Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Qualitative & Quantitative Analysis
From the deep stoichiometry of Kjeldahl back-titrations to the subtle mechanistic anomalies of the Tollens' and Iodoform tests. Uncover the hidden traps of organic identification.
Qualitative and Quantitative Analysis bridges the gap between theoretical structures and actual laboratory data. To conquer these JEE Advanced problems, you must not only memorize color changes but fundamentally understand the inorganic redox mechanisms and titration stoichiometry happening inside the flask.
View Solution
Step 1: The Chemistry of the Extract
During the preparation of Lassaigne's extract, the organic compound is fused with molten Sodium. For a standard organic compound containing Nitrogen, the Carbon and Nitrogen atoms fuse with Sodium to form Sodium Cyanide: $Na + C + N \rightarrow NaCN$.
Step 2: Analyzing the Anomalous Compounds
Neither Hydrazine ($NH_2NH_2$) nor Hydroxylamine ($NH_2OH$) contains any Carbon atoms in their molecular structure.
Step 3: The Mechanistic Failure
Because there is no Carbon available during the sodium fusion process, it is chemically impossible to form the $CN^-$ (cyanide) ion. Instead, these compounds decompose to release nitrogen gas or form sodamide ($NaNH_2$), which does not react with $FeSO_4$ to form the necessary ferrocyanide complex. Consequently, no Prussian Blue ($Fe_4[Fe(CN)_6]_3$) can be formed.
View Solution
Step 1: Nitro and Azo Compounds
In Nitrobenzene ($-NO_2$) and Azobenzene ($-N=N-$), the nitrogen atoms are already in higher oxidation states or bound in stable highly conjugated non-amine linkages. Boiling concentrated sulfuric acid is an oxidizing agent; it cannot effectively reduce these specific nitrogen functional groups into the $-3$ oxidation state required to form the $NH_4^+$ ion. Instead, the nitrogen often escapes as $N_2$ or nitrogen oxides during digestion.
Step 2: Pyridine (Ring Nitrogen)
In Pyridine, the nitrogen atom is an integral part of a highly stable, aromatic, six-membered ring. The aromatic ring is exceptionally resistant to oxidative cleavage by $H_2SO_4$. The ring fails to break apart entirely, meaning the nitrogen remains trapped in the stable heterocyclic structure and is never liberated as Ammonium Sulfate.
View Solution
Step 1: Calculate Pressure of Dry $N_2$
$P_{\text{dry}} = P_{\text{total}} - \text{Aqueous Tension}$
$P_{\text{dry}} = 715 - 15 = 700 \text{ mm Hg}$.
Step 2: Convert Volume to STP
Initial state: $P_1 = 700 \text{ mm}$, $V_1 = 50 \text{ mL}$, $T_1 = 300 \text{ K}$.
STP state: $P_2 = 760 \text{ mm}$, $T_2 = 273 \text{ K}$, $V_2 = ?$
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies \frac{700 \times 50}{300} = \frac{760 \times V_2}{273}$
$V_2 = \frac{700 \times 50 \times 273}{300 \times 760} = \frac{9555000}{228000} \approx 41.9 \text{ mL}$ at STP.
Step 3: Calculate Mass of $N_2$ and Percentage
$22400 \text{ mL}$ of $N_2$ at STP weighs $28 \text{ g}$.
Mass of $N_2$ = $\frac{28}{22400} \times 41.9 \approx 0.0524 \text{ g}$.
$\% \text{ N} = \left(\frac{\text{Mass of } N_2}{\text{Total Mass of Compound}}\right) \times 100$
$\% \text{ N} = \left(\frac{0.0524}{0.30}\right) \times 100 \approx 17.46\%$.
View Solution
Step 1: The Threat of Interfering Anions
If the organic compound contains Nitrogen and Sulfur, the sodium fusion extract will contain Sodium Cyanide ($NaCN$) and Sodium Sulfide ($Na_2S$).
Step 2: The False Positives
If $AgNO_3$ is added directly to this raw extract:
1. The $CN^-$ ions will react to form Silver Cyanide ($AgCN$), which is a highly insoluble white precipitate.
2. The $S^{2-}$ ions will react to form Silver Sulfide ($Ag_2S$), which is a highly insoluble black precipitate.
These precipitates will completely mask or mimic the white precipitate of $AgCl$, leading the chemist to falsely conclude halogens are present even if they are not.
Step 3: The Role of Boiling with $HNO_3$
Boiling the extract with concentrated Nitric Acid (a strong oxidizing acid) completely decomposes and drives off the interfering ions as volatile gases before the silver is added:
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$ (gas escapes)
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$ (gas escapes)
With the interferents destroyed, only the halide ions ($Cl^-$) remain to legitimately precipitate with $AgNO_3$.
View Solution
Step 1: The Fusion Chemistry
When Nitrogen and Sulfur are present together in the right ratios, they fuse with Carbon and Sodium to form Sodium Thiocyanate ($NaSCN$) instead of forming separate $NaCN$ and $Na_2S$ species:
$Na + C + N + S \rightarrow NaSCN$.
Step 2: Failure of Prussian Blue
Because the Nitrogen is locked up entirely in the thiocyanate ion ($SCN^-$), there are no free Cyanide ions ($CN^-$) available in the extract. Without free $CN^-$, the formation of the ferrocyanide complex required for Prussian Blue is impossible.
Step 3: Formation of the Blood Red Complex
When Ferric Chloride ($FeCl_3$) is added to the extract containing $SCN^-$, the $Fe^{3+}$ ions form a highly stable, intensely blood-red coordination complex with the thiocyanate ligands.
$Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}$
View Solution
Step 1: Total Acid Equivalents
$H_2SO_4$ is dibasic, so its n-factor = 2.
meq of $H_2SO_4$ taken = Molarity $\times$ n-factor $\times$ Volume(mL) = $0.5 \times 2 \times 50 = 50 \text{ meq}$.
Step 2: Excess Acid Neutralized
$NaOH$ is a monoacidic base, n-factor = 1.
meq of $NaOH$ used = $0.5 \times 1 \times 60 = 30 \text{ meq}$.
This means $30 \text{ meq}$ of $H_2SO_4$ was left unreacted.
Step 3: Acid Consumed by Ammonia
meq of acid reacting with $NH_3$ = Total - Excess = $50 - 30 = 20 \text{ meq}$.
Therefore, meq of $NH_3$ evolved = $20 \text{ meq}$.
Since $NH_3$ has an n-factor of 1, millimoles of $NH_3$ = $20 \text{ mmol}$.
Because each $NH_3$ contains one N atom, millimoles of Nitrogen = $20 \text{ mmol}$.
Step 4: Mass and Percentage Calculation
Mass of N = $20 \times 10^{-3} \text{ mol} \times 14 \text{ g/mol} = 0.28 \text{ g}$.
$\% \text{ N} = \left(\frac{0.28}{0.50}\right) \times 100 = 56\%$.
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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 acidic. 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. It does not dissolve in $NaOH$ to form a salt, but in its pure state or in solution, pseudonitroles inherently possess a deep Blue color.
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Step 1: The Role of Anhydrous $ZnCl_2$
The $-OH$ group is a terrible leaving group. Anhydrous $ZnCl_2$ is a powerful Lewis Acid. It coordinates with the lone pairs on the oxygen atom of the alcohol, drawing electron density away and converting the $-OH$ into an excellent leaving group ($[ZnCl_2OH]^-$), which departs to leave behind a carbocation.
Step 2: Standard Kinetics ($S_N1$)
Tertiary carbocations are highly stable, forming instantly (instant turbidity from insoluble alkyl chloride). Secondary carbocations take $\approx 5$ minutes. Primary carbocations are incredibly unstable, so the $S_N1$ reaction effectively does not occur at room temperature for standard alkanols.
Step 3: The Allylic Anomaly
Allyl alcohol, despite being primary, forms the Allyl carbocation ($CH_2=CH-CH_2^+$) upon leaving group departure. This carbocation is massively stabilized by resonance (conjugation with the adjacent $\pi$-bond), making its stability comparable to or greater than a tertiary carbocation. Therefore, it undergoes rapid $S_N1$ substitution, causing instant turbidity.
(A) 1-Phenyl-1-ethanol
(B) Acetic Acid ($CH_3COOH$)
(C) Acetyl Chloride ($CH_3COCl$)
(D) Lactic Acid ($CH_3-CH(OH)-COOH$)
View Solution
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.
View Solution
Step 1: Formic Acid vs Benzoic Acid
If you draw the structure of Formic Acid ($H-C(=O)-OH$), you will notice that one side of the molecule actually resembles an aldehyde group ($H-C=O$). Because of this unique $C-H$ bond on the carbonyl, Tollens' reagent can oxidize formic acid further into Carbonic Acid (which decomposes to $CO_2$ and $H_2O$), depositing a silver mirror. Benzoic acid ($Ph-COOH$) lacks this aldehydic hydrogen and cannot be oxidized.
Step 2: Fructose vs Sucrose
Fructose is a ketose sugar. Normally, ketones do not respond to Tollens'. However, fructose is an alpha-hydroxy ketone. In the basic medium of Tollens' reagent, it undergoes Lobry de Bruyn-van Ekenstein transformation (isomerization via an enediol intermediate) to form an equilibrium mixture containing Glucose and Mannose (which are aldoses). These aldoses then react to give a silver mirror. Sucrose is a non-reducing sugar because its anomeric carbons are locked in a glycosidic bond, preventing it from opening into a reactive open-chain aldehyde/ketone form.
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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 primary amine nitrogen attacks the empty p-orbital of the dichlorocarbene, forming a zwitterionic intermediate: $R-N^+H_2-C^-Cl_2$.
Step 3: The Steric/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.
View Solution
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, slowly hydrolyzing the reagent to sodium benzenesulfonate, which is water-soluble (hence the clear solution, but no reaction product).
View Solution
Step 1: Nitrosation
$NaNO_2$ and $H_2SO_4$ generate nitrous acid ($HNO_2$), which produces the nitrosonium electrophile ($NO^+$). Phenol undergoes electrophilic aromatic substitution at the para-position to form p-nitrosophenol.
Step 2: Tautomerization and Condensation
p-Nitrosophenol tautomerizes into the quinone monoxime. In the presence of concentrated acid, this condenses with a second, unreacted molecule of Phenol. This condensation yields an indophenol derivative (a protonated oxonium sulfate salt) which exhibits the initial deep green/blue/red colors depending on hydration/protonation.
Step 3: Alkalization (The Deep Blue Salt)
Upon adding excess $NaOH$, the acidic phenolic proton of the indophenol derivative is removed. The resulting highly conjugated anion is the Sodium salt of Indophenol. The extreme extended conjugation spanning two benzene rings and the central nitrogen atom produces an intense, deep blue color.
View Solution
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), electron density is donated into the carbonyl carbon. This makes the carbonyl carbon vastly less electrophilic and highly stabilizes the molecule against oxidation.
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.
View Solution
Step 1: Molar Mass of Precipitate
Molar mass of $BaSO_4$ = $137 (Ba) + 32 (S) + 4 \times 16 (O) = 137 + 32 + 64 = 233 \text{ g/mol}$.
Step 2: Calculate Mass of Sulfur
Since $233 \text{ g}$ of $BaSO_4$ contains exactly $32 \text{ g}$ of Sulfur:
Mass of Sulfur in the $0.233 \text{ g}$ precipitate = $\left( \frac{32}{233} \right) \times 0.233 = 0.032 \text{ g}$.
Step 3: Calculate Mass Percentage
$\% \text{ S} = \left(\frac{\text{Mass of Sulfur}}{\text{Mass of Organic Compound}}\right) \times 100$
$\% \text{ S} = \left(\frac{0.032}{0.156}\right) \times 100 \approx 20.51\%$.
View Solution
Step 1: Extreme Activation
In aqueous solution (Bromine water), phenol ionizes slightly to the phenoxide ion ($-O^-$), which is one of the most powerful activating groups known. The ortho and para positions possess immense electron density.
Step 2: Electrophilic Attack on Salicylic Acid (Ipso Attack)
Salicylic acid has a $-COOH$ group at the ortho position. When the highly reactive $Br^+$ electrophile attacks this specific ortho position (an "ipso" attack), a highly crowded intermediate Wheland complex (sigma complex) forms.
Step 3: Elimination (Decarboxylation)
Normally, EAS restores aromaticity by losing a proton ($H^+$). However, at the ipso position bearing the $-COOH$ group, the intermediate can restore aromaticity by expelling the massive carboxyl group entirely as Carbon Dioxide ($CO_2$) gas. This process is known as ipso-substitution or halodecarboxylation.
Step 4: Completion
Once the $-COOH$ group departs, the remaining ortho/para positions are rapidly brominated as usual, leading directly to 2,4,6-tribromophenol.
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Step 1: Reagent Functions
- Anhydrous $CaCl_2$: A powerful desiccant. It absorbs only moisture ($H_2O$) and ignores $CO_2$.
- Concentrated $KOH$: A strong base. It absorbs acidic $CO_2$ gas via the reaction $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$.
Step 2: The Critical Ordering
The gases must pass through $CaCl_2$ first. The $CaCl_2$ absorbs all the water generated by the combustion. The remaining dry $CO_2$ then passes into the $KOH$ tube, where it is exclusively absorbed.
Step 3: The Reversal Catastrophe
If the gases hit the $KOH$ tube first, a fatal error occurs. $KOH$ pellets are highly hygroscopic (they absorb water) AND they chemically absorb $CO_2$. Furthermore, the reaction of $KOH$ with $CO_2$ actually produces water ($H_2O$) as a byproduct! Thus, the $KOH$ tube would absorb both the original water from the combustion and the $CO_2$, rendering it impossible to distinguish how much mass came from C versus H. The subsequent $CaCl_2$ tube would be useless.
View Solution
Step 1: The Chemical Reaction
$R-OCH_3 + HI \rightarrow R-OH + CH_3I \uparrow$
The volatile Methyl Iodide is distilled off and reacts with $Ag^+$:
$CH_3I + AgNO_3 \rightarrow AgI \downarrow + CH_3NO_3$.
Every 1 mole of $AgI$ precipitate corresponds exactly to 1 mole of Methoxy ($-OCH_3$) groups originally present.
Step 2: Calculate Moles of $AgI$
Moles of $AgI = \frac{0.470 \text{ g}}{235 \text{ g/mol}} = 0.002 \text{ mol}$.
Therefore, moles of $-OCH_3$ groups = $0.002 \text{ mol}$.
Step 3: Calculate Mass and Percentage of Methoxy
Molar mass of $-OCH_3$ = $12 + 3 + 16 = 31 \text{ g/mol}$.
Mass of methoxy groups = $0.002 \text{ mol} \times 31 \text{ g/mol} = 0.062 \text{ g}$.
Percentage = $\left(\frac{0.062}{0.268}\right) \times 100 \approx 23.13\%$.
View Solution
Step 1: Identifying the Amine Class
The reaction described is the Hofmann Mustard Oil test. This test is given exclusively by Primary ($1^{\circ}$) Amines. Secondary and tertiary amines do not possess the two 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. Since the formula is $C_3H_9N$, the alkyl group must be propyl (or isopropyl).
Step 4: The Byproduct
The sulfur stripped from the intermediate forms Mercuric Sulfide ($HgS$), which deposits as a distinct black precipitate.
View Solution
Step 1: Analysis of the Reactant ($Na_2[Fe(CN)_5NO]$)
The complex anion is $[Fe(CN)_5NO]^{2-}$.
In Sodium Nitroprusside, the Nitroso ligand is notoriously present as the Nitrosonium cation ($NO^+$), donating three electrons.
Let oxidation state of Fe be $x$:
$x + 5(-1) + (+1) = -2 \implies x - 4 = -2 \implies x = +2$.
The Iron is in the $+2$ state (Low spin $d^6$, diamagnetic).
Step 2: Analysis of the Violet Complex
The sulfide ion ($S^{2-}$) from the extract acts as a nucleophile, directly attacking the highly electrophilic Nitrogen atom of the $NO^+$ ligand (NOT displacing a cyanide).
The new ligand formed is the thionitrosyl group ($NOS^-$).
The new complex is $Na_4[Fe(CN)_5(NOS)]$. The anion is $[Fe(CN)_5(NOS)]^{4-}$.
Step 3: Final Oxidation State
Let new oxidation state of Fe be $y$:
$y + 5(-1) + (-1) = -4 \implies y - 6 = -4 \implies y = +2$.
Step 4: The Anomaly
The deep color change suggests a massive electronic shift, usually indicating a redox change on the central metal. However, anomalously, the oxidation state of Iron remains strictly $+2$ throughout the entire intensely colored transformation!
View Solution
Step 1: Write the balanced reaction
$CH_3-CO-CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + CH_3COONa + 3NaI + 3H_2O$.
1 mole of acetone yields 1 mole of Iodoform.
Step 2: Calculate Moles of Reactant
Moles of acetone = $5.8 \text{ g} / 58 \text{ g mol}^{-1} = 0.1 \text{ moles}$.
Step 3: Calculate Theoretical Mass of Product
Molar mass of Iodoform ($CHI_3$) = $12(C) + 1(H) + 3 \times 127(I) = 13 + 381 = 394 \text{ g/mol}$.
Theoretical yield = $0.1 \text{ moles} \times 394 \text{ g/mol} = 39.4 \text{ g}$.
Step 4: Apply Percentage Yield
Actual yield = $39.4 \text{ g} \times 0.75 = 29.55 \text{ g}$.
View Solution
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 (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.
View Solution
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 to an aldehyde/ketone, reducing itself to colorless Cerium(III), and the red color fades. The test relies on observing the transient coordination intermediate.
View Solution
Step 1: Decode 2,4-DNP and Tollens'
- Orange ppt with 2,4-DNP means A is a carbonyl compound (aldehyde or ketone).
- Failure with Tollens' reagent proves A is strictly a Ketone (not an aldehyde).
Step 2: Decode Iodoform Test
- Yellow ppt with $I_2/NaOH$ (Iodoform test) proves the ketone possesses an active terminal methyl group adjacent to the carbonyl: meaning it is a Methyl Ketone ($R-CO-CH_3$).
Step 3: Reconcile with Molecular Formula
Formula is $C_8H_8O$. High carbon-to-hydrogen ratio strongly indicates a benzene ring ($C_6H_5-$) is present. Subtracting the ring ($C_6H_5 = C_6H_5$) from the formula leaves $C_2H_3O$, which matches perfectly with an acetyl group ($-CO-CH_3$).
Therefore, compound A is Acetophenone (Phenyl methyl ketone, $C_6H_5-CO-CH_3$).
Step 4: Oxidation to B
Vigorous oxidation of alkyl-substituted benzenes (like acetophenone) with hot $KMnO_4$ cleaves the alkyl side chain entirely at the benzylic position, leaving behind a Carboxyl group attached to the ring.
Therefore, compound B is Benzoic Acid ($C_6H_5COOH$). As a carboxylic acid, it readily effervesces with $NaHCO_3$ to release $CO_2$.
View Solution
Step 1: Analyze Hinsberg Test (Compound Y)
X reacts with benzenesulfonyl chloride (Hinsberg reagent), meaning it must be a $1^{\circ}$ or $2^{\circ}$ amine. The resulting sulfonamide Y is insoluble in $NaOH$. This absolutely proves that X is a Secondary ($2^{\circ}$) Amine (the product lacks the acidic N-H proton needed to form a soluble salt).
Step 2: Analyze Nitrous Acid Test (Compound Z)
Reaction with $HNO_2$ at low temp yields a yellow oil with NO nitrogen gas evolution. This confirms it is a secondary amine forming an N-Nitrosamine derivative ($R_2N-N=O$, which are notoriously toxic yellow oils). A primary amine would have evolved $N_2$ gas.
Step 3: Deduce Structure from Formula
Formula is $C_7H_9N$. A high C/H ratio implies a benzene ring ($C_6H_5-$). This leaves $1$ Carbon, $4$ Hydrogens, and $1$ Nitrogen. Since it must be a secondary amine, the nitrogen must be between the benzene ring and the remaining methyl group ($CH_3$).
Therefore, X is N-Methylaniline ($C_6H_5-NH-CH_3$).
Step 4: Identify Y and Z
Y is N-Methyl-N-phenylbenzenesulfonamide ($Ph-SO_2-N(CH_3)Ph$).
Z is N-Methyl-N-nitrosoaniline ($Ph-N(NO)-CH_3$).
Mastering the Laboratory in Theory
Congratulations on decoding these 25 ultra-challenging problems in Qualitative and Quantitative Analysis! Memorizing that "Tollens gives a silver mirror" is simply not enough for JEE Advanced. You must understand the why. You must trace the electron flow in the Lucas test, recognize the structural impossibilities in Kjeldahl digestion, and master the exact stoichiometric ratios of the Dumas method. Keep fusing your physical stoichiometry skills with your organic reaction mechanisms, and visit Chemca.in for more top-tier masterclasses!
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