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

The regioselectivity of the reaction is governed by the relative stability of the intermediate species, where the formation of the most stable carbocation dictates the course of the reaction. By following the principle that the reaction pathway leading to the most substituted and stable intermediate requires the lowest activation energy, the reaction preferentially proceeds through the route that minimizes the transition state energy. Thus, the structure depicted in choice D represents this most stable, kinetically and thermodynamically favored product, confirming it as the major product formed.

Sodium borohydride ($NaBH_4$) is a mild and selective reducing agent that specifically reduces aldehydes and ketones to primary and secondary alcohols, respectively. In the presence of less reactive functional groups like esters, $NaBH_4$ leaves them unaffected because it lacks the reducing power required to convert them, unlike stronger reducing agents such as $LiAlH_4$. Consequently, the ketone carbonyl group in the molecule is converted to a hydroxyl group while the ester functional group remains entirely intact in the final product.

Diisobutylaluminium hydride, commonly referred to as DIBAL-H, serves as a selective electrophilic reducing agent that facilitates the partial reduction of functional groups such as nitriles ($R-CN$), esters ($R-COOR'$), lactones, amides, and carboxylic acids. Because this reagent specifically targets the electrophilic center and terminates the reaction at the aldehyde stage rather than continuing to a primary alcohol, the transformation consistently yields an aldehyde as the final product. Identifying the correct chemical structure simply requires recognizing that the reactant's relevant group has been converted into an aldehyde moiety under the influence of this reagent.

The described chemical reactions are characteristic of aromatic aldehydes like benzaldehyde, which lack $\alpha$-hydrogen atoms and thus undergo the Cannizzaro reaction in the presence of a concentrated base, yielding benzyl alcohol as shown by $2 C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$. Furthermore, this compound reacts with acetophenone via Claisen-Schmidt condensation, a specific type of cross-aldol reaction, to form a single product, benzalacetophenone, following the equation $C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{OH^-} C_6H_5CH=CHCOC_6H_5 + H_2O$. The absence of $\alpha$-hydrogen atoms is the essential structural feature that prevents self-aldol condensation and enables these specific transformations.

Tollen's reagent, $\mathrm{[Ag(NH_3)_2]^{+}OH^{-}}$, acts as a mild oxidizing agent that selectively converts an aldehyde into a carboxylate ion, preserving other existing functional groups. Subsequent treatment with acidic methanol, $\mathrm{H}^{+}/\mathrm{CH_3OH}$, facilitates the conversion of the carboxylic acid into a methyl ester through Fischer esterification. Finally, the addition of the Grignard reagent, $\mathrm{CH_3MgBr}$, introduces a nucleophilic methyl group that attacks the electrophilic carbonyl center of the ester to produce the final tertiary alcohol.

The production of $C(CH_2OH)_4$ involves the reaction of an aldehyde having three acidic alpha-hydrogens, like acetaldehyde, with four equivalents of formaldehyde in an alkaline environment. The mechanism proceeds as three successive aldol condensations, which attach three hydroxymethyl groups to the alpha carbon of the starting aldehyde, creating an intermediate structure. This process concludes with a cross-Cannizzaro reaction between this intermediate and the final formaldehyde molecule, effectively reducing the carbonyl group to a primary alcohol and confirming that the synthesis requires exactly three aldol condensations and one Cannizzaro reaction.

Ozonolysis proceeds by the oxidative cleavage of a carbon-carbon double bond, which converts the alkene into two carbonyl-containing fragments, or into a single open-chain molecule with two carbonyl groups if the precursor is a cyclic alkene. To determine the specific cyclic precursor for a given product like 5-keto-2-methyl hexanal, one must identify the two carbonyl carbons-the aldehyde carbon and the ketone carbon-and theoretically reconnect them via a $C=C$ double bond to close the ring. This reverse-engineering process reconstructs the original cyclic alkene, ensuring that the ring size and the arrangement of substituents, such as the methyl group, match the carbon skeleton of the provided product.

The oxidation of toluene with $\text{KMnO}_4$ yields benzoic acid, which is subsequently converted into benzoyl chloride upon reaction with $\text{SOCl}_2$. This acid chloride ($B$) undergoes a Rosenmund reduction when treated with hydrogen gas in the presence of palladium on barium sulfate (\text{H}_2/\text{Pd}/\text{BaSO}_4$), a catalytic process that selectively reduces the acyl chloride group to an aldehyde. The use of$\text{BaSO}_4$poisons the catalyst, preventing further reduction of the aldehyde to a primary alcohol and resulting \in benzaldehyde (\text{C}_6\text{H}_5\text{CHO}$) as the final product $C$.

Reducing carboxylic acids to primary alcohols using lithium aluminum hydride is the initial step in this sequence, which transforms acetic acid into ethanol, $\mathrm{CH_3CH_2OH}$. Subsequent reaction with phosphorus pentachloride converts the alcohol into an alkyl halide by replacing the hydroxyl group with a chlorine atom, resulting in ethyl chloride, $\mathrm{CH_3CH_2Cl}$. In the final step, treating this alkyl halide with alcoholic potassium hydroxide induces a dehydrohalogenation reaction that removes hydrogen chloride to form the final product, ethylene, $\mathrm{CH_2=CH_2}$.

The reaction of sodium phenoxide with $CO_2$ at 125°C under pressure proceeds via the Kolbe-Schmitt reaction to form salicylic acid ($o$-hydroxybenzoic acid) as the intermediate product. Subsequent acetylation of the phenolic $-OH$ group in salicylic acid yields acetylsalicylic acid, which is commercially known as Aspirin.

The reduction of an aldehyde or ketone to a hydrocarbon can be performed using either the Clemmensen reduction or the Wolff-Kishner reduction, but the reaction conditions dictate the appropriate choice when sensitive functional groups are present. While $\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}$ is effective in the Clemmensen method, its acidic nature is unsuitable for substrates containing acid-sensitive moieties like alkenes or $-\mathrm{OH}$ groups. Consequently, the Wolff-Kishner reduction using hydrazine $\mathrm{NH}_2\ \mathrm{NH}_2$ and a strong base such as $\text{overset}{\ominus}{\mathrm{OH}}$ is required, as it successfully reduces the carbonyl group while maintaining the integrity of the rest of the molecule.

The acidity of carboxylic acids is primarily governed by the inductive effect, where electron-withdrawing substituents like chlorine pull electron density away from the $\mathrm{O-H}$ bond to facilitate the release of $\mathrm{H}^{+}$. Because this effect relies heavily on proximity, the chlorine atom in $\mathrm{CH_3CH_2CH(Cl)CO_2H}$ exerts its electron-pulling power most effectively by being attached to the alpha carbon, which is closest to the carboxyl group. This makes it significantly more effective at destabilizing the $\mathrm{O-H}$ bond compared to the other options, where the substituent is placed further along the carbon chain or is entirely absent.

Cannizzaro's reaction occurs when aldehydes lacking $\alpha$-hydrogen atoms are treated with a strong base, causing the molecules to undergo disproportionation through simultaneous oxidation and reduction. Because trichloroacetaldehyde, $\mathrm{CCl}_3\mathrm{CHO}$, has no hydrogen atoms attached to the alpha carbon, it reacts with $\mathrm{NaOH}$ to form the sodium salt of the corresponding carboxylic acid and the corresponding alcohol. This specific chemical transformation yields sodium trichloroacetate alongside 2,2,2-trichloroethanol, as illustrated by the reaction $\mathrm{CCl}_3\mathrm{CHO} + \mathrm{NaOH} \rightarrow \mathrm{CCl}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{CCl}_3\mathrm{COONa}$.

This reaction is a classic example of nucleophilic acyl substitution, where the ethoxide ion acts as a nucleophile to displace the chloride group from the acyl chloride. Specifically, the oxygen atom of the ethoxide group attacks the electrophilic carbonyl carbon of ethanoyl chloride, resulting in the formation of an ester known as ethyl ethanoate along with sodium chloride, as shown by the reaction $\text{CH}_3\text{COCl} + \text{NaOC}_2\text{H}_5 \rightarrow \text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaCl}$.

The Tollen's reagent test relies on the susceptibility of the aldehyde functional group to oxidation, which simultaneously reduces the silver diamine complex to metallic silver, creating a characteristic mirror effect. While ketones like acetone and benzophenone remain unreactive under these conditions, aldehydes readily participate in this redox reaction:
$RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag(s) + 4NH_3 + 2H_2O$
Although both formaldehyde and acetaldehyde are known to yield a silver mirror due to their reactive carbonyl structures, acetaldehyde serves as the representative compound in this context that facilitates the oxidation process to produce the reflective metallic deposit.

Ozonolysis involves the oxidative cleavage of a carbon-carbon double bond to form two carbonyl compounds. Given that the reaction yields two moles of an aldehyde with a molecular mass of 44 u, we can identify this product by analyzing the general aldehyde structure $R-CHO$. The functional group $CHO$ has a mass of $12 + 1 + 16 = 29$ u, implying that the alkyl group $R$ must have a mass of $44 - 29 = 15$ u, which corresponds to a methyl group ($CH_3$). This confirms the aldehyde is acetaldehyde, $CH_3CHO$. Combining two molecules of acetaldehyde by removing the oxygen atoms and creating a double bond between the former carbonyl carbons results in $CH_3CH=CHCH_3$, which is identified as 2-butene.

The production of a distinct, fruity odor in this chemical test indicates the formation of an ester, a process known as esterification that occurs when a carboxylic acid reacts with an alcohol in the presence of a concentrated acid catalyst. Since the reaction involves ethanol as the alcohol, the missing liquid must be a carboxylic acid to facilitate this transformation. Comparing the options, $\mathrm{CH}_3\mathrm{COOH}$ is the only choice that fits this requirement, leading to the following reaction: $\mathrm{CH}_3\mathrm{COOH} + \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \xrightarrow{\text{Conc. } \mathrm{H}_2\mathrm{SO}_4} \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5 + \mathrm{H}_2\mathrm{O}$.

The rate of a multi-step reaction is governed by the slowest stage in the mechanism, known as the rate-determining step. In the Cannizzaro reaction, an initial nucleophilic attack by $\mathrm{OH}^-$ on $\mathrm{PhCHO}$ creates a tetrahedral intermediate, which must then donate a hydride ion to another molecule of $\mathrm{PhCHO}$. This intermolecular transfer of the hydride ion is chemically demanding because it requires breaking a strong $\mathrm{C}-\mathrm{H}$ bond while simultaneously attacking the carbonyl carbon of the second aldehyde. Given this significant energetic barrier, the transfer of hydride to the carbonyl group constitutes the rate-determining step of the process.

When an alkyl benzene reacts with a strong oxidizing agent like alkaline $\mathrm{KMnO}_4$, the entire alkyl side chain is systematically converted into a carboxyl group attached directly to the benzene ring. This oxidative cleavage occurs regardless of the original side chain length, causing the ethyl group in ethylbenzene to be fully transformed into the functional group $-\mathrm{COOH}$. Consequently, the reaction consistently results in the formation of benzoic acid.

The acidity of carboxylic acids is fundamentally determined by the stability of their conjugate base, which is enhanced by electron-withdrawing substituents through the inductive effect and destabilized by electron-donating groups. Among these compounds, $(\mathrm{Me})_2 \mathrm{CH} \mathrm{COOH}$ (D) contains an electron-donating isopropyl group that makes it the least acidic, followed by the reference compound $\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}$ (A). Replacing a hydrogen in the methyl group with a methoxy substituent produces $\mathrm{MeOCH}_2 \mathrm{CO}_2 \mathrm{H}$ (B), which is more acidic due to the electron-withdrawing nature of oxygen, while the three highly electronegative fluorine atoms in $\mathrm{CF}_3 \mathrm{CO}_2 \mathrm{H}$ (C) exert the most powerful inductive effect to create the strongest acid. Consequently, the increasing order of acid strength is $D < A < B < C$.