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

The reaction is an instance of the Hofmann Bromamide Degradation, where a primary amide reacts with bromine in the presence of an aqueous base to yield a primary amine with one carbon atom fewer than the starting material. This transformation proceeds through the formation of an isocyanate intermediate, $R-N=C=O$, which subsequently hydrolyzes to the corresponding primary amine, $R-NH_2$, and carbon dioxide (as carbonate). Consequently, the structure $A$ corresponds to the amine produced by the removal of the carbonyl group from the starting amide substrate.

Sodium borohydride ($\mathrm{NaBH}_4$) acts as a chemoselective reducing agent that efficiently converts aldehydes and ketones into their respective primary and secondary alcohols. A critical limitation of this reagent is its inability to reduce carboxylic acid groups, which remain completely unaffected regardless of the reaction conditions. Consequently, when predicting the final products of such a reaction, one must ensure that any carboxylic acid moiety present in the starting material persists in its original form, while only the other susceptible functional groups undergo reduction.

The reaction produces 2-bromo-3-methylpentane, which has two chiral centers identified by the four distinct groups attached to each chiral carbon. In the structure $CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3$, the carbon at position 2 is bonded to a hydrogen atom, a bromine atom, a methyl group, and a sec-butyl group ($CH(CH_3)CH_2CH_3$), making it a chiral center. Similarly, the carbon at position 3 is bonded to a hydrogen atom, a methyl group, an ethyl group, and a 1-bromoethyl group ($CH(Br)CH_3$), confirming it as the second chiral center. Thus, the product contains a total of 2 chiral carbons.

The conversion of benzoic acid to benzaldehyde in one step is facilitated by passing the vapours of benzoic acid over heated manganous oxide ($\mathrm{MnO}$) at a temperature of $300^\circ\mathrm{C}$. This specific catalytic process promotes the decarboxylation and reduction of the carboxylic acid group directly into an aldehyde. Other reagents such as $\mathrm{LiAlH}_4$ and $\mathrm{NaBH}_4$ are reducing agents that typically convert carboxylic acids to primary alcohols, while $\mathrm{KMnO}_4$ is a strong oxidizing agent that cannot perform this transformation.

The Hell-Volhard-Zelinsky reaction involves the selective halogenation of carboxylic acids at the $\alpha$-carbon when carried out in the presence of catalytic red phosphorus. During this process, an $\alpha$-hydrogen atom is replaced by a halogen, transforming the starting material into an $\alpha$-halo carboxylic acid. Recognizing this specific substitution pattern allows for the identification of product P as the structure containing the halogen atom at the $\alpha$-position.

The reaction of benzene with ethyl chloride in the presence of anhydrous $AlCl_3$ proceeds via Friedel-Crafts alkylation to produce ethylbenzene ($C_6H_5CH_2CH_3$) rather than the target acetophenone. Achieving the synthesis of acetophenone requires Friedel-Crafts acylation, which involves the reaction of benzene with an acyl chloride, such as $CH_3COCl$, to effectively substitute a hydrogen atom with the $-COCH_3$ acetyl group.

Determining the degree of unsaturation reveals the number of double bonds in the hydrocarbon $X$. A 10 mg sample of the compound with a molar mass of 80 g/mol corresponds to $0.125 \text{ mmol}$ of the substance. Since 8.4 mL of $\mathrm{H}_2$ at STP is consumed, which represents $0.375 \text{ mmol}$ of hydrogen gas, the molar ratio of $\mathrm{H}_2$ to $X$ is $\frac{0.375}{0.125} = 3$. This confirms that the hydrocarbon contains three double bonds.

The structure of the compound with the molecular formula $\mathrm{C}_6\mathrm{H}_8$ and three double bonds is 1,3,5-hexatriene, which is written as $\mathrm{H}_2\mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$. Ozonolysis effectively cleaves each of these three double bonds, breaking the six-carbon chain into smaller segments. The cleavage of the two terminal double bonds yields two molecules of formaldehyde, while the cleavage of the internal double bond generates two molecules of glyoxal (a dialdehyde). Consequently, the total number of fragments produced from this ozonolysis reaction is four.

The nucleophilic addition of the Grignard reagent $\mathrm{CH}_3\mathrm{MgBr}$ to the cyano group of propionitrile ($\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CN}$) yields an imine salt intermediate, which upon acidic hydrolysis ($\mathrm{H}_3\mathrm{O}^+$) undergoes conversion to the ketone butanone ($\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}-\mathrm{CH}_3$). The final step involves the Clemmensen reduction of this ketone using an amalgam of zinc and mercury in the presence of hydrochloric acid, which selectively reduces the carbonyl group ($\mathrm{C=O}$) to a methylene group ($\mathrm{CH}_2$), resulting in the formation of n-butane ($\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$) as the final product $C$.

The photolytic decomposition of benzoyl peroxide, $(\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2)_2$, proceeds via the homolytic cleavage of the weak oxygen-oxygen bond to generate two benzoyloxy radicals, $\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^{\bullet}$. This species serves as the intermediate 'X', which subsequently undergoes decarboxylation to release carbon dioxide and form the phenyl radicals, $\mathrm{C}_6\mathrm{H}_5^{\bullet}$.

The stability of the enol form is significantly enhanced in $\beta$-dicarbonyl compounds like pentane-2,4-dione due to the formation of a stable six-membered chelated ring through intramolecular hydrogen bonding and extended conjugation. Unlike simple aldehydes or ketones where the keto form is thermodynamically more stable, the resonance stabilization and the chelation effect in the enol tautomer of these systems shift the equilibrium position substantially toward the enol form.

Acid-catalyzed reactions often proceed through a carbocation intermediate where the reaction pathway is governed by the relative stability of the intermediate and Zaitsev's rule for elimination products. Upon formation of the initial carbocation, a 1,2-hydride shift occurs to generate a more stable tertiary carbocation, ensuring the formation of the most substituted alkene ($C=C$) as the thermodynamic major product. This sequence of carbocation rearrangement followed by deprotonation is the fundamental principle determining the selectivity and the major product yield in these synthetic pathways.

Esterification is a fundamental nucleophilic acyl substitution where the alcohol acts as a nucleophile, attacking the carbonyl carbon of the carboxylic acid to displace the hydroxyl group and form an ester according to the general pathway $\mathrm{R-COOH} + \mathrm{R'-OH} \rightleftharpoons \mathrm{R-COOR'} + \mathrm{H_2O}$. The rate of this reaction is directly influenced by the electrophilicity of the carbonyl carbon, meaning the introduction of electron-withdrawing groups on the carboxylic acid increases the partial positive charge on that carbon, which in turn enhances the susceptibility to nucleophilic attack and accelerates the overall rate of esterification.

This reaction is governed by the principles of Aldol condensation, which occurs when carbonyl compounds possessing at least one acidic $\alpha$-hydrogen react in the presence of a base. The fundamental step involves the base abstracting an $\alpha$-proton to generate an enolate ion, which then functions as a nucleophile to attack the electrophilic carbonyl carbon of another molecule. This mechanism establishes a new carbon-carbon bond, leading to the formation of a $\beta$-hydroxy aldehyde or ketone structure consistent with the transformation displayed in the second option.

The starting material, $(\pm) \mathrm{Ph}(\mathrm{C}=\mathrm{O})\mathrm{C}(\mathrm{OH})(\mathrm{CN})\mathrm{Ph}$, is a racemic mixture containing one chiral center, and the reaction with $\mathrm{HCN}$ introduces a new chiral center at the carbonyl position. This new center can form in either $(R)$ or $(S)$ configuration, producing four potential combinations: $(R,R)$, $(S,S)$, $(R,S)$, and $(S,R)$. Since the final molecule features two equivalent chiral centers, the $(R,S)$ and $(S,R)$ forms represent the same meso compound, leaving a total of 3 distinct stereoisomers.

Lindlar's catalyst, which consists of palladium deposited on barium sulfate and poisoned with quinoline, is specifically engineered to perform the selective partial reduction of alkynes to cis-alkenes while remaining inert toward other functional groups such as carbonyls. Consequently, this poisoned catalyst cannot facilitate the reduction of an aldehyde like acetaldehyde ($CH_3CHO$) into a primary alcohol ($CH_3CH_2OH$), rendering the reaction represented in the option chemically impossible under standard conditions.

Fehling's reagent is characterized by the presence of a copper complex containing the metal in the $+2$ oxidation state. The $\text{Cu}^{2+}$ ion exhibits a $3d^9$ electronic configuration, which leaves a single unpaired electron in the $d$-orbitals. The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ represents the number of unpaired electrons. Substituting $n = 1$ leads to $\mu = \sqrt{3} \approx 1.73$ B.M., which rounds to 2 as the nearest integer.

Fehling's reagent, composed of $\mathrm{CuSO}_4$ and alkaline sodium potassium tartrate, acts as a specific test used to differentiate aliphatic aldehydes from other compounds. This mild oxidizing agent is insufficient to oxidize aromatic aldehydes, which remain unaffected when treated with it. As a result, any reaction sequence attempting to oxidize an aromatic aldehyde with Fehling's reagent cannot proceed.

The reaction proceeds through the formation of a carbocation intermediate, which undergoes rapid rearrangement-such as a hydride or alkyl shift-to achieve a more stable tertiary configuration. Following this rearrangement, the loss of a proton is governed by Zaitsev's rule, which dictates that the elimination favors the formation of the most substituted and, consequently, the most thermodynamically stable alkene. This energetic preference for the most stable transition state and product ensures that the rearrangement product corresponds to the major structure observed in option B.

The Etard reaction involves the oxidation of toluene using chromyl chloride ($CrO_2Cl_2$) in a non-polar solvent such as carbon disulfide, which forms an intermediate chromium complex that upon hydrolysis yields benzaldehyde. In the Gattermann-Koch reaction, benzene is treated with carbon monoxide ($CO$) and hydrogen chloride ($HCl$) in the presence of anhydrous aluminum chloride ($AlCl_3$) and cuprous chloride ($CuCl$) to produce benzaldehyde through direct formylation.

The reaction proceeds via an E2 elimination mechanism governed by Zaitsev's rule, which dictates that the major product is the most substituted and thermodynamically stable alkene. The strong base abstracts a proton from the $\beta$-carbon to facilitate the formation of the alkene with the highest degree of alkyl substitution, as this configuration possesses the lowest potential energy due to increased hyperconjugation and minimized steric hindrance. Consequently, the trisubstituted alkene is significantly more stable than less substituted isomers, making it the major product in the reaction mixture.