Question

In which case would a Williamson ether synthesis fail?

• A. Sodium ethoxide $+$ iodomethane
• B. Sodium ethoxide $+$ iodoethane
• C. Sodium ethoxide $+$ 2-iodopropane
• D. Sodium ethoxide $+$ 2-iodo-2-methylpropane

Answer 1

The correct option is D Sodium ethoxide $+$ 2-iodo-2-methylpropane

2-iodo-2-methylpropane is a tertiary alkyl halide.

This will undergo dehydrohalogenation on reaction with strong base such as sodium ethoxide to form 2-methyl-prop-1-ene which is an alkene and not an ether.

Williamson's reaction follows $S{N}^2$ mechanism.

Thus, Williamson ether synthesis will fail in this case.