How does Markovnikov's rule apply when there is equal numbers of Cs and Hs on each side of the double bond?

Question

For example, when 2-methyl-2-pentene (I think this is right, Im looking from a diagram) is reacted with sulfuric acid and water- which side of the bond would the hydroxyl group attach to?

Answer 1

But 2-methyl-2-pentene doesn't have equal numbers of Hs on each side.  2-Methyl-2-pentene is (CH3)2C=CHCH2CH3.

You should remember that Markovnikov's "rule", at least in the form in which it's usually taught (as opposed to how it was originally formulated) is just an emprical memory aid, with no actual reference to a rationale or guiding principle.  If you want to understand chemistry, you can't rely on empirical shortcuts like that, you need to grasp the underlying chemical basis for them -- that makes them easier to remember, easier to extrapolate to new examples, and actually makes more sense, because then organic chemistry just becomes the application of a few simple principles over and over, rather than the rote memorization of a thousand unrelated facts.  In this case, both Markovnikov and antiMarkovnikov additions occur with the regioselectivity they do because of the underlying mechanism, which generally involves a carbocationic intermediate.  If there's a choice, that carbocation will be more stable with the (+) on either one or the other C of the double bond, and the location of the (+) charge and the nature of the reagent you're using will determine the product.

But to answer the question you actually asked: it doesn't.  The empirical rule is based on differing numbers of Hs (or, if you're getting a bit more sophisiticated, differing electron donating properties of the R groups), so you cannot use it if their number is the same.