Read the Passage carefully and answer the Questions.
Carbonyl compounds such as aldehydes, ketones, and carboxylic acids play an important role in the biochemical process of plants and animal kingdom. Many aldehydes and ketones find applications in dye, perfumery and other industries. The carbonyl carbon atom is \( sp^2 \)-hybridized forming three sigma and one pi-bond with oxygen. C-O bonds are polar bond due to high electronegativity of oxygen than carbon. Hence the carbonyl carbon behaves as an electrophilic center and oxygen as a nucleophilic centre. Primary and secondary alcohols yield aldehydes and ketones respectively upon oxidation with variety of oxidizing agents. The boiling points of aldehydes and ketones differ from hydrocarbons and ethers. Aldehydes and ketones undergo nucleophilic addition reaction due to presence of electrophilic carbonyl carbon. Nucleophile and a proton add across the carbon oxygen double bond.
The alcohol which cannot be used to prepare carbonyl compound through oxidation is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-methylpropan-2-ol (\( \ce{(CH3)3C-OH} \))
Oxidation of an alcohol to a carbonyl compound forms a C=O bond by removing the hydrogen of the O–H group and a hydrogen on the carbon carrying the –OH group. A primary alcohol (two such hydrogens) gives an aldehyde and a secondary alcohol (one such hydrogen) gives a ketone. A tertiary alcohol has no hydrogen on that carbon, so it does not undergo this oxidation.
In 2-methylpropan-2-ol the carbon carrying the –OH group holds three methyl groups and no hydrogen, so no carbonyl compound can be formed from it. Under drastic conditions the molecule only breaks up into smaller fragments.
Option (1) butan-1-ol (\( \ce{CH3CH2CH2CH2OH} \)) — Can give a carbonyl compound. It is a primary alcohol and is oxidised to butanal:
\( \displaystyle \ce{CH3CH2CH2CH2OH} \xrightarrow{\text{PCC}} \ce{CH3CH2CH2CHO} \)
Option (2) 2-methylpropan-2-ol (\( \ce{(CH3)3C-OH} \)) — Cannot give a carbonyl compound. It is a tertiary alcohol: the carbon carrying –OH has no hydrogen to lose.
Option (3) butan-2-ol (\( \ce{CH3CH2CH(OH)CH3} \)) — Can give a carbonyl compound. It is a secondary alcohol and is oxidised to the ketone butanone:
\( \displaystyle \ce{CH3CH2CH(OH)CH3} \xrightarrow{\ce{CrO3}} \ce{CH3CH2COCH3} \)
Option (4) but-2-en-1-ol (\( \ce{CH3CH=CHCH2OH} \)) — Can give a carbonyl compound. It is a primary alcohol and is oxidised to the aldehyde but-2-enal:
\( \displaystyle \ce{CH3CH=CHCH2OH} \xrightarrow{\text{PCC}} \ce{CH3CH=CHCHO} \)
NCERT (2026-27 edition):
Oxidation – …
