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

This reaction sequence is a standard method for converting a nitrile into an aldehyde by utilizing an acid chloride intermediate. The initial acid-catalyzed hydrolysis of propanenitrile, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CN}$, transforms the cyano group into a carboxylic acid, yielding propanoic acid, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$. Treating this acid with thionyl chloride, $\mathrm{SOCl}_2$, substitutes the hydroxyl group with a chlorine atom to generate propanoyl chloride, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCl}$. Finally, the selective reduction of this acid chloride using hydrogen gas in the presence of a poisoned palladium catalyst, $\mathrm{Pd}/\mathrm{BaSO}_4$, follows the Rosenmund reduction mechanism, which limits the reaction to the aldehyde stage to produce propanaldehyde, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$.

The reaction sequence initiates with a Friedel-Crafts acylation, which introduces an acyl group onto the aromatic ring to yield a ketone as intermediate A. Subsequently, the Clemmensen reduction conditions specifically target the carbonyl group of this ketone, converting it into a $\mathrm{CH}_2$ group to produce the final alkylated product B.

The transformation of an alkyl benzene into benzoic acid is an oxidation reaction that specifically requires a strong oxidizing agent. Because $\mathrm{NaBH}_{4}$, $\mathrm{LiAlH}_{4}$, and the combination of $\mathrm{Zn}-\mathrm{Hg}$ with $\mathrm{HCl}$ function as reducing agents, they are chemically incapable of performing this type of conversion. Alkaline $\mathrm{KMnO}_{4}$ followed by $\mathrm{H}^{+}$ serves as the powerful oxidizing agent necessary to carry out this specific chemical change.

Reduction processes in organic chemistry require specific reagents capable of transferring hydrogen atoms or electrons to a functional group, a function served by catalytic systems like $\mathrm{Pt}-\mathrm{C} / \mathrm{H}_2$ and $\mathrm{Pd}-\mathrm{C} / \mathrm{H}_2$, or chemical reductants like $\mathrm{Zn} / \mathrm{H}_2\mathrm{O}$. In contrast, the combination of $\mathrm{Na} / \mathrm{H}_2$ fails to facilitate such a transformation because it simply does not function as a reducing agent. This lack of chemical reactivity makes it unsuitable for experimental reduction tasks where the other listed reagents would be effective.

Compounds able to liberate carbon dioxide when treated with sodium bicarbonate must be stronger acids than carbonic acid ($H_2CO_3$). Because $H_2CO_3$ readily dissociates into $H_2O$ and $CO_2$, the chemical equilibrium favors the evolution of gas only when the reactant acid is strong enough to drive the reaction forward. As compounds B and C possess the requisite acidity to displace carbonic acid, they successfully facilitate the release of $CO_2$, while other compounds in the set fail to push the equilibrium toward gas liberation.

The synthesis relies on the conversion of a carboxylic acid to an acyl chloride, followed by electrophilic aromatic substitution and nucleophilic addition to a carbonyl. First, acetic acid reacts with thionyl chloride to form acetyl chloride, $\mathrm{CH}_3\mathrm{COCl}$, which serves as the electrophile for the subsequent Friedel-Crafts acylation with benzene in the presence of $\mathrm{AlCl}_3$ to yield acetophenone, $\mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3$. Finally, the addition of the cyanide ion to the carbonyl group of acetophenone produces the cyanohydrin, 2-hydroxy-2-phenylpropanenitrile.

The reaction described is an aldol condensation, a fundamental process where a base like dil. $\mathrm{NaOH}$ abstracts an $\alpha$-hydrogen to generate a nucleophilic enolate ion. This enolate species subsequently attacks another carbonyl electrophile to yield a $\beta$-hydroxy ketone, which represents the intermediate product $X$. Upon further reaction, this aldol intermediate undergoes dehydration, eliminating a water molecule to produce the stable, conjugated enone known as product $Y$.

The reaction between a ketone and ethylene glycol is a classic example of nucleophilic addition, leading to the formation of a cyclic ketal. In this process, the diol acts as a nucleophile where the first hydroxyl group attacks the electrophilic carbonyl carbon to form a hemiketal intermediate. Subsequently, the second hydroxyl group from the same ethylene glycol molecule attacks the same carbon center, resulting in the elimination of a water molecule ($H_{2}O$) and the closing of the ring to form the stable cyclic ketal structure. This mechanism confirms that the sequential addition of two hydroxyl groups from a single diol molecule to a ketone yields the cyclic product represented in the reaction.

The synthesis sequence initiates with a Cannizzaro reaction where 4-methoxybenzaldehyde reacts with formaldehyde in the presence of $NaOH$ to produce $p$-methoxybenzyl alcohol. This alcohol is subsequently deprotonated by $NaH$ and coupled with bromoethane in the presence of $DMF$ to yield 1-ethoxymethyl-4-methoxybenzene. In the final step, the compound undergoes solvolysis followed by an $S_N2$ attack from $I^-$ ions, leading to the formation of 4-(iodomethyl) phenol as the product $A$.

The transformation occurs via electrophilic aromatic substitution, where the active electrophile, the nitronium ion ($NO_2^+$), attacks the electron-rich aromatic ring to form a resonance-stabilized carbocation intermediate known as the sigma complex. Subsequent loss of a proton from the tetrahedral carbon restores the aromaticity of the ring, yielding nitrobenzene as the final product.

Alkaline $KMnO_4$ acts as a potent oxidizing agent that targets alkyl groups directly attached to an aromatic ring, converting them into carboxylic acid groups. When 4-methoxy toluene undergoes this reaction, the methyl substituent is oxidized to a carboxyl moiety. This transformation results in the formation of 4-methoxy benzoic acid as the final product $X$.

The stability of keto-enol tautomers is fundamentally determined by the relative bond energies of the carbonyl group compared to the resulting double bond and hydroxyl group. In monocarbonyls like acetone, the keto form is significantly more stable because the $\mathrm{C=O}$ bond energy is much higher than the combined energy of the $\mathrm{C=C}$ and $\mathrm{O-H}$ bonds, keeping the enol concentration below $0.1\%$. Conversely, acetyl acetone exhibits an enol content of approximately $15\%$ because its enol tautomer is stabilized by the formation of a six-membered ring through intramolecular hydrogen bonding. This unique stabilizing effect, which is structurally impossible for simple acetone, directly accounts for the much higher enol proportion observed in the diketone.

Markovnikov's rule dictates that in the electrophilic addition of hydrogen halides to unsymmetrical alkenes, the hydrogen atom attaches to the carbon with more hydrogen substituents, while the halide attaches to the more substituted carbon. The reaction proceeds through the formation of a tertiary carbocation intermediate, which is significantly more stable than the alternative primary carbocation, thereby lowering the activation energy for the transition state leading to the final product. This regioselectivity ensures that the bromide ion attacks the more substituted position, resulting in 2-bromo-2-methylpropane as the dominant product.

The 2,4-DNP test relies on the chemical reactivity of Brady's reagent, known formally as 2,4-dinitrophenylhydrazine, which specifically targets the carbonyl group ($C=O$) present in aldehydes and ketones. When this reagent is added to a sample containing these functional groups, it undergoes a condensation reaction that produces a distinct reddish-orange precipitate. Because this color change is exclusive to the carbonyl-containing compounds, this test serves as a clear diagnostic indicator for identifying an aldehyde compared to non-carbonyl substances like amines, ethers, or halogenated compounds.

Determining the percentage yield requires comparing the actual mass of the product recovered to the theoretical maximum possible based on the reaction stoichiometry. Benzoic acid has a molar mass of 122 g/mol, and converting the 6.1 g sample into moles yields $\frac{6.1}{122} = 0.05$ mol. Since the product, m-bromo benzoic acid, has a molar mass of 201 g/mol, the theoretical yield is $0.05 \times 201 = 10.05$ g, resulting in a percentage yield of $\frac{7.8}{10.05} \times 100 \approx 77.61\%$, which rounds to 78%.

The reduction of esters using $\mathrm{LiAlH}_4$ serves as a fundamental chemical process to generate primary alcohols by cleaving the original ester linkage into two distinct alcohol components. Because compound A has the molecular formula $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_2$ and follows a straight-chain structure, it functions as an ester that reduces to form a four-carbon alcohol, which is subsequently oxidized to create a $\mathrm{C}_4$ carboxylic acid. This identifies the compound as ethyl butanoate, which splits into butanol and ethanol during the reduction step, allowing the resulting butanol to form the required four-carbon carboxylic acid in the final stage of the sequence.

The preparation of maleic anhydride is an example of intramolecular dehydration, which requires the carboxylic acid groups to be in close spatial proximity. When cis-but-2-enedioic acid is heated, the two adjacent carboxyl groups readily undergo a condensation reaction to lose a molecule of $H_2O$, thereby forming a stable five-membered cyclic anhydride. This cyclization is geometrically impossible for trans-but-2-enedioic acid, as its carboxylic acid groups are situated on opposite sides of the double bond and cannot reach each other to form the necessary ring structure.

The Tollen's test relies on a redox reaction where the aldehyde is oxidized to a carboxylate ion while the silver ions in the $[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$ complex are reduced to metallic $\mathrm{Ag}$. During this process, the aldehyde group effectively loses two electrons to satisfy the stoichiometry of the chemical transformation. Since each silver ion in the $[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$ complex has an oxidation state of $+1$, it requires one electron to be reduced to $\mathrm{Ag}$, meaning the two electrons released by the aldehyde are transferred to the silver species. Consequently, the total number of electrons transferred to the Tollen's reagent per aldehyde group is 2.

The Rosenmund reduction utilizes $\mathrm{H}_2 / \mathrm{Pd}-\mathrm{BaSO}_4$ to selectively convert acid chlorides ($R-\mathrm{COCl}$) to aldehydes ($R-\mathrm{CHO}$), while the Hell-Volhard-Zelinsky reaction employs $\mathrm{Cl}_2 / \text{Red P}$ for the $\alpha$-halogenation of carboxylic acids. Simultaneously, the Hofmann bromamide degradation converts amides ($R-\mathrm{CONH}_2$) to primary amines ($R-\mathrm{NH}_2$) using $\mathrm{Br}_2 / \mathrm{NaOH}$, and the Clemmensen reduction utilizes $\mathrm{Zn}(\mathrm{Hg}) / \text{conc. } \mathrm{HCl}$ to reduce ketones to their corresponding alkanes. These specific reagent conditions pair the listed transformations as A-2, B-4, C-1, and D-3.

The reaction yields two distinct products: a species featuring both alcohol and carboxylic acid functional groups, and a diol containing two alcohol groups. Because the transformation results exclusively in these specific molecular structures, a compound possessing two carboxylic acid functional groups is not produced.