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

The stoichiometric coefficient $x$ in a chemical reaction is determined by applying the law of conservation of mass, which necessitates that the number of atoms for every element must remain identical on both the reactant and product sides. To find the value of $x$ for compound 'D', the balanced chemical equation is derived by setting the total number of atoms and the total charge on the left-hand side equal to those on the right-hand side. By systematically applying the oxidation number method or the half-reaction method, one determines the molar ratios of all species involved in the reaction pathway. After ensuring that all elements, including the central metal atoms and the ligands, are stoichiometrically balanced across the entire reaction, the final coefficient $x$ for compound 'D' evaluates to 15.

The acidity of a compound is determined by the stability of its conjugate base, represented by the equilibrium $HA \rightleftharpoons H^+ + A^-$. Greater stability in the resulting conjugate base $A^-$ shifts the equilibrium more readily to the right, indicating a more acidic proton. This stability is governed by the electronic environment, where electron-withdrawing substituents disperse the generated negative charge through resonance or inductive effects, while electron-donating groups destabilize the conjugate base by increasing charge density. By evaluating the structural context at each position, one can rank these protons based on how effectively their respective conjugate bases delocalize negative charge.

Esters undergo a two-step nucleophilic reaction with an excess of Grignard reagent, where the first equivalent converts the ester into a ketone through nucleophilic acyl substitution, and the second equivalent performs a nucleophilic addition to this ketone to generate a tertiary alcohol. The methyl magnesium bromide nucleophile attacks the carbonyl carbon of the methyl propanoate to form the intermediate butan-2-one, which is then immediately attacked by another molecule of the Grignard reagent to produce a magnesium alkoxide species. Final acidic hydrolysis of this intermediate yields 2-methylbutan-2-ol as the major product 'P'.

Reaction mechanisms are primarily determined by the electronic distribution within the reactants, where the nucleophile attacks the electrophilic center to initiate bond formation. In processes governed by the stability of the products, factors such as resonance, hyperconjugation, and steric hindrance dictate the regioselectivity of the reaction pathway, such as the preference for specific substitution or addition modes. The correct outcome is identified by tracing the movement of electrons from the highest occupied molecular orbital of the nucleophile to the lowest unoccupied molecular orbital of the electrophile, ensuring the most stable intermediate or transition state is formed along the reaction coordinate.

Electrophilic aromatic substitution using $\mathrm{Br}_2/\mathrm{Fe}$ installs a bromine substituent on the aromatic ring, while the subsequent reduction of a nitro group to an amine is effectively achieved via $\mathrm{Fe}/\mathrm{H}^{+}$. The sequence proceeds with the diazotization of the amine using $\mathrm{HONO}$ to generate a diazonium salt intermediate, which is then converted into a chloro-substituted product through the Sandmeyer reaction with $\mathrm{CuCl}$. Finally, the side-chain oxidation of an alkyl group, such as a methyl group, into a carboxylic acid is accomplished using the strong oxidizing agent $\mathrm{KMnO}_4$.

Solubility in alkaline media is determined by a compound's intrinsic acidity or its susceptibility to base-promoted hydrolysis. Sodium bicarbonate acts as a weak base, reacting only with strong acids, while sodium hydroxide functions as a strong base capable of deprotonating phenols at room temperature. Compounds 2, 3, and 7 remain insoluble in both cold solutions because they lack acidic functional groups that can be readily removed under mild conditions, yet they become soluble upon heating with $\mathrm{NaOH}$ as the elevated temperature facilitates the hydrolysis of esters or amides into water-soluble carboxylate salts. Consequently, exactly three compounds from the set satisfy these specific reactivity criteria.

The formation of product 'A' in this reaction proceeds via the diazotization of a primary aromatic amine, a fundamental transformation for synthesizing aryl derivatives. By treating the amine with $NaNO_2$ and $HCl$ at low temperatures of $0-5^{\circ}C$, the nitrous acid generated in situ replaces the amino group to form the corresponding diazonium salt, represented by the structure $Ar-N_2^+ Cl^-$. This reaction is driven by the instability of the diazonium intermediate at higher temperatures, necessitating the maintenance of a cold environment to ensure the successful isolation and stabilization of 'A' as the primary product.

The conversion of salicylic acid to acetylsalicylic acid occurs via the acetylation of the phenolic hydroxyl group using acetic anhydride in the presence of an acid catalyst, denoted as $(CH_3CO)_2O / H^{+}$. Alternatively, the synthesis of methyl salicylate involves the esterification of the carboxylic acid group, which proceeds through a Fischer esterification reaction requiring methanol and an acid catalyst under heating conditions, denoted as $\mathrm{CH_3OH} / \mathrm{H}^{+}, \Delta$.

The compound represented in option A acts as the precursor for lactone formation through the selective oxidation of its terminal aldehyde group to a carboxylate ion using Tollens' reagent, followed by the acid-catalyzed cyclization of the intermediate hydroxy-acid. This intramolecular nucleophilic attack, where the hydroxyl oxygen attacks the carbonyl carbon of the carboxylic acid group to form an ester linkage, is thermodynamically favored due to the formation of a stable six-membered ring structure.

The reaction of glycine with chlorine in the presence of red phosphorus serves as an alpha-halogenation process that replaces one of the alpha-hydrogens to produce $\text{H}_2\text{N}-\text{CHCl}-\text{COOH}$. This modification results in a central carbon atom bonded to four distinct groups-a chlorine atom, an amino group, a carboxyl group, and a hydrogen atom-which defines a chiral center and makes the assertion true. The reason statement is incorrect because the presence of two chiral centers does not guarantee optical activity, as seen in meso compounds where an internal plane of symmetry renders the molecule optically inactive despite having stereogenic carbons.

The Ceric Ammonium Nitrate (CAN) test identifies alcohols by forming a red-colored complex, while the Iodoform test is specific for compounds containing either a methyl ketone group ($CH_3CO-$) or a secondary alcohol with a methyl group adjacent to the hydroxyl-bearing carbon ($CH_3CH(OH)-$). To satisfy both conditions, a compound must be an alcohol possessing a methyl carbinol moiety. Ethanol ($CH_3CH_2OH$), 2-propanol ($CH_3CH(OH)CH_3$), and 2-butanol ($CH_3CH_2CH(OH)CH_3$) fulfill both requirements, as they are alcohols that contain the necessary methyl carbinol structural unit. Methanol and 1-propanol fail the Iodoform test due to the absence of the required structural moiety, acetophenone is a ketone and thus fails the alcohol-specific CAN test, and phenol fails the Iodoform test while producing a brown-colored complex rather than the red coloration characteristic of alcohols with CAN. Therefore, exactly three compounds from the set provide positive results for both tests.

The reaction sequence involves the formation of a nucleophilic acetylide followed by an $S_N2$ alkylation and a stereoselective catalytic hydrogenation. First, sodium amide ($NaNH_2$) acts as a strong base to abstract the acidic terminal proton from $CH_3CH_2C \equiv C-H$, generating the nucleophilic acetylide anion, which then reacts with ethyl bromide ($CH_3CH_2Br$) to yield the disubstituted alkyne, 3-hexyne. Finally, the application of Lindlar's catalyst ($H_2, Pd/CaCO_3$) facilitates the syn-addition of hydrogen to the internal triple bond, ensuring the exclusive formation of the cis-alkene isomer, specifically cis-3-hexene.

Lithium borohydride ($LiBH_4$) functions as a specialized and selective reducing agent in organic synthesis, primarily valued for its specific reactivity toward certain functional groups. When this reagent interacts with molecules containing esters or lactones, it efficiently reduces the carbonyl moiety found within these structures. This process transforms the ester or lactone into the corresponding primary alcohol, which emerges as the major product of the reaction.

Fehling's solution acts as a mild oxidizing agent that reacts with aliphatic aldehydes of the form $R-CHO$ to produce a positive result, whereas aromatic aldehydes fail to undergo this oxidation. The positive result here dictates that the molecule must be an aliphatic aldehyde. Furthermore, the appearance of a blood-red color during the sodium nitroprusside test on the fusion extract confirms the presence of both nitrogen and sulfur in the compound, as this distinct coloration is only produced when both of these specific elements are identified.

Acetylation of the amino group in aniline using acetic anhydride in the presence of pyridine serves as a protective measure to reduce the strong electron-donating capability of the $-NH_2$ group, preventing uncontrolled oxidation or poly-substitution. The reaction involves the nucleophilic attack of the nitrogen lone pair on the carbonyl carbon of acetic anhydride, resulting in the formation of $N$-phenylethanamide, commonly known as acetanilide, as the intermediate product A. This modification effectively deactivates the amino group just enough to ensure that subsequent electrophilic aromatic substitution, such as nitration, occurs predominantly at the para-position.

The sequence initiates with the diazotization of aniline using sodium nitrite and hydrochloric acid at $273-278\ K$ to convert the amino group into a diazonium group, forming benzene diazonium chloride. Subsequent treatment of this diazonium salt with aqueous potassium iodide ($KI$) facilitates a nucleophilic substitution where the diazonium moiety is displaced by an iodine atom, yielding iodobenzene, which corresponds to the product B.

The molecular formula $\mathrm{C_9H_{10}O}$ corresponds to a degree of unsaturation of five, which indicates the presence of an aromatic benzene ring and one additional double bond. The inability of the compound to dissolve in $\mathrm{NaOH}$ or $\mathrm{HCl}$ rules out acidic phenolic groups and basic amine groups, while the negative response to 2,4-DNP indicates the absence of any aldehyde or ketone functionality. These chemical constraints point to an ether structure containing a carbon-carbon double bond, specifically methyl styryl ether with the structure $\mathrm{Ph-CH=CH-OCH_3}$.

Hydrogenation of this molecule adds one mole of hydrogen across the double bond to yield $\mathrm{C_9H_{12}O}$, confirming the presence of an alkene. Because the carbon-carbon double bond in this ether is non-terminal and maintains a rigid configuration, it allows for the formation of distinct cis and trans geometric isomers. Both of these stereoisomers satisfy all the stipulated chemical criteria, resulting in a total of two isomeric compounds.

The reaction sequence involves the diazotization of aniline with $NaNO_2/HCl$ at $273-278 K$ to form benzenediazonium chloride (A), followed by a nucleophilic substitution using $CuCN/KCN$ to yield benzonitrile (B). Finally, the acid-catalyzed hydrolysis of the nitrile group in (B) produces benzoic acid (C) as the terminal carboxylic acid product.

The degree of unsaturation ($DU$) for the compound $\mathrm{C}_{10}\mathrm{H}_{12}\mathrm{O}_2$ is calculated using the formula $DU = C + 1 - \frac{H}{2}$, which gives $10 + 1 - \frac{12}{2} = 5$. A benzene ring contributes 4 units of unsaturation, consisting of one ring and three $\pi$ bonds. Because the compound decolorizes alkaline $\mathrm{KMnO}_4$, it must possess an additional site of unsaturation in the side chain, such as an alkene group, which provides the remaining 1 unit of unsaturation. Therefore, the total number of $\pi$ bonds present in the compound is $3 + 1 = 4$.

A positive haloform test relies on identifying specific structural arrangements within a molecule, specifically the acetyl group, $\text{CH}_3\text{CO}-$, or the methyl carbinol group, $\text{CH}_3\text{CH(OH)}-$. By inspecting the provided chemical structures, you must look for a methyl group directly attached to a carbonyl carbon or a hydroxyl-bearing carbon. Each molecule containing at least one of these two motifs is capable of undergoing the reaction. Systematic verification of the entire set confirms that three distinct molecules possess these necessary structural features to yield a positive result.