Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Organic Reagents
From the delicate chemoselectivity of Luche reduction to the mechanistic anomalies of Swern oxidation and DIBAL-H. Master the catalysts of chemical transformation.
Knowing standard reactions is not enough for JEE Advanced. You must master Chemoselectivity (which functional group reacts first), Regioselectivity (where it reacts), and Stereospecificity (the 3D geometry of the product). These 25 problems will test your mastery of oxidizing agents, reducing agents, and organometallics under extreme, competing conditions.
View Solution
Step 1: Analyzing the Substrate
The molecule possesses an ester group ($-COOC_2H_5$) and a ketone group ($-C=O$). Generally, ketones are more easily reduced than esters by standard nucleophilic hydrides (like $NaBH_4$).
Step 2: The Action of DIBAL-H
DIBAL-H is an electrophilic reducing agent (due to the empty p-orbital on Aluminum). At $-78^{\circ}\text{C}$, it coordinates strongly with the more electron-rich oxygen. It typically reduces esters to aldehydes and stops there because it forms a stable tetrahedral hemiacetal-like aluminum complex that does not collapse until aqueous workup destroys the reagent.
Step 3: Chemoselectivity Limit
While DIBAL-H can reduce ketones, it is widely utilized for the controlled reduction of esters (and nitriles) to aldehydes. If $1 \text{ equivalent}$ is used, it will reduce the ketone (which is fundamentally more reactive) to a secondary alcohol. Wait, is a ketone more reactive to DIBAL-H than an ester? Yes! Electrophilic hydrides (like $AlH_3$ or DIBAL) actually reduce ketones faster than esters. However, if excess DIBAL-H is used at $-78^{\circ}\text{C}$, it reduces BOTH the ketone to an alcohol and the ester to an aldehyde.
Correction/Refinement: With exactly 1 equivalent, the most electrophilic group reacts first. The ketone is more electrophilic than the ester (which has resonance $+M$ from the alkoxy oxygen). Therefore, the ketone is reduced to a secondary alcohol, leaving the ester intact.
View Solution
Step 1: Identify Allylic Positions
Structure: $CH_3-C(CH_3)=CH-CH_3$.
There are three distinct allylic positions:
1. Two identical methyl groups on C2.
2. One methyl group on C4.
Step 2: $SeO_2$ Regioselectivity Rules
- Rule 1: Oxidation occurs preferentially at the more substituted end of the double bond.
- Rule 2: If the ends are equivalent, the order of preference is $CH_2 > CH_3 > CH$.
- Rule 3: The new double bond forms in the more stable (more substituted/conjugated) position (usually an (E)-alkene).
Step 3: Application
The more substituted end of the double bond is C2 (it bears two methyls, whereas C3 bears one methyl). Therefore, oxidation will occur at one of the methyl groups attached to C2.
The $SeO_2$ mechanism (Ene reaction followed by [2,3]-sigmatropic shift) targets the most sterically accessible proton at the more substituted end.
Step 4: The Product
One of the $CH_3$ groups on C2 is oxidized to a $CH_2OH$ group.
View Solution
Step 1: The Standard Problem
An $\alpha,\beta$-unsaturated ketone has two electrophilic sites: the carbonyl carbon (Hard electrophile) and the $\beta$-carbon (Soft electrophile). $NaBH_4$ is a borderline nucleophile that can attack both, leading to unwanted mixtures of saturated and unsaturated alcohols.
Step 2: The Role of $CeCl_3$ (Luche Conditions)
Cerium(III) is a highly oxophilic, "Hard" Lewis acid. It specifically and strongly coordinates to the carbonyl oxygen. This massive polarization drastically increases the electrophilicity (hardness) of the carbonyl carbon.
Simultaneously, the methanol solvent reacts with $NaBH_4$ to form alkoxyborohydrides (like $NaBH(OCH_3)_3$), which are "Harder" nucleophiles than $NaBH_4$ itself.
Step 3: The Resulting Attack
Because Hard likes Hard (HSAB principle), the newly hardened hydride nucleophile exclusively attacks the hardened carbonyl carbon via strict 1,2-addition.
Step 4: The Product
The ketone is cleanly reduced to an alcohol, leaving the alkene completely untouched.
View Solution
Step 1: The Nature of $MnO_2$
Activated Manganese dioxide is a very mild, heterogeneous oxidizing agent. It is not strong enough to oxidize standard, unactivated aliphatic primary or secondary alcohols under mild conditions.
Step 2: Chemoselectivity
$MnO_2$ is highly specific for the oxidation of allylic, benzylic, and propargylic alcohols. The adjacent $\pi$-system heavily stabilizes the transition state during the hydrogen abstraction/oxidation process on the solid $MnO_2$ surface.
Step 3: Product Formation
The primary aliphatic alcohol is left completely untouched. The secondary allylic alcohol is selectively oxidized to an $\alpha,\beta$-unsaturated ketone.
View Solution
Step 1: Activation of DMSO
DMSO reacts with oxalyl chloride at $-78^{\circ}\text{C}$ to form an unstable chlorodimethylsulfonium intermediate. During this activation, the oxalyl group violently breaks down, releasing Carbon Monoxide ($CO$) and Carbon Dioxide ($CO_2$) gases.
Step 2: Formation of the Alkoxysulfonium Ion
The primary alcohol (1-octanol) attacks the sulfur atom, displacing the chloride to form an alkoxysulfonium intermediate.
Step 3: Elimination (Addition of Base)
Triethylamine is added. It deprotonates a methyl group on the sulfur to form a sulfur ylide. This ylide then undergoes an intramolecular E2-like elimination, cleaving the $C-H$ bond of the alcohol to form the carbonyl double bond.
Step 4: The Final Products
The primary alcohol yields an aldehyde: Octanal.
The leaving group from the sulfur ylide is the notoriously foul-smelling, volatile liquid/gas Dimethyl Sulfide (DMS, $CH_3SCH_3$).
View Solution
Step 1: Analyze the Acetal Group
An acetal is an ether-like linkage that is highly stable in basic and neutral conditions. However, acetals are explicitly designed as protecting groups because they are highly sensitive to aqueous acids, which rapidly hydrolyze them back into aldehydes/ketones and alcohols.
Step 2: Evaluate Clemmensen Reduction
The Clemmensen reduction operates under brutally harsh acidic conditions (boiling concentrated $HCl$). These conditions will instantly hydrolyze and destroy the cyclic acetal protecting group.
Step 3: Evaluate Wolff-Kishner Reduction
The Wolff-Kishner reduction operates under strongly basic conditions (boiling $KOH$ in ethylene glycol). Since acetals are perfectly stable in base, the acetal will survive completely intact while the ketone is reduced.
View Solution
Step 1: The Initial Oxidation
Both reagents successfully oxidize the primary alcohol into the desired aldehyde, (E)-2-hexenal. The double bond is generally unreactive to both mild $Cr(VI)$ reagents under these conditions.
Step 2: The Aqueous Trap (Jones Reagent)
Jones reagent is an aqueous acidic solution. As soon as the aldehyde is formed, it reacts with the water present in the solvent to form an equilibrium mixture with its gem-diol (aldehyde hydrate): $R-CHO + H_2O \rightleftharpoons R-CH(OH)_2$.
Step 3: Over-oxidation
To a strong $Cr(VI)$ oxidant, a gem-diol looks and acts exactly like an alcohol. The Jones reagent immediately oxidizes the gem-diol further, converting it irreversibly into a carboxylic acid ((E)-2-hexenoic acid).
Step 4: The PCC Solution
PCC is used in strictly anhydrous (water-free) dichloromethane ($CH_2Cl_2$). Without water, the gem-diol cannot form. The reaction cleanly and permanently stops at the aldehyde stage.
View Solution
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 $CH_3-C(=O)-O-CH(CH_3)_2$.
View Solution
Step 1: Evaluate the Electrophilic Sites
- The Carbonyl carbon (C1) has a highly concentrated, localized partial positive charge ($\delta+$) due directly to the electronegative oxygen. It is a "Hard" electrophile.
- The $\beta$-carbon (C3) has a highly delocalized, diffuse partial positive charge generated by resonance. It is a "Soft" electrophile.
Step 2: Flask A (Grignard Reagent)
Grignard reagents ($RMgBr$) contain a highly polarized, charge-dense $C-Mg$ bond. The carbanion character is concentrated, making it a "Hard" nucleophile. Hard likes Hard. It attacks the carbonyl carbon directly (1,2-addition) to form an allylic alcohol: 1-methylcyclohex-2-en-1-ol.
Step 3: Flask B (Gilman Reagent)
Organocuprates (Gilman reagents) contain Carbon-Copper bonds. Copper is a large transition metal, making the bond highly polarizable and the electron cloud diffuse. It is a "Soft" nucleophile. Soft likes Soft. It attacks the $\beta$-carbon (1,4-conjugate addition), yielding an enolate which tautomerizes to 3-methylcyclohexan-1-one.
View Solution
Case (a): Cis-2-butene
Reagent: Hydrogen gas ($H_2$) over Lindlar's Catalyst ($Pd/CaCO_3$ poisoned with lead acetate/quinoline).
Mechanism: This is a heterogeneous catalytic hydrogenation. The alkyne adsorbs flat onto the solid palladium surface. Both hydrogen atoms are delivered simultaneously from the metal surface to the same face of the alkyne. This physical constraint strictly enforces a syn-addition, yielding the cis-alkene.
Case (b): Trans-2-butene
Reagent: Sodium or Lithium metal in liquid Ammonia ($Na/NH_{3(l)}$) - The Birch-type reduction conditions.
Mechanism: This is a dissolving metal reduction. A single electron adds to form a Radical Anion intermediate. To minimize the extreme electrostatic repulsion between the single electron cloud and the lone pair, the molecule rapidly adopts a trans-like (anti) geometry. Subsequent protonations lock this thermodynamically favored anti geometry in place, yielding the trans-alkene.
View Solution
Step 1: Flask A ($OsO_4$)
$OsO_4$ reacts via a concerted [3+2] cycloaddition. Two oxygen atoms from the same Osmium atom attack the $\pi$-bond simultaneously. Because they are tethered together, they must attach to the exact same face of the ring, forming a cyclic osmate ester. Reductive cleavage preserves this geometry, yielding strictly cis-1,2-cyclohexanediol (Syn-dihydroxylation).
Step 2: Flask B (mCPBA then $H_3O^+$)
mCPBA transfers one oxygen to form an epoxide ring. Upon protonation by acid, water acts as a nucleophile. Because the epoxide oxygen physically shields the top face, the water molecule must attack from the opposite face (backside) to open the ring. This $S_N2$-like inversion places the two hydroxyl groups on opposite sides, yielding strictly a racemic mixture of trans-1,2-cyclohexanediol (Anti-dihydroxylation).
View Solution
Step 1: Radical Generation and Abstraction
The NBS generates a Bromine radical ($Br^{\bullet}$). This radical abstracts a hydrogen atom from the weakest $C-H$ bond in 1-butene. The weakest bond is at the allylic carbon (C3), right next to the double bond.
Step 2: The Resonance-Stabilized Intermediate
Abstraction yields an allylic free radical: $CH_2=CH-C^{\bullet}H-CH_3$.
This radical is highly stable because the single electron delocalizes across the $\pi$-system via resonance, creating a secondary resonance structure: $^{\bullet}CH_2-CH=CH-CH_3$.
Step 3: Bromine Attack (Two Pathways)
Because the radical intermediate has high electron density at two different carbons (C3 and C1), the incoming $Br_2$ molecule (generated in trace amounts by NBS) can be attacked by either site.
- Attack at C3 yields: 3-bromo-1-butene (Kinetic product).
- Attack at C1 yields: 1-bromo-2-butene (Thermodynamic product, as the internal double bond is more stable).
View Solution
Step 1: The Structural Requirement
$HIO_4$ exclusively cleaves $C-C$ bonds where both adjacent carbons possess an oxygen-containing functional group (hydroxyl, carbonyl, or a mix). To cleave the bond, the Iodine atom must simultaneously coordinate with both oxygen atoms to form a 5-membered cyclic ester. If the groups are separated by a $CH_2$ group, the resulting 6-membered ring forms too slowly to trigger cleavage under standard conditions.
Step 2: Analyzing the Options
- (a) Ethylene glycol (1,2-ethanediol): Vicinal diol. Cleaves (yields 2 molecules of formaldehyde).
- (b) Propane-1,3-diol: The $-OH$ groups are separated by a $CH_2$ group. Fails to cleave.
- (c) 2-Hydroxypropanal: An $\alpha$-hydroxy aldehyde. The adjacent carbons have an $-OH$ and a $C=O$. Cleaves (yields acetaldehyde and formic acid).
- (d) 2,3-Butanedione: A vicinal diketone ($\alpha$-diketone). The hydrated carbonyls can coordinate. Cleaves (yields 2 molecules of acetic acid).
View Solution
Step 1: The Initial Cleavage
1-Methylcyclohexene is a ring with a double bond between C1 (bearing a methyl group) and C2 (bearing a hydrogen). Cleaving this bond opens the ring into a straight 6-carbon chain.
C1 becomes a ketone (because it is attached to a methyl and the ring chain).
C2 becomes an aldehyde (because it is attached to a hydrogen and the ring chain).
The raw skeleton is: $CH_3-C(=O)-CH_2-CH_2-CH_2-CH=O$ (6-oxoheptanal).
Step 2: Flask A (Reductive Workup with $Zn/H_2O$ or DMS)
Zinc acts as a reducing agent to destroy the highly explosive $H_2O_2$ byproduct formed during the hydrolysis of the ozonide. This prevents any further oxidation. The aldehyde group safely survives.
Product A: 6-oxoheptanal.
Step 3: Flask B (Oxidative Workup with $H_2O_2$)
Hydrogen peroxide is a strong oxidizing agent. The ketone group is stable, but the vulnerable aldehyde group is immediately oxidized into a carboxylic acid.
Product B: 6-oxoheptanoic acid.
View Solution
Case (a): $NaBH_4$
Sodium borohydride is a mild, chemoselective nucleophile. It easily attacks the highly electrophilic ketone, reducing it to a secondary alcohol. However, it is fundamentally too weak to attack resonance-stabilized (electron-rich) functional groups like esters or amides. The ester and amide remain completely untouched.
Case (b): $LiAlH_4$
Lithium aluminum hydride is a phenomenally aggressive, unselective hydride donor.
- The ketone is reduced to a secondary alcohol.
- The ester is completely cleaved and reduced to yield two primary alcohols.
- The primary amide is uniquely reduced, losing its carbonyl oxygen entirely, to yield a primary amine ($-CH_2NH_2$).
View Solution
Step 1: Reaction without Pyridine (The $S_Ni$ Mechanism)
Initially, the alcohol reacts with $SOCl_2$ to form an alkyl chlorosulfite intermediate ($-O-SO-Cl$) while releasing $HCl$. In ether without a base, the $HCl$ leaves as a gas. The intermediate collapses via a concerted internal nucleophilic substitution ($S_Ni$). The chlorine atom is delivered to the exact same face from which the $SO_2$ gas departs. Because the nucleophile attacks from the front, the result is complete Retention of Configuration (yielding the (S)-product).
Step 2: Reaction with Pyridine (The Switch to $S_N2$)
When pyridine (a base) is present, it immediately reacts with the generated $HCl$ to form pyridinium chloride ($C_5H_5NH^+ Cl^-$). This creates a massive concentration of free, highly reactive chloride ions ($Cl^-$) in the solution.
Step 3: External Backside Attack
Instead of waiting for the sluggish internal collapse, the abundant external $Cl^-$ ions aggressively attack the alkyl chlorosulfite intermediate from the back via a standard $S_N2$ mechanism, displacing the entire $-O-SO-Cl$ leaving group. Backside attack mandates Inversion of Configuration, yielding the (R)-product.
View Solution
Step 1: The Benzylic Carbon Vulnerability
The carbon atom directly attached to the benzene ring is 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.
View Solution
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).
View Solution
Step 1: Strong Acidic Medium ($Sn/HCl$)
In a strong acid, protons ($H^+$) are infinitely abundant. The reduction proceeds smoothly and entirely to completion, stripping both oxygen atoms and replacing them with hydrogens. The product A is Aniline ($Ph-NH_2$).
Step 2: Strong Alkaline Medium ($Zn/NaOH$)
In a basic medium, intermediate reduction products like nitrosobenzene ($Ph-NO$) and N-phenylhydroxylamine ($Ph-NH-OH$) are formed. Because the basic environment lacks the protons required to rapidly push them all the way to aniline, these two highly reactive intermediates build up and bump into each other. They condense together, linking their nitrogen atoms. Further reduction by $Zn/NaOH$ yields the fully symmetric hydrazo compound.
Product B is Hydrazobenzene (1,2-diphenylhydrazine, $Ph-NH-NH-Ph$).
View Solution
Step 1: The Failure of Standard $BH_3$
Because $BH_3$ is small and has three active hydrogens, it can add across the triple bond to form a vinylborane, and then immediately add across the resulting double bond a second time, undergoing double hydroboration. Oxidation of this mess yields complex mixtures, not a clean aldehyde.
Step 2: The Solution (Bulky Boranes)
Disiamylborane ($Sia_2BH$) has massive, sterically hindering alkyl groups and only one active $B-H$ bond. When it adds across the terminal alkyne, the massive steric bulk physically prevents a second borane molecule from approaching the resulting alkene. The reaction stops cleanly at the vinylborane stage.
Step 3: Anti-Markovnikov Regioselectivity
The sheer size of the $Sia_2B$ group forces it to attach exclusively to the least hindered terminal carbon. Subsequent oxidation ($H_2O_2 / OH^-$) replaces the boron with an $-OH$ group, yielding an enol at the terminal carbon, which tautomerizes cleanly to Hexanal.
View Solution
Step 1: Bypassing the Vinylic Carbocation
Protonating an alkyne directly would form an $sp$-hybridized vinylic carbocation ($H_2C=C^+-H$), which is highly unstable. The large, highly polarizable $Hg^{2+}$ ion acts as a potent electrophile, coordinating across the triple bond to form a stable, bridged mercurinium-like ion, avoiding the free carbocation entirely.
Step 2: Nucleophilic Attack and Demercuration
Water attacks the bridged intermediate. Following the loss of a proton, an organomercury enol is formed. The acidic environment then replaces the Mercury group with a proton via protonolysis, yielding a pure Enol intermediate ($CH_2=CH-OH$).
Step 3: Keto-Enol Tautomerization
The enol is thermodynamically unstable relative to its carbonyl counterpart. The pi-electrons shift to form a $C=O$ double bond, and the proton migrates to the adjacent carbon, rapidly tautomerizing into Acetaldehyde ($CH_3-CHO$).
View Solution
Tube A: Aniline (Primary $1^{\circ}$)
The $1^{\circ}$ amine reacts to form an aromatic diazonium salt ($Ph-N_2^+ Cl^-$). Because it is aromatic and kept cold ($0^{\circ}\text{C}$), orbital resonance from the benzene ring grants the $C-N$ bond partial double-bond character, making it stable. It yields a clear, stable, water-soluble solution. (If warmed, it would explosively release $N_2$ gas and form phenol).
Tube B: N-methylaniline (Secondary $2^{\circ}$)
The $2^{\circ}$ amine reacts with the $NO^+$ electrophile. It loses its one and only nitrogen proton to yield an N-nitroso compound ($Ph-N(CH_3)-N=O$). Because it has no more protons to lose, the dehydration sequence to form a triple bond cannot occur. The reaction stops here. These compounds are called N-nitrosamines. They are characteristically toxic, highly carcinogenic, water-insoluble yellow oils that separate from the aqueous solution.
View Solution
Step 1: The Initial Structure
Fructose contains a ketone group at C2 and a hydroxyl group at C1 (an $\alpha$-hydroxy ketone). Ketones generally cannot be oxidized by mild $Ag^+$ ions because they lack an abstractable hydrogen on the carbonyl carbon.
Step 2: Lobry de Bruyn-van Ekenstein Transformation
In the basic medium of Tollens' reagent, the mildly acidic alpha-proton of fructose is removed. The molecule tautomerizes into an enediol intermediate (a double bond between C1 and C2, with hydroxyl groups on both).
This enediol is highly unstable. When the equilibrium shifts back to the carbonyl form, the double bond can reform at C1 instead of C2. This converts the molecule into Glucose and Mannose, both of which are Aldoses (containing aldehyde groups).
Step 3: The Oxidation
Now that aldehyde groups are present in the basic solution, the $Ag^+$ ions are easily reduced to metallic silver ($Ag^0$), precipitating as the famous silver mirror.
View Solution
Step 1: The First Reduction
Palladium readily catalyzes the reduction of the highly reactive acyl chloride ($Ph-COCl$) into Benzaldehyde ($Ph-CHO$).
Step 2: The Threat of Over-Reduction
Aldehydes are highly electrophilic and easily hydrogenated. If pure, highly active Palladium metal were used, it would instantly reduce the newly formed Benzaldehyde further down into a primary alcohol: Benzyl alcohol ($Ph-CH_2OH$). This would completely ruin the synthesis.
Step 3: Poisoning the Catalyst
To stop the reaction exactly at the aldehyde stage, the Palladium catalyst must be "poisoned" (its activity must be throttled). Using Barium Sulfate as a dense support naturally lowers the surface activity. Adding a trace of sulfur, quinoline, or thiourea acts as a chemical poison, permanently binding to the most active catalytic sites. This leaves the catalyst just active enough to reduce the hyper-reactive acyl chloride, but too weak to reduce the moderately reactive aldehyde.
View Solution
Step 1: Formation of B (Friedel-Crafts Alkylation)
Benzene reacts with methyl chloride to yield Toluene ($C_6H_5CH_3$). Compound B is Toluene.
Step 2: Formation of C (Benzylic Oxidation)
Hot $KMnO_4$ oxidizes any alkyl side chain with a benzylic hydrogen completely down to a carboxyl group. Toluene becomes Benzoic Acid ($C_6H_5COOH$). Compound C is Benzoic acid.
Step 3: Formation of D (Chlorination)
Treatment with $SOCl_2$ converts the carboxylic acid into an acyl chloride. Compound D is Benzoyl Chloride ($C_6H_5COCl$).
Step 4: Formation of E (Curtius Rearrangement)
Reaction of an acyl chloride with $NaN_3$ yields an acyl azide ($Ph-CO-N_3$). Upon heating, it undergoes the Curtius rearrangement: expelling $N_2$ gas and undergoing a 1,2-aryl shift to form an Isocyanate ($Ph-N=C=O$). The aqueous medium instantly hydrolyzes the isocyanate, releasing $CO_2$ and yielding a primary amine. Compound E is Aniline ($C_6H_5NH_2$).
Mastering the Catalysts of Synthesis
Congratulations on conquering these 25 ultra-challenging problems on Reagents in Organic Chemistry! The secret to mastering reagents for JEE Advanced is understanding chemoselectivity. Memorizing that DIBAL-H reduces esters is useless unless you know it stops at the aldehyde at $-78^{\circ}\text{C}$, but reduces it fully to an alcohol at room temperature. Whether it is the Hard-Soft dynamics of Gilman vs Grignard, or the strict syn/anti geometry of OsO4 vs mCPBA, the reaction mechanism is the true key. Keep honing your intuition, and visit Chemca.in for more elite organic chemistry masterclasses!
No comments:
Post a Comment