Aldehydes, Ketones and Carboxylic Acids MCQs for UPSC Prelims

Correct answer: B - Ethanoic acid

Ethanoic acid molecules pair up through two hydrogen bonds into a dimer, so the strongest intermolecular attraction of the four must be broken before it boils. Propan-1-ol hydrogen bonds as well, but each molecule offers only one $\text{O-H}$ hydrogen, so its attraction is weaker. Propanal is polar yet has no $\text{O-H}$ and manages only dipole-dipole attraction, and butane is held together by weak dispersion forces alone.

Correct answer: A - Benzoyl chloride, $\text{C}_6\text{H}_5\text{COCl}$

The Rosenmund reduction turns an acyl chloride $\text{RCOCl}$ into $\text{RCHO}$, so the starting material must carry chlorine on a carbonyl carbon, as benzoyl chloride does. In benzyl chloride the chlorine sits on a $\text{CH}_2$ group, so hydrogenation cannot build a $\text{CHO}$ group. Chlorobenzene holds its chlorine on the ring and is untouched, and a carboxylic acid is not reduced by hydrogen over poisoned palladium.

Correct answer: C - Propanone

Acetone is a three-carbon chain with the carbonyl on carbon 2, so its IUPAC name is propanone. Propanal has the same three carbons but the oxygen at the end of the chain, which makes it an aldehyde rather than a ketone. Ethanal is the IUPAC name of acetaldehyde, a different common name. Butanone has four carbons and is the IUPAC name of methyl ethyl ketone.

Correct answer: C - Propan-2-ol

Sodium borohydride delivers hydride to the carbonyl carbon, so a ketone becomes a secondary alcohol with the $-\text{OH}$ on the middle carbon. A primary alcohol would need the carbonyl at the end of the chain, as in propanal. Stripping the oxygen away to leave propane needs Clemmensen or Wolff-Kishner conditions, and a reducing agent cannot produce an acid.

Correct answer: C - Propan-2-ol

The test needs a $\text{CH}_3\text{CO}-$ group, or a $\text{CH}_3\text{CH(OH)}-$ group that alkaline iodine can first oxidise to one. Propan-2-ol has $\text{CH}_3\text{CH(OH)}-$ and is oxidised to propanone, which then yields $\text{CHI}_3$. Propan-1-ol is oxidised only to propanal, and in propanal and propanoic acid alike the carbonyl carbon carries an ethyl group rather than a methyl group.

Correct answer: C - Propanone

A dialkylcadmium hands its alkyl group to the acyl carbon in place of chlorine, so $\text{CH}_3\text{COCl}$ becomes $\text{CH}_3\text{COCH}_3$. Ethanal would need a reduction, which cadmium reagents do not carry out. Propanal has its carbonyl at the end of the chain, which attack at the acyl carbon cannot produce, and no water is supplied to give an acid.