Which hydrogen does the nucleophile attack in E2 reaction?
Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2.

As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options.

In future similar examples, you may come across reactants with multiple possible reactive positions that are not equivalent. For example, consider 3-chlorohexane (instead of 3-chloropentane, as in the original problem). In this case, again there are two possible sites to be deprotonated (Ha and Hb). Looking at the two possible products, both are disubstituted alkenes, so they will be close in stability. There isn't much difference in the groups nearby Ha and Hb, so the kinetics of the deprotonation are probably similar. So which is the "correct" answer? In reality, both are correct. They may not be formed in exactly equal amounts, but they will probably both be formed in substantial amounts.



There are a couple of key lessons here for those studying organic chemistry:

1) Draw out all of the possible products that come from mechanisms that you think make sense. This might be lots of drawing! Nevertheless, it is much easier to realize the differences if everything is drawn out in front of you.

2) Organic chemistry is messy. Often times multiple mechanistic pathways will be viable and multiple products will form. Much of the creativity in organic chemistry is finding ways to ensure that only one mechanism can occur or one (desired) compound is formed as the major product.