When 2-methylpropene is bubbled into dilute sulfuric acid at room temperature, it appears to dissolve. What ne?

1. in the reaction of 2-methylpropene with water in the presence of sulfuric acid, the first step is the attack of the electrons of the pi bond on a proton (usually shown on a hydronium ion).
2. this gives a tertiary carbocation. In the next step of the reaction, a lone pair of electrons from the oxygen of water attack the carbocation to give another intermediate which bears a positive charge on oxygen, this is called an oxonium ion
3.in the final step of the reaction, a proton is removed from the oxygen by water to regenerate the hydronium ion, and to give the alcohol product.