Why tertiary alkyl halides doesn't take part in Wurtj reaction?

1 Answer
Apr 26, 2018

It's because a critical step in the reaction involves an #"S"_text(N)2# substitution.

Explanation:

The general equation for a Wurtz reaction is

#"2R-X + 2Na" color(white)(l)stackrelcolor(blue)("dry ether"color(white)(mm))(→) "R-R + 2NaX"#

The reaction occurs through a free radical mechanism.

Step 1. Electron transfer to halogen

The #"Na"# atom transfers an electron to the halogen, producing an alkyl halide and an alkyl radical.

#"R-X + ·Na → R· + Na"^"+""X"^-"#

Step 2. Electron transfer to alkyl radical

Another sodium transfers an electron to the alkyl radical, forming a carbanion.

#"R· + ·Na → R:"^"-" + "Na"^"+"#

Step 3. Nucleophilic attack on alkyl halide

The carbanion attacks the alkyl halide in an #"S"_text(N)2# displacement reaction.

#"R:"^"-" + "R-X → R-R + :X"^-"#

Nucleophilic attack on a tertiary alkyl halide is extremely slow because of steric hindrance.

That's why the Wurtz reaction does not work with tertiary alkyl halides.