Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Functional Group Tests
From the deep kinetic mechanisms of the Lucas Test to the complex color chemistry of Liebermann's Nitroso Reaction. Master the logic of Practical Organic Chemistry.
In JEE Advanced, simply memorizing that "Tollens gives a silver mirror" is insufficient. You must understand why aromatic aldehydes fail Fehling's test, how the Iodoform intermediate is cleaved, and what specific structural features govern the solubility rules in the Hinsberg test. Prepare to push your diagnostic logic to the absolute limit.
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Step 1: The Role of $ZnCl_2$
Anhydrous $ZnCl_2$ acts as a powerful Lewis acid. It coordinates with the oxygen atom of the alcohol, turning the poor $-OH$ leaving group into an excellent leaving group ($[ZnCl_2OH]^-$).
Step 2: Carbocation Formation
Upon departure of the leaving group, allyl alcohol forms the Allyl carbocation ($CH_2=CH-C^+H_2$). While technically located on a primary carbon, this is not a standard primary carbocation.
Step 3: Resonance Stabilization
The empty p-orbital of the carbocation is perfectly adjacent to the $\pi$-bond. The $\pi$-electrons extensively delocalize into the empty orbital via resonance: $[CH_2=CH-C^+H_2 \leftrightarrow C^+H_2-CH=CH_2]$. This immense delocalization stabilizes the allyl carbocation to a level comparable to, or greater than, a standard tertiary carbocation.
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Step 1: Primary Alcohols (The Red Path)
$R-CH_2OH \rightarrow R-CH_2I \rightarrow R-CH_2NO_2$ (Primary nitroalkane).
Reaction with $HNO_2$: The primary nitroalkane has two $\alpha$-hydrogens. It reacts with $O=N-OH$ to form Nitrolic Acid ($R-C(=NOH)-NO_2$).
Nitrolic acid is highly acidic due to the withdrawing groups. When dissolved in $NaOH$, it forms a sodium salt which is intensely Blood Red in color.
Step 2: Secondary Alcohols (The Blue Path)
$R_2CH-OH \rightarrow R_2CH-I \rightarrow R_2CH-NO_2$ (Secondary nitroalkane).
Reaction with $HNO_2$: The secondary nitroalkane has only ONE $\alpha$-hydrogen. It reacts with $HO-N=O$ to form a Pseudonitrole ($R_2C(NO)-NO_2$).
A pseudonitrole contains no acidic hydrogen on the central carbon. Because it lacks an acidic proton, it cannot react with the added base ($NaOH$) to form a soluble salt. It remains as an insoluble oil, which in its pure state or in solution inherently possesses a deep Blue color.
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Step 1: The Initial Contact
When an aliphatic alcohol is added to the yellow Ceric Ammonium Nitrate solution ($ (NH_4)_2[Ce(NO_3)_6] $), an immediate ligand exchange reaction occurs.
Step 2: The Ligand Substitution
The oxygen atom of the alcohol acts as a nucleophilic ligand, displacing one or more of the nitrate ($NO_3^-$) ligands from the coordination sphere of the Cerium(IV) heavy metal core.
$[Ce(NO_3)_6]^{2-} + R-OH \rightleftharpoons [Ce(NO_3)_5(OR)]^{2-} + HNO_3$
Step 3: The Color Generation
This new intermediate alkoxy-cerium complex is highly unstable but intensely Red colored due to Ligand-to-Metal Charge Transfer (LMCT) transitions. If left standing, the Cerium(IV) will eventually oxidize the alcohol, reducing itself to colorless Cerium(III), and the red color fades. The test relies on observing the transient coordination intermediate.
(A) 1-Phenyl-1-ethanol
(B) Acetic Acid ($CH_3COOH$)
(C) Acetyl Chloride ($CH_3COCl$)
(D) Lactic Acid ($CH_3-CH(OH)-COOH$)
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Compound A: 1-Phenyl-1-ethanol ($Ph-CH(OH)-CH_3$)
This contains the $CH_3-CH(OH)-$ group. The reagent oxidizes it to acetophenone ($Ph-CO-CH_3$), which possesses an active methyl group adjacent to a carbonyl. This will undergo halogenation and cleavage to yield $CHI_3$ (yellow ppt). Positive.
Compound B & C: Acetic Acid and Acetyl Chloride
Despite having a $CH_3-CO-$ group, the carbonyl carbon is attached to a heteroatom possessing a lone pair ($-OH$ or $-Cl$). The strong $+M$ (resonance) effect heavily suppresses the electrophilicity of the carbonyl carbon and stabilizes the enolate poorly. Furthermore, the $OH^-$ base simply attacks the acidic proton or performs acyl substitution on the chloride. Negative.
Compound D: Lactic Acid ($CH_3-CH(OH)-COOH$)
It contains the required $CH_3-CH(OH)-$ moiety. It oxidizes to pyruvic acid ($CH_3-CO-COOH$). The $CH_3$ group is active and will form $CI_3-CO-COOH$, which base cleaves into Iodoform and oxalate. Positive.
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Step 1: Evaluate Reagent Strength
Tollens' reagent (ammoniacal $Ag^+$) is a relatively strong mild oxidizing agent. Fehling's solution (a tartrate complex of $Cu^{2+}$) is a significantly weaker oxidizing agent than Tollens'.
Step 2: Analyze Aromatic Aldehydes
In Benzaldehyde ($Ph-CHO$), the carbonyl group is directly conjugated with the electron-rich $\pi$-system of the benzene ring. Through resonance ($+M$ effect from the ring), massive electron density is donated into the carbonyl carbon. This makes the carbonyl carbon vastly less electrophilic and highly stabilizes the molecule against oxidative attack.
Step 3: The Verdict
Because Benzaldehyde is exceptionally stable due to aromatic conjugation, it requires a reasonably strong oxidant to break that stability. Tollens' reagent is just strong enough to accomplish this. Fehling's reagent, being much weaker, lacks the thermodynamic oxidizing power to overcome the resonance stabilization of the aromatic aldehyde.
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Step 1: The Decolorization
Rosaniline is a highly conjugated triphenylmethane dye. When sulfur dioxide ($SO_2$ / sulfurous acid) is added, it forms an addition product (a sulfonic acid derivative) at the central carbon. This converts the central $sp^2$ carbon into an $sp^3$ carbon, breaking the extended conjugation across the three rings and rendering the molecule colorless.
Step 2: The Aldehyde Reaction
Aldehydes are highly reactive electrophiles. When added to the colorless Schiff's reagent, the aldehyde reacts with the free amine ($-NH_2$) groups on the periphery of the dye molecule, forming imine/Schiff base complexes (or bis-hydroxysulfonate adducts depending on the exact mechanism proposed by researchers).
Step 3: Restoration of Conjugation
This nucleophilic attack by the dye's amino groups onto the aldehyde pulls electron density outward. This triggers the elimination of the sulfite group from the central carbon, converting it back to an $sp^2$ hybridized state. The massive conjugated $\pi$-network is completely restored, resulting in the return of the intense magenta color.
Note: Ketones generally fail this test because they are sterically hindered and not electrophilic enough to form the required adducts efficiently.
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Step 1: The Standard Mechanism
The nucleophile is the terminal $-NH_2$ group of 2,4-DNP. The electrophile is the carbonyl carbon. Mild acid protonates the carbonyl oxygen, enhancing its electrophilicity without destroying the nucleophile.
Step 2: Failure in Strong Acid ($pH \approx 1$)
If the acid is too strong, the lone pair on the terminal $-NH_2$ group of the 2,4-DNP reagent gets completely protonated to form $-NH_3^+$. A protonated amine has no lone pairs left, meaning it is no longer a nucleophile. The reaction completely stops.
Step 3: Failure in Strong Base ($pH \approx 10$)
If the medium is basic, there are no protons available to protonate the carbonyl oxygen. The unactivated carbonyl carbon is a poor electrophile, resulting in extremely slow or zero reaction. Furthermore, in strong base, the acidic protons on the 2,4-DNP ring itself could be abstracted, forming unreactive anionic species.
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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 give effervescence.
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 give 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 give violent effervescence.
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Step 1: Structure of the Complex
The violet color is due to the formation of a hexacoordinated Ferric phenoxide complex:
$6 PhOH + FeCl_3 \rightleftharpoons H_3[Fe(OC_6H_5)_6] + 3 HCl$.
Step 2: The Role of Acidity
Ferric chloride ($FeCl_3$) is a salt of a strong acid ($HCl$) and a weak base ($Fe(OH)_3$). In water, it undergoes massive cationic hydrolysis, making the solution naturally highly acidic (high $H^+$ concentration). Phenol is a weak acid ($PhOH \rightleftharpoons PhO^- + H^+$).
Step 3: Le Chatelier's Suppression
If the $FeCl_3$ solution is highly acidic, the massive excess of $H^+$ ions strongly suppresses the ionization of phenol via the Common Ion Effect. Without a sufficient concentration of the active nucleophilic phenoxide ions ($PhO^-$) to coordinate with the Ferric ion, the colored complex cannot form effectively.
Step 4: Neutralization
By neutralizing the excess acid with $NH_4OH$, we remove the common ion suppression, allowing phenol to ionize sufficiently to form the deep violet complex.
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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 yields an indophenol derivative (a protonated oxonium sulfate salt) which exhibits the initial red/green colors.
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: Alkene Observation
Alkenes undergo simple electrophilic addition across the double bond to form a vicinal dibromide (or bromohydrin in water). The orange/brown color of $Br_2$ disappears, leaving a colorless solution.
Step 2: Phenol Observation
Phenol undergoes substitution. The orange color disappears, but a dense White Precipitate instantly drops out of the solution.
Step 3: The Runaway Mechanism in Phenol
Water is highly polar and strongly hydrogen-bonds. It forces phenol to heavily ionize into the phenoxide ion ($-O^-$). The phenoxide ion is an aggressively powerful activating group, flooding the ring with electron density via the $+M$ effect. 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 at all ortho and para positions.
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Step 1: Generation of the Intermediate
Alcoholic $KOH$ acts as a strong base, stripping a proton from $CHCl_3$ to form the trichloromethyl carbanion ($:CCl_3^-$). This carbanion rapidly expels a chloride ion (alpha-elimination) to generate Dichlorocarbene ($:CCl_2$). Dichlorocarbene is an electron-deficient, highly reactive neutral electrophile.
Step 2: Amine Attack
The lone pair on the amine nitrogen attacks the empty p-orbital of the dichlorocarbene, forming a zwitterionic intermediate: $R-N^+H_2-C^-Cl_2$.
Step 3: The Protonic Requirement
To form the final product (an Isocyanide, $R-N \equiv C$), the nitrogen atom must lose two protons to the base, while the carbon atom expels the remaining two chloride ions via successive eliminations. A primary amine ($R-NH_2$) has exactly the two required protons on the nitrogen.
A secondary amine ($R_2NH$) only has one proton, halting the mechanism halfway. A tertiary amine ($R_3N$) has zero protons, making the reaction completely impossible.
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Outcome 1: Ethylamine (Primary Amine)
Reaction yields N-ethylbenzenesulfonamide ($Ph-SO_2-NH-Et$). Because the highly electron-withdrawing sulfonyl group ($-SO_2$) is attached directly to the nitrogen, the remaining hydrogen atom on the nitrogen becomes highly acidic. It reacts with $NaOH$ to form a soluble sodium salt ($Ph-SO_2-N^--Et \ Na^+$).
Outcome 2: Diethylamine (Secondary Amine)
Reaction yields N,N-diethylbenzenesulfonamide ($Ph-SO_2-N(Et)_2$). The nitrogen atom is completely substituted with alkyl groups; there are zero acidic protons remaining. Therefore, it cannot react with $NaOH$ to form a salt and remains as an insoluble precipitate.
Outcome 3: Triethylamine (Tertiary Amine)
Tertiary amines have no hydrogens on the nitrogen to begin with. They cannot substitute the chloride from the Hinsberg reagent to form a stable sulfonamide. They merely act as a base/catalyst, slowly hydrolyzing the reagent to sodium benzenesulfonate, which is water-soluble (hence the clear solution, but no reaction product).
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Step 1: Identifying the Amine Class
This test is given exclusively by Primary ($1^{\circ}$) Amines. Secondary and tertiary amines do not possess the required N-H protons to complete the sequence.
Step 2: The Reaction Mechanism
The primary amine ($R-NH_2$) acts as a nucleophile, attacking $CS_2$ to form an unstable dithiocarbamic acid intermediate ($R-NH-C(=S)-SH$).
Step 3: Action of Mercuric Chloride
When heated with heavy metal salts like $HgCl_2$, the intermediate is desulfurized. It loses $H_2S$ (which reacts with Mercury) to form an Alkyl Isothiocyanate ($R-N=C=S$). This specific molecule is responsible for the aggressive mustard-like smell.
Step 4: The Byproduct
The sulfur stripped from the intermediate forms Mercuric Sulfide ($HgS$), which deposits as a distinct black precipitate.
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Step 1: The Activation Requirement (Mildly Alkaline)
The neutral $\beta$-Naphthol molecule is not nucleophilic enough to successfully attack the weak diazonium electrophile. By making the solution mildly basic, it is deprotonated to form the naphthoxide ion ($-O^-$). The full negative charge pushes a massive amount of electron density into the ring via resonance, hyper-activating it for coupling.
Step 2: Failure in Acidic Medium
If the medium is acidic, the molecule remains as neutral $\beta$-Naphthol. The ring is insufficiently activated, and the incredibly weak diazonium electrophile is completely unable to perform the substitution. The reaction fails.
Step 3: Failure in Strongly Basic Medium
If the medium is made too strongly basic ($pH > 11$), the nucleophile is fine, but the electrophile is destroyed! The diazonium ion reacts with excess hydroxide ions ($OH^-$) to form a diazohydroxide ($Ar-N=N-OH$), which further deprotonates to form the diazotate anion ($Ar-N=N-O^-$). The diazotate anion is completely unreactive as an electrophile.
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Step 1: 1-nitropropane (Primary)
It has two $\alpha$-hydrogens. It reacts with $HNO_2$ to form a Nitrolic acid. Nitrolic acids contain an acidic proton, so they dissolve in $NaOH$ to form a salt with an intense Blood Red color.
Step 2: 2-nitropropane (Secondary)
It has only one $\alpha$-hydrogen. It reacts with $HNO_2$ to form a Pseudonitrole. Pseudonitroles have no acidic protons, so they do not dissolve in $NaOH$, but they inherently possess a Blue color in solution or as an oil.
Step 3: 2-methyl-2-nitropropane (Tertiary)
It is a tertiary nitroalkane and possesses zero $\alpha$-hydrogens. It cannot react with Nitrous acid at all. The solution remains Colorless.
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Step 1: The Mild Reduction
Normally, strong acidic reduction ($Sn/HCl$) reduces nitrobenzene all the way to aniline. However, $Zn / NH_4Cl$ provides a neutral, mild reducing environment.
Step 2: The Intermediate
This mild reduction halts halfway. The nitro group ($-NO_2$) is reduced to a hydroxylamine group ($-NHOH$). The product is N-Phenylhydroxylamine ($C_6H_5-NHOH$).
Step 3: Tollens' Reaction
N-Phenylhydroxylamine is highly susceptible to oxidation. When Tollens' reagent ($Ag^+$) is added, it easily oxidizes the hydroxylamine back to Nitrosobenzene ($C_6H_5-NO$), simultaneously reducing the silver ions to metallic silver (the mirror).
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Step 1: The Acidity Factor
1-butyne is a terminal alkyne. The hydrogen is attached to an $sp$-hybridized carbon. Because $sp$-orbitals have $50\%$ s-character, they hold electrons very tightly to the nucleus, rendering that terminal hydrogen unusually acidic ($pK_a \approx 25$). 2-butyne is internal and completely lacks this acidic proton.
Step 2: The Reaction with Heavy Metals
The basic ammoniacal medium ($OH^-$ from $NH_4OH$) provides enough basicity to facilitate the removal of the acidic proton, while the transition metal cations ($Ag^+$ or $Cu^+$) strongly coordinate with the resulting alkynide anion.
Step 3: Precipitation
This forms Silver butynide ($CH_3CH_2-C \equiv C^- Ag^+$), which is highly insoluble and instantly drops out of solution as a heavy white precipitate. Similarly, Copper(I) butynide precipitates as a red solid.
Conclusion: 2-butyne has no acidic proton, cannot form the alkynide anion, and therefore remains unreactive.
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Step 1: Baeyer's Reagent (KMnO4)
The permanganate ion ($MnO_4^-$) adds to the $\pi$-bond via a concerted, cyclic 5-membered manganate ester intermediate. Because both oxygen atoms are delivered simultaneously from the same massive central Manganese atom, they must attach to the same face of the cyclopentene ring. Subsequent hydrolysis yields a strictly cis-1,2-cyclopentanediol (Syn-addition).
Step 2: Bromine Water ($Br_2 / H_2O$)
Bromine attacks the $\pi$-bond to form a rigid, 3-membered cyclic bromonium ion intermediate. This bulky bridge physically blocks one face of the ring. The nucleophile (water) is forced to execute an $S_N2$-like backside attack from the opposite face. This yields strictly trans-2-bromocyclopentanol (Anti-addition).
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Step 1: Nucleophilic Acyl Substitution
The ester ($R-CO-OR'$) is heated with Hydroxylamine ($NH_2OH$) in an alkaline medium. The basic medium makes the hydroxylamine a potent nucleophile, which attacks the carbonyl carbon, kicking out the alkoxide leaving group ($-OR'$).
The product formed is a Hydroxamic acid ($R-CO-NHOH$).
Step 2: Complexation with Iron(III)
The reaction mixture is acidified, and a few drops of Ferric Chloride ($FeCl_3$) are added. The hydroxamic acid acts as a powerful bidentate ligand. It chelates the $Fe^{3+}$ ion using the carbonyl oxygen and the hydroxylamine oxygen.
Step 3: Color Generation
This chelation forms a highly stable, octahedral Ferric hydroxamate complex. Like many Iron-phenoxide-like complexes, it exhibits intense Ligand-to-Metal Charge Transfer (LMCT) transitions, producing the characteristic deep red-violet/burgundy color.
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Step 1: Structural Analysis
- Pentan-2-one: $CH_3-CO-CH_2-CH_2-CH_3$. It possesses a terminal methyl group directly attached to the carbonyl (a methyl ketone).
- Pentan-3-one: $CH_3-CH_2-CO-CH_2-CH_3$. The carbonyl is flanked by two ethyl groups. It lacks a terminal methyl ketone group.
Step 2: The Iodoform Test
Treat both tubes with Iodine ($I_2$) and aqueous Sodium Hydroxide ($NaOH$).
Pentan-3-one will not react. Pentan-2-one will undergo sequential alpha-halogenation followed by base-catalyzed cleavage.
Step 3: The Equation
$CH_3-CO-CH_2CH_2CH_3 + 3I_2 + 4NaOH \rightarrow \mathbf{CHI_3 \downarrow} + CH_3CH_2CH_2COONa + 3NaI + 3H_2O$.
The product $CHI_3$ is Iodoform, an insoluble, bright yellow precipitate with a distinct antiseptic, hospital-like odor.
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Step 1: The Initial Imine Formation
Both sugars react with the first equivalent of phenylhydrazine. Glucose forms a phenylhydrazone at C1. Fructose forms a phenylhydrazone at C2.
Step 2: Amadori-like Rearrangement/Oxidation
A second equivalent of phenylhydrazine acts as an oxidizing agent. For Glucose, it oxidizes the adjacent C2 hydroxyl group into a ketone. For Fructose, it oxidizes the adjacent C1 primary alcohol into an aldehyde. (The phenylhydrazine is reduced to aniline and ammonia).
Step 3: The Final Condensation
A third equivalent of phenylhydrazine condenses with the newly formed carbonyl group. The result is a molecule with two phenylhydrazone groups located at C1 and C2.
Step 4: The Convergence
Because Glucose and Fructose differ structurally only at C1 and C2 (they have identical stereochemistry from C3 to C6), the Osazone reaction completely wipes out their only structural difference. Thus, they yield the exact same Glucosazone (or Fructosazone) crystals.
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Step 1: Action on Ethylamine hydrochloride
When $NaOH$ is added to the amine salt at room temperature, an immediate acid-base neutralization occurs: $R-NH_3^+Cl^- + NaOH \rightarrow R-NH_2 + NaCl + H_2O$. The free ethylamine is released. It has a strong fishy/ammonia-like odor and will turn damp red litmus paper blue. However, this happens instantly in the cold.
Step 2: Action on Acetamide
Amides are highly stable due to resonance ($O=C-N \leftrightarrow ^-O-C=N^+$). Simply adding cold $NaOH$ does nothing. However, if the solution is boiled vigorously, the strong base forces a nucleophilic acyl substitution (alkaline hydrolysis).
$CH_3CONH_2 + NaOH \xrightarrow{\Delta} CH_3COONa + NH_3 \uparrow$.
Step 3: The Distinguishing Factor
Only the Amide requires prolonged heating/boiling to release Ammonia gas ($NH_3$), which then turns the damp red litmus blue. The amine salt releases its basic gas instantly without any heating required.
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Step 1: Oxidative Decarboxylation
Ninhydrin (which exists as a hydrate of an aromatic vicinal triketone) oxidizes the $\alpha$-amino acid. The amino acid is violently degraded, losing Carbon Dioxide ($CO_2$) and decomposing into an aldehyde, while the nitrogen is liberated as a free molecule of Ammonia ($NH_3$). The Ninhydrin is simultaneously reduced to Hydrindantin.
Step 2: The Condensation Step
The newly liberated Ammonia ($NH_3$) acts as a nucleophile. It attacks a second molecule of unreacted (oxidized) Ninhydrin and the molecule of reduced Hydrindantin.
Step 3: Formation of Ruhemann's Purple
A massive condensation reaction links the two Ninhydrin derivatives together through the single Nitrogen atom provided by the ammonia. The resulting bis-ninhydrin-enamine complex is highly conjugated across multiple aromatic rings, yielding the intense purple color.
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Step 1: Analyze Compound X ($C_3H_8O$)
Reaction with Sodium metal strictly proves it is an Alcohol (contains an acidic $-OH$ group), eliminating the possibility of it being an ether.
A positive Iodoform test for an alcohol strictly requires a secondary methyl carbinol group: $CH_3-CH(OH)-$.
Since it has 3 carbons, the structure must be $CH_3-CH(OH)-CH_3$. Therefore, X is Propan-2-ol (Isopropanol).
Step 2: Analyze Compound Y (Dehydration)
Heating an alcohol with concentrated $H_2SO_4$ triggers an E1 elimination (dehydration).
$CH_3-CH(OH)-CH_3 \xrightarrow{\Delta} CH_3-CH=CH_2 + H_2O$.
The gas Y is Propene.
Step 3: Analyze Compound Z (Baeyer's Test)
Cold, dilute, alkaline $KMnO_4$ is Baeyer's reagent. It tests for unsaturation by performing a syn-dihydroxylation across the double bond of propene. The purple permanganate is reduced to a brown precipitate of $MnO_2$.
$CH_3-CH=CH_2 + [O] + H_2O \rightarrow CH_3-CH(OH)-CH_2(OH)$.
The resulting vicinal diol is a sweet-tasting, viscous liquid due to extensive hydrogen bonding.
Mastering the Analytical Logic
Congratulations on decoding these 25 ultra-challenging problems in Practical Organic Chemistry! In the laboratory, functional group tests are the only way to physically "see" the invisible molecular structures we draw on paper. For JEE Advanced and NEET, mastering these tests means going beyond memorizing the colors. You must understand the pH limitations of the 2,4-DNP test, the steric/protonic requirements of the Hinsberg and Carbylamine tests, and the redox thermodynamics separating Tollens' from Fehling's. Keep sharpening your deductive logic, and visit Chemca.in for more elite organic chemistry masterclasses!
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