Animated Solution for Chemistry - Organic Chemistry: The major product [B] in the following reactions is ${\text{CH}}_3-{\text{CH}}_2-\underset{\stackrel{|}{{\text{CH}}_3}}{\mathrm{CH}}-{\text{CH}}_2-{\text{OCH}}_2{\text{CH}}_3\underset{\text{Heat}}{\overset{\text{HI}}{\to }}[\text{A}]\text{ alcohol}\underset{\Delta }{\overset{{\text{H}}_2{\text{SO}}_4}{\to }}[\text{B}]$

Ether Cleavage

The problem presents us with a classic two-step organic synthesis. We begin with an unsymmetrical ether, specifically 1-ethoxy-2-methylbutane, and subject it to heating with hydrogen iodide ($\text{HI}$).

This is a textbook ether cleavage reaction. When an ether is treated with a strong acid like $\text{HI}$, the first step is always the protonation of the oxygen atom. This turns the ether into a good leaving group.

A Primary Alcohol

Once protonated, the iodide ion (${\text{I}}^{-}$) must decide where to attack. Since both alkyl groups attached to the oxygen are primary, the reaction proceeds via an ${\text{S}}_{\text{N}}2$ mechanism.

In an ${\text{S}}_{\text{N}}2$ reaction, the nucleophile prefers the path of least resistance. It attacks the less sterically hindered carbon. Here, the ethyl group is less hindered than the 2-methylbutyl group.

Therefore, the iodide ion attacks the ethyl group, breaking the carbon-oxygen bond to form ethyl iodide (${\text{CH}}_3{\text{CH}}_2\text{I}$) and our intermediate [A], which is the primary alcohol 2-methylbutan-1-ol.

Dehydration and Rearrangement

Now, we take our intermediate alcohol [A] and heat it with concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$). This is the standard condition for the dehydration of an alcohol.

The acid protonates the hydroxyl group, creating a water molecule that acts as an excellent leaving group. As water departs, it leaves behind a primary carbocation.

However, primary carbocations are notoriously unstable. Nature always seeks a lower energy state. Right next to our primary carbocation is a tertiary carbon holding a hydrogen atom.

This sets the stage for a 1,2-hydride shift. The hydrogen atom, along with its electron pair, migrates to the adjacent positively charged carbon. This rearrangement transforms the unstable primary carbocation into a highly stable tertiary carbocation.

Saytzeff's Rule

With our stable tertiary carbocation formed, the final step is the elimination of a proton (${\text{H}}^{+}$) to form a double bond.

According to Saytzeff's Rule, the elimination will occur in a way that yields the most highly substituted, and therefore most thermodynamically stable, alkene.

A base (like water or bisulfate ion) removes a proton from the adjacent $\beta$-carbon. Removing the proton from the internal ${\text{CH}}_2$ group results in a tri-substituted double bond.

This gives us our final major product [B], which is 2-methylbut-2-ene (${\text{CH}}_3-\text{CH}=\text{C}({\text{CH}}_3)-{\text{CH}}_3$).