Aldehydes, Ketones and Carboxylic Acids MCQs for UPSC Prelims
105 practice questions covering Aldehydes, Ketones and Carboxylic Acids, organised into 1 topics.
105 questions include a written explanation.
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Butane, propanal, propan-1-ol and ethanoic acid all have molar masses between $58$ and $60\ \text{g mol}^{-1}$. Which of them has the highest boiling point?
AButane
BEthanoic acid
CPropanal
DPropan-1-ol
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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.
Q2
easy
Benzaldehyde, $\text{C}_6\text{H}_5\text{CHO}$, is to be made by the Rosenmund reduction. Which compound should be hydrogenated over $\text{Pd/BaSO}_4$?
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.
Q3
easy
Acetone is the common name of the simplest ketone, $\text{CH}_3\text{COCH}_3$. What is its IUPAC name?
APropanal
BButanone
CPropanone
DEthanal
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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.
Q4
easy
Propanone is treated with sodium borohydride. What is the product?
APropan-1-ol
BPropane
CPropan-2-ol
DPropanoic acid
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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.
Q5
medium
Which compound gives a yellow precipitate when warmed with iodine and sodium hydroxide solution?
APropan-1-ol
BPropanal
CPropan-2-ol
DPropanoic acid
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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.
Q6
medium
Ethanoyl chloride is treated with dimethylcadmium, $(\text{CH}_3)_2\text{Cd}$. What is the organic product?
AEthanoic acid
BEthanal
CPropanone
DPropanal
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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.
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