Thioester Reactions

Thioester formation and hydrolysis

A thioester has the linkage R–C(=O)–S–R′. An acid chloride can acylate a thiol to form this bond; hydrolysis breaks the bond between the acyl carbon and sulfur. The net products retain a carbonyl C=O group.

Acetyl chloride, CH3COCl. Ethanethiol with its sulfur and ethyl carbons blue.

pyridine

S-Ethyl ethanethioate, CH3C(=O)SCH2CH3, with the ethanethiol fragment blue.
Acetyl chloride and ethanethiol give ethyl thioacetate. Blue traces the thiol's heavy atoms; pyridine accepts the released proton.

At a glance

  • Formation: Acid chloride + thiol, with a base such as pyridine.
  • Hydrolysis: Aqueous hydroxide gives a carboxylate; acid workup gives the carboxylic acid.
  • Mechanism class: Nucleophilic acyl substitution.
  • Key exam point: Track the acyl C–S bond separately from the sulfur–alkyl bond.

Mechanism

Formation: addition, elimination and proton removal

In the direct thiol-attack pathway shown here, ethanethiol sulfur attacks the acid chloride's carbonyl carbon. Collapse of the tetrahedral intermediate restores C=O and releases chloride. Pyridine then removes the sulfur-bound proton, giving the neutral thioester and pyridinium chloride. The panels separate the electron-pair movements; blue tracks the intact S–ethyl fragment.

Ethanethiol sulfur attacks acetyl chloride at the carbonyl carbon. The C=O pi pair moves to oxygen, giving a tetrahedral intermediate with O− and S+H; chloride is still attached. Blue follows sulfur and both ethyl carbons.
An oxygen lone pair reforms C=O and the acyl C–Cl pair leaves with chloride. The thioester is initially protonated on sulfur; proton removal follows in the next panel.
Pyridine nitrogen takes the explicit S-bound proton, and the S–H bond pair returns to sulfur. The products are neutral ethyl thioacetate and pyridinium; chloride from the previous step is the counterion. This direct thiol-attack teaching sequence omits possible acyl-pyridinium participation.

Pyridine is drawn as the proton acceptor in this teaching pathway; acyl-pyridinium participation is not shown.

Hydrolysis: cleave the acyl C–S bond

Hydroxide attacks the carbonyl carbon. In the addition–elimination representation below, the tetrahedral intermediate collapses and releases ethanethiolate. Proton transfer then gives acetate and ethanethiol overall for one equivalent of hydroxide. The sulfur–ethyl bond stays intact throughout.

In the addition–elimination teaching representation, hydroxide attacks the carbonyl carbon of ethyl thioacetate. C=O pi donation to oxygen gives a tetrahedral alkoxide. Sulfur remains attached to its ethyl carbon.
The alkoxide reforms C=O while the acyl C–S pair leaves with ethanethiolate. The leaving sulfur remains attached to its original ethyl group. Ethanethiolate then removes the carboxylic-acid proton; O–H electrons stay on oxygen, giving acetate and ethanethiol for one equivalent of hydroxide. Excess strong base can deprotonate the thiol further.

Excess strong base can leave more of the sulfur fragment as ethanethiolate. Acid workup gives acetic acid and protonates the sulfur fragment; it does not convert that fragment into ethanol.

Scope, limits and traps

  • Keep C=O: a thioester contains sulfur beside the carbonyl carbon, not in place of the carbonyl oxygen.
  • Before acid workup, draw the carboxylate rather than the neutral carboxylic acid for base hydrolysis.
  • The sulfur fragment's protonation depends on conditions. The one-equivalent hydroxide net equation can be written with RSH; excess strong base can give RS⁻. Acid workup protonates the sulfur fragment.
  • Do not split the sulfur–ethyl bond when predicting hydrolysis. The ethyl group leaves together with sulfur, not as ethanol.

Worked examples

Example 1: Acetyl chloride + ethanethiol → ethyl thioacetate

The product is S-ethyl ethanethioate, commonly called ethyl thioacetate, CH₃C(=O)SCH₂CH₃. Its two acetyl-fragment carbons come from acetyl chloride, and its two ethyl carbons come from ethanethiol. Pyridinium chloride accounts for the proton and chloride removed during formation.

Ethyl thioacetate with a C=O group and a sulfur-linked ethyl group.
Ethyl thioacetate is C₄H₈OS; the blue sulfur is bonded to both the acyl carbon and its original ethyl carbon.

Example 2: Ethyl thioacetate → acetate and ethanethiol

Aqueous hydroxide hydrolyzes the thioester. The following shows the net products for one hydroxide equivalent; the RSH/RS⁻ balance changes with excess base. After acidic workup, the acyl product is acetic acid.

Ethyl thioacetate. Aqueous hydroxide hydrolysis. Acetate anion and ethanethiol, with the sulfur-linked ethyl fragment blue.
The carbonyl acyl fragment becomes acetate. The intact blue S–ethyl fragment becomes ethanethiol in this net equation.

Tools

  • For Example 1, open Reaction Solver Pro and manually enter acetyl chloride and ethanethiol, with pyridine and a standard workup. This generic link does not preload the example.
  • Open the IUPAC Namer for naming practice.

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