Alkylation of 1-naphthol in trifluoroethanol versus in DMSO
The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:



Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.

In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.

EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."