Organic Chemistry · Carbonyl Condensation Reactions
Mixed Aldol Reactions
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A Mixed (crossed) aldol Aldol reaction between two different carbonyl compounds Full entry → reaction joins two different carbonyl compounds: an enolate from one partner adds to the carbonyl of the other, giving a β-Hydroxy carbonyl The initial aldol adduct (alcohol β to the carbonyl) Full entry → that usually dehydrates to an Enone α,β-Unsaturated carbonyl (C=C–C=O) Full entry →. The catch is that if both partners have α-hydrogens, each can serve as either enolate source or electrophile, so up to four products can form. Chemists tame this with two tactics: (1) use a partner with no α-hydrogens (only one compound can enolize), or (2) pre-form the enolate of one partner with a strong base such as LDA before adding the second. This topic shows how to predict which product a mixed aldol delivers — the skill behind syntheses ranging from cinnamaldehyde in flavor chemistry to the aldol steps of the Robinson annulation (topic 12).
Why this matters
- Mixed aldols build one specific C–C bond between two chosen fragments — exactly what real syntheses need, and what self-aldols cannot provide.
- The four-product trap is a classic exam question: counting products and explaining why a benzaldehyde partner eliminates three of them is high-yield reasoning.
- Directed aldols (LDA enolates) are routine in drug and natural-product synthesis, and the regiochemical thinking here recurs in mixed Claisen condensations (topic 08) and the Michael/Robinson reactions (topics 10–12).
The college version
Core Concepts
The four-product problem
Mix acetaldehyde and propanal with hydroxide and both enolates form, each able to attack either carbonyl. The result is up to four adducts — two self-aldols and two crossed:
\[ \text{enolate A + A} \quad \text{enolate A + B} \quad \text{enolate B + A} \quad \text{enolate B + B} \]
The crossed product you wanted is only a fraction of the mixture. Two structural features rescue the reaction.
Strategy 1: a partner with no α-hydrogens
If one carbonyl has no α-H, it cannot enolize — it can only act as the electrophile. Benzaldehyde (C₆H₅CHO) is the classic example: the ring carbon bears no hydrogen. Formaldehyde (HCHO) has no α carbon at all. With benzaldehyde plus acetone under hydroxide, only acetone enolizes, and its enolate has only benzaldehyde to attack:
\[ \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{CH(OH)CH}_2\text{COCH}_3 \rightarrow \text{C}_6\text{H}_5\text{CH}=\text{CHCOCH}_3 \]
The β-hydroxy adduct dehydrates in hot base (topic 03) to the conjugated enone — here benzylideneacetone, a single product.
Strategy 2: a preformed enolate (directed aldol)
When both partners have α-H's, form the enolate of one partner before adding the other. LDA (lithium diisopropylamide) Strong, bulky, non-nucleophilic base (conjugate acid pKa ≈ 36) Full entry →, a strong, bulky, non-nucleophilic base, deprotonates a ketone completely and irreversibly at −78 °C. The enolate solution is then added to the second carbonyl, which never gets a chance to enolize — one product, complete regiochemical control. The equilibrium argument from Chapter 22 explains why LDA works: for deprotonation of a ketone (pKa ≈ 19) by LDA (conjugate acid pKa ≈ 36),
\[ K_{\text{eq}} = 10^{(pK_a(\text{amine}) - pK_a(\text{ketone}))} = 10^{(36 - 19)} = 10^{17} \]
Formula first, then substitution — the equilibrium lies so far right that enolate formation is essentially quantitative.
Regiochemistry: which α-carbon enolizes?
If the enolizable partner has two different α positions (e.g., 2-methylcyclohexanone), the Directed aldol Enolate of one partner pre-formed with a strong base before adding the other Full entry → can give two regioisomers. Kinetic control (LDA, −78 °C, fast and irreversible) removes the less hindered, more hydrogen-rich α-H; thermodynamic control (weaker base, longer time) favors the more substituted, more stable enolate. Predict the enolate first, then the aldol product.
Dehydration completes the sequence
Aldol adducts are β-hydroxy carbonyls that typically dehydrate to α,β-unsaturated carbonyls under the reaction conditions. With benzaldehyde, dehydration is especially favorable because it extends conjugation into the aromatic ring — which is why cinnamaldehyde (C₆H₅CH=CHCHO) forms so readily from benzaldehyde and acetaldehyde.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Mixed aldol | Self-aldol | Mixed uses two different carbonyls; self-aldol uses two molecules of the same one |
| Benzaldehyde (no α-H) | Acetone (has α-H) | Only the partner with α-H can be the enolate; the no-α-H partner is always the electrophile |
| Kinetic enolate | Thermodynamic enolate | Kinetic forms at the less hindered α-H under LDA/cold conditions; thermodynamic is the more substituted enolate from equilibration |
| Aldol adduct | Enone product | The adduct is the β-hydroxy carbonyl formed first; the enone is its dehydration product |
| Directed aldol (LDA) | Equilibrium aldol (OH⁻) | LDA pre-forms one enolate irreversibly; hydroxide generates both enolates reversibly, inviting all four products |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Mixing two different carbonyls in an aldol is like a dance where both partners can lead — you'd get four different couples. To force one specific couple, you either pick a partner that can't lead (like benzaldehyde, which has no loose hydrogens to grab), or you grab the hydrogen off your chosen leader first with a very strong base so only it can move. Then the dance has exactly one outcome.
Worked example
Example 1: Acetaldehyde + formaldehyde
Problem: Acetaldehyde and aqueous formaldehyde are treated with dilute NaOH. Predict the product.
Reasoning: Formaldehyde (HCHO) has no α-H, so only acetaldehyde can enolize. Its enolate attacks formaldehyde's carbonyl:
\[ \text{CH}_2=\text{CHO}^- \text{(from CH}_3\text{CHO)} + \text{HCHO} \rightarrow \text{HOCH}_2\text{CH}_2\text{CHO} \]
Answer: 3-Hydroxypropanal (HOCH₂CH₂CHO) — a single crossed product; under forcing conditions it dehydrates to acrolein (CH₂=CHCHO).
Example 2: Benzaldehyde + acetone → benzylideneacetone
Step 1 — Identify the enolate source: benzaldehyde has no α-H, so only acetone enolizes: CH₃COCH₂⁻.
Step 2 — Attack: the acetone enolate adds to the benzaldehyde carbonyl carbon:
\[ \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_2^- \rightarrow \text{C}_6\text{H}_5\text{CH(O}^-\text{)CH}_2\text{COCH}_3 \]
Step 3 — Protonation and dehydration: the alkoxide picks up a proton and the β-hydroxy ketone dehydrates to the conjugated enone:
\[ \text{C}_6\text{H}_5\text{CH(OH)CH}_2\text{COCH}_3 \rightarrow \text{C}_6\text{H}_5\text{CH}=\text{CHCOCH}_3 \]
Answer: 4-Phenylbut-3-en-2-one (benzylideneacetone). With excess benzaldehyde, a second condensation at the remaining α position gives dibenzylideneacetone — evidence that each α-H enolizes independently.
Example 3: Directed aldol with LDA — why one product
Problem: You need the crossed aldol of cyclohexanone and isobutyraldehyde, both with α-H's. Why does the LDA procedure succeed where hydroxide fails?
Step 1 — Count products with hydroxide: each enolate can attack either partner → up to four adducts, all interconverting through enolate equilibria.
Step 2 — Pre-form the enolate: LDA removes the less hindered α-H of cyclohexanone completely at −78 °C:
\[ K_{\text{eq}} = 10^{(pK_a(\text{amine}) - pK_a(\text{ketone}))} = 10^{(36 - 19)} = 10^{17} \]
essentially all ketone is enolate before isobutyraldehyde is added.
Step 3 — Add the electrophile: the preformed enolate attacks isobutyraldehyde's carbonyl; with no base present, isobutyraldehyde never enolizes.
Answer: 2-(1-Hydroxy-2-methylpropyl)cyclohexanone — one product, exactly as designed.
Key takeaways
- Four products form when two enolizable carbonyls are mixed under basic aldol conditions; the crossed product is only one of them.
- A partner without α-H (benzaldehyde, formaldehyde) cannot enolize — it is a pure electrophile, and the product count drops to one.
- Directed aldol: pre-form one enolate with LDA at −78 °C, then add the second carbonyl — one product with complete control.
- K_eq for enolate formation = 10^(pKa(conjugate acid of base) − pKa(substrate)); LDA (pKa ≈ 36) deprotonates ketones (pKa ≈ 19) essentially completely.
- Predict regiochemistry from the enolate: kinetic (less hindered α-H, LDA, cold) vs thermodynamic (more substituted enolate, equilibration).
- Aldol adducts are β-hydroxy carbonyls and usually dehydrate to enones; conjugation with an aromatic ring makes dehydration especially favorable.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why does mixing acetaldehyde with propanal under NaOH give several products, not one?
Show answer
Both compounds have α-H's, so each can act as both enolate source and electrophile — up to four adducts (two self, two crossed) form simultaneously.
Which of these can serve as the "no-enolate" partner: benzaldehyde, acetone, formaldehyde, ethyl acetate? Explain.
Show answer
Benzaldehyde and formaldehyde only. Acetone and ethyl acetate both have α-H's and can enolize.
Predict the product of benzaldehyde + cyclohexanone with NaOH (after dehydration).
Show answer
2-Benzylidenecyclohexanone: the cyclohexanone enolate attacks benzaldehyde, and the adduct dehydrates to the conjugated enone.
Calculate K_eq for deprotonation of acetone (pKa 19.3) by LDA (conjugate acid pKa 36). Is enolate formation complete?
Show answer
K_eq = 10^(36 − 19.3) = 10^16.7 ≈ 5 × 10¹⁶ — essentially complete deprotonation; this is why LDA gives a clean, stoichiometric enolate.
2-Methylcyclohexanone is enolized with LDA at −78 °C, then treated with benzaldehyde. Which regioisomeric product forms, and why?
Show answer
The kinetic enolate forms at the less hindered α-H (the CH₂ away from the methyl group), so benzaldehyde adds at the unsubstituted α position: 2-(hydroxyphenylmethyl)-6-methylcyclohexanone.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Mixed (crossed) aldol
- Aldol reaction between two different carbonyl compounds
- Non-enolizable partner
- Carbonyl with no α-hydrogen (benzaldehyde, formaldehyde)
- Directed aldol
- Enolate of one partner pre-formed with a strong base before adding the other
- LDA (lithium diisopropylamide)
- Strong, bulky, non-nucleophilic base (conjugate acid pKa ≈ 36)
- β-Hydroxy carbonyl
- The initial aldol adduct (alcohol β to the carbonyl)
- Enone
- α,β-Unsaturated carbonyl (C=C–C=O)
Sources & references
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