Answer:
2-bromopentane
Explanation:
In this case, we have a substitution. An Sn2 reaction. With this in mind, we have to identify the leaving group in this case "OH" from the 2-pentanol. The nucleophile is a "Br" atom provided by the PBr3. So, the "OH" would be replaced by "Br". If we have an Sn2 reaction we will have an inversion in the configuration. If we start with a wedge bond for "OH" we will have a dashed bond for "Br", if we have a dashed bond for "OH" we will have a wedge bond for "Br". In the 2-pentanol we have any special configuration because we don't have a chiral carbon, so we can have both options (wedge and dashed) for the product (See figure 1).