Carbonyl addition, acyl substitution and enolate chemistry

Carbonyl addition moves two pairs together

In a selected polar carbonyl addition, a nucleophile’s pair forms a bond to carbonyl carbon while the C=O π pair moves to oxygen. These coupled changes reduce C=O bond order and keep the ordinary second-row carbon within an octet. Adding another complete bond while retaining the original C=O double bond would instead give five bond pairs around carbon: an invalid ten-electron product in this model.

For CH₃CHO plus OH⁻, the selected tetrahedral adduct is CH₃CH(OH)O⁻. The original carbonyl oxygen bears the negative charge in this contributor, and the entering OH oxygen retains its hydrogen. Both sides contain C₂H₅O₂, total charge −1 and 26 valence electrons. The ledger constrains the proposal; it does not establish an experimental rate, equilibrium fraction or unique pathway.

A later proton transfer can regenerate hydroxide

In a symbolic base-catalyzed hydration sequence, water transfers a proton to the adduct’s O⁻, producing CH₃CH(OH)₂ and OH⁻. The proton-transfer ledger uses the adduct oxygen pair to make O–H while the donating water O–H pair stays with its oxygen. Both sides of this separate step contain C₂H₇O₃, charge −1 and 34 valence electrons.

Sum the two steps and cancel the intervening adduct and hydroxide appearing on opposite sides. The net is CH₃CHO + H₂O ⇌ CH₃CH(OH)₂. Hydroxide is regenerated in the stated catalytic sequence, while the added water appears in the net transformation. Reverse paths are required for the equilibrium description. Catalyst regeneration alone gives neither a particular hydration fraction nor a practical preparation.

An acyl substitution restores the carbonyl after addition

A selected ester addition–elimination model first forms a tetrahedral anion. A pair on the original carbonyl oxygen then restores C=O while the carbon–leaving-group bond pair goes to the departing group. This is different from retaining an aldehyde or ketone addition product, and from a single concerted SN2 displacement.

For the symbolic ethyl ethanoate/OH⁻ ledger, addition gives a tetrahedral C₄H₉O₃⁻ adduct. Elimination can be written as ethanoic acid plus ethanolate, followed by proton transfer to give ethanoate plus ethanol. The resulting net CH₃COOCH₂CH₃ + OH⁻ → CH₃COO⁻ + CH₃CH₂OH retains C₄H₉O₃, charge −1 and 44 valence electrons. Hydroxide is consumed in this net, unlike the selected catalytic hydration ledger.

The tetrahedral intermediate has four ordinary bonds around that carbon, rather than five complete bonds. Which groups depart, how proton transfers occur and whether a particular intermediate persists depend on substrate and medium. The ledger does not turn a qualitative leaving-group trend into an unconditional rate ranking or a guarantee of product formation.

An enolate’s contributors keep every nucleus attached

For the ethanal-derived enolate, a carbon-centered contributor ⁻CH₂CH=O and an oxygen-centered contributor CH₂=CH–O⁻ share C₂H₃O⁻, 18 valence electrons and the same nuclear attachments. Redrawing the carbon pair into C=C and the C=O π pair onto O relates contributors of one delocalized species. It does not create two independently exchanging populations.

Neutral ethanal CH₃CHO and ethenol CH₂=CHOH instead have C₂H₄O with a hydrogen attached to different atoms. Their interconversion is tautomerism and needs an actual proton-transfer pathway. Neither enolate contributor is that neutral enol. Describing both C and O as possible reactive sites does not by itself predict which will react under every set of conditions.

Locate the neighboring carbon before counting alpha hydrogens

An alpha carbon in this context is a carbon directly adjacent to the carbonyl carbon. Formaldehyde H₂C=O has two hydrogens on the carbonyl carbon and no alpha carbon: its conventional alpha-H count is zero. Ethanal has three alpha hydrogens on its methyl group. Propanone has six across its two neighboring methyl groups.

These are structural counts, not statements that all individual hydrogens have equal acidity or reactivity in every medium. Conventional alpha enolization needs an appropriate alpha H and compatible conditions. Other acidic positions or reaction types can exist; the term alpha does not mean every hydrogen anywhere near a carbonyl.

Worked examples

Try it yourself

1. Why must the C=O π pair move in the stated OH⁻ addition?

   Show the answer and reasoning

   To reduce carbonyl bond order while the new bond forms.

   Keeping the double bond and adding a complete fifth bond would give ten electrons around ordinary second-row carbon in this model.

2. Does hydroxide regeneration in hydration imply it is also regenerated in the stated ester-to-carboxylate net?

   Show the answer and reasoning

   No.

   Summing each separate ledger shows regeneration in the first case and consumption in the second.

3. Does drawing charge on C versus O in the two enolate contributors move a hydrogen?

   Show the answer and reasoning

   No.

   Nuclear attachments remain unchanged; the drawings describe one electronic structure.

4. Do formaldehyde’s two carbonyl hydrogens provide conventional alpha-H enolization?

   Show the answer and reasoning

   No.

   It has no alpha carbon. Carbonyl H and alpha H refer to different locations.

What to carry forward

• Move the carbonyl π pair when adding a new bond.
• Sum steps before assigning catalyst regeneration.
• Retain leaving-group departure and proton transfers in an acyl ledger.
• Keep enolate resonance separate from neutral tautomerism.
• Count alpha hydrogens from the actual carbon framework.