Final answer:
The SN₂ reaction between 1-bromo-2-methylpropane and methoxide is unlikely due to steric hindrance, as the substrate is tertiary. The SN1 or E2 mechanism is more plausible for this reaction, where substitution might lead to a racemic mixture or elimination to an alkene product.
Step-by-step explanation:
Understanding SN₂ Reactions
When considering the SN₂ reaction between 1-bromo-2-methylpropane and methoxide, one must be attentive to the reaction mechanism occurring within this context. The SN₂ mechanism involves a bimolecular process where the nucleophile attacks the substrate and the leaving group departs simultaneously. Specifically, in the transition state of an SN₂ reaction, the methoxide ion would approach the 1-bromo-2-methylpropane from the opposite side of the leaving bromide ion, forming a bond with the central carbon atom as the bromide leaves.
However, it's important to note that 1-bromo-2-methylpropane is a tertiary halide, which typically does not undergo SN₂ reactions due to steric hindrance. Instead, it's more likely to react via an SN1 or E2 mechanism. The SN1 mechanism involves the formation of a carbocation intermediate where the nucleophile can attack either side of the planar intermediary, possibly leading to a racemic mixture if the original molecule was chiral.
In the case of 1-bromo-2-methylpropane, an E2 elimination reaction might compete with substitution, yielding an alkene as the product, which could further react with HBr. This elimination reaction is favored by the steric hindrance that prevents the nucleophile from approaching the carbon in an SN₂ fashion. Nevertheless, the initial question relates to an SN₂ reaction, despite the unlikelihood of its occurrence in this particular case. The curved arrows in such a reaction would show the methoxide attacking the central carbon and the bromide departing in one concerted step, though this is not the favored pathway for this substrate.