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

Let's dive into this fascinating organic chemistry problem. We are given 3-iodo-2,2-dimethylbutane and it is reacting with tert-butanol ($t$-BuOH) under heating conditions.

The first thing we must do is analyze our reagents. Tert-butanol is a bulky, weak base and a polar protic solvent. When you combine a weak base with heat, the reaction is heavily biased towards an E1 (Elimination Unimolecular) mechanism. Unlike E2 reactions which happen in a single concerted step, E1 reactions are a multi-step journey that begins with the formation of a carbocation.

Carbocation Formation

In the first step, the carbon-iodine bond breaks heterolytically. Iodine, being highly electronegative and a great leaving group, takes the bonding electrons and departs as an iodide ion ($I^{-}$).

This leaves behind a secondary ($2^{\circ }$) carbocation. This step is the slow, rate-determining step of the reaction. Now, whenever you form a carbocation, you must immediately ask yourself: "Can this carbocation rearrange to become more stable?"

Look closely at the carbon adjacent to our newly formed positive charge. It is a quaternary carbon, meaning it is bonded to three other methyl groups. Nature always seeks the lowest energy state, and a tertiary carbocation is significantly more stable than a secondary one due to increased hyperconjugation and inductive effects.

To achieve this stability, one of the methyl groups on the adjacent quaternary carbon migrates over to the positively charged carbon, taking its bonding pair of electrons with it. This is known as a 1,2-methyl shift.

We now have a highly stable tertiary ($3^{\circ }$) carbocation.

Saytzeff's Rule

Now for the final act of elimination. Our weak base ($t$-BuOH) will abstract a proton ($H^{+}$) from a carbon adjacent to the positive charge to form a double bond.

But which proton will it take? According to Saytzeff's Rule, elimination will preferentially occur to form the most highly substituted, and therefore most thermodynamically stable, alkene.

By removing a proton from the adjacent $-CH$ group, we form a double bond between two carbons that are each attached to two methyl groups. This yields a tetrasubstituted alkene, which is incredibly stable.

The final major product is 2,3-dimethyl-2-butene, which perfectly matches option (b). This problem is a beautiful reminder to always watch out for carbocation rearrangements!