Organic Chemistry · Aldehydes and Ketones: Nucleophilic Addition Reactions

Nucleophilic Addition Reactions of Aldehydes and Ketones

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

The carbonyl group (C=O) is the most important functional group in organic chemistry, and nucleophilic addition is its signature reaction. Because oxygen is more electronegative than carbon, the C=O bond is polarized: carbon carries a partial positive charge (an ), and oxygen a partial negative charge. A attacks the carbonyl carbon, the π electrons move up onto oxygen, and a forms; protonation of that oxygen completes the addition:

$R 2 C = O + Nu - \to R 2 C (Nu)(O -) H + ⟶ R 2 C (Nu)(OH)$

Aldehydes react more readily than ketones for two compounding reasons: steric (their carbonyl carbon is less crowded) and electronic (alkyl groups push electron density toward the carbonyl carbon, making it less positive). Because the carbonyl carbon carries no good leaving group, the reaction stops at addition. This one mechanism explains dozens of reactions — hydration, cyanohydrin formation, alcohol formation, imine formation — so mastering it turns the rest of the chapter into variations on a theme.

Why this matters

Nucleophilic addition is where organic chemistry stops being about naming compounds and starts being about building them: every carbon–carbon bond-forming reaction that constructs a drug, polymer, or natural product relies on a nucleophile attacking an electrophilic carbonyl carbon. Biochemistry runs the same chemistry constantly — sugars cyclize through intramolecular hemiacetal formation, enzymes form imines with substrates via lysine side chains, and NADH delivers hydride to carbonyls in metabolic redox reactions. On exams, the aldehyde-versus-ketone reactivity comparison is a perennial question, and knowing whether an addition is reversible or irreversible determines which conditions a synthesis requires.

The college version

Core Concepts

The carbonyl carbon is the electrophilic site

Oxygen's greater electronegativity pulls electron density away from carbon, leaving carbon electron-poor (δ⁺) and oxygen electron-rich (δ⁻); resonance — a structure with the π electrons entirely on oxygen — reinforces this. The carbonyl carbon is therefore the site a nucleophile always attacks; attacking oxygen would be wrong, since oxygen already has excess electron density.

The mechanism: attack, form alkoxide, protonate

Nucleophilic addition proceeds in two stages. First, the nucleophile's lone pair forms a new bond to the carbonyl carbon while the C=O π electrons move up onto oxygen, converting the planar sp² carbon into a tetrahedral sp³ carbon bearing an alkoxide (C–O⁻). shows this as one arrow from the nucleophile to carbon and a second from the C=O π bond to oxygen, drawn together because bond formation and π-bond breaking happen in the same step. Second, a proton source (H₂O, H₃O⁺, or the acidic reagent) transfers H⁺ to the alkoxide, giving the neutral product. The alkoxide is a strong base, so this protonation is fast.

Aldehydes are more reactive than ketones

Two effects make aldehydes more reactive. Sterically, an aldehyde carbonyl carbon carries one carbon substituent and one small hydrogen; a ketone carries two, and bulky groups block the nucleophile's perpendicular approach. Electronically, alkyl groups donate electron density, so a ketone's two alkyl groups partially satisfy the carbon's positive charge, making it less attractive to nucleophiles. Formaldehyde, with two hydrogens, is the most reactive of all; electron-withdrawing substituents (the three chlorines in chloral, Cl₃CCHO) boost reactivity for the same reason in reverse.

Reversible vs. irreversible addition

Weak nucleophiles — water, alcohols, neutral amines — add reversibly, the equilibrium depending on product stability and conditions (why acetal and imine formation need acid catalysis or water removal). Strong nucleophiles — hydride (H⁻), cyanide (CN⁻), Grignard reagents (R–MgX) — add essentially irreversibly: the new C–Nu bond is far stronger than the C=O π bond lost.

Attack can come from either face

Because the carbonyl carbon is sp² and planar, the nucleophile can attack from either face. For simple aldehydes and ketones both faces are equivalent, so a new stereocenter forms as a racemic mixture; bulky or chiral substituents can block one face and make attack selective — a concept central to asymmetric synthesis later in the course.

Common Confusions

Do Not Confuse	With	Difference
Nucleophile attacking carbon	Nucleophile attacking oxygen	Attack on carbon is correct; oxygen is already electron-rich (δ⁻) and repels nucleophiles.
Aldehyde reactivity	Ketone reactivity	Aldehydes react faster: one small H substituent vs. two bulky, electron-donating R groups.
Addition reaction	Substitution reaction	Addition forms a new bond without expelling a leaving group; substitution swaps one group for another.
Reversible addition	Irreversible addition	Weak nucleophiles (H₂O, ROH) add reversibly; strong ones (H⁻, CN⁻, R⁻) add irreversibly.
The alkoxide intermediate	The final product	The alkoxide (C–O⁻) is transient; the neutral product forms only after protonation.
Steric effect alone	Electronic effect alone	Both operate together; exams often test whether you can name both reasons.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a playground seesaw (the C=O bond) where the carbon side is slightly magnetized to attract things and the oxygen side is not. When a ball (the nucleophile) rolls onto the carbon side, the seesaw tips — the oxygen side grabs the extra electrons, and then a friend (a proton) jumps on to balance it. Aldehydes are easier to hit because there's less furniture blocking the carbon side, while ketones have two big chairs in the way.

Worked example

Example 1: How much of a carbonyl is reacted at equilibrium?

The hydration equilibrium constant measures how readily a carbonyl accepts a water nucleophile:

$K eq = [gem-diol] [carbonyl]$

For acetone, K_eq ≈ 1.4 × 10^-3. What fraction of acetone exists as the hydrated gem-diol in pure water?

Solve for the fraction f converted to hydrate:

$f = K eq 1 + K eq = 1.4 \times 10 -3 1 + 1.4 \times 10 -3 = 1.4 \times 10 -3 1.0014 = 1.4 \times 10 -3$

As a percentage:

$f \times 100 % = 1.4 \times 10 -3 \times 100 % = 0.14 %$

So in pure water, roughly 0.14% of acetone exists as the hydrate — essentially no reaction. The units cancel (concentration over concentration), so the fraction is dimensionless — a good check that the algebra is correct.

Example 2: Why does formaldehyde react so much faster?

Repeat the calculation for formaldehyde, whose hydration constant is approximately 2 × 10³:

$f = K eq 1 + K eq = 2 \times 103 1 + 2 \times 103 = 20002001 = 0.9995$

$f \times 100 % = 99.95 %$

Formaldehyde is essentially fully hydrated, while acetone is essentially untouched — and the only structural difference is two small hydrogens instead of two methyl groups. This single comparison explains why aldehydes are the "fast" carbonyls: less steric bulk and less electron donation make their carbonyl carbon far more attractive to nucleophiles.

Key takeaways

The carbonyl carbon is electrophilic (δ⁺); nucleophiles always attack carbon, never oxygen.
Mechanism: nucleophile attacks C → tetrahedral alkoxide → protonation gives the product.
Aldehydes react faster than ketones: less steric hindrance + less electron donation (the small H substituent).
Formaldehyde is the most reactive simple carbonyl; electron-withdrawing groups boost reactivity, donors suppress it.
No good leaving group on the carbonyl carbon → addition stops at the tetrahedral product.
Strong nucleophiles (H⁻, CN⁻, R⁻) add irreversibly; weak ones (H₂O, ROH, amines) add reversibly.
The planar carbonyl allows attack from either face; new stereocenters form as racemic mixtures unless the environment is chiral.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Draw (in words) the two curved arrows of the first step of nucleophilic addition to a carbonyl.
Show answer
Arrow 1: from the nucleophile's lone pair to the carbonyl carbon (new C–Nu bond). Arrow 2: from the C=O π bond to oxygen (forming the alkoxide). Both arrows are drawn simultaneously in one curved-arrow step.
Why is the carbonyl carbon, not the oxygen, the site of nucleophilic attack?
Show answer
Oxygen is more electronegative, so the C=O bond is polarized with δ⁺ on carbon; carbon is electron-poor and accepts the nucleophile's electron pair, while oxygen already has excess electron density.
Give two structural reasons aldehydes are more reactive than ketones toward nucleophilic addition.
Show answer
(1) Steric: a ketone has two alkyl substituents that crowd the carbonyl carbon and hinder nucleophile approach; an aldehyde has one small hydrogen. (2) Electronic: alkyl groups donate electron density that partially satisfies the carbon's positive charge, making ketones less electrophilic.
Why does nucleophilic addition stop at the tetrahedral product instead of continuing to substitution?
Show answer
The carbonyl carbon has no good leaving group (H⁻, R⁻, and OH⁻ are all poor leaving groups). Without a leaving group, the tetrahedral intermediate cannot collapse by expulsion, so the reaction ends at addition.
Is the addition of a Grignard reagent reversible or irreversible? What about addition of water?
Show answer
Grignard addition is irreversible (a strong carbon nucleophile forming a strong C–C bond); water addition is reversible (a weak nucleophile, with the hydrate often reverting to the carbonyl).

Keep learning

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Nucleophile: An electron-rich species that donates a lone pair to form a new bond.
Electrophile: An electron-poor atom that accepts an electron pair.
Tetrahedral alkoxide intermediate: The negatively charged C–O⁻ species formed after nucleophilic attack.
Curved-arrow notation: A convention showing where electron pairs move during a step.
Reversible addition: An addition whose equilibrium lies partway; product can revert to starting materials.
Irreversible addition: An addition that goes essentially to completion.
Steric hindrance: Blocking of a reaction site by bulky neighboring groups.

Sources & references

openstax.org — Organic Chemistry

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