Aldol Condensation
Aldehydes and ketones undergo a number of reactions due to the acidic nature of α-hydrogen. Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes or β-hydroxy ketones, respectively. This is known as aldol reaction. The aldol and ketol readily lose water to give α, β-unsaturated carbonyl compounds, which are aldol condensation products; the reaction is called aldol condensation. When aldol condensation is carried out between two different aldehydes and/or ketones, it is called cross aldol condensation. Ethanal and propanal undergo this reaction to form mixture of products. However, benzaldehyde and acetophenone react to give only one product . Moreover, aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction ( disproportionation) reaction on heating with concentrated alkali. The reaction is known as Cannizzaro reaction. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
Which of the following pairs will not undergo an aldol condensation reaction?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → One mole of methanal and one mole of benzaldehyde
Aldol condensation starts with the removal of an α-hydrogen by the base to form a carbanion. So at least one of the two carbonyl compounds must have an α-hydrogen. If neither has one, no carbanion can form and no aldol condensation is possible; such aldehydes undergo the Cannizzaro reaction with concentrated alkali instead.
Option (1) 2 moles of ethanal — Undergoes aldol condensation. Ethanal has three α-hydrogens. Its carbanion adds to a second ethanal molecule to give the aldol 3-hydroxybutanal, which loses water on heating to give but-2-enal.
\( \displaystyle \ce{2CH3CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3CH(OH)CH2CHO} \xrightarrow[-\ce{H2O}]{\Delta} \ce{CH3CH=CHCHO} \)
Option (2) 1 mole of methanal and 1 mole of propanone — Undergoes (cross) aldol condensation. Methanal has no α-hydrogen, but propanone has six. The carbanion of propanone adds to the C=O group of methanal, giving 4-hydroxybutan-2-one, which loses water to give but-3-en-2-one.
\( \displaystyle \ce{HCHO + CH3COCH3} \xrightarrow{\text{dil. NaOH}} \ce{HOCH2CH2COCH3} \xrightarrow[-\ce{H2O}]{\Delta} \ce{CH2=CHCOCH3} \)
Option (3) two moles of acetophenone — Undergoes aldol condensation. The \( \ce{CH3} \) group of acetophenone carries three α-hydrogens, so acetophenone condenses with itself; the ketol loses water to give 1,3-diphenylbut-2-en-1-one.
\( \displaystyle \ce{2C6H5COCH3} \xrightarrow{\text{base}} \ce{C6H5C(OH)(CH3)CH2COC6H5} \xrightarrow[-\ce{H2O}]{\Delta} \ce{C6H5C(CH3)=CHCOC6H5} \)
Option (4) One mole of methanal and one mole of benzaldehyde — Does not undergo aldol condensation. Methanal (HCHO) has no carbon next to the C=O group, and in benzaldehyde (\( \ce{C6H5CHO} \)) the C=O group is attached to a ring carbon that has no hydrogen. Neither has an α-hydrogen, so no carbanion can form. With concentrated alkali this pair undergoes a cross Cannizzaro reaction instead: methanal is oxidised to formate and benzaldehyde is reduced to benzyl alcohol.
\( \displaystyle \ce{HCHO + C6H5CHO} \xrightarrow{\text{conc. NaOH}} \ce{HCOONa + C6H5CH2OH} \)
NCERT (2026-27 edition):
Other Reactions
