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

The addition of $\mathrm{HCN}$ to carbonyl compounds is a nucleophilic addition reaction governed by the electrophilic nature of the carbonyl carbon and the extent of steric hindrance surrounding it. Electron-donating groups reduce the electrophilic character of the carbonyl carbon through the $+I$ effect, while bulky groups impede the approach of the incoming nucleophile, causing the reaction rate to drop as the size and number of these substituents increase. Consequently, formaldehyde is the most reactive due to the absence of such groups, whereas benzophenone is the least reactive due to the presence of two bulky phenyl groups, leading to an increasing order of reaction rates expressed as $\mathrm{PhCOPh} < \mathrm{PhCOCH}_3 < \mathrm{CH}_3 \mathrm{COCH}_3 < \mathrm{HCHO}$.

The Reimer-Tiemann reaction of $p$-cresol with chloroform in an alkaline medium introduces a formyl group at the ortho position relative to the hydroxyl group, yielding $2$-hydroxy-$5$-methylbenzaldehyde. This aldehyde undergoes nucleophilic addition with hydrogen cyanide to produce a cyanohydrin intermediate, which, upon acidic hydrolysis, converts the nitrile moiety into a carboxylic acid group, resulting in $2$-hydroxy-$5$-methylmandelic acid. The final product is chiral because the alpha-carbon atom is bonded to four distinct substituents-a hydrogen atom, a hydroxyl group, a carboxyl group, and the substituted phenyl ring-thereby creating a stereocenter.

The acidity of carboxylic acids is primarily governed by the electronic effects of the attached groups, with stronger acids exhibiting a higher acid dissociation constant, $\mathrm{K}_a$. Because the relationship is defined by $\mathrm{pK}_a = -\log \mathrm{K}_a$, the compound with the highest $\mathrm{K}_a$ necessarily corresponds to the lowest $\mathrm{pK}_a$ value. Among the given options, $\mathrm{HCOOH}$ is the strongest acid because it lacks the electron-donating alkyl groups that destabilize the conjugate bases of the other compounds. Consequently, this maximum acidity results in the lowest $\mathrm{pK}_a$ value in the set.

The reaction between a ketone, such as cyclohexanone, and a secondary amine like dimethylamine ($HN(CH_3)_2$) in the presence of an acid catalyst proceeds via the formation of a carbinolamine intermediate which then undergoes dehydration to form an iminium ion. Because the nitrogen atom in a secondary amine lacks the hydrogen atom required to form a stable $C=N$ double bond (imine), the system eliminates a proton from the $\alpha$-carbon to restore neutrality, resulting in an $\alpha,\beta$-unsaturated amine known as an enamine. The continuous removal of water serves to shift the chemical equilibrium in favor of this enamine product.

Mixing ethyl acetate with an aqueous solution of sodium chloride does not initiate any chemical reaction because neither substance possesses the chemical properties required to modify the other under these conditions. Since no transformation takes place, the individual compounds remain chemically distinct and coexist in the mixture as $\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5 + \mathrm{NaCl}$.

The Clemmensen reduction utilizes zinc amalgam ($Zn-Hg$) in the presence of concentrated hydrochloric acid to effectively reduce the carbonyl group found in aldehydes and ketones into a methylene group, thereby producing an alkane. Because butan-2-one is a ketone, it reacts under these specific conditions to yield the hydrocarbon butane according to the process $\text{CH}_3\text{COCH}_2\text{CH}_3 \xrightarrow[\text{Conc. HCl}]{\text{Zn-Hg}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$. Other candidates such as acetic acid, acetamide, and ethyl acetate contain distinct functional groups that do not participate in this particular reduction pathway to form hydrocarbons.

Aldehydes that lack alpha-hydrogen atoms undergo a specific disproportionation process known as the Cannizzaro reaction when exposed to a strong base like $50\%$ sodium hydroxide. Benzaldehyde meets this structural requirement, and during this reaction, it simultaneously oxidizes and reduces to yield two distinct products: sodium benzoate and benzyl alcohol. This unique behavior of converting a single reactant into both an acid salt and an alcohol is characteristic of this specific transformation.

Lithium aluminum hydride ($\mathrm{LiAlH}_4$) functions as a highly selective reducing agent that targets carboxylic acid groups to form primary alcohols while leaving carbon-carbon double bonds untouched. Consequently, when $\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH}$ is treated with this reagent, the carboxyl group is reduced to an alcohol while the vinyl group remains intact, producing $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\mathrm{OH}$ according to the reaction:
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\mathrm{OH}$

The reaction between a primary amine, such as p-toluidine, and chloroform in the presence of alcoholic potassium hydroxide constitutes the carbylamine reaction, which proceeds through the formation of a highly reactive dichlorocarbene intermediate ($\mathrm{:CCl_2}$). This intermediate electrophilically attacks the nitrogen atom of the amine, leading to the synthesis of p-tolyl isocyanide ($\mathrm{CH_3C_6H_4NC}$), which is identified by its characteristically unpleasant odor.

The reaction sequence begins with the Hell-Volhard-Zelinsky reaction, which specifically targets the $\alpha$-carbon of a carboxylic acid to introduce a halogen atom. When propanoic acid is treated with chlorine in the presence of red phosphorus, a chlorine atom replaces a hydrogen at the $\alpha$-position, transforming the molecule into 2-chloropropanoic acid, $\mathrm{CH}_3\mathrm{CHClCOOH}$, which serves as intermediate A. Subsequent treatment with alcoholic KOH promotes a dehydrohalogenation reaction. In this elimination process, the strong base removes a hydrogen atom from the $\beta$-carbon alongside the chlorine atom from the $\alpha$-carbon, establishing a carbon-carbon double bond to yield $\mathrm{CH}_2=\mathrm{CHCOOH}$, also known as acrylic acid, as the final product B.

A redox reaction occurs when there is a net transfer of electrons between reactants, which is identifiable by a change in the oxidation states of one or more elements. While the first three options represent double displacement or acid-base processes where oxidation numbers remain entirely unchanged, the transformation $\mathrm{Zn} + 2\,\mathrm{AgCN} \rightarrow 2\,\mathrm{Ag} + \mathrm{Zn(CN)_2}$ involves zinc shifting from an oxidation state of 0 to +2 and silver moving from +1 to 0. This distinct shift in oxidation values confirms that electrons have been exchanged, marking it as a true redox process.

Strong oxidizing agents like permanganate induce the oxidative cleavage of carbon-carbon double bonds and the oxidation of terminal aldehyde groups into carboxylic acids. Applying this to $(\mathrm{CH}_3)_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO}$ causes the double bond to split, transforming the disubstituted alkene carbon into acetone $(\mathrm{CH}_3)_2 \mathrm{C}=\mathrm{O}$ and the remaining fragment, which includes the terminal aldehyde group, into malonic acid $\mathrm{HOOC}-\mathrm{CH}_2-\mathrm{COOH}$.

Picric acid is the common chemical name used to identify the compound 2, 4, 6-trinitrophenol. This aromatic substance is a derivative of phenol, featuring three strongly electron-withdrawing nitro groups substituted at the second, fourth, and sixth positions of the benzene ring, represented by the chemical formula $C_6H_2(OH)(NO_2)_3$.

The Cannizzaro reaction involves the base-catalyzed disproportionation of aldehydes lacking $\alpha$-hydrogens, where two equivalents of benzaldehyde react in the presence of $\mathrm{NaOD}$ and $\mathrm{D}_2\mathrm{O}$ to produce a benzoate ion and benzyl alcohol. During the reaction mechanism, a hydride ($\mathrm{H}^-$) is transferred from the carbonyl carbon of one benzaldehyde molecule to the other, forming the methylene group ($\mathrm{CH_2}$) of the benzyl alcohol, while the hydroxyl group of the alcohol is formed through exchange with the deuterated solvent, resulting in the final product $\mathrm{Ph-CH_2OD}$.

The reaction between propanal and methanal in a basic medium begins with an aldol condensation, where the enolate of propanal acts as a nucleophile to attack the electrophilic carbonyl carbon of methanal. This process constructs the backbone of the molecule, which is subsequently transformed through a series of steps-including the addition of cyanide followed by acid hydrolysis-that introduces a carboxylic acid functional group. The final molecule, identified with the molecular formula $\mathrm{C}_5\mathrm{H}_8\mathrm{O}_3$, possesses a chiral carbon atom. Because the synthesis does not involve any stereoselective reagents, the product is generated as a racemic mixture, meaning it contains equal amounts of both enantiomers. Furthermore, the presence of the carboxylic acid group makes the compound acidic, allowing it to react with a saturated $\mathrm{NaHCO}_3$ solution to liberate carbon dioxide gas.

Ozonolysis, involving treatment with $O_3$ followed by a reductive workup with $\mathrm{Zn-H_2O}$, cleaves carbon-carbon double bonds to form carbonyl groups at the original site of unsaturation. Cyclohex-1,3-diene undergoes simultaneous cleavage at both its double bonds to yield ethane-1,2-dicarbaldehyde from the internal carbon chain and glyoxal from the remaining two-carbon segment.

Similarly, 1-methylcyclopent-1-ene undergoes ring-opening ozonolysis because the endocyclic double bond breaks to form a linear chain where the carbon originally bearing the methyl group becomes a ketone and the other double-bonded carbon becomes a terminal aldehyde, resulting in 5-oxohexanal.