Thiol Disulfide Redox

Thiol oxidation and disulfide reduction

Oxidation joins two thiol groups, R–SH, to form a disulfide, R–S–S–R. Reduction cleaves that S–S bond and restores two thiols. Iodine is a common oxidant for this pair; zinc in acidic medium provides a reductive route back. The carbon–sulfur bonds remain intact in both directions.

Two ethanethiol molecules, A black and B blue, react with iodine to give diethyl disulfide plus two hydrogen iodide molecules. — Oxidation forms one S–S bond from two thiols. Blue tracks the heavy atoms of copy B into one half of the disulfide; the new S–S bond is black. HI denotes the formal acid product.

At a glance

Oxidation: Two thiols plus I₂ give one disulfide and two equivalents of acid/iodide.
Reduction: Zinc in acidic protic medium can cleave a disulfide to two thiols; aqueous acid with an alcohol-containing medium is a supported formulation.
Bond changed: S–S forms or breaks; the C–S framework is retained.
Stoichiometry: Two RSH molecules correspond to one RSSR molecule.
Key exam point: Zinc is consumed as a reductant, becoming Zn²⁺; it is not a catalyst in the balanced reduction below.

Mechanism

The oxidation panels show a selected iodine pathway through a sulfenyl iodide, R–S–I. Thiol sulfur reacts with iodine; solvent proton transfer gives the sulfenyl iodide. A second thiol then forms the S–S bond, and another proton transfer gives the neutral disulfide. These formal charged drawings explain the electron flow; they do not imply that the intermediates are isolated or that every thiol oxidation uses this route. Red arrows show electron-pair movement. Blue tracks the heavy atoms of thiol copy B, and the new S–S bond is black.

A sulfur lone pair attacks I₂, and the I–I bond pair moves to the leaving iodine. The resulting S-iodinated thiol is drawn as a formal cation with iodide, not as an isolated preparative species.

Solvent removes the S–H proton, and the bond pair returns to sulfur, giving the sulfenyl iodide Et–S–I. The iodide from the first step is retained in the balance.

Copy B attacks sulfur of Et–S–I while iodide leaves. The new S–S bond is black; copy B heavy atoms remain blue. Hydronium and iodide already formed are carried through explicitly.

Solvent removes the remaining sulfur proton. The neutral product contains both original ethyl–sulfur units. In aqueous bookkeeping, the two lost thiol protons appear as two hydronium ions, with two iodide ions.

Zinc/acid reduction is a different, reagent-driven process. The final panel keeps the net electron and proton balance without inventing an elementary mechanism at the metal surface.

Zinc/acid reduction is shown as a balanced net redox process, not an elementary metal-surface mechanism. One disulfide returns two thiols; zinc supplies two electrons and is consumed as Zn²⁺. Total charge is +2 on both sides.

Oxidation removes two protons and two electrons overall: 2 RSH → RSSR + 2 H⁺ + 2 e⁻. Iodine accepts the electrons: I₂ + 2 e⁻ → 2 I⁻. The explicit solvent drawings use H₃O⁺; the equivalent formal balance is 2 RSH + I₂ → RSSR + 2 H⁺ + 2 I⁻, often written with 2 HI.

For reduction, Zn → Zn²⁺ + 2 e⁻ combines with RSSR + 2 H⁺ + 2 e⁻ → 2 RSH. Thus RSSR + Zn + 2 H⁺ → 2 RSH + Zn²⁺. Chloride from HCl is an unchanged spectator.

Scope, limits and traps

Do not draw only one thiol molecule as the balanced reduction product of one disulfide. Each sulfur retains its carbon group and gains H, giving two RSH molecules.
A disulfide contains two sulfur atoms joined together. A sulfide, R–S–R′, contains one sulfur between two carbon groups and is a different functional group.
Iodine oxidation does not add oxygen. A sulfoxide or sulfone is not the disulfide product in this scheme.
Mixing two different thiols can produce symmetrical and mixed disulfides. Selective R–S–S–R′ formation is not guaranteed merely by mixing RSH and R′SH.
Zinc/acid compatibility depends on the rest of the substrate. Cleaving a disulfide selectively is not a promise that every other reducible or acid-sensitive group survives.

Worked examples

Example 1: Ethanethiol to diethyl disulfide

Two molecules of ethanethiol, CH₃CH₂SH, react with iodine to give diethyl disulfide, CH₃CH₂–S–S–CH₂CH₃.

Two ethanethiol molecules labeled A and B, plus iodine. Copy B heavy atoms are blue. — 2 C₂H₆S + I₂ → C₄H₁₀S₂ + 2 HI. Both sulfur atoms are retained; the two S–H hydrogens account for the acid products.

Diethyl disulfide with its copy B half blue, plus a coefficient of two for hydrogen iodide. — 2 C₂H₆S + I₂ → C₄H₁₀S₂ + 2 HI. Both sulfur atoms are retained; the two S–H hydrogens account for the acid products.

The organic product has one fewer pair of hydrogen atoms than the two starting thiols combined. Neither ethyl group is removed or transferred to iodine.

Example 2: Reduce diethyl disulfide back to ethanethiol

Diethyl disulfide gives two ethanethiol molecules with zinc and acid. The ionic equation makes the electron donor and charge balance explicit.

Diethyl disulfide plus neutral zinc and two formal aqueous H+ equivalents; chloride spectator omitted. — C₄H₁₀S₂ + Zn + 2 H⁺ → 2 C₂H₆S + Zn²⁺. Total charge is +2 on both sides; chloride is omitted consistently.

Two ethanethiol molecules labeled A and B plus Zn2+, with copy B heavy atoms still blue. — C₄H₁₀S₂ + Zn + 2 H⁺ → 2 C₂H₆S + Zn²⁺. Total charge is +2 on both sides; chloride is omitted consistently.

The S–S bond breaks, while both C–S bonds remain. The two product structures are chemically identical ethanethiol molecules; A/B labels distinguish their atom origins rather than different products.

Tools

Oxidation partners: Open Reaction Solver Pro. Enter ethanethiol and iodine manually as separate participants, with the standard workup setting. Use the balanced equation to track two thiols per disulfide.
Explore reaction mechanisms with the Mechanism Solver.
Explore names with the IUPAC Namer.

Study Tools