Aldehydes & ketones & carboxylic acids practice · Medium

The correct answer is B (D > C > B > E > A) because the acidity increases with the electron-withdrawing effect of the substituents, with fluorine being the strongest electron-withdrawing group among the given options.

Option C (3-pentanone) is correct because 3-pentanol is oxidized to 3-pentanone using aluminium tertiary butoxide in the presence of acetone (Oppenauer oxidation).

The correct answer is Ethyl methyl ketone because the reaction between a dialkylcadmium reagent and an acyl chloride leads to the formation of a ketone.

Option B is correct because the reaction with $NaHSO_3$ is a simple nucleophilic addition, without the elimination of any molecule like water.

Option C ($\mathrm{N}{\mathrm{H}_2}\mathrm{CONHN}{\mathrm{H}_2}$) is correct because it is the correct chemical formula for semicarbazide.

The correct answer is $C{H_3}COOH$ (Option B) because the reaction sequence involves ammonolysis, dehydration, and hydrolysis, ultimately converting the starting material to acetic acid.

Option D (Benzyl alcohol) is correct because it lacks the required $CH_3CO-$ or $CH_3CH(OH)-$ group to give a positive iodoform test.

Option D (Aldol condensation) is correct because ethanal reacting with alkali to form 3-hydroxybutanal is a classic example of aldol condensation.

Option B ($\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$) is correct because Cannizzaro reaction is given by aldehydes which do not have alpha-hydrogen atoms, and benzaldehyde fits this criterion.

Option A (I $>$ II $>$ III $>$ IV) is correct because the rate of esterification decreases with increasing steric hindrance around the alcohol's hydroxyl group.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}$) is correct because only aldehydes react with both Tollen’s reagent and Fehling’s solution.

Option B (Oxalic acid) is correct because oxalic acid (COOHCOOH) is a dibasic acid, meaning it can donate more than one proton, thus it is a polyprotic acid.

Option B ($\mathrm{NaHC}{\mathrm{O}_3}$) is correct because the carbon dioxide ($\mathrm{C}{\mathrm{O}_2}$) released in the reaction comes from the bicarbonate ion ($\mathrm{HC}{\mathrm{O}_3}^-$) in $\mathrm{NaHC}{\mathrm{O}_3}$.

Option D ($5 \;\mathrm{\% }\;\mathrm{NaHC}{\mathrm{O}_3}$) is correct because benzoic acid reacts with sodium bicarbonate while phenol does not, allowing for separation.

Option A ($R - C{H_2} - OH$) is correct because $LiAl{H_4}$ reduces carboxylic acids to primary alcohols.

Option D ($HCOONa + {C_6}{H_5}C{H_2}OH$) is correct because formaldehyde undergoes oxidation to form formate, while benzaldehyde undergoes reduction to form benzyl alcohol in the Cannizzaro reaction.

Option D is correct because ozonolysis of alkenes typically produces aldehydes, ketones, or carboxylic acids, but not alcohols directly. The other methods (Grignard reaction, alkyl halide hydrolysis, and hydroboration-oxidation) are valid alcohol synthesis methods.

The correct answer is D (Formaldehyde and benzaldehyde) because Cannizzaro reaction requires aldehydes without alpha-hydrogens, and both formaldehyde and benzaldehyde meet this criterion.

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

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.