Which reaction cannot occur with 2 bromo 2 methyl butane? and why?

Think about what you have here. The leaving group (Br) of this reaction is bonded to a tertiary carbon. This actually favors radicals, sn1, e1, and is forced E2 when there is a strong base present. It favors radicals because of inductive effects. The bonds to adjacent carbons donate electron density. They don't give electrons. It's a stabilizing effect due to proximity. It stabilizes carbocation formation for the same reasons. The only one not favored is Sn2. Sn2 reactions have to occur via a nucleophile coming in opposite the leaving group. In this case, the tertiary carbon is too sterically hindered for a nucleophile to come in opposite the leaving group. That is the absolute reason as to why Sn2 won't occur. I hope this helped you.

The concern here is that you're making an attempt to convert an alcohol into an alkyl halide. To do that you want an Sn2 response, which in flip requires a excellent leaving team. So here's the main issue: -OH, as it is, shouldn't be a just right leaving group. So including HBr gives no reaction, seeing that the OH cannot go away within the first situation. There are three methods to override this quandary: 1) Use an Sn1 mechanism: area the preliminary reactant to powerful acid. This may protonate the OH crew, making it into a excellent leaving group. As soon as the OH2+ goes off, you're left with a carbocation, and this conveniently reacts with HBr to type the preferred product. 2) Use an Sn2 mechanism: Add Tosylate chloride (TsCl for short) to your alcohol. This transforms the unhealthy leaving crew OH into thegood leaving staff O-Ts, and the Br can then be with ease brought, forming the favored product. Three) Brominate your alcohol immediately via adding PBr3.