Aldehydes, Ketones and Carboxylic Acids · Lesson 8 of 15
Crossed Aldol, Cannizzaro and Aromatic Substitution
“Choose between enolate chemistry, disproportionation and aromatic-ring reactions from structure and conditions.”
• Enumerate donor–acceptor combinations in crossed aldol reactions. • Predict Cannizzaro products and explain hydride transfer. • Distinguish aldehydes without α-H from ketones without α-H. • Explain meta substitution of directly attached carbonyl groups.
Two enolisable aldehydes give several constitutional products
A crossed aldol uses two different carbonyl compounds. If each can form an enolate and each can accept nucleophilic attack, both self-reactions and both cross-reactions are possible. For two simple aldehydes with one distinct α site each, there are four donor–acceptor combinations. This count concerns constitutional products and does not separately count E/Z or other stereoisomers. More complicated ketones can have multiple enolate sites, so “always exactly four” is not a universal rule.
Use the donor as the framework containing the carbonyl that survives. Its α-carbon makes the new bond to the acceptor carbonyl carbon, which becomes the β-OH carbon. On dehydration, donor α-carbon and former acceptor carbonyl carbon form C=C. This tracing method works even when product names become complicated.
| Enolate donor | Carbonyl acceptor | Dehydrated product |
|---|---|---|
| Ethanal | Ethanal | CH₃CH=CHCHO; but-2-enal |
| Propanal | Propanal | CH₃CH₂CH=C(CH₃)CHO; 2-methylpent-2-enal |
| Propanal | Ethanal | CH₃CH=C(CH₃)CHO; 2-methylbut-2-enal |
| Ethanal | Propanal | CH₃CH₂CH=CHCHO; pent-2-enal |
Problem
List products by donor and acceptor rather than guessing names.
- 1.Propanal donor plus propanal acceptor gives OHC–C(CH₃)=CH–CH₂CH₃: 2-methylpent-2-enal.
- 2.Butanal donor plus butanal acceptor gives OHC–C(CH₂CH₃)=CH–CH₂CH₂CH₃: 2-ethylhex-2-enal.
- 3.Propanal donor plus butanal acceptor gives OHC–C(CH₃)=CH–CH₂CH₂CH₃: 2-methylhex-2-enal.
- 4.Butanal donor plus propanal acceptor gives OHC–C(CH₂CH₃)=CH–CH₂CH₃: 2-ethylpent-2-enal. Self-products have 6 or 8 carbons; cross-products each have 7. Stereoisomers may add further complexity.
A non-enolisable aldehyde can still accept an enolate
Benzaldehyde lacks α-H but its carbonyl remains electrophilic. With an enolisable partner such as acetophenone, the partner supplies the enolate. Base-catalysed reaction followed by dehydration yields C₆H₅COCH=CHC₆H₅, 1,3-diphenylprop-2-en-1-one, commonly chalcone. Benzaldehyde plus propanal similarly gives PhCH=C(CH₃)CHO. A non-enolisable component reduces some competing donor choices, although conditions and stoichiometry are still needed to control self-reaction or further condensations.
Cannizzaro transfers reducing power between aldehyde molecules
Aldehydes without an enolisable α-H, such as methanal, benzaldehyde and 2,2-dimethylbutanal, can undergo Cannizzaro disproportionation with concentrated alkali. One aldehyde molecule is oxidised to carboxylate and another is reduced to a primary alcohol. This is a redox reaction without an external oxidant or reductant; the two substrate molecules have different fates. The base conditions differ from dilute-alkali aldol conditions.
Hydroxide first adds to an aldehyde carbonyl, making RCH(OH)O⁻. In the key redox step, the C–H bond of this intermediate transfers hydride character to the carbonyl carbon of a second aldehyde. Oxygen at the donor reforms C=O; the acceptor oxygen becomes alkoxide. Subsequent proton transfer gives alcohol and carboxylate. The transferred species is not a freely released H⁻ ion swimming in water; the bond-to-bond transfer is part of the mechanism.
Problem
Predict benzaldehyde and methanal with concentrated alkali.
- 1.For benzaldehyde: 2C₆H₅CHO + NaOH → C₆H₅COONa + C₆H₅CH₂OH. One molecule gives benzoate and the other benzyl alcohol.
- 2.For methanal: 2HCHO + KOH → HCOOK + CH₃OH. Methanal has no carbon neighbour and therefore no α-carbon bearing H.
- 3.The carboxyl product remains a salt until acid work-up. No new C–C bond is created, unlike aldol chemistry.
| Substrate | Relevant structural fact | Ordinary choice in this scope |
|---|---|---|
| Methanal | No α-carbon | Cannizzaro |
| Benzaldehyde | No enolisable α-H | Cannizzaro; also crossed-aldol acceptor |
| 2,2-Dimethylbutanal | Adjacent carbon has no H | Cannizzaro |
| Cyclohexanone | Ketone with α-H | Aldol |
| Phenylacetaldehyde, PhCH₂CHO | CH₂ supplies α-H | Aldol |
| Benzophenone | Ketone without α-H | Neither ordinary self-aldol nor Cannizzaro |
| Butan-1-ol | Not a carbonyl compound | Neither without prior conversion |
A carbonyl directly attached to a ring directs electrophiles meta
In ArCHO and ArCOR, the group attached to the ring withdraws electron density by induction and resonance. Electrophilic aromatic substitution is slower than in benzene, and the resonance pattern particularly destabilises intermediates from ortho and para attack. Meta substitution is consequently favoured. Nitration of benzaldehyde with concentrated HNO₃/H₂SO₄ gives mainly 3-nitrobenzaldehyde under suitable controlled conditions. Acetophenone likewise gives mainly m-nitroacetophenone.
The directing effect depends on direct attachment: a CH₂ spacer can interrupt conjugation, and the rest of the substituent then matters. “Contains a carbonyl somewhere” is insufficient to declare a ring meta-directing. Nor does a meta-directing group mean every reaction produces exclusively one isomer. Product prediction should state the major pattern within the stated conditions.
Quick Quiz
Quiz
How many donor–acceptor combinations exist for ethanal plus propanal before stereoisomers are counted?
Explanation: Two donor choices multiplied by two acceptor choices include both self- and cross-reactions.
Which compound undergoes ordinary Cannizzaro reaction?
Explanation: It is an aldehyde with no α-H on the adjacent quaternary carbon.
What is oxidised in benzaldehyde Cannizzaro reaction?
Explanation: Another benzaldehyde molecule is reduced to benzyl alcohol.
What role does benzaldehyde play in a crossed aldol with acetophenone?
Explanation: Acetophenone supplies the enolate because benzaldehyde has no α-H.
Where does nitration of acetophenone occur mainly?
Explanation: The directly attached acyl group is deactivating and meta-directing.
Practice Problems
- Write the product of benzaldehyde plus propanal under crossed-aldol condensation conditions.
- Does “no α-H” imply Cannizzaro for every carbonyl compound?
- Write Cannizzaro products of 2,2-dimethylbutanal.
- Identify the donor and acceptor for 2-ethylpent-2-enal from propanal/butanal.
- Contrast electron flow and carbon skeleton outcomes of aldol and Cannizzaro.
- Plan m-nitroacetophenone from benzene and explain the order.
Propanal enolate carbon bonds to benzaldehyde carbonyl carbon. The addition product is OHC–CH(CH₃)–CH(OH)Ph; dehydration gives OHC–C(CH₃)=CHPh, 2-methyl-3-phenylprop-2-enal. The surviving aldehyde is from propanal, not benzaldehyde.
Key Takeaways
• Crossed aldol products follow donor–acceptor carbon tracing. • Four combinations apply to two simple partners with one donor site each, before stereoisomers. • Non-enolisable benzaldehyde can accept an enolate from another compound. • Cannizzaro disproportionates aldehydes without α-H into alcohol and carboxylate. • Hydride transfer does not require a free hydride ion in water. • Ketones without α-H are not automatically Cannizzaro substrates. • Directly attached CHO and COR groups generally direct aromatic electrophiles meta.