49 questions on this topic of Organic Compounds Containing Oxygen (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (B) Ethanal
Dry distillation of calcium formate alone yields methanal, and calcium acetate alone yields propanone. When an equimolar mixture of calcium acetate, (CH₃COO)₂Ca, and calcium formate, (HCOO)₂Ca, is distilled, the cross-product ethanal (CH₃CHO) is obtained along with CaCO₃.
Answer: (A) Both Statement I and Statement II are correct
In carboxylic acids, resonance donation from the lone pair of the hydroxyl oxygen into the carbonyl carbon greatly decreases its positive character compared to aldehydes and ketones. Consequently, carboxylic acids do not undergo characteristic nucleophilic additions like forming NaHSO₃ bisulphite adducts.
Answer: (D) Cyclopentylidene-(2,4-dinitrophenyl)hydrazine
The reaction of cyclopentanone with 2,4-dinitrophenylhydrazine (Brady's reagent) is a nucleophilic addition followed by elimination of a water molecule, forming a hydrazone named cyclopentylidene-(2,4-dinitrophenyl)hydrazine.
Answer: (B) 3-Nitrobenzaldehyde
The aldehyde group (-CHO) on a benzene ring is strongly deactivating and meta-directing due to its electron-withdrawing nature via resonance and inductive effects. Electrophilic aromatic substitution (nitration) occurs predominantly at the meta position, giving 3-nitrobenzaldehyde.
Answer: (A) A hemiacetal
Aldehydes react with one equivalent of a monohydric alcohol in the presence of dry HCl gas to yield alkoxyalcohol intermediate known as a hemiacetal. Reaction with another equivalent of alcohol produces a gem-dialkoxy compound termed an acetal.
Answer: (A) Propanoic acid
Compound 'X' is a ketone that gives a positive iodoform test, so it must be butan-2-one (CH₃COCH₂CH₃). It does not react with NaHSO₃ easily due to steric factors and does not reduce Fehling's solution. Cleavage during the haloform reaction gives CHI₃ and sodium propanoate ('Y'). Acidification of 'Y' yields propanoic acid.
Answer: (C) Ketones have two alkyl groups that cause steric hindrance and reduce the electrophilicity of the carbonyl carbon
Two electron-releasing alkyl groups in ketones reduce the positive charge on carbonyl carbon and hinder nucleophilic attack sterically.
NCERT Class 12
Answer: (A) Cannizzaro reaction (self oxidation-reduction)
In the Cannizzaro reaction one molecule is reduced to the alcohol and another oxidised to the carboxylate salt.
NCERT Class 12
Answer: (D) 2-Ethylcyclohexanone
Cyclohexanone reacts with pyrrolidine (a secondary amine) to yield an enamine (compound P). Enamines act as nucleophiles at the α-carbon. Alkylation with ethyl iodide introduces an ethyl group at the C-2 position. Subsequent acidic hydrolysis regenerates the carbonyl group, yielding 2-ethylcyclohexanone (Stork enamine synthesis).
Answer: (D) Sodium formate and benzyl alcohol
In a crossed Cannizzaro reaction involving formaldehyde, OH⁻ preferentially attacks the more electrophilic and sterically unhindered carbonyl carbon of HCHO. Consequently, formaldehyde is oxidized to sodium formate, and benzaldehyde is reduced to benzyl alcohol.
Answer: (C) 2,2-Dimethylpropanoic acid and 2,2-dimethylpropan-1-ol
2,2-Dimethylpropanal [(CH₃)₃C-CHO] lacks α-hydrogens, so it undergoes the Cannizzaro reaction on heating with concentrated KOH. Disproportionation gives potassium 2,2-dimethylpropanoate (which on acidification yields 2,2-dimethylpropanoic acid) and 2,2-dimethylpropan-1-ol.
Answer: (C) Propan-2-one
Reaction of a ketone with a Grignard reagent yields a tertiary alcohol after hydrolysis. Formaldehyde yields primary alcohols, and all other aldehydes yield secondary alcohols. Propan-2-one is a ketone and therefore gives 2-methylpropan-2-ol, a tertiary alcohol.