Chemca
Master Organic Chemistry
20 Advanced Word Problems
Dive into the intricate world of organic synthesis, deductive reasoning, and chemical testing. These 20 highly detailed problems cover distinction tests (Tollens, Iodoform, Hinsberg), isomerism, and reaction mechanisms. Click any problem to reveal an exhaustive, step-by-step logical deduction accompanied by professional chemical typesetting.
Q1. An alkene A ($\ce{C5H10}$) on ozonolysis gives a mixture of two compounds B and C. Compound B gives a positive Tollens' test but a negative Iodoform test. Compound C gives a positive Iodoform test but a negative Tollens' test. Identify A, B, and C with logical reasoning.
For compound A with the molecular formula $\ce{C5H10}$, the Degree of Unsaturation (DU) is calculated as: $DU = C + 1 - \frac{H}{2} = 5 + 1 - \frac{10}{2} = 1$. Since it undergoes ozonolysis, this single degree of unsaturation corresponds exactly to one carbon-carbon double bond ($\ce{C=C}$). A is an alkene.
Compound B gives a positive Tollens' test (silver mirror formation). Tollens' reagent ($\ce{[Ag(NH3)2]+}$) selectively oxidizes aldehydes, meaning B must be an aldehyde. However, B gives a negative Iodoform test. The Iodoform test requires a methyl ketone group ($\ce{CH3-CO-}$). The only aldehyde that contains this group is acetaldehyde ($\ce{CH3CHO}$). Since B is negative for Iodoform, B cannot be acetaldehyde.
Compound C gives a negative Tollens' test, meaning it is not an aldehyde (it is a ketone). It gives a positive Iodoform test, meaning it must be a methyl ketone.
Since the total carbon count of A is 5, the carbons in B + C = 5.
If C is the simplest methyl ketone, Propanone ($\ce{CH3-CO-CH3}$, 3 carbons), then B must have $5 - 3 = 2$ carbons. If B has 2 carbons and is an aldehyde, it would be Ethanal ($\ce{CH3CHO}$). But we already established B cannot be Ethanal (because B fails the iodoform test).
Therefore, let's reverse the carbon counts. Let B have 3 carbons and C have 2 carbons. A 3-carbon aldehyde is Propanal ($\ce{CH3CH2CHO}$). Propanal gives a positive Tollens' and negative Iodoform test. This perfectly fits B!
Wait, if C has 2 carbons and is a ketone... a 2-carbon ketone doesn't exist! Let's re-evaluate. What if B has 1 carbon? A 1-carbon aldehyde is Methanal (Formaldehyde, $\ce{HCHO}$). Methanal gives positive Tollens' and negative Iodoform. This fits!
If B is Methanal (1C), then C must have 4 carbons. A 4-carbon methyl ketone is Butan-2-one ($\ce{CH3-CO-CH2-CH3}$). Butan-2-one gives a positive Iodoform test and negative Tollens' test. This is the perfect logical fit.
We connect the carbonyl carbons of B ($\ce{HCHO}$) and C ($\ce{CH3-CO-CH2-CH3}$) to form the original double bond. Removing the oxygen atoms gives: $\ce{CH2=C(CH3)(CH2CH3)}$. The IUPAC name is 2-Methylbut-1-ene.
Q2. An organic compound A ($\ce{C4H10O}$) reacts with Lucas reagent causing turbidity within 5 minutes. On oxidation with acidified $\ce{K2Cr2O7}$, it gives compound B ($\ce{C4H8O}$). Compound B gives a yellow precipitate with 2,4-DNP but does not reduce Tollens' reagent. Identify A and B.
The formula $\ce{C4H10O}$ ($\ce{C_nH_{2n+2}O}$) suggests a saturated alcohol or ether. The reaction with Lucas reagent (anhydrous $\ce{ZnCl2}$ + conc. $\ce{HCl}$) confirms it is an alcohol. The appearance of turbidity (due to the formation of insoluble alkyl chloride) within 5 minutes is the classical diagnostic observation for a Secondary ($2^\circ$) Alcohol. (Tertiary alcohols react immediately; primary alcohols do not react at room temp).
The only secondary alcohol containing four carbon atoms is Butan-2-ol ($\ce{CH3-CH(OH)-CH2-CH3}$).
Oxidation of a secondary alcohol (Butan-2-ol) with a strong oxidizing agent like acidified potassium dichromate ($\ce{K2Cr2O7 / H+}$) yields a ketone. The resulting ketone is Butan-2-one ($\ce{CH3-CO-CH2-CH3}$), which perfectly matches the molecular formula $\ce{C4H8O}$.
Compound B (Butan-2-one) is a ketone. It reacts with 2,4-Dinitrophenylhydrazine (Brady's reagent) to form a yellow/orange 2,4-dinitrophenylhydrazone precipitate, proving the presence of a carbonyl group. Crucially, as a ketone, it resists further mild oxidation and thus fails to reduce Tollens' reagent. This confirms B is a ketone and not an aldehyde.
Q3. An amine A ($\ce{C3H9N}$) reacts with Hinsberg reagent to form a solid product completely insoluble in aqueous KOH. When A reacts with nitrous acid, it yields an oily yellow liquid. Identify A and detail the mechanisms.
The molecular formula $\ce{C3H9N}$ points to an aliphatic amine. The Hinsberg test (using benzene sulphonyl chloride, $\ce{C6H5SO2Cl}$) differentiates amine classes:
- $1^\circ$ amines form sulphonamides with an acidic proton on nitrogen, making them soluble in alkali ($\ce{KOH}$).
- $2^\circ$ amines form N,N-dialkylbenzene sulphonamides lacking an acidic proton, making them insoluble in alkali.
- $3^\circ$ amines do not react.
Since the product is insoluble in aqueous $\ce{KOH}$, Compound A is definitively a Secondary Amine.
Secondary amines react with nitrous acid ($\ce{HNO2}$, generated from $\ce{NaNO2 + HCl}$) to form N-nitrosamines. These N-nitrosamines characteristically separate out of the aqueous mixture as yellow oily liquids. This visual confirmation solidifies that A is a secondary amine.
A secondary amine containing exactly 3 carbon atoms must have one methyl group and one ethyl group attached to the nitrogen atom. Therefore, the structure is $\ce{CH3-NH-CH2CH3}$. The IUPAC name is N-Methylethanamine.
$$ \text{Nitrous Acid Reaction:} $$ $$ \ce{CH3-NH-C2H5 + HNO2 -> CH3-N(NO)-C2H5 + H2O} $$ $$ \text{(N-Nitroso-N-methylethanamine - Yellow oil)} $$
Q4. An alkyl halide A ($\ce{C4H9Br}$) reacts with alcoholic KOH to give a major product B. Compound B on reductive ozonolysis gives only one organic compound, Ethanal. Deduce structures A and B, and state the governing elimination rule.
Compound B (an alkene) undergoes reductive ozonolysis to yield only Ethanal ($\ce{CH3CHO}$). Because only a single type of product is formed, the parent alkene must be perfectly symmetrical. Joining two molecules of Ethanal at their carbonyl carbons creates the double bond: $\ce{CH3-CH=O} + \ce{O=CH-CH3} \rightarrow \ce{CH3-CH=CH-CH3}$.
Therefore, Compound B is But-2-ene.
Compound B (But-2-ene) is the major product formed by the dehydrohalogenation of alkyl halide A ($\ce{C4H9Br}$) using alcoholic KOH (a strong base facilitating E2 elimination).
If A were 1-Bromobutane, the only possible elimination product would be But-1-ene. For But-2-ene to be the major product, the bromine atom must be located on the second carbon. Thus, A must be 2-Bromobutane ($\ce{CH3-CHBr-CH2-CH3}$).
The reaction proceeds via an E2 (Bimolecular Elimination) mechanism. The formation of the disubstituted But-2-ene as the major product over the monosubstituted But-1-ene is dictated by Saytzeff's Rule (Zaitsev's Rule). This rule states that in $\beta$-elimination reactions, the highly substituted alkene is formed as the major product because it is thermodynamically more stable due to hyperconjugation.
$$ \text{Ozonolysis:} $$ $$ \ce{CH3-CH=CH-CH3 ->[1. O3][2. Zn/H2O] 2CH3CHO} \quad \text{(Ethanal)} $$
Q5. An aromatic compound A ($\ce{C7H8O}$) is sparingly soluble in water but highly soluble in aqueous NaOH. It gives a violet colouration with neutral $\ce{FeCl3}$. On treatment with bromine water, it yields a tribromo derivative. Identify A.
The molecular formula $\ce{C7H8O}$ has a Degree of Unsaturation of 4, corresponding to a benzene ring. Its solubility in aqueous $\ce{NaOH}$ indicates it is an acidic compound, but it is not a carboxylic acid (only one oxygen). This points to a Phenol.
The characteristic violet color with neutral $\ce{FeCl3}$ is the definitive confirmatory test for the phenolic $-\ce{OH}$ group, which forms a colored iron complex.
Subtracting the phenol core ($\ce{C6H5OH}$) from the formula ($\ce{C7H8O}$) leaves one $-\ce{CH3}$ group. The compound is a methylphenol (cresol). The three position isomers are o-Cresol, m-Cresol, and p-Cresol. (Anisole is eliminated as it doesn't dissolve in $\ce{NaOH}$ or react with $\ce{FeCl3}$).
The problem states A yields a tribromo derivative with aqueous bromine. The $-\ce{OH}$ group is strongly activating and ortho/para directing.
- In o-Cresol, one ortho position is blocked by the methyl group, leading to a dibromo product.
- In p-Cresol, the para position is blocked, also leading to a dibromo product.
- In m-Cresol (3-Methylphenol), both ortho positions and the para position (relative to the $-\ce{OH}$) are free and unoccupied. Because the weak activation of the methyl group aligns cooperatively with the strong activation of the $-\ce{OH}$ group, bromination rapidly occurs at positions 2, 4, and 6 to form 2,4,6-tribromo-3-methylphenol.
Therefore, Compound A is m-Cresol.
Q6. An aromatic compound A ($\ce{C7H6O}$) reduces Tollens' reagent but does not undergo Aldol condensation. On heating with 50% NaOH, it forms compounds B and C. B on acidification gives a solid D ($\ce{C7H6O2}$). Identify A, B, C, D and the reaction.
Reduction of Tollens' reagent to a silver mirror proves A is an aldehyde. The formula $\ce{C7H6O}$ (DU = 5) indicates an aromatic ring. Thus, Compound A is Benzaldehyde ($\ce{C6H5CHO}$). Benzaldehyde lacks any $\alpha$-hydrogen atoms (the carbon adjacent to the carbonyl group is part of the aromatic ring and has no hydrogens), which perfectly explains its inability to undergo Aldol condensation.
Aldehydes lacking $\alpha$-hydrogens, when heated with concentrated alkali (50% $\ce{NaOH}$), undergo a classic disproportionation redox process known as the Cannizzaro reaction. One molecule is oxidized to the sodium salt of a carboxylic acid, and another is reduced to a primary alcohol.
- Reduction product (C): Benzyl alcohol ($\ce{C6H5CH2OH}$)
- Oxidation product (B): Sodium benzoate ($\ce{C6H5COONa}$)
Compound B is a water-soluble salt. Acidification with mineral acid ($\ce{HCl}$) protonates the carboxylate ion, causing the precipitation of a white crystalline solid, Compound D. This solid is Benzoic Acid ($\ce{C6H5COOH}$), matching the final formula $\ce{C7H6O2}$.
Q7. An organic acid A ($\ce{C3H6O2}$) is treated with $\ce{NH3}$ and heated to form B. Compound B with $\ce{Br2}$ and alcoholic KOH produces C ($\ce{C2H7N}$). C reacts with $\ce{CHCl3}$ and ethanolic KOH to form intensely foul-smelling D. Identify A, B, C, D.
The molecular formula $\ce{C3H6O2}$ conforms to $\ce{C_nH_{2n}O2}$, indicating an aliphatic carboxylic acid. With 3 carbons, Compound A is Propanoic acid ($\ce{CH3CH2COOH}$).
Reaction of a carboxylic acid with ammonia first yields an ammonium salt ($\ce{CH3CH2COONH4}$). Strong heating dehydrates the salt to form a primary amide. Compound B is Propanamide ($\ce{CH3CH2CONH2}$).
The reagents $\ce{Br2}$ + alcoholic $\ce{KOH}$ act on amides to perform the Hoffmann bromamide degradation. This mechanism features the migration of the alkyl group from the carbonyl carbon to the nitrogen atom, followed by the expulsion of the carbonyl group as carbonate. The result is a primary amine with one carbon less than the parent amide. Propanamide (3 carbons) yields Ethanamine ($\ce{CH3CH2NH2}$, 2 carbons), which perfectly matches the formula $\ce{C2H7N}$. Compound C is Ethanamine.
Primary amines (like Ethanamine) react with chloroform ($\ce{CHCl3}$) in the presence of strong alkali (ethanolic $\ce{KOH}$) to form isocyanides. This is the Carbylamine test. The product, Compound D, is Ethyl isocyanide ($\ce{CH3CH2NC}$), famous for its intolerable, foul odor.
Q8. An alkene A ($\ce{C6H12}$) exhibits geometrical isomerism. Upon catalytic hydrogenation, it yields n-hexane. Ozonolysis of A yields two different aldehydes, B and C. Identify A, B, and C with stereochemical reasoning.
Hydrogenation of A yields n-hexane, proving that A has a straight, unbranched 6-carbon chain.
Ozonolysis cleaves the double bond to form aldehydes.
- If A is Hex-1-ene, products are Methanal (1C) + Pentanal (5C).
- If A is Hex-2-ene, products are Ethanal (2C) + Butanal (4C).
- If A is Hex-3-ene, products are two molecules of Propanal (3C).
The problem states A yields two different aldehydes, eliminating Hex-3-ene.
Compound A must exhibit cis-trans (geometrical) isomerism. This requires each carbon of the double bond to be attached to two different groups.
- Hex-1-ene ($\ce{CH2=CH-R}$) has two identical hydrogen atoms on Carbon-1. It cannot show geometrical isomerism.
- Hex-2-ene ($\ce{CH3-CH=CH-CH2CH2CH3}$) has one H and one methyl on C-2, and one H and one propyl on C-3. It does show geometrical isomerism.
Conclusion: Compound A is Hex-2-ene. Compounds B and C are Ethanal and Butanal.
Q9. Compound A ($\ce{C4H10O}$) does not react with sodium metal. On heating with excess HI, it yields two moles of an alkyl iodide B. B reacts with aqueous KOH to produce C, which gives a positive iodoform test. Identify A, B, and C.
The formula $\ce{C4H10O}$ is either an alcohol or an ether. The failure to react with sodium metal (no effervescence of $\ce{H2}$ gas) proves the absence of an acidic hydroxyl proton. Thus, A is an Ether.
Heating with excess $\ce{HI}$ cleaves the ether linkage to form alkyl iodides. Producing exactly two moles of the same alkyl iodide (B) indicates the ether is symmetrical. A symmetrical 4-carbon ether is Diethyl ether (Ethoxyethane, $\ce{CH3CH2-O-CH2CH3}$).
Consequently, Compound B is Iodoethane ($\ce{CH3CH2I}$).
Iodoethane (B) reacts with aqueous $\ce{KOH}$ via an $S_N2$ mechanism, substituting the iodine for a hydroxyl group. This forms Compound C, Ethanol ($\ce{CH3CH2OH}$).
Does Ethanol give a positive iodoform test? Yes. The mild oxidizing environment of the iodoform reagent ($\ce{NaOI}$) oxidizes ethanol to ethanal ($\ce{CH3CHO}$), which possesses the requisite methyl ketone group, yielding the yellow $\ce{CHI3}$ precipitate.
Q10. A biomolecule A ($\ce{C6H12O6}$) reduces Fehling's solution. Prolonged heating with HI forms n-hexane. Oxidation with bromine water gives a monocarboxylic acid; with nitric acid, a dicarboxylic acid (saccharic acid). Map its functional groups.
1. Chain Structure: Reduction with powerful $\ce{HI}$ to n-hexane proves all 6 carbons are arranged in a straight, unbranched linear chain.
2. Aldehyde Group: $\ce{Br2}$ water is a mild oxidant that oxidizes aldehydes but not ketones. Formation of a monocarboxylic acid (gluconic acid) proves the carbonyl group is an aldehyde at C-1.
3. Primary Alcohol: Nitric acid ($\ce{HNO3}$) is a strong oxidant that oxidizes both the C-1 aldehyde and the terminal primary alcohol into carboxyl groups. The resulting dicarboxylic acid (saccharic acid) proves a primary alcohol ($\ce{-CH2OH}$) exists at C-6.
Conclusion: Compound A is an aldohexose, specifically D-Glucose.
Q11. An aromatic primary amine A undergoes diazotization at 273-278 K to form B. Compound B reacts with phenol in a mildly alkaline medium to form an intensely coloured orange dye C. Identify A, B, C and the reaction mechanism.
Diazotization: Compound A is Aniline ($\ce{C6H5NH2}$). At ice-cold temperatures (0-5°C), it reacts with nitrous acid ($\ce{NaNO2 + HCl}$) to form a diazonium salt, Benzene diazonium chloride (Compound B, $\ce{C6H5N2+Cl-}$). Maintaining low temperature is critical as diazonium salts are highly unstable and decompose to phenol at higher temperatures.
Coupling Reaction: The diazonium ion is a weak electrophile. In mildly alkaline medium, phenol is converted to the highly activated phenoxide ion. The electrophilic nitrogen attacks the para-position of the phenol ring (Electrophilic Aromatic Substitution), creating an extended conjugated azo system. Compound C is p-Hydroxyazobenzene, an orange dye.
Q12. An optically active alkyl halide A ($\ce{C4H9Cl}$) reacts with aqueous KOH. Kinetics show a mix of $S_N1$ and $S_N2$ pathways, yielding a racemic mixture and inverted product B. Name A and explain the stereochemistry.
Identification: For the halide $\ce{C4H9Cl}$ to be optically active, it must possess a chiral carbon. 2-Chlorobutane is the only isomer where the central carbon is attached to four distinct groups (H, Cl, Methyl, Ethyl).
Mechanistic Duality: As a secondary halide, it undergoes hydrolysis via both $S_N1$ and $S_N2$ mechanisms. The $S_N1$ pathway involves a planar secondary carbocation intermediate, leading to attack from both faces and subsequent racemization. The concurrent $S_N2$ pathway involves concerted backside attack by the nucleophile, leading exclusively to Walden inversion. The net result is a partially racemized mixture with an excess of the inverted Butan-2-ol (Compound B).
Q13. A hydrocarbon A ($\ce{C4H6}$) gives a white precipitate with Tollens' reagent. On hydration with dilute $\ce{H2SO4}$ and $\ce{HgSO4}$, it gives compound B ($\ce{C4H8O}$) which shows positive iodoform test. Identify A and B.
Terminal Alkyne Test: The reaction with ammoniacal silver nitrate (Tollens') to form a white precipitate indicates an acidic terminal alkyne hydrogen ($\ce{\equiv C-H}$). Thus, A is But-1-yne ($\ce{CH3CH2C\equiv CH}$).
Kucherov Hydration: Hydration catalyzed by $\ce{Hg^2+}$ follows Markovnikov's rule. The $\ce{OH}$ adds to the more substituted internal carbon, forming an unstable enol which rapidly tautomerizes into a ketone. The product, Compound B, is Butan-2-one ($\ce{CH3-CO-CH2-CH3}$), which confirms the final clue by giving a positive Iodoform test.
Q14. Alkyl halide A ($\ce{C3H7Br}$) reacts with Mg in dry ether to form B. Compound B reacts with dry ice followed by acid hydrolysis to form C. C turns blue litmus red. Identify A, B, C.
Grignard Synthesis: Reaction with Mg in dry ether forms a Grignard reagent (B). $\ce{C3H7Br + Mg -> C3H7MgBr}$.
Carboxylation: Grignard reagents act as strong nucleophiles, attacking the electrophilic carbon in carbon dioxide ($\ce{O=C=O}$). Acid hydrolysis yields a carboxylic acid (C) with one extra carbon atom. Thus, C is a 4-carbon acid (turns litmus red).
Structures: If A is 1-Bromopropane, C is Butanoic acid. If A is 2-Bromopropane, C is 2-Methylpropanoic acid. Both pathways perfectly satisfy the chemical logic of the problem.
Q15. How would you distinguish chemically between Aniline and N-methylaniline in the laboratory? Detail the mechanism of selectivity.
The Carbylamine Test: This is the definitive test. Aniline is a primary ($1^\circ$) aromatic amine. N-methylaniline is a secondary ($2^\circ$) amine.
Observation: Heating the compounds with chloroform ($\ce{CHCl3}$) and ethanolic $\ce{KOH}$ produces an intolerably foul odor (phenyl isocyanide) for Aniline, while N-methylaniline produces no such odor.
Mechanistic Selectivity: The reaction involves the generation of a highly reactive dichlorocarbene ($:\ce{CCl2}$) intermediate. To successfully form the terminal triple bond of the isocyanide ($-\ce{N#C}$), the amine nitrogen must lose two protons during the mechanism. Primary amines have two protons on nitrogen; secondary amines only have one, causing the reaction to fail.
Q16. Benzene reacts with methyl chloride in the presence of anhydrous $\ce{AlCl3}$ to give A. Compound A on vigorous oxidation with alkaline $\ce{KMnO4}$ followed by acidification gives B. B on heating with soda lime yields C. Identify A, B, C.
Friedel-Crafts Alkylation: Anhydrous $\ce{AlCl3}$ (Lewis acid) generates the methyl electrophile ($\ce{CH3+}$). Electrophilic attack on benzene yields Toluene (Compound A, $\ce{C6H5CH3}$).
Side-Chain Oxidation: Vigorous oxidation of alkylbenzenes with alkaline $\ce{KMnO4}$ oxidizes the entire alkyl side chain (provided it has benzylic hydrogens) directly into a carboxyl group. Compound B is Benzoic Acid ($\ce{C6H5COOH}$).
Decarboxylation: Heating a carboxylic acid with soda lime ($\ce{NaOH + CaO}$) removes the carboxyl group as carbon dioxide. This regenerates the unsubstituted aromatic ring. Compound C is Benzene.
Q17. Compare the acidic character of Phenol and Ethanol using resonance structures. Give one chemical test that proves Phenol is more acidic.
Theoretical Comparison: Acidity is determined by the stability of the conjugate base.
In Phenol, loss of a proton yields the phenoxide ion ($\ce{C6H5O-}$). The negative charge on the oxygen is delocalized through resonance into the pi-system of the aromatic ring, stabilizing the ion and driving the dissociation forward.
In Ethanol, the ethoxide ion ($\ce{CH3CH2O-}$) lacks resonance. Furthermore, the electron-donating $+I$ effect of the ethyl group intensifies the negative charge on the oxygen, destabilizing the ion and making ethanol a very weak acid (weaker than water).
Chemical Test: Phenol dissolves in aqueous $\ce{NaOH}$ to form sodium phenoxide and water, proving it is acidic enough to react with a strong base. Ethanol does not react with aqueous $\ce{NaOH}$.
Q18. Explain the mechanistic pathway for the acid-catalyzed dehydration of 3,3-dimethylbutan-2-ol, including the crucial carbocation rearrangement.
Step 1 (Protonation): The acid protonates the hydroxyl group, forming a good leaving group ($\ce{-OH2+}$).
Step 2 (Carbocation Formation): Water departs, leaving a secondary ($2^\circ$) carbocation at C-2: $\ce{CH3-C(CH3)2-C+H-CH3}$.
Step 3 (Wagner-Meerwein Rearrangement): Secondary carbocations are less stable. A methyl group from the adjacent quaternary C-3 migrates to C-2 with its bonding pair (a 1,2-methyl shift). This shifts the positive charge to C-3, creating a highly stable tertiary ($3^\circ$) carbocation: $\ce{CH3-C+(CH3)-CH(CH3)-CH3}$.
Step 4 (Elimination): A base removes a proton to form the double bond. Following Saytzeff's rule to form the most substituted alkene, elimination yields 2,3-dimethylbut-2-ene (a tetrasubstituted alkene) as the major product, rather than the expected 3,3-dimethylbut-1-ene.
Q19. You are provided with Pentan-2-one and Pentan-3-one. How will you distinguish them? Describe the test and the structural requirement.
The Iodoform Test: This test is specific for the methyl ketone group ($\ce{CH3-CO-}$).
Analysis:
Pentan-2-one ($\ce{CH3-CO-CH2CH2CH3}$) possesses the required terminal methyl group attached to the carbonyl. Heating it with Iodine and $\ce{NaOH}$ produces a pale yellow precipitate of Iodoform ($\ce{CHI3}$).
Pentan-3-one ($\ce{CH3CH2-CO-CH2CH3}$) has two ethyl groups flanking the carbonyl. It lacks the methyl ketone motif, hence it produces no yellow precipitate.
Q20. Compound A ($\ce{C8H8O}$) forms a precipitate with 2,4-DNP and a yellow precipitate with iodine/NaOH. It doesn't reduce Tollens' or decolorise bromine water. Drastic oxidation gives carboxylic acid B ($\ce{C7H6O2}$). Identify A and B.
Deductive Reasoning:
- Positive 2,4-DNP: Contains a carbonyl group (aldehyde/ketone).
- Negative Tollens': Not an aldehyde, therefore it is a ketone.
- Positive Iodoform: It is a methyl ketone ($\ce{-CO-CH3}$).
- Negative Bromine water: No aliphatic double/triple bonds.
- Formula analysis ($\ce{C8H8O}$): DU = 5 (Benzene ring + carbonyl). Subtracting the acetyl group ($\ce{CH3CO}$, 2 carbons) leaves $\ce{C6H5}$ (phenyl group).
Thus, Compound A is Acetophenone ($\ce{C6H5-CO-CH3}$).
Oxidation: Drastic oxidation of acetophenone with chromic acid cleaves the methyl group, oxidizing the ring-attached carbon to a carboxyl group. Compound B is Benzoic Acid ($\ce{C6H5COOH}$), matching formula $\ce{C7H6O2}$.
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