chloride, the chlorine is attached to an $sp^3$ hybridized carbon. When the $Cl^-$ leaves, it forms a benzylic carbocation ($C_6H_5C^+H_2$).
Step 3: Resonance Stabilization
The empty p-orbital of this benzylic carbocation is perfectly aligned with the $\pi$-cloud of the adjacent benzene ring. The $\pi$-electrons from the ring delocalize into the empty orbital, spreading the positive charge over the ortho and para positions of the ring. This massive resonance stabilization lowers the activation energy tremendously, allowing the rapid $S_N1$ mechanism to proceed.
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Step 1: The Bond Strength Barrier
The $C-Cl$ bond in chlorobenzene is significantly shorter and stronger than the $C-Br$ bond in bromobenzene due to greater orbital overlap and stronger partial double-bond character. It is incredibly difficult for the Magnesium metal to insert into this strong bond.
Step 2: The Role of the Solvent
To form a Grignard reagent, the solvent must coordinate with the newly formed, highly electron-deficient Magnesium atom to stabilize the transition state and the final complex. Standard diethyl ether provides weak stabilization, which is sufficient to assist the cleavage of the weaker $C-Br$ bond, but fails to provide enough thermodynamic driving force to break the tough $C-Cl$ bond.
Step 3: The Power of THF
Tetrahydrofuran (THF) is a cyclic ether. Its oxygen atom is highly exposed and sterically unhindered compared to the floppy ethyl chains of diethyl ether. Therefore, THF is a vastly superior, highly aggressive coordinating ligand. The intense stabilization energy provided by THF coordinating to the Magnesium atom pays the thermodynamic penalty required to break the stubborn $C-Cl$ bond.
View Solution
Step 1: Selective Grignard Formation
Magnesium selectively inserts into the weaker Carbon-Bromine bond, leaving the stronger Carbon-Fluorine bond intact. This forms the Grignard reagent: 2-fluorophenylmagnesium bromide.
Step 2: Intramolecular Elimination
The $C-Mg$ bond is highly polarized, acting essentially as a carbanion on the ring carbon. This carbanion is located strictly ortho to the highly electronegative Fluorine atom (an excellent leaving group in this specific highly basic environment). The carbanion electron pair kicks in to form a triple bond, violently expelling the Fluoride ion ($F^-$) and the $[MgBr]^+$ complex.
Step 3: The Intermediate
This rapid elimination generates the highly strained, incredibly reactive Benzyne intermediate, which immediately undergoes a Diels-Alder $[4+2]$ cycloaddition with the furan diene trap.
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Step 1: The Transition State
The rate-determining step is the attack of the nucleophile, forming an anionic Meisenheimer complex. The negative charge specifically delocalizes to the ortho and para positions relative to the attack site.
Step 2: Additive Resonance Stabilization
- With one $-NO_2$ group (para), the charge delocalizes into one oxygen sink.
- With two $-NO_2$ groups (ortho/para), the charge delocalizes into two separate oxygen sinks, drastically lowering the energy of the intermediate.
- With three $-NO_2$ groups (ortho/para/ortho), every single carbon that bears a negative charge in the resonance hybrid is directly attached to a powerful electron-withdrawing nitro group. The negative charge is perfectly dispersed outside the ring onto six highly electronegative oxygen atoms.
Step 3: Conclusion
This perfect, synergistic resonance stabilization makes the activation energy barrier for nucleophilic attack nearly non-existent, allowing the reaction to proceed with mere warm water.
View Solution
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 ($-O^-$), which is an aggressively powerful activating group, 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 is non-polar. 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 hindered para position.
View Solution
Step 1: Ring Activation
In aqueous bromine, the phenolic $-OH$ group acts as a powerful $+M$ director, driving immense electron density to the two ortho positions and the one para position.
Step 2: The Ipso Attack
The highly reactive $Br^+$ electrophile easily attacks the two empty ortho positions. However, the para position is currently occupied by the $-SO_3H$ group. Because the ring is hyper-activated, the $Br^+$ attacks this specific carbon anyway, forming a crowded Wheland intermediate containing both Br and $SO_3H$ on the same carbon. This is an ipso attack.
Step 3: Restoring Aromaticity
Normally, the ring restores aromaticity by losing a proton ($H^+$). However, the $SO_3H$ group is an excellent leaving group (as a weak electrophile, $SO_3$ gas and $H^+$). The intermediate collapses, violently expelling the entire sulfonic acid group to restore aromaticity, leaving the Bromine atom securely attached.
View Solution
Step 1: Map the Positions
In 1,4-dichlorobenzene, Chlorine A is at C1, and Chlorine B is at C4. All remaining positions (C2, C3, C5, C6) are chemically equivalent due to the high symmetry of the molecule.
Step 2: Apply Directing Rules
- Chlorine A (at C1) directs the incoming nitronium ion ($NO_2^+$) to its ortho positions (C2 and C6). It cannot direct para because C4 is blocked.
- Chlorine B (at C4) directs the electrophile to its ortho positions (C3 and C5). It cannot direct para because C1 is blocked.
Step 3: Conclusion
Because all open positions are exactly equivalent, nitration at any of the four open carbons yields the exact same molecule. The electrophile simply attaches to one of the open positions adjacent to a chlorine atom.
View Solution
Step 1: The Initial Elimination
BHC is a cyclohexane ring with a chlorine atom on every single carbon. Alcoholic $KOH$ initiates an E2 elimination. A base removes a proton, kicking out an adjacent, anti-periplanar chloride ion, forming a double bond.
Step 2: The Blockade
Once the first double bond is formed, the two carbons involved become $sp^2$ hybridized. They no longer possess a hydrogen and a chlorine atom capable of anti-periplanar alignment with their remaining adjacent neighbors. Furthermore, forming cumulated double bonds (an allene inside a 6-membered ring) introduces impossible strain.
Step 3: Spacing the Eliminations
Therefore, the second elimination MUST occur at an isolated part of the ring, separated from the first double bond by an $sp^3$ carbon. The base abstracts a proton from a carbon separated by one position, forming an alternating (conjugated) double bond system. This pattern repeats for the third elimination, strictly mandating the formation of alternating, symmetrical double bonds (restoring the aromatic benzene core).
View Solution
Step 1: Formation of A
Aniline reacts with cold nitrous acid to undergo diazotization.
Compound A is Benzenediazonium chloride ($C_6H_5N_2^+Cl^-$).
Step 2: Formation of B
Treatment of a diazonium salt with $CuCN$ is the Sandmeyer reaction, substituting the diazonium group with a cyanide group.
Compound B is Cyanobenzene (Benzonitrile, $C_6H_5CN$).
Step 3: Formation of C
Complete acidic hydrolysis of a nitrile group ($-C \equiv N$) converts it into a carboxylic acid group ($-COOH$).
Compound C is Benzoic Acid ($C_6H_5COOH$).
Step 4: Formation of D
Treatment of a carboxylic acid with thionyl chloride ($SOCl_2$) replaces the $-OH$ with $-Cl$.
Compound D is Benzoyl Chloride ($C_6H_5COCl$).
Step 5: Formation of E
Benzoyl chloride acts as an acylating agent in a Friedel-Crafts acylation with Benzene and Lewis acid $AlCl_3$. The electrophile is the acylium ion ($C_6H_5C \equiv O^+$), which attacks the benzene ring.
Compound E is Benzophenone (Diphenylketone, $C_6H_5-CO-C_6H_5$).
Mastering the Unbreakable Bonds
Congratulations on conquering these 25 ultra-challenging problems on Haloarenes! You have successfully navigated the stringent stereochemical constraints of the Benzyne mechanism, the paradoxical directing effects of halogens, and the extreme kinetic barriers of unactivated Nucleophilic Aromatic Substitution. Remember, in JEE Advanced, the aromatic ring is a thermodynamic fortress—to break it, you must understand exactly how substituents manipulate resonance and inductive pathways to lower transition state energies. Keep honing your mechanistic intuition, and visit Chemca.in for more elite masterclasses!
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