Animated Solution for Chemistry - Organic Chemistry: Acid catalysed hydration of alkenes except ethene leads to the formation of

Imagine you are in a chemistry lab, and you have a flask full of an alkene. You want to convert it into an alcohol. One of the most classic ways to do this is through acid-catalyzed hydration. But here is the million-dollar question: what kind of alcohol will you get? Primary, secondary, or tertiary?

To answer this, we need to dive into the microscopic world and look at the reaction mechanism. The reaction doesn't happen in one magic step. It's a carefully choreographed dance of electrons.

Acid-Catalyzed Hydration

When you add water to an alkene, nothing happens. Water is a weak nucleophile, and the alkene's pi-electron cloud repels it. That's why we need an acid catalyst. The acid provides a proton ($H^{+}$), which is a strong electrophile. The pi-electrons of the double bond eagerly attack this proton.

This attack breaks the double bond and leaves one of the carbon atoms electron-deficient, creating a positively charged intermediate known as a carbocation.

Now, nature is lazy. It always wants to take the path of least resistance, which means forming the most stable intermediate possible. Carbocations are stabilized by the presence of electron-donating alkyl groups through two main effects: inductive effect and hyperconjugation.

A tertiary carbocation (attached to three alkyl groups) is more stable than a secondary carbocation (attached to two), which in turn is much more stable than a primary carbocation (attached to only one). This preference for forming the most stable carbocation is the physical basis of Markovnikov's Rule.

The Lone Exception

Let's look at the simplest alkene: ethene ($C{H}_2=C{H}_2$). It is perfectly symmetrical. When it grabs a proton, the positive charge has no choice but to land on a terminal carbon, forming an ethyl carbocation ($C{H}_3-C{H}_2^{+}$). This is a primary carbocation.

When water attacks this primary carbocation and subsequently loses a proton, the final product is ethanol ($C{H}_3-C{H}_2-OH$), which is a primary alcohol. Because ethene only has two carbon atoms, it is physically impossible for it to form anything other than a primary carbocation.

The Rule Followers

Now, what happens when we move to any other alkene, like propene ($C{H}_3-CH=C{H}_2$)? Propene is unsymmetrical. When it reacts with a proton, it has a choice. It can form a primary carbocation at the end of the chain, or a secondary carbocation in the middle.

Because of the stabilizing effects we discussed earlier, it overwhelmingly chooses to form the secondary carbocation ($C{H}_3-C{H}^{+}-C{H}_3$). When water attacks this more stable intermediate, the resulting product is isopropyl alcohol ($C{H}_3-CH(OH)-C{H}_3$), a secondary alcohol.

The same logic applies to even more substituted alkenes like isobutylene, which will form a highly stable tertiary carbocation, leading to a tertiary alcohol.

The pattern is crystal clear. Because any alkene other than ethene has at least one alkyl group attached to the double bond, protonation will always occur to place the positive charge on the more substituted carbon. This guarantees the formation of a secondary or tertiary carbocation.

Consequently, the hydration of any alkene—except ethene—will always yield a secondary or tertiary alcohol, depending on the specific structure of the starting alkene. Ethene stands alone as the only alkene capable of producing a primary alcohol through this method.