Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Imagine you are looking at a molecular battlefield. On one side, we have our substrate: 2-bromo-3-methylbutane. This is a secondary ($2^{\circ }$) alkyl halide. On the other side, we have our reagent and solvent: methanol ($C{H}_{3}OH$).

Methanol is a weak nucleophile and a polar protic solvent. This specific combination is a massive neon sign pointing towards the $S_{N}1$ mechanism. Weak nucleophiles don't have the aggressive "push" required for a concerted $S_{N}2$ backside attack, especially on a sterically hindered secondary substrate. Instead, they patiently wait for the leaving group to depart on its own.

Carbocation Formation

The very first step of an $S_{N}1$ reaction is the rate-determining step (RDS). The carbon-bromine bond breaks heterolytically. Bromine, being highly electronegative and a good leaving group, takes the bonding electrons and leaves as a bromide ion ($B{r}^{-}$).

This leaves behind a secondary ($2^{\circ }$) carbocation.

Now, I know this intermediate looks fine, but let's take a breath and look closer. Carbocations are highly reactive, electron-deficient species. They are always looking for ways to become more stable.

Carbocation Rearrangement

Look at the carbon atom directly adjacent to our positively charged carbon. It is a tertiary ($3^{\circ }$) carbon, and crucially, it has a hydrogen atom attached to it.

Carbocations will spontaneously rearrange if a simple shift can lead to a more stable state. Here, a 1,2-hydride shift takes place. The hydrogen atom migrates with its entire electron pair to the positively charged carbon.

Following this shift, the positive charge moves to the tertiary carbon. We have now formed a tertiary ($3^{\circ }$) carbocation.

This new intermediate is significantly more stable due to increased hyperconjugation (more adjacent $\alpha$-hydrogens donating electron density into the empty p-orbital).

Nucleophilic Attack

Now our highly stable electrophile is ready. The oxygen atom in methanol swoops in and uses its lone pair to attack the positively charged tertiary carbon.

After the attack, the oxygen atom bears a positive charge because it formed three bonds. To regain neutrality, it quickly loses a proton ($H^{+}$) to the surrounding solvent.

This final deprotonation gives us our major product: 2-methoxy-2-methylbutane.

This perfectly matches option (c). It's an elegant mechanism that beautifully demonstrates why you must always check for carbocation rearrangements in $S_{N}1$ reactions!