Wurtz reaction template

Sodium gives an electron to bromomethane (methyl radical), a second Na turns the radical into methylsodium, and CH₃⁻ displaces bromide from another CH₃Br to give ethane.

Mechanism step by step

Two alkyl halides and sodium in dry ether join into one alkane

1. Electron transfer – Na gives one electron; Br⁻ leaves and a methyl radical forms
2. Organosodium – A second Na reduces the radical to CH₃⁻ Na⁺
3. C–C bond (SN2) – The carbanion displaces Br⁻ from a second CH₃Br

Key points

• Shown: the ionic (organosodium) pathway, R–X + 2 Na → R⁻Na⁺ + NaX, then R⁻ + R–X → R–R. Some books show two R• radicals pairing instead; both start with electron transfer from Na.
• Gives symmetrical alkanes with an even number of carbons (2 CH₃Br → C₂H₆). Methane cannot be made.
• Two different halides give a mixture (R–R, R–R′, R′–R′), so it is not used for unsymmetrical alkanes.
• Ether must be dry: water would protonate R⁻ to R–H. Side products are an alkene and alkane (disproportionation), mainly with 2° and 3° halides.

Drawn with 23 atoms, 10 arrows, 8 shapes – every one editable in MolDraw.

Sources

• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020 Textbook