Nucleophilic Acyl Substitution

Updated Apr 17, 2026

Nucleophilic acyl substitution (NAS) is the fundamental mechanism for all carboxylic acid derivatives. It has two phases: (1) addition of a nucleophile to the carbonyl C, then (2) elimination of a leaving group to restore the carbonyl. The net result is that one group on the acyl carbon has been replaced by another - substitution at the carbonyl.

This mechanism is the key distinction between carboxylic acids/derivatives and aldehydes/ketones. Aldehydes and ketones CANNOT do NAS because they have no leaving group attached to the carbonyl carbon (just H or R). Carboxylic acids and their derivatives all have a leaving group (OH, Cl, OR, NR₂) that can be kicked out.

The Two-Phase Mechanism

Phase 1: Addition. Exactly like the aldehyde/ketone addition mechanism of Chapter 6:

Nucleophile attacks the carbonyl C.
Pi bond breaks, electrons flow to oxygen.
Result: a tetrahedral alkoxide intermediate.

Phase 2: Elimination. Different from aldehyde/ketone chemistry:

The alkoxide oxygen reforms the C=O double bond.
Electrons flow back to kick out whatever leaving group is attached to the carbon (Cl, OR, OH, NR₂).
Result: the original carbonyl is restored, but now with a new substituent in place of the leaving group.

Net: R-CO-LG + Nu⁻ → R-CO-Nu + LG⁻. Substitution of Nu for LG at the acyl carbon.

Why Aldehydes and Ketones Don’t Do NAS

In an aldehyde (R-CHO), the groups on the carbonyl are R and H. Neither is a leaving group (H⁻ is an incredibly strong base; R⁻ is too). After nucleophilic addition, the tetrahedral intermediate can only protonate the alkoxide - it cannot eliminate H or R. So the reaction is stuck at the tetrahedral alcohol stage (the addition product), and NAS does not happen.

Carboxylic acid derivatives, on the other hand, always have at least one good leaving group (Cl, OR, OCOR’, OH after protonation, NR₂ as a poor LG). The tetrahedral intermediate collapses by kicking out the LG, restoring the C=O and giving the substitution product.

The Reactivity Ladder

The order of reactivity in NAS is:

acyl halide > anhydride > aldehyde > ketone > ester > carboxylic acid > amide

Wait - “aldehyde > ester”? Yes, aldehydes are more electrophilic than esters (because ester’s OR donates by resonance, reducing electrophilicity). But aldehydes cannot DO NAS (no leaving group), so they only react via addition.

For the actual NAS mechanism (which requires a leaving group), the order is:

acyl halide > anhydride > ester > amide

With carboxylic acid sitting roughly between ester and anhydride in reactivity - it can be protonated to activate further, but OH⁻ is a poor LG.

Why Chloride Is a Great Leaving Group in NAS

In the tetrahedral intermediate after nucleophile attack, the leaving group has to depart with the bond electrons. Chloride is a weak base (stable as Cl⁻), so it leaves easily. OR⁻ (alkoxide) is a strong base (~pKa 16 for its conjugate acid), so it leaves much more reluctantly - that is why esters react more slowly than acid chlorides.

Acid vs. Base Catalysis

NAS can work under acidic or basic conditions, with slightly different mechanisms:

Base-catalyzed NAS: The nucleophile is already anionic (hydroxide, alkoxide, amide). No acid catalysis needed. Mechanism: attack → tetrahedral intermediate → LG leaves → restore C=O.
Acid-catalyzed NAS: Neutral nucleophile (water, alcohol, amine). Acid protonates the carbonyl oxygen first, activating it. Then the neutral nucleophile attacks; a proton shuffle converts the alcohol-like OH to a better leaving group; that LG leaves with acid assistance. This is the Fischer esterification pattern from Section 8.8.

Tetrahedral Intermediate Stability

In many NAS reactions, the tetrahedral intermediate can be detected or even isolated. Its stability determines whether the reaction proceeds cleanly:

Stable intermediate → reaction works well.
Unstable intermediate → reaction collapses back to starting material, giving low yield.

Temperature, base strength, and substrate steric/electronic properties all influence this balance.

Common NAS Reactions

Ester hydrolysis (saponification): R-COOR’ + OH⁻ → R-COO⁻ + R’-OH.
Ester + amine: R-COOR’ + R”-NH₂ → R-CO-NH-R” + R’-OH (amide formation).
Transesterification: R-COOR’ + R”-OH → R-COOR” + R’-OH (swap OR groups).
Amide hydrolysis: R-CO-NR’₂ + H₂O → R-COOH + HNR’₂. Slow under mild conditions; requires strong acid or base.
Acid chloride + nucleophile: R-COCl + Nu → R-CO-Nu + Cl⁻. Very fast with any nucleophile.

Why is a ketone NOT a substrate for nucleophilic acyl substitution, even though it has an electrophilic carbonyl carbon that can be attacked by a nucleophile?

Click to reveal answer

A ketone's carbonyl carbon has two R (alkyl) substituents. Neither R nor H is a leaving group (R⁻ and H⁻ are extremely strong bases that cannot leave). After a nucleophile attacks, the tetrahedral intermediate cannot eliminate anything - it just protonates to give the tetrahedral alcohol (nucleophilic ADDITION product, not substitution). Carboxylic acid derivatives have Cl, OR, NR₂, or OCOR' attached, all of which CAN leave, allowing the carbonyl to reform with a new nucleophile - substitution.