In aldehydes, the carbonyl group is bonded to a carbon and hydrogen, while in ketones it is bonded to two carbon atoms. The compounds in which the carbonyl group is bonded to oxygen are known as carboxylic acids and their derivatives.
Arrange the following steps of Aldol condensation in the correct order of their occurrence.
A. Abstraction of proton
B. Protonation
C. Formation of Enolate
D. Nucleophilic Addition
E. Elimination
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → ACDBE
Note: The question is from beyond the NCERT textbook.
In base-catalysed aldol condensation (shown here for ethanal) the steps follow one after another:
A. Abstraction of proton — \( \ce{OH-} \) removes an acidic α-hydrogen from the aldehyde or ketone:
\( \ce{CH3CHO + OH- -> ^-CH2CHO + H2O} \)
C. Formation of enolate — the carbanion formed is stabilised by resonance as the enolate ion.
D. Nucleophilic addition — the enolate attacks the carbonyl carbon of a second molecule, giving an alkoxide ion:
\( \ce{^-CH2CHO + CH3CHO -> CH3CH(O^-)CH2CHO} \)
B. Protonation — the alkoxide takes a proton from water to give the β-hydroxy aldehyde (aldol):
\( \ce{CH3CH(O^-)CH2CHO + H2O -> CH3CH(OH)CH2CHO + OH-} \)
E. Elimination — on heating, the aldol loses water to give the α,β-unsaturated carbonyl compound:
\( \ce{CH3CH(OH)CH2CHO} \xrightarrow{\Delta} \ce{CH3CH=CHCHO + H2O} \)