Which one of the following product is NOT formed when mixture of ethanal and propanal heated with dil NaOH?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Pent-3-enal
Ethanal and propanal both have α-hydrogens, so with dilute NaOH each can form a carbanion and add to either aldehyde (cross aldol condensation). The aldols lose water to give α,β-unsaturated aldehydes, with the C=C between C-2 and C-3. Four products are possible:
Ethanal + ethanal → but-2-enal, \( \ce{CH3CH=CHCHO} \)
Propanal + propanal → 2-methylpent-2-enal, \( \ce{CH3CH2CH=C(CH3)CHO} \)
Propanal carbanion + ethanal → 2-methylbut-2-enal, \( \ce{CH3CH=C(CH3)CHO} \)
Ethanal carbanion + propanal → pent-2-enal, \( \ce{CH3CH2CH=CHCHO} \)
Pent-3-enal (\( \ce{CH3CH=CHCH2CHO} \)) has its double bond between C-3 and C-4, not conjugated with C=O, so it is not an aldol condensation product.
For example, the self condensation of ethanal:
\( \ce{2CH3CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3CH(OH)CH2CHO} \xrightarrow[-\ce{H2O}]{\Delta} \ce{CH3CH=CHCHO} \)
Option (1) But-2-enal — Formed in this reaction. By the self condensation of ethanal.
Option (2) Pent-3-enal — Not formed. Its C=C is not conjugated with C=O, so it is not an aldol condensation product.
Option (3) 2-Methylpent-2-enal — Formed in this reaction. By the self condensation of propanal.
Option (4) 2-Methylbut-2-enal — Formed in this reaction. A cross product, from the propanal carbanion adding to ethanal.
NCERT (2026-27 edition):
Cross Aldol Condensation – “When aldol condensation is carried out between two different aldehydes and / or ketones, it is called cross aldol condensation. If both of them contain α-hydrogen atoms, it gives a mixture of four products.”