Draw the product of the reaction between 2-pentanol and PBr3. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include counter-ions, e. g., Na , I-, in your answer. If no reaction occurs, draw the organic starting material.

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).

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