Question

Account for the following : 
t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butyl methyl ether.

Answer 1

Sodium methoxide is a strong base hence elimination predominates over substitution 
\[\begin{array}{cc}
\ce{CH3}\phantom{.............................}\\
|\phantom{...............................}\\
\ce{CH3-C-Cl +CH3O^-Na^+->CH3-C=CH3OH + NaCI}\\
|\phantom{...............................}||\\
\phantom{....}\ce{CH3}\phantom{..............}\phantom{..............}\ce{CH2}
\end{array}\]

William synthesis

\[\ce{R - X + R -\underset{\bullet\bullet}{{}^{\bullet\bullet}_{O}}  - Na->R - \underset{\bullet\bullet}{{}^{\bullet\bullet}_{O}} - R + NaX}\]

The alkyl halide must be 1⁰ as the reaction involves the S_N2 reaction pathway.
If alkyl halide is 3⁰ then the alkene will be the major product. Hence, 2-methylpropene is formed instead of t-butylmethylether.