Aldehydes Ketones And Carboxylic Acids – JEE Main ChemistryPractice Questions & PYQs

The structure of the hydrocarbon 'X' is determined by reversing the oxidative cleavage process, where the carbonyl groups of the propan-2-one and propanoic acid fragments are joined to reconstruct the backbone of the original molecule. Because the compound absorbs 2 moles of hydrogen, we account for the degree of unsaturation that defines the location of the multiple bonds. By systematically counting each carbon-carbon single bond and every carbon-hydrogen bond in the fully constructed molecule, and recognizing that each double or triple bond contributes exactly one $\sigma$ bond to the structure, the total count is established. Summing these individual covalent interactions throughout the entire molecular framework results in exactly 27 $\sigma$ bonds.

The position of the hydration equilibrium for carbonyl compounds is primarily determined by the electrophilicity of the carbonyl carbon and the extent of steric hindrance. Formaldehyde possesses small hydrogen substituents that minimize steric hindrance, facilitating an effective nucleophilic attack that leads to an equilibrium constant $K$ of 2280. For trichloroacetaldehyde, the strong inductive (-I) effect of the three chlorine atoms drastically enhances the electrophilicity of the carbonyl carbon, which significantly drives the equilibrium toward the hydrated form with an equilibrium constant $K$ of 2000. As these factors accurately explain the high reactivity observed in both species, both provided statements are correct.

Fehling's solution, which consists of copper(II) ions in an alkaline medium, functions as a specific reagent for identifying aliphatic aldehydes and $\alpha$-hydroxy ketones by undergoing a characteristic redox reaction. In this process, the aldehyde group is oxidized to a carboxylate anion while the blue $Cu^{2+}$ ions are reduced to form a distinctive brick-red precipitate of copper(I) oxide ($Cu_2O$). Compounds (C), (D), and (E) yield a positive result because they contain these necessary functional groups capable of reducing the copper species, whereas other structures lacking this specific reactivity, such as (A) and (B), fail to produce the required precipitate.

Ozonolysis of a cyclic alkene involves the oxidative cleavage of the carbon-carbon double bond, yielding a dicarbonyl compound through the formation of an ozonide intermediate. The subsequent intramolecular aldol condensation proceeds via the formation of an enolate from one of the resulting carbonyl groups, which performs a nucleophilic attack on the other carbonyl to form the target cyclic structure $x$. The cyclic alkene corresponding to option C is the unique precursor that possesses the appropriate carbon chain and functional group positioning to generate the specific dicarbonyl intermediate required for this cyclization.

The reactivity of a carbonyl compound toward nucleophilic addition is directly proportional to the positive charge density on the electrophilic carbon atom, where electron-withdrawing substituents increase this charge and electron-donating groups or steric hindrance reduce it. Acetophenone ranks lowest in reactivity because the additional methyl group provides both steric hindrance and electron donation, while p-tolualdehyde follows as the next least reactive because the electron-donating $-\text{CH}_3$ group reduces the electrophilicity of the carbonyl carbon compared to unsubstituted benzaldehyde. At the top of the scale, p-nitrobenzaldehyde displays the highest reactivity because the strongly electron-withdrawing $-\text{NO}_2$ group significantly increases the positive charge on the carbonyl carbon, thereby accelerating the nucleophilic attack.

The percentage yield of a chemical reaction is defined as the ratio of the actual mass of product obtained to the theoretical mass calculated from stoichiometry, multiplied by 100. We first determine the moles of benzaldehyde by evaluating $\frac{5.3}{106} = 0.05$. Since the reaction stoichiometry requires 2 moles of benzaldehyde to produce 1 mole of dibenzalacetone, 0.05 moles of reactant will yield 0.025 moles of the product. The theoretical yield is found by $0.025 \times 234 = 5.85$ g, and the final percentage yield is calculated as $\frac{3.51}{5.85} \times 100 = 60$ %.

The synthesis relies on a logical progression of functional group interconversions starting from an alkyl halide. The optically active compound, sec-butyl bromide (\text{CH}_3-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$), undergoes base-catalyzed elimination with hot alcoholic$\text{KOH}$to form the more stable internal alkene, but-2-ene (\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$). Adding bromine to this alkene leads to 2,3-dibromobutane, which then undergoes double dehydrohalogenation using alcoholic $\text{NaNH}_2$ to yield the corresponding alkyne, but-2-yne (\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$). Finally, the hydration of this alkyne by adding 18 grams of water-which corresponds to one mole-\in the presence of mercuric sulphate and acid at$333\text{ K}$adds a water molecule across the triple bond to form the ketone, butan-2-one (\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3$).

The iodoform reaction serves as a diagnostic test for compounds containing a methyl ketone unit, $R-CO-CH_3$, or a secondary alcohol with an adjacent methyl group, $R-CH(OH)-CH_3$. Molecules missing these specific structural patterns cannot form the yellow $CHI_3$ precipitate when treated with iodine and a base. Ethanol, isopropyl alcohol, bromoacetone, 2-butanol, 2-butanone, and 2-pentanone all possess these required motifs, whereas butanal, 3-pentanone, pentanal, and 3-pentanol do not. This confirms there are four molecules in the list that cannot undergo the iodoform reaction.

The reaction sequence begins with an acid-catalyzed Pinacol-Pinacolone rearrangement, where the protonation of a diol followed by dehydration and a 1,2-alkyl shift transforms the molecule ($P$) into a stable ketone ($Q$). Subsequent ozonolysis generates reactive carbonyl intermediates, which undergo an intramolecular aldol condensation during alkaline reflux to yield the cyclic conjugated enone system designated as ($R''$).

The reaction with 2,4-dinitrophenylhydrazine confirms that the compound possesses a carbonyl functional group, while the lack of reaction with Tollens reagent indicates that this group is not an aldehyde, establishing it as a ketone. Given the molecular formula $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}$, the compound contains a single degree of unsaturation that accounts for this carbonyl moiety. Since the compound must be optically active, the carbon chain is branched in a way that introduces a chiral center, distinguishing the correct molecular arrangement.

When aromatic compounds with alkyl side chains are subjected to vigorous oxidation using hot $\mathrm{KMnO}_4$, the side chains are cleaved and the benzylic carbon is oxidized to a carboxylic acid group, yielding benzoic acid. This transformation is strictly dependent on the presence of at least one benzylic hydrogen atom, defined as a hydrogen atom attached to the carbon bonded directly to the aromatic ring. Substituents lacking this essential hydrogen remain unreactive under these oxidizing conditions, meaning only molecules with suitable benzylic protons will successfully convert. Applying this structural criterion to the set confirms that 5 compounds meet the requirement and result in the formation of benzoic acid.

Amino acids behave as amphoteric molecules whose protonation state shifts based on the acidity or alkalinity of their environment. At a pH of 2, the acidic medium provides enough protons to ensure the amino group exists as $-\mathrm{NH}_3^+$, creating structure A. Alternatively, when the pH reaches 10, the basic medium facilitates the loss of a proton from the carboxyl group, resulting in the $-\mathrm{COO}^-$ form of structure B.

Clemmensen reduction utilizes a zinc-mercury amalgam ($Zn(Hg)$) in the presence of concentrated hydrochloric acid ($HCl$) to selectively transform aldehyde or ketone groups into methylene groups ($>CH_2$). A key characteristic of this method is its chemoselectivity, meaning it functions efficiently while leaving other functional groups entirely untouched. Specifically, an ester group remains resistant to these strongly acidic reducing conditions and does not undergo reduction, ensuring that the final structure retains the ester functionality exactly as it appeared in the reactant.

Distinguishing between isomers with the molecular formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$ relies on their chemical behavior, as ketones do not oxidise readily while aldehydes are easily converted to carboxylic acids, allowing us to identify the compound $(X)$ as acetone ($\mathrm{CH}_3\mathrm{COCH}_3$). Reducing acetone yields the secondary alcohol propan-2-ol ($\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$) as $(Y)$, which subsequently reacts with hydrogen bromide to replace the hydroxyl group with bromine, forming the alkyl bromide $(Z)$ known as 2-bromopropane ($\mathrm{CH}_3\mathrm{CH}(\mathrm{Br})\mathrm{CH}_3$). Finally, treating the Grignard reagent derived from 2-bromopropane with acetone initiates a nucleophilic addition to the carbonyl carbon, followed by hydrolysis, to produce 2,3-dimethylbutan-2-ol and complete the sequence.

Intramolecular aldol condensation proceeds via the formation of a stable enolate ion from an $\alpha$-carbon, which subsequently undergoes an intramolecular nucleophilic attack on the second carbonyl group to generate a cyclic $\beta$-hydroxy ketone intermediate. The reaction concludes with the dehydration of this intermediate to yield a stable, conjugated $\alpha, \beta$-unsaturated cyclic ketone, where the major product is dictated by the preference for forming five- or six-membered rings due to favorable entropic factors and minimal ring strain. The isomer provided in option B represents the thermodynamically favored product because it optimizes internal orbital overlap and minimizes steric congestion within the newly formed cyclic framework.

The fundamental principle governing this reaction is that a stronger acid will displace a weaker acid from its corresponding salt. Sodium bicarbonate ($\mathrm{NaHCO}_3$) effectively acts as a source of the weak acid carbonic acid ($\mathrm{H_2CO}_3$), and for gas evolution to occur, the reactant must be acidic enough to protonate the bicarbonate ion, which then decomposes into water and $\mathrm{CO}_2$. Since compounds A, C, and D are inherently stronger acids than $\mathrm{H_2CO}_3$, they are capable of initiating this transformation, whereas compounds that are weaker than carbonic acid will not produce $\mathrm{CO}_2$ under these conditions.

Thermal decomposition of sodium bicarbonate yields anhydrous sodium carbonate ($Na_2CO_3$) as the solid residue (A). This residue subsequently reacts with dilute hydrochloric acid ($HCl$) via an acid-base neutralization process to produce sodium chloride ($NaCl$) as compound (B), water, and carbon dioxide, according to the balanced chemical equation:

$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$

The reaction sequence systematically modifies the substituent on the benzene ring, beginning with the conversion of acetophenone into a geminal dichloride using $\mathrm{PCl}_5$. Treatment of this intermediate with three equivalents of $\mathrm{NaNH}_2$ facilitates a double dehydrohalogenation to generate the phenylacetylide anion, which upon acidic workup yields phenylacetylene. The final hydroboration-oxidation step involves the anti-Markovnikov addition of water across the terminal triple bond, creating an enol intermediate that rapidly tautomerizes to form phenylacetaldehyde.

To determine the number of $\mathrm{sp}^2$ hybridized carbon atoms in the final product, phenylacetaldehyde, one must examine the hybridization state of each carbon atom in the structure. The six carbon atoms forming the aromatic benzene ring each possess a pi bond and are consequently $\mathrm{sp}^2$ hybridized. Additionally, the carbonyl carbon within the aldehyde group is double-bonded to an oxygen atom, making it $\mathrm{sp}^2$ hybridized as well. The remaining carbon, which is the methylene group linking the phenyl ring to the aldehyde, is bonded through four single bonds and is therefore $\mathrm{sp}^3$ hybridized. Combining these values, the molecule contains a total of 7 $\mathrm{sp}^2$ hybridized carbon atoms.

Both acetaldehyde and acetone feature an acetyl group ($CH_3CO-$) that facilitates the Iodoform reaction, and the presence of $\alpha$-hydrogens in both molecules allows them to undergo the Aldol condensation. They both respond to the Clemmensen reduction, which converts the carbonyl moiety into a methylene group ($CH_2$), making this process compatible with both reactants. Conversely, the Tollen's test selectively identifies aldehydes rather than ketones, and the Cannizzaro reaction is reserved for aldehydes lacking $\alpha$-hydrogens, meaning neither of these reactions applies to both substances.

The synthesis of carboxylic acids often relies on specific chemical transformations capable of generating the $-COOH$ functional group. The reaction of a Grignard reagent, $R-MgX$, with carbon dioxide followed by an acidic workup constitutes a standard route for preparing these acids, as the nucleophilic alkyl group attacks the electrophilic carbon of $CO_2$. Separately, bromine water serves as an oxidizing agent that directly converts aldehydes into their corresponding carboxylic acids, confirming that both of these pathways are effective for achieving the target product.