NEET Aldehydes, Ketones and Carboxylic Acids Previous Year Questions with Solutions

Benzene reacts with acetyl chloride in the presence of anhydrous aluminium chloride (Lewis acid catalyst). An acylium ion (CH₃-C⁺=O) is generated, which attacks the benzene ring to yield acetophenone.

• • Electrophile: CH₃-C⁺=O (resonance-stabilized acylium cation).
• • Product: Acetophenone (methyl phenyl ketone).

Acetophenone contains the characteristic -COCH₃ group attached to carbon. Sodium hypochlorite (NaOCl), an aqueous mixture of chlorine and sodium hydroxide, oxidizes and halogenates the methyl ketone. Tri-alpha chlorination occurs followed by hydroxide attack and cleavage of the C-C bond to yield Sodium Benzoate (P) and Chloroform (Q).

• • Product P: Sodium benzoate (C₆H₅COONa).
• • Product Q: Chloroform / trichloromethane (CHCl₃).

Acidification of sodium benzoate (P) yields Benzoic Acid (C₆H₅COOH). When chloroform (Q) is exposed to atmospheric oxygen in the presence of sunlight, it undergoes slow photochemical oxidation to form carbonyl chloride (Phosgene, COCl₂), an extremely poisonous gas.

• • Acidification: C₆H₅COONa + H⁺ → C₆H₅COOH (Benzoic acid).
• • Safety protocol in labs: Chloroform is stored in dark amber bottles filled to the brim with 1% ethanol added as a stabilizer.

2,2-dimethylcyclopentan-1-one has two α-carbons: - C2: fully substituted with two methyl groups (quaternary carbon, 0 α-hydrogens). - C5: secondary carbon with two acidic α-hydrogens. Therefore, deprotonation by hydroxide base occurs exclusively at C5 to form a single regioselective enolate.

• • Benzaldehyde (PhCHO) has zero α-hydrogens and cannot form an enolate.

1. Cross-Aldol Condensation with PhCHO: C5 enolate attacks the carbonyl of PhCHO. Subsequent dehydration eliminates the remaining C5 α-hydrogen and the β-hydroxyl group to give an exocyclic conjugated C=CH-Ph double bond. This yields the E-isomer (Structure 2) and Z-isomer (Structure 4). 2. Self-Aldol Condensation: C5 enolate attacks the carbonyl of another 2,2-dimethylcyclopentanone molecule. Dehydration of the resulting β-hydroxyketone occurs between C5 and C1' of the newly attached ring to yield the fully conjugated enone (Structure 3).

• • Structures 2, 3, and 4 are standard feasible products.

In Structure (1), the double bond is endocyclic in the second ring (between C1' and C2' or C2' and C3'), leaving the inter-ring bond as a C-C single bond. This fails to eliminate the C5 α-hydrogen, producing a non-conjugated, thermodynamically unstable β,γ-unsaturated framework. Elimination in aldol condensation always proceeds toward the carbonyl to achieve resonance stabilization of the α,β-unsaturated system.

• • Matches Option (1).

Chromyl chloride (CrO₂Cl₂) in a non-polar solvent like CS₂ oxidizes the methyl group of toluene into a brown chromium complex [C₆H₅CH(OCrOHCl₂)₂]. Hydrolysis of this intermediate yields benzaldehyde.

• • Compound P is Benzaldehyde (C₆H₅CHO).

Rosenmund reduction involves catalytic hydrogenation of an acyl chloride (benzoyl chloride) over palladium catalyst supported on barium sulfate (Pd/BaSO₄), partially poisoned with sulfur or quinoline to prevent over-reduction to an alcohol.

• • This independently synthesizes Compound P (Benzaldehyde).
• • Statement (4) is scientifically correct.

1. Statement (1): Benzaldehyde is an aldehyde, not a carboxylic acid; it does NOT release CO₂ gas with NaHCO₃ (no brisk effervescence). 2. Statement (2): Benzene + CH₃COCl/AlCl₃ is Friedel-Crafts acylation, producing Acetophenone (PhCOCH₃), not benzaldehyde. 3. Statement (3): Bromine water gives white precipitates with phenol (2,4,6-tribromophenol) or aniline, not benzaldehyde.

• • Matches Option (4).

Carboxylic acids dissociate in water to generate carboxylate anions and hydronium ions:

• • Any factor that disperses the negative charge stabilizes the carboxylate anion and increases acidity.
• • Any factor that concentrates or intensifies the negative charge destabilizes the anion and decreases acidity.

Alkyl groups are electron-releasing (+I) groups. As the size and branching of the alkyl group increases, its +I electron-donating power increases dramatically:

• • In (CH₃)₃CCOOH, three electron-releasing methyl groups pump electron density into the carboxylate group, creating maximum destabilization of the anion (weakest acid).
• • In HCOOH, there is no alkyl group attached to the carboxyl carbon, so there is no +I destabilization (strongest acid).

Invert the +I hierarchy to obtain the decreasing acidity series:

• • pK_a values: HCOOH (3.75) < CH₃COOH (4.76) < (CH₃)₂CHCOOH (4.86) < (CH₃)₃CCOOH (5.05).
• • Matches Option (3) perfectly.

The methyl Grignard reagent (CH₃MgBr) acts as a strong nucleophile, attacking the electrophilic carbonyl carbon of the ketone:

• • Subsequent protonation during acidic workup (H₃O⁺) affords a tertiary alcohol: Ph-C(CH₃)(OH)-.

Nitriles react with Grignard reagents by nucleophilic attack on the carbon-nitrogen triple bond to form an imine intermediate salt:

• • The negative charge on the imine nitrogen prevents a second Grignard addition.
• • Upon acidic hydrolysis (H₃O⁺), the imine bond is cleaved to liberate a methyl ketone group:
• • -\text{CH}_2-\text{C}(\text{CH}_3)=\text{N-MgBr} \xrightarrow{\text{H}_3\text{O}^+} -\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 + \text{NH}_4^+ + \text{Mg}^{2+} + \text{Br}^-.

Combine the modified functionalities at both ends of the molecule:

• • This structure corresponds exactly to Option (2).

Diisobutylaluminium hydride, [(CH₃)₂CH-CH₂]₂AlH or AlH(iBu)₂, delivers exactly one hydride ion to the carbonyl carbon of the ester at low temperature (-78 °C):

• • The tetrahedral intermediate is stable at low temperatures, preventing further hydride attack.

Upon addition of water (H₂O workup), the tetrahedral aluminium intermediate collapses, releasing the aldehyde and methanol:

• • The reaction stops cleanly at benzaldehyde without over-reduction.

Verify why the other reagents fail for this specific transformation:

• • NaBH₄ does not react with esters at a measurable rate.
• • H₂/Pd-BaSO₄ requires an acid chloride (Ph-COCl), not an ester.
• • Therefore, (i) AlH(iBu)₂ (ii) H₂O is the unique correct reagent, matching Option (2).

Bicyclohexylidene is an alkene where two cyclohexyl rings are joined by an endocyclic/exocyclic double bond. Reductive ozonolysis cleaves the double bond with ozone followed by zinc dust reduction, yielding two equivalents of cyclohexanone:

Benzene reacts with benzoyl chloride (C6H5COCl) in the presence of anhydrous aluminum chloride (Lewis acid catalyst) to form diphenylmethanone (benzophenone):

Cyclohexanol is a secondary (2°) alcohol. Oxidation with chromium trioxide (CrO3 in aqueous acetone / Jones-type conditions) cleanly yields the corresponding cyclic ketone, cyclohexanone:

Regardless of the length of the alkyl side chain (provided it contains at least one benzylic hydrogen), vigorous oxidation with alkaline KMnO4 cleaves the side chain, converting it into the potassium salt of benzoic acid:

Reaction of 3-cyclohexylprop-1-ene with borane (BH3) proceeds with syn-addition across the double bond, with boron attaching to the less hindered terminal carbon. Subsequent oxidation with alkaline hydrogen peroxide (H2O2 / OH⁻) replaces boron with a hydroxyl group with retention of stereochemistry, yielding the primary alcohol 3-cyclohexylpropan-1-ol:

Primary alcohols are easily over-oxidized to carboxylic acids in aqueous acidic or alkaline media. Pyridinium chlorochromate (PCC, C5H5NH⁺ ClCrO3⁻) in anhydrous dichloromethane (CH2Cl2) selectively oxidizes primary alcohols to aldehydes without further oxidation:

Option (1) uses alkaline KMnO4 which oxidizes primary alcohols completely to carboxylic acids (-COOH). Options (2) and (3) use H2O/H⁺ which follows Markovnikov addition to yield a secondary alcohol, which upon oxidation produces a ketone rather than an aldehyde.

Fehling’s test is a standard diagnostic test used to detect aliphatic aldehydes and reducing sugars. The reagent comprises two separate solutions: - Fehling’s solution A: Aqueous copper(II) sulphate (CuSO4 · 5H2O), imparting a bright blue colour. - Fehling’s solution B: An alkaline solution containing sodium potassium tartrate (Rochelle’s salt, KOOC-CH(OH)-CH(OH)-COONa) and sodium hydroxide (NaOH).

When solutions A and B are mixed just prior to use, tartrate acts as a bidentate/chelating ligand for Cu²⁺ ions. This chelation prevents the premature precipitation of copper(II) hydroxide (Cu(OH)2) in strongly alkaline medium:

Upon heating with an aliphatic aldehyde, Cu²⁺ is reduced to Cu⁺, forming a characteristic red-brown precipitate of copper(I) oxide (Cu2O):

Acidic potassium permanganate (KMnO4 / H⁺) is a potent, vigorous oxidizing agent that causes oxidative cleavage of carbon-carbon double bonds. Unlike cold dilute alkaline KMnO4 (Baeyer’s reagent) which performs syn-dihydroxylation, hot acidic KMnO4 completely ruptures both the σ and π components of the alkene:

The products of oxidative cleavage depend on the substitution of the alkene carbons: 1. Disubstituted carbon (=CR2) yields a ketone (R2C=O). 2. Monosubstituted carbon (=CHR) is oxidized past the aldehyde stage directly to a carboxylic acid (R-COOH). 3. Terminal methylene (=CH2) is oxidized to formic acid, which is rapidly oxidized further to CO2 and H2O.

The substrate possesses two identical monosubstituted alkene carbons bonded to cyclohexyl rings. Cleavage of the central double bond produces two equivalents of cyclohexanecarboxylic acid:

Cyanide ion (:CN⁻) attacks the electrophilic carbonyl carbon of cyclohexanone, followed by protonation of the alkoxide intermediate to yield the cyanohydrin, 1-hydroxycyclohexanecarbonitrile [B]:

In the presence of concentrated sulfuric acid and water at elevated temperature, the nitrile group undergoes complete hydrolysis through an amide intermediate to a carboxylic acid (-COOH):

Under the hot, strongly acidic conditions of concentrated H₂SO₄, the tertiary alcohol hydroxyl group is protonated to form -OH₂⁺ and departs. Concurrently, a proton is eliminated from an adjacent ring carbon (C-2) to yield a thermodynamically stabilized α,β-unsaturated carboxylic acid:

The structure of product [C] consists of a cyclohexene ring with the carboxylic acid group directly attached to the sp²-hybridized alkene carbon, matching Option (1).

The reaction utilizes zinc amalgam in concentrated hydrochloric acid (Zn–Hg / conc. HCl), which is the standard protocol for the Clemmensen reduction. This heterogeneous single-electron transfer reduction converts carbonyl groups (>C=O) of aldehydes and ketones directly into methylene groups (>CH₂):

The acetyl group attached to the cyclohexyl ring is converted into an ethyl group:

Similarly, the acetyl group attached to the meta position of the benzene ring is reduced to an ethyl group:

Both ketone groups are fully reduced, producing 2 moles of H₂O as explicitly stated in the balanced equation. The final product features ethyl groups at both positions on the unchanged ring skeletons, corresponding identically to Option (2).

The reactant is 2-acetylbenzaldehyde, containing two different carbonyl functional groups on adjacent positions of a benzene ring: an aldehyde (–CHO) and a methyl ketone (–COCH₃):

Ammoniacal silver nitrate solution [Ag(NH₃)₂]⁺ in the presence of base (OH⁻) is Tollens’ reagent, a classic mild oxidizing agent:

Tollens’ reagent readily oxidizes the formyl group (–CHO) to the corresponding carboxylate anion (–COO⁻) while metallic silver is precipitated as a silver mirror:

Under these mild basic oxidizing conditions, ketones do not have an oxidizable hydrogen atom on the carbonyl carbon and remain completely unreacted. The –COCH₃ group persists unchanged in the product:

The oxidation of toluene to benzaldehyde using chromyl chloride ($CrO_2Cl_2$) dissolved in carbon disulfide ($CS_2$) or carbon tetrachloride ($CCl_4$), followed by aqueous hydrolysis, is the Étard Reaction.

Chromyl chloride is a mild oxidizing agent that selectively oxidizes the benzylic methyl group without over-oxidizing the ring to benzoic acid. Two moles of $CrO_2Cl_2$ react with one mole of toluene to form a distinctive brown precipitated chromium complex:

Hydrolysis of this complex in acidic water ($H_3O^+$) cleaves the $Cr-O$ and $C-O$ bonds, releasing chromium(III) inorganic salts and furnishing benzaldehyde:

When sodium salts of carboxylic acids are heated strongly with soda lime, they undergo decarboxylation (loss of carbon dioxide) to form an alkane containing one less carbon atom than the parent carboxylic acid: $$R-COO^-Na^+ + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} R-H + Na_2CO_3$$

Here, sodium propanoate ($CH_3CH_2COONa$, 3 carbons) is heated to produce ethane ($CH_3CH_3$, 2 carbons) and sodium carbonate:

In soda lime ($3\text{NaOH} : 1\text{CaO}$): 1. $\text{CaO}$ keeps the reaction mixture dry by absorbing atmospheric moisture ($\text{NaOH}$ is deliquescent and would turn into a sticky liquid). 2. It raises the fusion temperature, allowing the reaction mixture to be heated to high temperatures without melting glass containers.

When benzene or its derivative is treated with carbon monoxide ($CO$) and hydrogen chloride ($HCl$) in the presence of anhydrous aluminium chloride ($AlCl_3$) and cuprous chloride ($CuCl$), benzaldehyde is formed. This is the Gattermann-Koch reaction: $$C_6H_6 + CO + HCl \xrightarrow{\text{Anhyd. } AlCl_3 / CuCl} C_6H_5CHO + HCl \implies \mathbf{\text{(a)-(ii)}} delivery$$

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones, $RCOCH_3$) are oxidized by sodium hypohalite ($NaOX$) to sodium salts of corresponding carboxylic acids having one carbon atom less, while the methyl group is converted to haloform ($CHX_3$). This is the Haloform reaction: $$R-CO-CH_3 + 3\,NaOX \longrightarrow RCOONa + CHX_3\downarrow + 2\,NaOH \implies \mathbf{\text{(b)-(iii)}}$$

Carboxylic acids react with alcohols in the presence of a mineral acid catalyst (such as concentrated sulfuric acid) to form esters with elimination of water. This is Esterification: $$R'COOH + R-CH_2-OH \xrightleftharpoons{\text{Conc. } H_2SO_4} R'COOCH_2R + H_2O \implies \mathbf{\text{(c)-(iv)}}$$

Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus, followed by hydrolysis, yielding $\alpha$-halocarboxylic acids. This is the Hell-Volhard-Zelinsky reaction: $$R-CH_2-COOH \xrightarrow[\text{(ii) } H_2O]{\text{(i) } X_2 / \text{Red P}} R-CH(X)-COOH \implies \mathbf{\text{(d)-(i)}}$$

The reactant molecule contains two carbonyl-containing functional groups: 1. A cyclic ketone group located on the cyclohexane ring. 2. An aliphatic methyl ester group ($-CH_2COOCH_3$) on the side chain. The carbonyl carbon of a ketone is significantly more electrophilic than the carbonyl carbon of an ester because the ester oxygen atom donates electron density through resonance into the carbonyl $\pi^*$ orbital ($R-C(=O)-OCH_3 \leftrightarrow R-C(O^-)=O^+CH_3$).

Sodium borohydride ($NaBH_4$) is a mild hydride donor. Due to the lower nucleophilicity of the $[BH_4]^-$ ion compared to $[AlH_4]^-$ ($LiAlH_4$): - $NaBH_4$ readily reduces aldehydes and ketones to primary and secondary alcohols, respectively. - Under standard conditions in alcoholic solvent ($C_2H_5OH$), $NaBH_4$ does not reduce esters, carboxylic acids, or amides.

- The cyclic ketone ($C=O$) is smoothly converted to a secondary alcohol group ($-OH$). - The ester group ($-CH_2COOCH_3$) survives untouched. - This chemoselective transformation corresponds precisely to Structure (4).

Under photochemical conditions (light/heat), toluene undergoes free radical substitution at the benzylic methyl group rather than electrophilic ring chlorination: $$\text{C}_6\text{H}_5\text{-CH}_3 + 2\text{Cl}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_5\text{-CHCl}_2 + 2\text{HCl}$$ The compound formed is benzal chloride (benzylidene chloride, $$X$$).

Nucleophilic substitution of the two geminal chlorine atoms by water at 373 K yields an unstable gem-diol intermediate: $$\text{C}_6\text{H}_5\text{-CHCl}_2 + 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} [\text{C}_6\text{H}_5\text{-CH(OH)}_2] + 2\text{HCl}$$

Geminal diols attached to the same carbon are thermodynamically unstable and spontaneously undergo elimination of a water molecule: $$[\text{C}_6\text{H}_5\text{-CH(OH)}_2] \xrightarrow{-\text{H}_2\text{O}} \text{C}_6\text{H}_5\text{-CHO (Benzaldehyde)}$$

Option (1) is Benzotrichloride ($$C_6H_5-CCl_3$$, which would give Benzoic acid upon hydrolysis); Option (2) is Benzyl chloride ($$C_6H_5-CH_2Cl$$, which yields Benzyl alcohol); Option (3) is Benzal chloride ($$C_6H_5-CHCl_2$$). Hence, X is Option (3).

Benzaldehyde ($$\text{C}_6\text{H}_5\text{CHO}$$) is an aromatic aldehyde with zero $$\alpha$$-hydrogens on the formyl carbon. Acetophenone ($$\text{C}_6\text{H}_5\text{COCH}_3$$) is an aromatic methyl ketone with three active $$\alpha$$-hydrogens on the methyl group.

Dilute $$\text{NaOH}$$ abstracts an $$\alpha$$-proton from the methyl group of acetophenone to generate a resonance-stabilized carbanion/enolate: $$\text{C}_6\text{H}_5-\text{CO}-\text{CH}_2^- + \text{C}_6\text{H}_5-\text{CHO} \to \text{C}_6\text{H}_5-\text{CH(O}^-) - \text{CH}_2-\text{CO}-\text{C}_6\text{H}_5$$ Proton transfer yields a $$\beta$$-hydroxy ketone: $$\text{C}_6\text{H}_5-\text{CH(OH)}-\text{CH}_2-\text{CO}-\text{C}_6\text{H}_5$$

The $$\beta$$-hydroxy ketone readily undergoes dehydration (loss of $$\text{H}_2\text{O}$$) to form an $$\alpha,\beta$$-unsaturated ketone with an extended conjugated system, 1,3-diphenylprop-2-en-1-one (benzalacetophenone / chalcone): $$\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{CO}-\text{C}_6\text{H}_5$$