Animated Solution for Chemistry - Organic Chemistry: The main product of the following reaction is ${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH(OH)CH(CH}}_3{)}_2\stackrel{{\text{Conc. H}}_2{\text{SO}}_4}{\to }\text{?}$

The Power of Conjugation

Let's dive into a classic organic chemistry problem involving the dehydration of an alcohol. We are given a secondary alcohol, specifically ${\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH(OH)CH(CH}}_3{)}_2$, and we are treating it with concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$).

Concentrated sulfuric acid is a powerful dehydrating agent. Its primary role is to provide protons (${\text{H}}^{+}$). The very first step in any acid-catalyzed dehydration is the protonation of the hydroxyl ($-\text{OH}$) group. The lone pairs on the oxygen atom grab a proton, transforming the poor leaving group ($-\text{OH}$) into an excellent leaving group, an oxonium ion ($-{\text{OH}}_2^{+}$).

Once the oxonium ion is formed, it is highly unstable due to the positive charge on the electronegative oxygen atom. The water molecule quickly departs, taking its bonding electrons with it. This departure leaves behind a positive charge on the carbon atom, generating a secondary carbocation intermediate:

Which Beta-Hydrogen to Eliminate?

To form an alkene, the molecule must eliminate a proton from an adjacent carbon, known as a $\beta$-carbon. Looking closely at our carbocation, we have two distinct $\beta$-carbons available for elimination. Let's label them ${\beta }_1$ and ${\beta }_2$.

${\beta }_1$ is the $-{\text{CH}}_2-$ group located between the benzene ring and the carbocation. ${\beta }_2$ is the $-\text{CH}-$ group of the isopropyl moiety.

If we remove a proton from the ${\beta }_2$ position, the resulting double bond forms between the secondary and tertiary carbons. This yields a tri-substituted alkene: ${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{CH}={\text{C(CH}}_3{)}_2$. While tri-substituted alkenes are generally stable according to Zaitsev's rule, we must look at the bigger picture.

Conjugation

What happens if we remove a proton from the ${\beta }_1$ position instead? The double bond forms directly adjacent to the benzene ring. This creates a conjugated system:

Conjugation allows the $\pi$ electrons of the new double bond to delocalize with the $\pi$ electron cloud of the benzene ring. This extensive delocalization provides immense thermodynamic stability to the molecule, far outweighing the hyperconjugative stabilization of the isolated tri-substituted alkene.

Therefore, the conjugated alkene is the overwhelmingly favored major product. When we examine the given options, this structure perfectly matches option (a).