Learn · Organic Chemistry

Nucleophilic Acyl Substitution

The addition–elimination mechanism and the reactivity ladder of carboxylic acid derivatives.

Quick answer A nucleophile adds to the sp2 acyl carbon to give a tetrahedral intermediate, which collapses and expels the leaving group — net substitution. Reactivity runs acid chloride > anhydride > ester ≈ acid > amide, and you walk down that ladder, not up.

Mechanism · Nucleophilic Acyl Substitution2 steps
Step 1 — the nucleophile adds to the acyl carbon.
OCH3Clδ+NuOCH3ClNutetrahedral
Unlike a ketone, an acyl group carries a leaving group (here Cl). The nucleophile adds to the δ+ carbonyl carbon, and the C=O π electrons move to oxygen, giving a tetrahedral intermediate.
Step 2 — the intermediate collapses, expelling the leaving group.
OCH3ClNuOCH3Nu+ Clsubstitution product
The alkoxide reforms the C=O and pushes out the leaving group (chloride). Net result: the LG is substituted, not just added — this is why acyl chlorides convert readily to esters, amides, etc.

Give a carbonyl carbon a heteroatom leaving group and its tetrahedral intermediate gains an exit — addition becomes substitution, uniting the whole derivative family under one mechanism.

Acid chloride
Anhydride
Ester
Amide

The five members of the family share one acyl carbon (a carboxylic acid, , rounds out the set) — they differ only in the leaving group hanging off it.

1. The Defining Feature Is a Leaving Group Attached to the Carbonyl Carbon

Each derivative pairs an electrophilic sp2 acyl carbon with a leaving group — –Cl, –OCOR, –OR, –OH, or –NR2 — the license aldehydes and ketones lack.

Acyl carbon + leaving group (Cl)
Aldehyde — no leaving group

2. The Mechanism Is Addition–Elimination Through a Tetrahedral Intermediate

This is not one-step SN2; it runs in two stages:

  1. Addition. The nucleophile attacks the carbonyl carbon; C=O π electrons shift onto oxygen, giving the sp3, O tetrahedral intermediate.
  2. Elimination. The oxygen lone pair re-forms C=O and ejects the leaving group, restoring the carbonyl with its new group.
Start: the acyl carbon bears a leaving group (Cl).
Nucleophile adds → sp3 tetrahedral intermediate.
Leaving group is expelled, C=O reforms → net substitution.

3. Reactivity Follows the Leaving Group: Acid Chloride > Anhydride > Ester ≈ Acid > Amide

One clean scale predicts most of the chemistry — memorize it in structure form:

Acid chloride (most reactive)
Anhydride
Ester
Amide (least reactive)

The spread is enormous: acid chlorides react violently with water, while amides need hours of hot acid or base to hydrolyze.

4. Two Reinforcing Factors Explain the Order — Leaving-Group Ability and Resonance Donation

Two effects point the same way — a good leaving group is also a poor donor:

  • Leaving-group ability. The better the leaving group (conjugate base of a stronger acid), the more reactive: chloride > carboxylate > alkoxide/hydroxide > –NR2.
  • Resonance donation. The more the attached atom donates into C=O, the less electrophilic and reactive: N (amides) > O (esters/acids) > Cl (acid chlorides).
Amide: N donates strongly → unreactive
Acid chloride: Cl barely donates → reactive

5. Interconversions Run Downhill — You Step Down the Ladder, Not Up

A more-reactive derivative converts into a less-reactive one, not the reverse — so an acid chloride plus an amine gives an amide fast and irreversibly:

Acetyl chloride + methylamine → N-methylacetamide, drawn live.

To climb up, synthesis first activates an acid (e.g. with SOCl2) to the acid chloride, then steps down.

6. The Same Mechanism Explains Ester Hydrolysis and Transesterification

In ester hydrolysis, water is the nucleophile and –OR the leaving group; ester and acid sit close, so it is an equilibrium:

Ethyl acetate hydrolyzes to acetic acid (plus ethanol).

Swap water for an alcohol and one –OR replaces another — transesterification, same two steps:

Methyl acetate + ethanol → ethyl acetate (transesterification).

7. Summary

Add–eliminate through a tetrahedral intermediate · leaving group expelled for net substitution · reactivity acid chloride > anhydride > ester ≈ acid > amide · move down the ladder, activate to move up.

Worked example

Problem. What is the product of acetyl chloride + ammonia (excess)?
  1. Ammonia adds to the electrophilic carbonyl carbon → a tetrahedral intermediate.
  2. The intermediate collapses, expelling chloride (a good leaving group) and re-forming the C=O.
  3. A second equivalent of NH3 neutralises the HCl by-product.

Answer. Acetamide. Acyl chlorides are the most reactive carboxylic-acid derivatives.

Quiz yourself

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The derivatives have a leaving group on the carbonyl carbon (–Cl, –OCOR, –OR, –NR2). After the nucleophile adds and the tetrahedral intermediate forms, that leaving group can be expelled to reform the C=O — giving substitution. Aldehydes and ketones are bonded only to C and H, which cannot leave, so their tetrahedral intermediate has no exit and stays as the addition product.

It is the sp3, negatively charged species formed the instant the nucleophile adds — the carbon holds four groups: the acyl chain, the nucleophile, an O, and the leaving group. It is the branch point: its collapse (oxygen lone pair pushing down to reform C=O) is what expels the leaving group and completes the substitution. Every member of the family passes through it.

Acid chloride > anhydride > ester > amide. Two reinforcing factors: leaving-group ability (chloride leaves best, –NR2 worst) and resonance donation into the C=O (chlorine donates least so the carbonyl stays electrophilic; nitrogen donates most so amides are stabilized and unreactive).

No — that runs up the ladder. It would require expelling the poor leaving group (chloride is the good one you would need to install) and giving up the strongly stabilized amide carbonyl, which is unfavorable. Interconversions go downhill: you activate an acid (e.g., with SOCl2) to reach the acid chloride, then step down to esters or amides — never the other way under mild conditions.

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