Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

The correct answer is CO2 because CO2 reacts with Grignard reagent (CH3MgI) to form an adduct, which upon hydrolysis yields CH3COOH and Mg(OH)I.

Option A (Only Statement I is correct) is correct because benzoic acid is indeed stronger than acetic acid due to the resonance stabilization of the benzoate ion, while the acetate ion is less stabilized.

The correct answer is A because Fehling's solution can distinguish benzaldehyde from acetaldehyde, and the C-H bond in benzaldehyde's CHO group is indeed stronger than in acetaldehyde's.

Option C (ethylene) is correct because the reaction sequence involves reduction of carboxylic acid to alcohol, chlorination to alkyl halide, and then elimination to alkene.

Option B (Cannizaro’s reaction) is correct because benzaldehyde, having no alpha-hydrogen, undergoes Cannizaro's reaction to yield benzyl alcohol and a salt of benzoic acid.

Option C (Aldehydic group) is correct because Fehling's solution is a reagent used to detect the presence of aldehydes, which are oxidized to carboxylic acids.

Option B (Acetone and Ethanol) is correct because both acetone and ethanol give a positive iodoform test, so I₂/NaOH cannot distinguish them.

Option A ((+) tartaric acid) is correct because tamarind contains dextrorotatory or (+) tartaric acid.

Option D (All of these) is correct because a mixture of calcium acetate and calcium formate on heating gives acetone, formaldehyde, and acetaldehyde due to different possible reaction pathways.

Option C (Esters) is correct because carboxylic acids react with diazomethane to form methyl esters and nitrogen gas.

Option A (Hell-Volhard-Zelinsky reaction) is correct because it involves the alpha-halogenation of carboxylic acids.

Option C (Reduces solution of ammoniacal silver nitrate or Fehling's solution) is correct because formic acid, unlike acetic acid, can reduce Tollen's reagent and Fehling's solution due to its aldehyde-like structure.

Option C (Both (a) and (b)) is correct because both formulas represent a saturated carboxylic acid.

Option B (MeCHO) is correct because aldehydes are more susceptible to nucleophilic attack at the carbonyl group than acid derivatives due to less steric hindrance and the absence of a leaving group.

Option B (2-methyl-2-butene) is correct because its oxidative cleavage with KMnO4/H+ yields a ketone (acetone) and a carboxylic acid (acetic acid).

Option A (Acids containing even number of carbon atoms) is correct because even-numbered carbon acids pack more efficiently in the crystal lattice, leading to higher melting points.

Option D (Acetamide) is correct because the reaction sequence involves hydrolysis of methyl cyanide to acetic acid, followed by amidation to acetamide.

Option A is correct because the reactivity of acid derivatives towards nucleophilic acyl substitution depends on the leaving group ability, with acid chlorides being the most reactive and amides the least.

Option A (Baeyer-Villiger oxidation) is correct because the reaction of acetophenone with perbenzoic acid to form phenyl acetate is a Baeyer-Villiger oxidation, which involves the insertion of an oxygen atom into a carbonyl compound.

Option D (Tollen’s reagent) is correct because benzaldehyde gives a positive Tollen's test (silver mirror), while acetone does not.

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

Option D (Acetaldehyde) is correct because Cannizzaro's reaction requires the absence of α-hydrogen atoms, and acetaldehyde possesses three α-hydrogen atoms.

Option C (Acetylation) is correct because the reaction of acetyl chloride with an amine results in the replacement of a hydrogen atom of the amine with an acetyl group, which is known as acetylation.

Option B (Oxygen) is correct because Nessler's reagent is $K_2[HgI_4]$, which does not contain oxygen.

Option D (lodoform test) is correct because both $\mathrm{CH}_3 \mathrm{CHO}$ and $\mathrm{CH}_3 \mathrm{COCH}_3$ give a positive iodoform test, so it cannot be used to distinguish them.

Option D (Saponification) is correct because alkaline hydrolysis of an ester is called saponification, which produces soap and alcohol.

Option B (Both electrophiles and nucleophiles) is correct because acetaldehyde has an electrophilic carbonyl carbon and can also undergo reactions involving the acidic alpha-hydrogens.

Option A (Ethanol) is correct because the oxidation of ethanol yields acetaldehyde, which further oxidizes to ethanoic acid (acetic acid), containing two carbon atoms.

The correct answer is B (I < IV < III < II < V) because boiling points increase with increasing molecular weight and strength of intermolecular forces, with alcohols having the highest due to hydrogen bonding.

Option A (70) is correct because the equivalent weight of the acid is calculated using the equivalent weights of the silver salt and silver.