To determine the relationship between $\Delta H^{\circ}$ and $\Delta E^{\circ}$ (internal energy change), we first write the balanced thermochemical equation for the combustion of isobutylene (2-methylpropene, $C_{4}H_{8}$). Based on general combustion principles provided in the sources, the reaction is:

$C_{4}H_{8}(g) + 6O_{2}(g) \rightarrow 4CO_{2}(g) + 4H_{2}O(l)$

According to the sources, the relationship between enthalpy and internal energy is $\Delta H = \Delta U + \Delta n_{g}RT$ . Here, $\Delta n_{g}$ is the change in the number of moles of gaseous products and reactants .

1. Gaseous products: 4 moles of $CO_{2}(g)$. (Note: $H_{2}O$ is produced as a liquid under standard conditions).
2. Gaseous reactants: 1 mole of $C_{4}H_{8}(g)$ and 6 moles of $O_{2}(g)$, totaling 7 moles.
3. Calculate $\Delta n_{g}$: $\Delta n_{g} = 4 - 7 = -3$ .

Substituting this into the formula yields: $\Delta H^{\circ} = \Delta E^{\circ} + (-3)RT$, which simplifies to $\Delta H^{\circ} = \Delta E^{\circ} - 3RT$.

Since $R$ and $T$ are positive values, subtracting $3RT$ from $\Delta E^{\circ}$ results in a value that is smaller. Therefore, $\Delta H^{\circ} < \Delta E^{\circ}$, which matches Option D.