Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Name Reactions
From the thermodynamic limits of the Claisen Condensation to the rigorous stereochemistry of the Beckmann Rearrangement. Master the language of organic synthesis.
Name reactions are not just trivia; they are the standardized mechanisms by which complex molecules are assembled. To conquer JEE Advanced, you must move beyond memorizing the reagents and delve into the kinetic vs thermodynamic control, migratory aptitudes, and stereospecific transition states defining each transformation.
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Step 1: Identify Enolate Sites
2,5-hexanedione ($CH_3-CO-CH_2-CH_2-CO-CH_3$) has two distinct types of $\alpha$-hydrogens:
- The terminal methyl groups (C1 or C6).
- The internal methylene groups (C3 or C4).
Step 2: Attack from C3 (Internal Enolate)
If the base deprotonates C3, the carbanion must bend backward to attack the carbonyl at C5. This connects C3 directly to C5, forming a tiny 3-membered ring. A 3-membered ring possesses catastrophic angle strain ($60^{\circ}$ vs ideal $109.5^{\circ}$). This pathway has an impossibly high activation energy and is thermodynamically forbidden.
Step 3: Attack from C1 (Terminal Enolate)
If the base deprotonates the terminal C1 methyl group, the carbanion reaches across the molecule to attack the far carbonyl at C5. This connects C1 to C5, cleanly looping five carbon atoms together to form a highly stable 5-membered ring. Following E1cB dehydration, the product is an $\alpha,\beta$-unsaturated cyclic ketone.
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Step 1: Evaluate Electrophilicity
Formaldehyde ($HCHO$) has absolutely no bulky or electron-donating alkyl/aryl groups. It is completely unhindered sterically and is the most fiercely electrophilic carbonyl compound possible. Benzaldehyde is sterically blocked by the massive phenyl ring, and its electrophilicity is heavily dampened by resonance ($+R$ effect of the ring).
Step 2: The Kinetic Attack
Because it is a vastly superior target, the hydroxide ion ($OH^-$) preferentially and almost exclusively attacks the Formaldehyde molecule. This transforms the Formaldehyde into the tetrahedral intermediate $[H_2C(O^-)(OH)]$.
Step 3: The Hydride Transfer
This formaldehyde intermediate now collapses, forcing one of its hydrogen atoms to act as a Hydride ($H^-$) donor. It ejects the hydride to attack the unreacted Benzaldehyde. By donating the hydride, Formaldehyde oxidizes to Formate. By accepting the hydride, Benzaldehyde reduces to Benzyl alcohol.
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Step 1: Structural Analysis
Ethyl acetate ($CH_3-COOEt$) possesses three $\alpha$-hydrogens.
Ethyl isobutyrate ($(CH_3)_2CH-COOEt$) possesses only one $\alpha$-hydrogen.
Step 2: The Attack
$NaOEt$ removes the single $\alpha$-hydrogen from Ethyl isobutyrate to form an enolate. This enolate attacks a second molecule, forming the tetrahedral intermediate, which collapses to expel ethoxide and yield the $\beta$-keto ester product: $(CH_3)_2CH-CO-C(CH_3)_2-COOEt$.
Step 3: The Thermodynamic Trap (The Failure)
Every step of the Claisen condensation is unfavorable and highly reversible ($K_{eq} \ll 1$). The only way the reaction succeeds is if the final $\beta$-keto ester possesses an incredibly acidic proton between the two carbonyls. If it does, the ethoxide base instantly rips it off, forming a highly stabilized enolate anion, dragging the entire equilibrium forward irreversibly.
Look at the product from Ethyl isobutyrate: $(CH_3)_2CH-CO-C(CH_3)_2-COOEt$. The central $\alpha$-carbon has zero protons remaining!
Without a final highly acidic proton to trap the product as an anion, the unfavorable equilibrium dominates, and the entire system instantly reverts back to the starting ester materials.
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Step 1: Diethyl Adipate (6 carbons)
Structure: $EtOOC-CH_2^{(1)}-CH_2^{(2)}-CH_2^{(3)}-CH_2^{(4)}-COOEt$.
The base removes a proton from the $\alpha$-carbon (C1) forming an enolate. The enolate attacks the carbonyl carbon on the opposite end. This loops C1 directly to the carbonyl carbon, incorporating the four internal carbons plus the carbonyl carbon, forming a highly stable 5-membered ring (Ethyl 2-oxocyclopentanecarboxylate).
Step 2: Diethyl Succinate (4 carbons)
Structure: $EtOOC-CH_2-CH_2-COOEt$. Enolate at the $\alpha$-carbon attacks the far carbonyl. This would form a 3-membered ring. The angle strain required to form a cyclopropanone derivative is absolutely catastrophic. The reaction completely fails to cyclize, opting instead for intermolecular polymerization or decomposition.
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Step 1: Enolate Formation and Attack
The weak base (acetate ion) deprotonates the $\alpha$-carbon of the acetic anhydride. The resulting enolate attacks the electrophilic carbonyl carbon of benzaldehyde. After an internal acyl transfer and subsequent aqueous hydrolysis, an intermediate $\beta$-hydroxy acid is formed.
Step 2: Dehydration (E1cB Mechanism)
Under the applied heat, the intermediate undergoes an E1cB dehydration to form the $C=C$ double bond, establishing deep conjugation between the benzene ring and the carboxyl group.
Step 3: Thermodynamic Stereocontrol
Because the dehydration is driven by heat and establishes an extended $\pi$-system, it is under strict thermodynamic control. The molecule will adopt the configuration that minimizes steric repulsion.
If the massive Phenyl ring and the Carboxyl ($-COOH$) group were on the same side (cis/Z), their electron clouds would severely clash, violently twisting the molecule out of planarity and destroying the stabilizing resonance.
Therefore, the molecule locks into the Trans (E) configuration, placing the bulky groups on opposite sides, allowing the entire system to lie perfectly flat and maximize resonance stabilization.
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Step 1: The Grignard Trap
If you react an $\alpha$-halo ester ($Br-CH_2-COOEt$) with Magnesium, it will rapidly form a Grignard reagent ($BrMg-CH_2-COOEt$). However, Grignard reagents are phenomenally powerful, ferocious nucleophiles. They do not wait politely to be added to a ketone. The freshly formed carbanion will instantly attack the ester group of another molecule of the starting material. This triggers uncontrollable, violent self-condensation and polymerization, completely destroying the reagent.
Step 2: The Zinc Solution
Zinc is a much less electropositive metal than Magnesium. When Zinc inserts into the Carbon-Bromine bond, it forms an organozinc compound ($BrZn-CH_2-COOEt$). Because the $C-Zn$ bond has significantly more covalent character and less ionic carbanion character, it is a much weaker, milder nucleophile.
Step 3: The Chemoselectivity
This organozinc reagent is completely inert towards the relatively unreactive ester group, meaning it will not self-condense. It sits stably in solution until a highly electrophilic ketone or aldehyde is introduced, at which point it attacks cleanly to yield the desired $\beta$-hydroxy ester.
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Step 1: Unstabilized Ylides (Kinetic Control)
$Ph_3P=CH_2$ is an unstabilized ylide. It has a highly localized negative charge on the carbon, making it incredibly reactive. When it attacks the aldehyde, the [2+2] cycloaddition to form the 4-membered oxaphosphetane ring is extremely fast and essentially irreversible. To minimize steric repulsion between the massive triphenylphosphine group and the aldehyde's phenyl ring during the attack, they approach in a puckered, folded geometry that rigidly locks the final substituents on the same side, yielding the (Z)-alkene (Kinetic product).
Step 2: Stabilized Ylides (Thermodynamic Control)
$Ph_3P=CH-COOEt$ is a stabilized ylide because its negative charge is delocalized into the adjacent ester carbonyl group via resonance. It is much less reactive. The formation of the oxaphosphetane ring becomes highly reversible. Because the ring can open and close, the intermediate naturally equilibrates into its most sterically relaxed, lowest-energy trans-conformation before finally eliminating triphenylphosphine oxide. This perfectly yields the (E)-alkene (Thermodynamic product).
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Step 1: The Intermediate
The peroxyacid attacks the carbonyl carbon to form a tetrahedral Criegee adduct. As the weak $O-O$ peroxide bond breaks, one of the alkyl groups attached to the carbonyl carbon must migrate over to the electron-deficient oxygen atom.
Step 2: Migratory Aptitude
3-methyl-2-butanone is unsymmetrical. The carbonyl carbon is attached to a Methyl group ($1^{\circ}$-like) on one side, and an Isopropyl group ($2^{\circ}$) on the other.
In the transition state, the migrating group acquires significant carbocation character. Because a secondary group stabilizes a positive charge much better than a methyl group, the Isopropyl group migrates exclusively. The general order is: Tertiary > Secondary > Primary > Methyl.
Step 3: The Final Structure
The oxygen atom is inserted strictly between the carbonyl carbon and the Isopropyl group. The resulting structure is $CH_3-C(=O)-O-CH(CH_3)_2$.
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Step 1: Activation of the Leaving Group
The strong acid protonates the oxime hydroxyl group ($-OH \rightarrow -OH_2^+$), transforming it into an excellent leaving group (water).
Step 2: The Concerted Anti-Migration
As the water molecule begins to depart, a massive electron deficiency develops on the nitrogen atom. Simultaneously, an alkyl or aryl group from the adjacent carbon detaches and migrates to the nitrogen.
To execute this migration, the shifting group must attack the back-lobe of the breaking $N-O$ bond. Therefore, it is an absolute stereoelectronic requirement that the migrating group must be Anti (trans) to the departing hydroxyl group.
Step 3: Applying to the Substrate
The problem states the $-OH$ group is syn (on the same side) to the Methyl group. Therefore, the $-OH$ group is anti to the Phenyl ring.
Following the anti-migration rule, the Phenyl ring must migrate from the carbon to the nitrogen. The carbon, having lost the phenyl group, is attacked by water and tautomerizes to a carbonyl group.
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Step 1: Selective Protonation and Dehydration
The diol has a primary $-OH$ (at C1) and a tertiary $-OH$ (at C2). Loss of the primary $-OH$ would yield a highly unstable primary carbocation. Loss of the tertiary $-OH$ yields a highly stable tertiary carbocation: $CH_2(OH)-C^+(CH_3)_2$. Therefore, the acid selectively dehydrates the tertiary position.
Step 2: The 1,2-Shift (Migration)
The molecule now possesses a tertiary carbocation adjacent to a carbon bearing a hydroxyl group. Nature seeks a deeper energy well. A hydride ion ($H^-$) from the C1 carbon undergoes a 1,2-shift, migrating to the carbocation center at C2.
Step 3: The Thermodynamic Driving Force (Resonance)
This shift moves the positive formal charge directly onto the carbon bearing the oxygen atom: $C^+H(OH)-CH(CH_3)_2$. This new intermediate is spectacularly stabilized by resonance. The lone pair on the oxygen drops down to form a pi-bond, satisfying every atom's octet ($[CH(=O^+H)-CH(CH_3)_2]$).
Step 4: Deprotonation
Loss of the proton from the oxygen yields the final stable ketone, Isobutyraldehyde ($2$-methylpropanal). Wait, let's re-read the structure. The migrating group was a hydride, shifting from C1 to C2. C1 becomes the carbonyl. It forms an aldehyde, not a ketone! Let's re-verify: $CH_2(OH)-C^+(CH_3)_2$. A methyl shift from C2 to C1 would yield $CH_2(CH_3)-C^+(OH)(CH_3)$? No, the carbocation is at C2. The adjacent carbon C1 has $-OH$ and $H$. The migration must be from C1 to C2? No, migration goes FROM the carbon with the $-OH$ TO the carbocation.
Correction: The carbocation is at C2. The groups at C1 must migrate. C1 has $-OH$, $H$, $H$. A hydride ($H^-$) migrates from C1 to C2. The carbocation moves to C1. $C^+H(OH)-CH(CH_3)_2$. Oxygen lone pair drops down. Product is 2-methylpropanal. Yes, it's an aldehyde.
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Step 1: Nucleophilic Addition
Benzil ($Ph-CO-CO-Ph$) has no $\alpha$-hydrogens, preventing Aldol reactions. The strong hydroxide ion ($OH^-$) attacks one of the highly electrophilic carbonyl carbons, forming a tetrahedral alkoxide intermediate: $Ph-CO-C(O^-)(OH)-Ph$.
Step 2: The 1,2-Aryl Shift
To re-establish the strong $C=O$ double bond, the oxygen's negative charge pushes back down. To avoid breaking the octet rule, the attached Phenyl group detaches with its electrons and migrates to the adjacent carbonyl carbon. This internal nucleophilic attack breaks the second carbonyl's pi-bond, pushing its electrons onto oxygen.
Step 3: Rapid Proton Transfer
The migration yields an intermediate containing both a carboxylic acid group ($-COOH$) and a strongly basic alkoxide ion ($-O^-$) on the adjacent carbon. An instant, irreversible intramolecular acid-base reaction occurs: the proton shifts from the acid to the alkoxide, yielding a highly stable carboxylate salt: $Ph_2C(OH)-COO^- K^+$.
Step 4: Acidification
Acidic workup protonates the carboxylate, yielding the final product: Benzilic Acid (2-hydroxy-2,2-diphenylacetic acid).
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Step 1: Enolate Formation
The methoxide base removes an $\alpha$-proton from the carbon opposite to the chlorine-bearing carbon (C6), forming a resonance-stabilized enolate anion.
Step 2: Intramolecular $S_N2$ (Cyclopropanone Formation)
The enolate carbon (C6) reaches across and performs an intramolecular $S_N2$ backside attack on the chlorine-bearing carbon (C2), expelling the chloride ion. This closes a highly strained 3-membered ring fused to the cyclohexane ring, forming a bicyclohexane-cyclopropanone intermediate.
Step 3: Nucleophilic Ring Opening
The highly strained cyclopropanone ring is fiercely reactive. A methoxide ion ($CH_3O^-$) attacks the carbonyl carbon, forming a tetrahedral alkoxide. The oxygen pushes its electrons back down to reform the carbonyl, breaking the 3-membered ring.
Step 4: Ring Contraction
The ring cleaves to yield the most stable carbanion intermediate. The bond breaks to separate the former C2 and C6 carbons. The resulting structure is a 5-membered cyclopentane ring attached to a $-CH_2^-$ carbanion (which rapidly protonates from solvent) and an ester group. The 6-membered ring has contracted to a 5-membered ring.
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Step 1: Formation of the Isocyanate
The amide reacts with $Br_2/NaOH$ to form an N-bromoamide, which loses a proton to form a highly unstable nitrene-like intermediate (acyl nitrene). To stabilize, the entire alkyl group (the sec-butyl group) migrates from the carbonyl carbon to the electron-deficient nitrogen atom, forming an Isocyanate.
Step 2: The Stereochemistry of Migration
Crucially, the 1,2-alkyl shift is a completely concerted, intramolecular process. The migrating chiral carbon never fully detaches from the molecule; its bonding electron pair simply swoops over from the carbon to the nitrogen in a single, fluid motion.
Step 3: Retention of Configuration
Because the migrating group never becomes a free planar carbocation or carbanion, its internal 3D geometry is perfectly preserved. The bond to the new nitrogen atom is formed on the exact same face from which the bond to the carbonyl carbon was broken.
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Step 1: Nucleophilic Acyl Substitution
The azide ion ($N_3^-$) attacks the benzoyl chloride, displacing the chloride ion to form Benzoyl Azide ($Ph-CO-N_3$). This is intermediate A.
Step 2: The Concerted Rearrangement
Heating benzoyl azide initiates the rearrangement. The $-N_3$ group is primed to lose the incredibly stable $N_2$ molecule. As the $N-N_2$ bond breaks, an intense electron deficiency develops on the remaining nitrogen atom.
Simultaneously, the Phenyl ring takes its bonding electrons and migrates (1,2-aryl shift) from the carbonyl carbon strictly to the electron-deficient nitrogen atom. A lone pair from nitrogen drops down to form a double bond with the carbonyl carbon.
Step 3: The Isocyanate
This concerted dance expels $N_2$ gas and perfectly forms Phenyl Isocyanate ($Ph-N=C=O$). This is intermediate B.
Step 4: Hydrolysis
Hydrolysis of the isocyanate yields a carbamic acid ($Ph-NH-COOH$), which is highly unstable and instantly decarboxylates (loses $CO_2$) to yield Aniline.
View Solution
Step 1: The Reagents
The reaction requires treating the carboxylic acid ($R-COOH$) with Hydrazoic acid ($HN_3$) in the presence of concentrated Sulfuric acid ($H_2SO_4$).
Step 2: Formation of the Intermediate
The acid protonates the carboxylic acid, making it vulnerable to nucleophilic attack by $HN_3$. Following the loss of water, an acyl azide intermediate is formed ($R-CO-N_3$).
Step 3: The Rearrangement
Under acidic and thermal conditions, the acyl azide violently expels stable Nitrogen gas ($N_2$). As the nitrogen departs, the molecule undergoes a concerted 1,2-alkyl shift. The $R$ group migrates from the carbonyl carbon directly to the electron-deficient nitrogen atom, forming an Isocyanate intermediate ($R-N=C=O$).
Step 4: Hydrolysis
The highly reactive Isocyanate is immediately hydrolyzed by the aqueous acidic medium, releasing $CO_2$ gas and yielding the final Primary Amine ($R-NH_2$).
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Step 1: Analyze the Substrate
The substrate is $R-CO-NH-O-CO-R'$. The nitrogen is bonded to a carbonyl group on one side and an oxygen atom on the other. That oxygen is attached to a bulky acyl group ($-O-CO-R'$), which is an ester-like linkage.
Step 2: Activation and Deprotonation
The nitrogen possesses an acidic proton. A base removes this proton to form a nitrogen anion: $R-CO-N^--O-CO-R'$.
Step 3: The Concerted Rearrangement
The $-O-CO-R'$ group (a carboxylate anion) is a phenomenal leaving group because it is highly resonance stabilized. As this group departs, it creates an intense electron deficiency on the nitrogen.
Exactly as in the Hofmann degradation, the $R$ group migrates from the carbonyl carbon to the nitrogen, while the nitrogen's lone pair pushes down to form the $C=N$ double bond.
Step 4: The Driving Force
The reaction is thermodynamically driven by the expulsion of the stable Carboxylate anion ($R'-COO^-$).
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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.
View Solution
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, forming a non-aromatic cyclohexadienone intermediate.
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 rapid tautomerization to restore aromaticity, the final product is a phenol with the inverted allyl group at the ortho position.
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Step 1: The Activation Step
Carboxylic acids have extremely low enol content because the carboxyl $-OH$ group donates resonance into the carbonyl, stabilizing it. Phosphorus reacts with the halogen to form $PCl_3$, which converts a small amount of the carboxylic acid into the corresponding acyl chloride ($CH_3CH_2COCl$).
Step 2: Enolization and Attack
Acyl chlorides lack this stabilizing resonance (due to poor $3p-2p$ orbital overlap of Cl). Consequently, the acyl chloride easily tautomerizes to its enol form ($CH_3-CH=C(OH)Cl$). The pi-electrons of this enol double bond act as a nucleophile, attacking the $Cl_2$ molecule. This specifically halogens the alpha-carbon (C2), yielding an alpha-halo acyl chloride.
Step 3: Exchange and Propagation
The newly formed alpha-halo acyl chloride undergoes a rapid acyl exchange with an unreacted molecule of propanoic acid, transferring the chloride group to the new acid and yielding the final alpha-halo carboxylic acid product.
Step 4: Regioselectivity
The reaction strictly requires an enol intermediate, which involves only the alpha-carbon. The terminal methyl group (beta-carbon) cannot participate in enolization, making it completely immune to HVZ halogenation.
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Step 1: The Core Mechanism
The potassium salt of phthalimide produces a bulky, highly resonance-stabilized nitrogen nucleophile. To form the amine, this anion must perform a strict $S_N2$ backside attack on an alkyl halide ($R-X$).
Step 2: Failure for Aniline (Aromatic Amines)
To synthesize aniline, the phthalimide anion would have to attack an Aryl Halide (e.g., Chlorobenzene). Aryl halides violently resist $S_N2$ substitution because the $C-Cl$ bond has strong partial double-bond character (due to resonance with the ring), making it incredibly strong. Furthermore, the massive aromatic $\pi$-cloud electrostatically and sterically repels the massive incoming nucleophile.
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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: Mechanism of Coupling
The final step of the Corey-House synthesis is the reaction: $R_2CuLi + R'-X \rightarrow R-R' + R-Cu + LiX$.
Mechanistically, this is fundamentally an $S_N2$ substitution reaction. The $R$ group from the Gilman reagent acts as a nucleophile and attacks the carbon bearing the halogen in $R'-X$.
Step 2: Steric Requirements of $S_N2$
Because $S_N2$ reactions require backside attack, they are extremely sensitive to steric hindrance. Therefore, the alkyl halide ($R'-X$) must be primary ($1^{\circ}$) or a methyl halide. If a secondary ($2^{\circ}$) or tertiary ($3^{\circ}$) halide is used, elimination (E2) will dominate over substitution, ruining the yield.
Step 3: Assigning the Roles
We have 1-bromopropane ($1^{\circ}$) and 2-bromobutane ($2^{\circ}$).
To satisfy the $S_N2$ requirement, the $1^{\circ}$ halide (1-bromopropane) MUST be kept as the alkyl halide ($R'-X$).
The Gilman reagent is immune to these steric constraints when acting as the nucleophile source. Therefore, the $2^{\circ}$ halide (2-bromobutane) must be converted into the Gilman reagent (Lithium di-sec-butylcuprate).
View Solution
Step 1: The Concerted Mechanism
The diene (4 pi electrons) and the dienophile (2 pi electrons) approach each other in parallel planes. The electrons shift in a cyclic transition state to form two new sigma bonds and one new pi bond simultaneously.
Step 2: The Stereospecificity Rule
The defining rule of the Diels-Alder reaction is that it is absolutely stereospecific with respect to the dienophile. Whatever geometric relationship exists in the dienophile is perfectly conserved in the final cyclic adduct.
Step 3: Conclusion
Maleic anhydride is a cis-dienophile (the two carbonyl groups are locked on the same side of the double bond by the anhydride ring). Therefore, in the final cyclohexene adduct, those two groups must remain strictly cis to each other.
Mastering the Lexicon of Synthesis
Congratulations on conquering these 25 ultra-challenging problems on Name Reactions! In JEE Advanced, name reactions are rarely tested as direct memory recall. Instead, they are woven into complex multi-step syntheses where you must predict migratory aptitudes (like in Pinacol or Baeyer-Villiger) and stereospecific outcomes (like anti-migration in Beckmann or retention in Hofmann). Understanding the fundamental electronic and steric constraints of each named mechanism is the absolute key to organic mastery. Keep honing your intuition, and visit Chemca.in for more elite organic chemistry masterclasses!
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