Opening the book…
After completing this section, you should be able to
- write an equation to illustrate an intramolecular aldol reaction.
- identify the product formed when a given dicarbonyl compound undergoes an intramolecular aldol condensation.
- identify the dicarbonyl compound which, when treated with a suitable base, could be used to prepare a given cyclic enone by an intramolecular aldol condensation.
“Intramolecular ” means “within the same molecule.” You have already seen some examples of intramolecular reactions in previous chapters. Another term for intramolecular aldol reaction is “internal aldol reaction.”
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. The term “Intramolecular” means “within the same molecule.” In this case, it means that the enolate donor and the electrophilic acceptor of an aldol reaction are contained in the same molecule such as dialdehydes, keto aldehydes, or diketones. In these cases, the small distance between the donor and acceptor leads to faster reaction rates for intramolecular condensations making intermolecular condensations (which require two molecules to collide in solution) less favorable. For example, base treatment of a 1,4-diketone such as 2,5-hexanedione yields a cyclopentenone product, and base treatment of a 1,5-diketone such as 2,6-heptanedione yields a cyclohexenone.
The mechanism of intramolecular aldol reactions is similar to that of intermolecular reactions. The only difference is that both the nucleophilic carbonyl anion donor and the electrophilic carbonyl acceptor are now in the same molecule. One complication, however, is that intramolecular aldol reactions might lead to a mixture of products, depending on which enolate ion is formed. For example, 2,5-hexanedione might yield either the five-membered-ring product 3-methyl-2-cyclopentenone or the three-membered-ring product (2-methylcyclopropenyl) ethanone (Figure \(\PageIndex{1}\)). In practice, though, only the cyclopentenone is formed.
The selectivity observed in the intramolecular aldol reaction of 2,5-hexanedione is due to the fact that all steps in the mechanism are reversible, so an equilibrium is reached. Thus, the relatively strain-free cyclopentenone product is considerably more stable than the highly strained cyclopropene alternative. For similar reasons, intramolecular aldol reactions of 1,5-diketones lead only to cyclohexenone products rather than to the more strained acylcyclobutenes.
Please draw the expected product of an intramolecular aldol condensation with the following molecule:
- Answer
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Analysis: 6-Oxoheptanal has three unique sets of alpha-hydrogens which could be deprotonated to form an enolate. Selecting the correct set involves analyzing the carbonyl reactivity’s along with the possible ring sizes of the products.
Solution: Once the preferred alpha-hydrogens are determined go through the steps discussed in the previous sections to determine the aldol condensation product. Remember to form the aldol intermediate first.
Treatment of a 1,3-diketone such as 2,4-pentanedione with base does not give an aldol condensation product. Explain.
- Answer
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2,4-Pentanedione like other 1,3-Diketones have particularly acidic alpha-hydrogens between the two carbonyl. When base is applied these will be the first to deprotonate and form an enolate. For an intramolecular aldol condensation to occur an alpha-hydrogen one of the methyl groups would have to be removed which would be difficult.
The CH2 position between the two carbonyl groups is so acidic that it is completely deprotonated to give a stable enolate ion.
What product would you expect to obtain from base treatment of 1,6-cyclodecanedione?

- Answer
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This page titled 9.6: Intramolecular Aldol Reactions was last modified on Mon, 17 Mar 2025 22:17:08 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sol Parajon Puenzo (Cañada College) .
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