$\mathop {C{H_3} - \mathop {\mathop {C\,\, = \,}\limits^{|\,\,\,\,\,\,} }\limits^{C{H_3}} C{H_2}}\limits_{{\text{Isobutylene}}} + C{H_3}OH + NaBr$
$C{H_3}ONa \to C{H_3}{O^ - } + N{a^ + }$
methoxide ion is a $(C{H_3}{O^ - })$strong base, therefore it abstract proton from $3^\circ$alkyl halide and favours elimination reaction.
$2-$ મીથાઈલબ્યુટેન $\xrightarrow{{B{r_2},\,hv}}$ $2-$ બ્રોમો $-3-$ મીથાઈલબ્યુટેન
(not the major product)