What are the common products of the reaction: 2-bromobutane and NaOH
A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them.

An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution ($\mathrm{S_N1}$ and $\mathrm{S_N2}$) and elimination.

Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions. 

Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a $\mathrm{S_N2}$ reaction, a secondary halide – $\mathrm{S_N2}$ or elimination depending upon the strength of base/nucleophile and a tertiary halide – $\mathrm{S_N1}$ or elimination depending upon the stability of carbocation or the more substituted alkene.

So, I think preferably in aqueous NaOH or KOH there should be formation of alcohol but that too by $\mathrm{S_N1}$ if there's a tertiary; and, if primary halide, then it would follow $\mathrm{S_N2}$. It's bit tricky for secondary; again it depends on the reagents and solvents used.