$\mathop {C{H_3} - \mathop {\mathop {C\,\, = \,}\limits^{|\,\,\,\,\,\,} }\limits^{C{H_3}} C{H_2}}\limits_{{\text{Isobutylene}}} + C{H_3}OH + NaBr$
$C{H_3}ONa \to C{H_3}{O^ - } + N{a^ + }$
methoxide ion is a $(C{H_3}{O^ - })$strong base, therefore it abstract proton from $3^\circ$alkyl halide and favours elimination reaction.



${C_2}{H_5}OH + SOC{l_2}\xrightarrow{{{\text{Pyridine}}}}{C_2}{H_5}Cl + S{O_2} + HCl$