What are $X$, $Y$ and $Z$ in the following reactions? $\begin{aligned} & \left(\mathrm{H}_3…
What are $X$, $Y$ and $Z$ in the following reactions?
$\begin{aligned}
& \left(\mathrm{H}_3 \mathrm{C}\right)_3 \mathrm{C}-\stackrel{\circ}{\mathrm{O}} \mathrm{\oplus} \mathrm{Na}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \longrightarrow X+\mathrm{NaBr} \\
& \left(\mathrm{H}_3 \mathrm{C}\right)_3 \mathrm{C}-\mathrm{Br}+\mathrm{CH}_3 \mathrm{CH}_2 \stackrel{\ominus}{\mathrm{O}} \mathrm{\oplus} \mathrm{a} \longrightarrow Y+\mathrm{Z}
\end{aligned}$
Solution
\(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{ONa}^{+}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \longrightarrow\left(\mathrm{CH}_3\right)_3 \underset{(X)}{\mathrm{COCH}_2 \mathrm{CH}_3}+\mathrm{NaBr}\)
These reactions are known as Williamson synthesis where the product (ether or alkene) depends upon the type of halide and alkoxide ion. The reaction (I) involves \(\mathrm{SN}_2\) attack of \(3^{\circ}\) alkoxide ion on \(1^{\circ}\) halide giving ether whereas in reaction (II), a tertiary alkyl halide and primary alkoxide ion are present which gives alkene and alcohol.
