Animated Solution for Chemistry - Organic Chemistry: In the given reaction, 3-bromo-2, 2-dimethyl butane $\stackrel{C_2{H}_{5}OH}{\to }$ (A) Major product. Product A is

Welcome to a classic organic chemistry puzzle! We are given 3-bromo-2,2-dimethyl butane and asked to react it with ethanol ($C_2{H}_{5}OH$).

The first step in solving any organic reaction is identifying the nature of the reactants. Our substrate is a secondary alkyl halide. Our reagent, ethanol, is a weak nucleophile and a polar protic solvent. This specific combination is the perfect recipe for an $S_{N}1$ (Substitution Nucleophilic Unimolecular) reaction.

In an $S_{N}1$ mechanism, the reaction does not happen all at once. It is a stepwise process. The very first step—and the rate-determining step—is the departure of the leaving group.

The bromide ion ($B{r}^{-}$) takes its bonding electrons and leaves the molecule. This heterolytic cleavage leaves behind a positively charged carbon atom, creating a secondary ($2^{\circ }$) carbocation.

Carbocation Rearrangement

Now, you must always pause when a carbocation is formed. Carbocations are highly reactive and will always try to rearrange themselves into a more stable configuration if possible.

Look closely at our secondary carbocation. Right next to the positively charged carbon is a quaternary carbon (a carbon bonded to four other carbons). If one of those methyl groups shifts over to the positively charged carbon—taking its bonding electrons with it—the positive charge moves to the tertiary carbon.

This is called a 1,2-methyl shift. The resulting tertiary ($3^{\circ }$) carbocation is significantly more stable due to increased hyperconjugation and inductive effects. This rearrangement is the critical "catch" in this problem!

Now that we have our highly stable tertiary carbocation, the nucleophile can finally make its move. Ethanol uses the lone pairs on its oxygen atom to attack the positively charged carbon.

This forms a new carbon-oxygen bond, but it leaves the oxygen atom with a positive charge (since it is now bonded to three things). This intermediate is a protonated ether.

To regain neutrality and stability, the oxygen atom quickly loses a proton ($H^{+}$). This final deprotonation step yields our neutral product.

The final product is an ether where the ethoxy group ($-O{C}_2{H}_5$) is attached to the tertiary carbon. Naming this according to IUPAC rules gives us 2-ethoxy-2,3-dimethyl butane.

Comparing this to our options, it perfectly matches Option (c). Always remember: in $S_{N}1$ reactions, the carbocation will rearrange if it can find a more stable home!