Which of the following will not form propionaldehyde?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → propan-1-ol with KMnO4/H+
Propionaldehyde (propanal, \( \ce{CH3CH2CHO} \)) is formed by reactions that stop at the aldehyde stage: partial reduction of an ester or an acid chloride, or ozonolysis of a suitable alkene. A strong oxidising agent such as acidified \( \ce{KMnO4} \) does not stop at the aldehyde: it oxidises a primary alcohol right up to the carboxylic acid.
Option (1) methyl propanoate with (i) DIBAL-H (ii) H2O — Forms propionaldehyde. DIBAL-H reduces an ester only as far as the aldehyde.
\( \ce{CH3CH2COOCH3} \xrightarrow[\ce{(ii) H2O}]{\ce{(i) DIBAL-H}} \ce{CH3CH2CHO + CH3OH} \)
Option (2) 2-methylpent-2-ene with (i) O3 (ii) Zn/H2O — Forms propionaldehyde. Ozonolysis breaks the C=C bond and puts a C=O on each carbon. The left fragment is propanal and the right one is acetone.
\( \ce{CH3CH2CH=C(CH3)CH3} \xrightarrow[\ce{(ii) Zn/H2O}]{\ce{(i) O3}} \ce{CH3CH2CHO + CH3COCH3} \)
Option (3) propan-1-ol with KMnO4/H+ — Does not form propionaldehyde. Acidified \( \ce{KMnO4} \) oxidises the primary alcohol; any aldehyde formed is oxidised further, so the product is propanoic acid.
\( \ce{CH3CH2CH2OH + 2[O]} \xrightarrow{\ce{KMnO4/H+}} \ce{CH3CH2COOH + H2O} \)
Option (4) propanoyl chloride with H2/Pd — Forms propionaldehyde. This is the Rosenmund reduction (Pd on \( \ce{BaSO4} \)), which reduces an acid chloride to the aldehyde.
\( \ce{CH3CH2COCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{CH3CH2CHO + HCl} \)
NCERT (2026-27 edition):
Methods of Preparation of Carboxylic Acids – …
