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

The Gattermann-Koch reaction serves as an essential method for the direct formylation of benzene or its substituted derivatives to synthesize aromatic aldehydes. By subjecting the aromatic substrate to carbon monoxide ($CO$) and hydrogen chloride ($HCl$) in the presence of anhydrous $AlCl_3$ and $CuCl$ as catalysts, the process generates a formyl cation that acts as an electrophile to substitute a hydrogen atom on the aromatic ring. This transformation results in the formation of benzaldehyde, or its respective derivative, as the major product.

The reaction between two moles of benzaldehyde and one mole of acetone in the presence of dilute aqueous $\mathrm{NaOH}$ proceeds via a Claisen-Schmidt condensation to form dibenzylideneacetone, which has the structure $C_6H_5CH=CHCOCH=CHC_6H_5$. This molecule consists of two benzene rings, two conjugated carbon-carbon double bonds, and one carbonyl group. The total number of $\pi$ bonds is the sum of 3 $\pi$ bonds from each benzene ring, 1 $\pi$ bond from each alkene unit, and 1 $\pi$ bond from the carbonyl group, yielding $3 \times 2 + 2 + 1 = 9$ $\pi$ bonds.

The reaction follows the mechanism of electrophilic addition to alkenes, where the protonation of the double bond creates the most stable carbocation intermediate. The initial secondary carbocation undergoes a 1,2-hydride shift to form a more stable tertiary carbocation, which is the driving force for the reaction. The nucleophilic bromide ion then attacks this tertiary carbocation to form the thermodynamically favored product, 2-bromo-2-methylbutane.

The reaction proceeds through the principle of Neighboring Group Participation, where an adjacent phenyl ring acts as an internal nucleophile to facilitate the substitution process. The electron-rich pi system of the phenyl ring attacks the carbon bearing the leaving group, effectively displacing the $\mathrm{Br}$ atom through an intramolecular mechanism. This distinct mechanistic pathway leads to the formation of product 'P' via a bridged intermediate, which accelerates the overall reaction rate compared to standard substitution routes.

The Claisen-Schmidt condensation proceeds through a stoichiometric relationship where one molecule of acetone reacts with two molecules of benzaldehyde to form one molecule of dibenzalacetone. Even though 87 g of acetone is provided as a reactant, the stoichiometric requirement for the production of 351 g of dibenzalacetone determines the mass of benzaldehyde needed. Given that the molar mass of dibenzalacetone is 234 g/mol, the target mass corresponds to 1.5 moles of product. The reaction requires two moles of benzaldehyde for every mole of product formed, which necessitates a total of 3 moles of benzaldehyde. Multiplying these 3 moles by the molar mass of benzaldehyde, which is 106 g/mol, is calculated as:

$106 \times 3 = 318$

Thus, 318 g of benzaldehyde is required for the synthesis.

The synthesis relies on the selective reduction of a carboxylic acid derivative to an aldehyde using $\text{DIBAL-H}$, which is preferred over $\text{LiAlH}_4$ because it prevents over-reduction to a primary alcohol. Alcoholic $\text{NaOH}$ is subsequently employed to facilitate $E2$ dehydrohalogenation for the introduction of unsaturation, whereas the use of $\text{Zn/HCl}$ performs a Clemmensen reduction to convert carbonyl groups into methylene units. This specific reagent sequence is essential to navigate the necessary chemoselectivity required to transform the starting material into the target hydrocarbon while preserving specific functional group configurations.

The Hell-Volhard-Zelinsky reaction specifically facilitates the halogenation of the $\alpha$-carbon in a carboxylic acid that possesses at least one $\alpha$-hydrogen atom. When the carboxylic acid reacts with a halogen such as $Br_2$ or $Cl_2$ in the presence of catalytic red phosphorus, the mechanism proceeds through an acyl halide intermediate that significantly acidifies the $\alpha$-protons, allowing for controlled electrophilic substitution at that site. This chemical pathway results in the halogen being attached directly to the carbon adjacent to the carboxyl group, yielding the structure depicted in option A.

Catalytic hydrogenation of propanenitrile using $H_2$ and $Ni$ reduces the cyano group to a primary amine, yielding propan-1-amine as $A$. Subsequent treatment of this primary aliphatic amine with $HNO_2$ proceeds through the formation of a highly unstable alkyl diazonium salt, which rapidly undergoes substitution by water upon the evolution of nitrogen gas to produce propan-1-ol as $B$.

The synthesis of salicylaldehyde from phenol is a classic transformation known as the Reimer-Tiemann reaction, which involves treating phenol with chloroform (\text{CHCl}_3$) \in the presence of an aqueous base like$\text{NaOH}$. This reaction proceeds through the generation of a highly reactive dichlorocarbene intermediate (\text{:CCl}_2$), which acts as an electrophile to attack the ortho position of the phenoxide ion. Following this electrophilic aromatic substitution, hydrolysis of the resulting intermediate yields the final salicylaldehyde product.

The Gatterman-Koch reaction is an electrophilic aromatic substitution process in which benzene or substituted benzenes react with carbon monoxide ($CO$) and hydrogen chloride ($HCl$) in the presence of anhydrous aluminium chloride ($AlCl_3$) and cuprous chloride ($CuCl$) to synthesize benzaldehyde or its derivatives. This transformation involves the generation of a reactive formyl cation ($HCO^+$) intermediate, which facilitates the direct introduction of a formyl group ($-CHO$) onto the aromatic ring as shown by the reaction $C_6H_6 + CO + HCl \xrightarrow{AlCl_3/CuCl} C_6H_5CHO$.

Diamond features a covalent network structure where each carbon atom undergoes $sp^3$ hybridization, bonding to four other carbon atoms to form a rigid, three-dimensional tetrahedral arrangement. This geometry, characterized by bond angles of $109.5^{\circ}$, creates the highly stable lattice structure inherent to this specific allotrope.

Ozonolysis involves the oxidative cleavage of carbon-carbon double bonds, where the molecule is effectively severed at the alkene linkage and each vinylic carbon is transformed into a carbonyl group. For 2-hexene, which has the structure $\mathrm{CH_3-CH_2-CH_2-CH=CH-CH_3}$, the double bond is cleaved to split the six-carbon chain into a four-carbon segment and a two-carbon segment. Because each of the carbons involved in the double bond is bonded to a hydrogen atom, this reaction produces aldehydes rather than ketones. Consequently, the four-carbon fragment is oxidized into butanal ($\mathrm{CH_3-CH_2-CH_2-CHO}$) and the two-carbon fragment is oxidized into acetaldehyde ($\mathrm{CH_3-CHO}$).

The reaction sequence initiates with the conversion of cyclohexylamine to cyclohexanol (P) via the formation of an unstable diazonium salt using nitrous acid ($HNO_2$). Treatment of this secondary alcohol with PCC oxidizes it to cyclohexanone (Q). Subjecting cyclohexanone to dilute sodium hydroxide ($NaOH$) induces a self-aldol condensation followed by dehydration to produce 2-(1-cyclohexenyl)cyclohexanone (R).

Aniline undergoes diazotization when treated with nitrous acid, generated in situ from $NaNO_2$ and $HCl$ at $0-5^{\circ}C$, to form benzenediazonium chloride as the intermediate 'A'. The subsequent reaction with hypophosphorous acid ($H_3PO_2$) and water results in the reductive deamination of the diazonium group, replacing the functional group with a hydrogen atom to yield benzene as the final product 'B'.

The reaction proceeds via an elimination mechanism where the formation of the major product is dictated by Saytzeff's rule. This principle states that the most highly substituted alkene is the major product because it possesses the highest thermodynamic stability due to hyperconjugation and inductive effects. Consequently, the base abstracts a proton from the $\beta$-carbon that results in the most substituted double bond, ensuring that the isomer corresponding to option A is the most stable and predominant species formed.

The oxidation of cyclohexanecarbaldehyde ($C_7H_{12}O$) by Tollen's reagent is represented by the chemical equation
$C_7H_{12}O + 2[Ag(NH_3)_2]OH \rightarrow C_7H_{11}O_2NH_4 + 2Ag + 3NH_3 + H_2O$
which indicates that one mole of aldehyde produces three moles of ammonia. The molar mass of the aldehyde is calculated as $(7 \times 12) + (12 \times 1) + 16 = 112\,\text{g/mol}$. The number of moles in $131.8\,\text{kg}$ is $\frac{131.8\,\text{kg}}{112\,\text{g/mol}} = 1.1768\,\text{kmol}$. Applying the stoichiometric ratio, the amount of ammonia produced is $3 \times 1.1768 = 3.5304\,\text{kmol}$. Using the molar mass of ammonia ($17\,\text{g/mol}$), the mass generated is $3.5304 \times 17 = 60.0168\,\text{kg}$, which is $60\,\text{kg}$ to the nearest integer.

The Clemmensen reduction, which employs $\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$, is a reliable method for transforming a carbonyl group into a $-\mathrm{CH}_2-$ group, making the reason statement accurate. However, the assertion is false because this reaction medium is strongly acidic, rendering it incompatible with substrates that possess acid-sensitive functional groups such as alcohols. These groups react readily with the $\mathrm{HCl}$ present in the reagent, leading to undesirable side products instead of the target reduction.

Benedict's solution, containing alkaline copper sulphate and sodium citrate, detects reducing sugars by oxidizing aldehyde or alpha-hydroxy ketone functional groups to produce a characteristic reddish precipitate of $\mathrm{Cu_2O}$. Neutral $\mathrm{FeCl}_3$ solution identifies phenols through the formation of a distinct violet-colored coordination complex, while the reaction with alkaline chloroform serves as the carbylamine test to confirm the presence of primary amines via the generation of a foul-smelling isocyanide. Potassium iodide combined with sodium hypochlorite facilitates the haloform reaction to detect the methyl ketone functional group, which is evidenced by the appearance of a yellow precipitate of iodoform.

The oxidation of D-glucose using bromine water selectively converts the terminal aldehyde group into a carboxylic acid, yielding gluconic acid as product A (\text{COOH}(\text{CHOH})_4\text{CH}_2\text{OH}$). Subsequent reaction with concentrated nitric acid acts as a strong oxidizing agent that converts both the terminal carboxylic acid group and the primary alcohol group into carboxylic acid groups, resulting \in the formation of product E, saccharic acid (\text{COOH}(\text{CHOH})_4\text{COOH}$).

Analyzing an organic reaction requires identifying the reactant, the specific reagents, and the reaction conditions to determine the appropriate pathway, such as electrophilic addition, nucleophilic substitution, or rearrangement. As the chemical reaction and options are missing from your request, please provide the full question text or an image of the reaction. Once the details are supplied, the mechanism, regioselectivity, and the rationale behind the correct product formation will be thoroughly explained based on established electronic and steric principles.