Question
Using the Wurtz reaction, how do we synthesise butane from bromomethane? Also, what is the major limitation of this reaction when we use two different alkyl halides?
Solution — Step by Step
The Wurtz reaction couples two alkyl halide molecules using sodium metal in dry ether:
Two sodium atoms donate electrons, forming two carbanion-like species that couple together. The driving force is formation of the very stable sodium halide salt.
We start with bromomethane (CH₃Br). Two molecules react with 2 Na:
Wait — that gives ethane, not butane. To get butane (C₄H₁₀), we need bromoethane (CH₃CH₂Br) as our starting material.
The product is n-butane. This is a clean reaction because both starting molecules are identical — only one coupling product is possible.
Now suppose we use CH₃Br and C₂H₅Br together. Three products form:
| Coupling | Product |
|---|---|
| CH₃ + CH₃ | Ethane |
| C₂H₅ + C₂H₅ | Butane |
| CH₃ + C₂H₅ | Propane (the one we wanted) |
We get a mixture that’s hard to separate. This is why the mixed Wurtz is impractical for targeted synthesis.
Why This Works
Sodium is a powerful reducing agent (low ionisation energy, readily gives up its electron). When it reacts with an alkyl halide, it forms a highly reactive organo-sodium intermediate — essentially a carbanion — that immediately attacks another alkyl halide molecule.
The reaction works best with primary alkyl halides. Secondary and tertiary halides give poor yields because bulky groups cause steric hindrance at the reaction site, and elimination competes with coupling.
This is one of the earliest named reactions in organic chemistry — Charles-Adolphe Wurtz developed it in 1855. For JEE, the reaction is typically tested in two ways: identifying the correct starting material to make a target alkane, or explaining why mixed Wurtz fails.
Alternative Method
For odd-carbon alkanes (propane, pentane), the Wurtz reaction using two different halides is officially impractical. Instead, consider the Kolbe’s electrolysis of appropriate carboxylate salts — it also couples two carbon fragments but gives a cleaner product. In NCERT problems, you’ll mostly be asked about even-carbon products where Wurtz works perfectly.
If we specifically need propane via a Wurtz-type strategy, the trick is to use a symmetrical dihalide approach — but that’s beyond Class 11 scope. For boards, just remember: Wurtz = even-carbon alkanes cleanly, odd-carbon = mixed product problem.
Common Mistake
Counting carbons wrong. Students often write: “to make butane (C₄), use bromomethane (C₁)” — thinking the Na somehow adds carbons. It doesn’t. Each reactant molecule contributes its full carbon chain. Two CH₃Br molecules give ethane (C₂), not butane. Two C₂H₅Br molecules give butane (C₄). The rule is simple: product carbons = 2 × reactant carbons. Always double-check this before writing your answer in boards.
- Works best with: primary alkyl halides
- Solvent: dry ether (moisture destroys the reaction)
- Product carbons: double the reactant carbons
- Mixed Wurtz: always gives 3 products — not useful for pure compound synthesis