Aldehydes & ketones & carboxylic acids practice · Medium

Option B (HCOOH) has the lowest pKa value because it is the strongest acid among the given options.

Option D ($\alpha -$ hydroxy ketone) is correct because benzoin is an $\alpha$-hydroxy ketone formed from benzaldehyde.

Option C ($\mathrm{HCOOH}$) is correct because formic acid (HCOOH) is the strongest acid among the given options, and stronger acids have lower $\mathrm{p}{K_a}$ values.

Option D (Clemmensen reduction) is correct because the reaction described is the Clemmensen reduction, which reduces a ketone to an alkane using zinc amalgam and concentrated HCl.

The correct answer is B because both A and R are true, but R is not the correct explanation of A. The acidity is due to the -I effect of chlorine, not the +R effect.

Option D (Assertion (A) and Reason (R) both are false) is correct because formic acid is not the strongest monocarboxylic acid, and trifluoroacetic acid is stronger than formic acid.

Option C is correct. The assertion is correct because electron-withdrawing groups increase acidity, but the reason is incorrect because chlorine has a -I effect, not a +M effect.

Option B (${Cl}_3 {C} . {COOH}>{Cl}_2 {CH} \cdot {COOH}> Cl. {CH}_2 {COOH}>{CH}_3 {COOH}$) is correct because the acidity increases with the number of chlorine atoms due to the electron-withdrawing inductive effect (-I effect) of chlorine.

Option C ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{COCH}_3$) is correct because both compounds give a positive iodoform test.

The correct answer is D (All of the above) because the formation of A is a nucleophilic addition, B is optically active, and the intermediate in step I is a hydroxy alkoxide ion.

Option D (It does not reduce Tollens' reagent) is correct because formic acid (HCOOH) *does* reduce Tollens' reagent due to the presence of an aldehyde-like group.

Option C (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CHO}$) is correct because benzaldehyde undergoes Cannizzaro's reaction, reduces Schiff's reagent, but does not reduce Fehling's solution.

Option A (C6H5CHO) is correct because benzaldehyde (C6H5CHO) lacks alpha-hydrogens and undergoes Cannizzaro's reaction with 50% NaOH to give alcohol and acid.

Option D (Ester) is correct because a silver salt of a fatty acid reacts with an alkyl halide to form an ester in the Hunsdiecker reaction.

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the reason for benzaldehyde's lower reactivity is due to resonance stabilization, not solely the hybridization of its carbon atoms.

Option D (Urotropine) is correct because the reaction of ammonia and formaldehyde produces hexamethylenetetramine, also known as urotropine.

Option A (I > II > III) is correct because the rate of decarboxylation depends on the stability of the carbanion formed, which is influenced by the -I effect of the halogens.

Option A (3 – Pentanone) is correct because 3-pentanone (a ketone) forms phenyl hydrazone, gives negative Tollen's and iodoform tests, and yields n-pentane upon reduction.

Option D (All of the above) is correct because carbonyl compounds are used in the production of fabrics, flavorings, plastics, and drugs.

Option B is correct because the reduction of benzoyl chloride with Pd and BaSO4 (Rosenmund reduction) yields benzaldehyde.

Option A is correct. Propanol-2 (isopropyl alcohol) gives a positive iodoform test due to the presence of a methyl ketone structure after oxidation, while propanol-1 does not. Acetaldehyde (CH3CHO) *does* give a positive iodoform test.

Option C is correct. The assertion is correct that isobutanal does not give iodoform test, but the reason is incorrect because it does have alpha hydrogens.

Option D (Trifluoroacetic acid) is correct because the presence of three highly electronegative fluorine atoms increases the acidity due to their strong -I effect.

Option A (Hexamethylene tetramine) is correct because formaldehyde and ammonia react to form hexamethylene tetramine.

Option D (A and R are false) is correct because aldehydes, not ketones, react with Tollen's reagent, and the reason is that aldehydes have a hydrogen attached to the carbonyl group, while ketones do not.

Option B (Claisen condensation) is correct because the reaction of two molecules of ethyl acetate in the presence of sodium ethoxide to form acetoacetic ester is a Claisen condensation.

Option D ($\mathrm{Al}^{3+}$) is correct because, according to the Hardy-Schulze rule, ions with higher charges have greater coagulating power for colloids.

Option D ($\mathrm{CCl}_3-\mathrm{COOH}$) is correct because the presence of three highly electronegative chlorine atoms increases the acidity, thus lowering the pKa value.

Option B (Hydrolysis) is correct because methyl cyanide (acetonitrile) undergoes hydrolysis to form acetic acid.

Option D (An aromatic aldehyde or HCHO) is correct because only aldehydes without alpha-hydrogens undergo Cannizzaro reaction, which involves disproportionation into an alcohol and a carboxylic acid.