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

The reaction sequence involves the side-chain oxidation of Toluene (A) using an oxidizing agent such as $KMnO_4$ to yield Benzoic acid, followed by its conversion to Benzoyl chloride through the application of $SOCl_2$. The intermediate Benzoyl chloride undergoes Rosenmund reduction with $H_2/Pd-BaSO_4$ to produce Benzaldehyde, which is subsequently reduced by $NaBH_4$ to provide Benzyl alcohol (E).

Organic acids, such as carboxylic acids, react with aqueous $NaHCO_3$ to evolve $CO_2$ gas, while chiral carbons are identified as atoms bonded to four distinct substituents; therefore, compound $(X)$ satisfies Statement I if it possesses a carboxylic acid functional group and two such stereogenic centers. Hybridization is determined by the number of sigma bonds and lone pairs, where $sp^3$ carbons possess four sigma bonds, $sp^2$ carbons possess three, and $sp$ carbons possess two; thus, Statement II is true if compound $(Y)$ exhibits the specific distribution of hybrid states described in the problem.

The iodoform test detects the presence of a methyl ketone group or a methyl carbinol moiety, which reacts with sodium hypoiodite ($NaOI$) to produce a characteristic yellow precipitate of iodoform ($CHI_3$). Since compound $Z$ contains one of these functional groups, it is chemically verified to produce the observed precipitate, making Statement I correct. Similarly, the ratio of aromatic to aliphatic hydrogen atoms is calculated by comparing the number of hydrogen atoms directly attached to the benzene ring with those present in the alkyl side chains; identifying this ratio as $1:3$ for compound $Q$ confirms that the structural features of the molecule satisfy the conditions specified in Statement II.

The 2, 4-dinitrophenylhydrazine (2, 4-DNP) test acts as a definitive diagnostic tool for identifying carbonyl groups, as both aldehydes and ketones readily undergo nucleophilic addition with this reagent to produce distinct precipitates. Because the hydrolysis products (Y) and (Z) both participate in this addition reaction, it confirms that each molecule must contain a carbonyl moiety within its structure.

Furthermore, the condition that (Y) and (Z) share the same molar mass of 58-which corresponds to the molecular weight of isomers like acetone or propanal ($C_{3}H_{6}O$)-is consistent with them being structural isomers. While they share an identical molar mass, their differing electronic and steric environments mean they exhibit distinct reactivity profiles toward other reagents, supporting the conclusion that only the statements regarding their molar mass and 2, 4-DNP reactivity are correct.

The selective reduction of carbonyl groups is governed by the use of $\mathrm{NaBH_4}$ in methanol, which converts aldehydes and ketones into alcohols without affecting more stable functional groups like esters or carboxylic acids. Subsequent treatment with aqueous $\mathrm{NaOH}$ at elevated temperatures facilitates an intramolecular aldol condensation of the intermediate, driving the formation of a stable cyclic or conjugated product through dehydration. The structure shown in option C reflects the final outcome of this specific sequence of chemoselective reduction followed by base-catalyzed cyclization.

The Gattermann-Koch reaction of benzene with carbon monoxide and hydrogen chloride in the presence of cuprous chloride produces benzaldehyde, which is $x$. Friedel-Crafts acylation of benzene with ethanoyl chloride in the presence of anhydrous $AlCl_3$ yields acetophenone, which is $y$.

The base-catalyzed Claisen-Schmidt condensation of benzaldehyde and acetophenone results in the formation of 1,3-diphenylprop-2-en-1-one, commonly known as chalcone, which corresponds to $z$.

Each of the two benzene rings in chalcone contains 3 $\pi$ bonds, contributing 6 $\pi$ electrons per ring for a total of 12. The alkene linkage provides 2 $\pi$ electrons, and the carbonyl group adds another 2 $\pi$ electrons. Summing these values gives the total number of $\pi$ electrons:
$6 + 6 + 2 + 2 = 16$

The Stephen reduction of propanenitrile ($CH_3CH_2CN$) using stannous chloride ($SnCl_2$) and hydrochloric acid yields propanal ($CH_3CH_2CHO$) as product 'x', while the ozonolysis of but-2-ene ($CH_3CH=CHCH_3$) yields two molecules of acetaldehyde ($CH_3CHO$) as product 'y'. When these products react in the presence of an alkali, they undergo cross-aldol condensation and dehydration to form $\alpha,\beta$-unsaturated aldehydes, specifically producing 2-methylpent-2-enal from the self-condensation of propanal, pent-2-enal from the attack of the acetaldehyde enolate on propanal, and 2-methylbut-2-enal from the attack of the propanal enolate on acetaldehyde. 3-Methylbut-2-enal is not formed in this reaction because its structure contains a branching pattern that cannot be derived from the carbon skeletons of propanal and acetaldehyde.

Reactions involving the electrophilic addition of an unsymmetrical reagent to an unsymmetrical alkene are governed by Markovnikov's rule, which states that the nucleophilic part of the reagent adds to the carbon atom with the fewer number of hydrogen atoms to form the more stable carbocation intermediate. Option A violates this fundamental regioselectivity principle by placing the substituent on the less substituted carbon in the absence of peroxides or radical initiators, which is chemically invalid. This representation fails to account for the formation of the more stable intermediate, making it an incorrect description of the reaction pathway.

Acidic hydrolysis of enol ethers generates carbonyl compounds, typically resulting in a mixture of an aldehyde and a ketone. Fehling solution, consisting of alkaline copper(II) tartrate, selectively oxidizes aliphatic aldehydes to carboxylates while forming a brick-red precipitate of $Cu_2O$, whereas ketones remain unaffected by this reagent. Other choices like 2,4-DNP react with both functional groups and Lucas reagent is used to identify alcohols, making Fehling solution the specific test required to distinguish between the aldehyde and ketone products.

Each reagent set represents a distinct named chemical reaction specifically utilized for the preparation of aldehydes. The reduction of acid chlorides using $H_2$ in the presence of $Pd-BaSO_4$ constitutes the Rosenmund reduction, whereas the reduction of nitriles using $SnCl_2$ and $HCl$ is defined as the Stephen reaction. Furthermore, the oxidation of toluene to benzaldehyde using $CrO_2Cl_2$ in $CS_2$ is known as the Etard reaction, and the direct formylation of benzene using $CO$ and $HCl$ with anhydrous $AlCl_3$ is identified as the Gattermann-Koch reaction.

Cross aldol condensation between two aldehydes with $\alpha$-hydrogens theoretically yields four products, but if one of the reactants lacks $\alpha$-hydrogens, self-aldol condensation for that component is impossible, making the claim that it always produces four products incorrect. Furthermore, nucleophilic addition of semicarbazide to a mixture of carbonyl compounds is governed by the electrophilicity of the carbonyl carbon, where aldehydes are significantly more reactive than ketones due to reduced steric hindrance and electronic stabilization; consequently, semicarbazide reacts preferentially with the aldehyde, benzaldehyde, rather than the ketone, acetophenone, rendering the statement that it reacts with acetophenone only false.

Grignard reagents ($RMgBr$) react with water to produce alkanes ($RH$) and with carbon dioxide followed by acid hydrolysis to yield carboxylic acids ($RCOOH$). The number of moles of gas (Q) at STP is calculated as $n = 1.4 / 22.4 = 0.0625\ \text{mol}$. Given the mass of 1 g, the molar mass of the gas is $M = 1 / 0.0625 = 16\ \text{g/mol}$, which identifies the alkane as methane ($CH_4$) where the alkyl group $R$ is a methyl group ($CH_3$).

Consequently, compound (Z) is acetic acid ($CH_3COOH$), which has a molar mass of $M = (12 \times 2) + (1 \times 4) + (16 \times 2) = 60\ \text{g/mol}$. The final weight of $0.1\ \text{mol}$ of (Z) is calculated by the expression $\text{mass} = 0.1 \times 60 = 6\ \text{g}$.

The acidic strength of carboxylic acids is determined by the stability of the conjugate carboxylate anion, which is enhanced by electron-withdrawing substituents and destabilized by electron-donating groups attached to the carboxyl carbon. The major products formed in the reactions are formic acid ($\mathrm{HCOOH}$) for C, benzoic acid ($\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}$) for A, phenylacetic acid ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{COOH}$) for D, and propanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$) for B. Comparing their acid dissociation constants, formic acid is the strongest acid ($\mathrm{p}K_a \approx 3.75$), followed by benzoic acid ($\mathrm{p}K_a \approx 4.20$) where the phenyl group exerts a strong inductive withdrawal, phenylacetic acid ($\mathrm{p}K_a \approx 4.31$) where the electron-withdrawing effect of the phenyl ring is reduced by the $\mathrm{CH}_2$ group, and lastly propanoic acid ($\mathrm{p}K_a \approx 4.89$) which is the weakest due to the electron-donating inductive effect of the ethyl group.

The haloform reaction relies on the presence of a methyl ketone ($CH_3-C=O$) or a secondary alcohol with a methyl group ($CH_3-CH(OH)-$), where sodium hypochlorite serves as an oxidizing agent that converts iodide to the active halogenating species, hypoiodite ($OI^-$), making statement A true and statement B false because hypoiodite is an oxidizing agent rather than a reducing one. $\alpha, \beta$-unsaturated methyl ketones possess the necessary intact methyl ketone moiety to react, and isopropyl alcohol is successfully oxidized to acetone, which yields a positive test, establishing statement C as true and statement D as false. Conversely, methanoic acid lacks the required methyl group adjacent to the carbonyl carbon, meaning it cannot participate in the haloform reaction, which renders statement E incorrect.

Intramolecular aldol condensation is governed by the thermodynamic stability of the resulting cyclic structure, which strongly favors the formation of $5$- or $6$-membered rings due to minimal ring strain and favorable entropy. Products that require the creation of highly strained $3$- or $4$-membered rings or unstable bridgehead systems are generally not formed because the prohibitive activation energy barriers render these cyclizations kinetically and thermodynamically unfavorable. Consequently, the structure represented in option D is not a viable product because it violates these fundamental geometric limitations inherent in the condensation mechanism.

Vanillin contains an aldehyde group directly attached to an aromatic ring, allowing it to provide a positive Tollen's reagent test and undergo a Cannizzaro reaction in the presence of $\text{NaOH}$, which confirms the first statement. The ability to undergo self-aldol condensation requires the presence of at least one $\alpha$-hydrogen atom on the carbon adjacent to the carbonyl group. Because the aldehyde carbon in vanillin is bonded to the aromatic ring and lacks an $\alpha$-hydrogen, it cannot participate in this reaction, making the second statement incorrect.

The formation of the major product $(P)$ is governed by the relative stability of the possible reaction intermediates, where the pathway proceeding through the more stable carbocation or transition state possesses the lower activation energy and thus occurs at a faster rate. Electronic factors, such as resonance and inductive effects, further stabilize this preferred intermediate, ensuring that the final structure adheres to fundamental regioselective principles. These thermodynamic and kinetic considerations ultimately determine the selectivity and yield of the reaction.

Organic name reactions are fundamental transformations defined by the specific reagents required to convert precursors into aldehyde functional groups. The Stephen reaction involves the reduction of nitriles using $SnCl_2$ and $HCl$, which corresponds to the description labeled (IV). Rosenmund reduction utilizes the selective hydrogenation of acid chlorides to aldehydes over a $Pd/BaSO_4$ catalyst, aligning with entry (III). The Etard reaction uses chromyl chloride ($CrO_2Cl_2$) to oxidize a toluene methyl group to an aldehyde, mapping to entry (I). Finally, the Gatterman-Koch reaction introduces a formyl group onto a benzene ring using carbon monoxide and $HCl$ in the presence of anhydrous $AlCl_3$, matching entry (II).

The reaction sequence proceeds via an initial aldol condensation followed by a cross-Cannizzaro reaction, utilizing the acidic $\alpha$-hydrogens of the propanal. In the presence of $\mathrm{OH}^{-}$ ions and excess $\mathrm{HCHO}$, the $\alpha$-hydrogens of the propanal are substituted with hydroxymethyl groups to form 2,2-dihydroxy methyl propanal. This intermediate then undergoes a Cannizzaro reaction with the remaining $\mathrm{HCHO}$ to reduce the aldehyde group to an alcohol, yielding the final product, 2,2-hydroxymethylpropan-1-ol.

The iodoform test is specific to compounds containing the methyl ketone group ($\mathrm{CH_3CO-}$) or secondary alcohols containing a methyl group adjacent to the hydroxyl carbon ($\mathrm{CH_3CH(OH)-}$), which undergo oxidation and subsequent cleavage by $\mathrm{NaOI}$ to form a yellow precipitate of iodoform ($\mathrm{CHI_3}$). Acetone possesses the requisite methyl ketone structure, and ethanol, although a primary alcohol, is oxidized to acetaldehyde during the reaction, which subsequently satisfies the structural requirement for the haloform reaction. Compounds like diethyl ketone, acetic acid, and methanol lack the necessary methyl-bearing functional groups, ensuring that only the substances represented by (A) and (C) yield the positive test result.