Question

Draw the product you expect from the reaction of (R) - 2 -bromooctane with CH_3CO_2. Use the wedge/hash bond tools to indicate stereochemistry. Include H atoms at chiral centers only. If a group is achiral, do not use wedged or hashed bonds on it.

Answer 1

The first step is to figure out what kind of mechanism that will work with this reaction. Your choices are an Sn1 and Sn2. This reaction is an Sn2 because -CN is a strong nucleophile which favors the Sn2 mechanism.

In an Sn2 mechanism, its a one step process where one bond breaks while another bond forms. So the bromine (because its a great leaving group) just comes right off. A characteristic about the Sn2 mechanism is that it only does the backside attack meaning that the nucleophile can only creep up from behind the Br as the Br leaves.

Another important characteristic about the Sn2 mechanism is that it does the inversion of configuration which is when the leaving group is "wedged" then the nucleophile that comes in will be dashed instead of wedged, the opposite. So because there is a stereogenic center, it is allowed to do the inversion of configuration.

Here's a picture

CN CN + Br

This is an SN2 reaction, so Br will be replaced with CN. Also, since it's a backside attack, the stereocenter will be converted from R to S. The product will be (S)-2-cyanooctane.