In Williamson synthesis of mixed ether having a primary and a tertiary alkyl group if tertiary halide is used, then :
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- Rate of reaction will be slow due to slow cleavage of carbon-halogen bond.
- Alkene will be the main product.
- Simple ether will form instead of mixed ether.
- Expected mixed ether will be formed.
The Correct Option is B
Solution and Explanation
With a tertiary alkyl halide, backside attack is blocked by steric hindrance, while the alkoxide ion acts as a strong base.
So the alkoxide abstracts a \(\beta\)-hydrogen (E2 elimination), giving an alkene, e.g. \(\mathrm{(CH_3)_3C{-}Br + C_2H_5O^- \rightarrow (CH_3)_2C{=}CH_2}\).
Hence the alkene is the main product, not the ether.
Answer: (2) Alkene will be the main product.
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