Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option B ($\mathrm{pd - CaC}{\mathrm{O}_3} + \mathrm{BaS}{\mathrm{O}_4}$) is correct because Lindlar's catalyst is palladium supported on calcium carbonate, poisoned with barium sulfate or quinoline.

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

Option A (122) is correct because benzoic acid is a monobasic acid, so its molecular weight equals its neutralization equivalent.

Option B (Formic acid) is correct because the stings of bees, red ants, and wasps contain formic acid.

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

The correct answer is D (NH3) because HCHO forms hexamethylenetetramine, CH3CHO forms acetaldehyde ammonia trimer, and CH3COCH3 forms ketimine, showing different reaction patterns with NH3.

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.

The correct answer is D (Reaction of acid chloride with Grignard reagents) because this reaction leads to a tertiary alcohol, not a ketone.

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 D is correct because the carboxyl group deactivates the benzene ring and the catalyst (AlCl3) coordinates with the carboxyl group, preventing it from acting as a Lewis acid.

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 C (Formic acid) is correct because formic acid is obtained by the distillation of red ants.

Option B (Chloral) is correct because chloral (CCl3CHO) reduces Fehling's solution and gives a monocarboxylic acid on oxidation, and it can be obtained by chlorination of ethanol.

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 A (Only H-atom) is correct because the reaction of an acid with a metal involves the replacement of the hydrogen atom from the -COOH group with the metal, releasing hydrogen gas.

Option A (primary and secondary alcohols) is correct because primary alcohols oxidize to aldehydes, and secondary alcohols oxidize to ketones.

Option B ($\mathrm{CHCl}_3, \mathrm{NaOH}$) is correct because salicylaldehyde is formed from phenol via the Reimer-Tiemann reaction using chloroform and sodium hydroxide.

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 A is correct because Tollen's reagent is an ammoniacal silver nitrate solution, specifically [Ag(NH3)2]NO3.

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.

Option C (It cannot be used for all types of crops and soils) is correct because urea can be used for all types of crops and soils.

Option A (2,4 - DNP) is correct because acetaldehyde reacts with 2,4-Dinitrophenylhydrazine (2,4-DNP) to form an orange-colored precipitate of 2,4-Dinitrophenylhydrazone.

The correct answer is C ($\mathrm{CH}_3-\mathrm{C}(\mathrm{Cl}_2)-\mathrm{CH}_2-\mathrm{CH}_3$) because it forms a methyl ketone upon reaction with aqueous acetone, which then undergoes the iodoform reaction.

Option B (Partially deactivated palladised charcoal) is correct because Lindlar's catalyst is palladium on calcium carbonate, poisoned with lead or quinoline, used for the partial reduction of alkynes to cis-alkenes.

Option C (CHO · CHO) is correct because acetaldehyde oxidation with $Se{O_2}$ yields glyoxal.

Option C (${\left[ {Ag{{(N{H_3})}_2}} \right]^ + }$) is correct because silver chloride reacts with excess ammonia to form the diamminesilver(I) complex ion, ${\left[ {Ag{{(N{H_3})}_2}} \right]^ + }$.