Organic Chemistry, "Change" - Part 19; Dehydrolysis of an Alcohol to form an Alkene (Final Steps)

 Greetings,

The last post, Part 18, focused on the start of an Alcohol Dehydrolysis Reaction to form an Alkene. Specifically, protonation of the hydroxyl functional group was explored. In this post, the final steps of the reaction is discussed.

The protonation of the hydroxyl group occurs because a good leaving group is formed, namely water. Because the formation of water is a reaction product, a greater concentration of free water at the reaction start will sequester (hinder) the overall change.

There are three steps involved with the mechanism of this reaction: 1) Protonation of the hydroxyl group to form a positively charged water group, 2) Loss of water to form the unstable carbocation intermediate, and 3) Loss of hydrogen ion (a proton) from the adjacent carbon atom to form the alkene product. Each of  the three steps are reversible because each step is an equilibrium reaction. Because of the instability of the carbocation intermediate, its formation occurs at the lowest rate of the mechanism. Therefore, step 2), the loss of water reaction, is the overall reaction's rate determining step. The rate of the dehydrolysis reaction only depends on the speed of step 2), the loss of water. Also, it is only step 2) which involves water, directly (as an equilibrium reaction product).

That's all for this post. The next post will feature a detailed diagram with molecular structures to clearly show exactly what is happening at each of the three mechanism steps.

As always, Thank you for reading!