An Example Organic Chemistry Mechanism Problem

Suppose you’re working an organic chemistry assignment, and you're asked to propose a mechanism for the conversion of the alcohol shown in the following figure to the alkene shown in the same figure.

An example mechanism problem.
An example mechanism problem.

First, you want to identify what kind of mechanism you’re dealing with. Is it an acid or a base mechanism? Is it a nucleophile or an electrophile mechanism? Is it a free radical mechanism? Because the reagent is sulfuric acid (H2SO4) plus ∆ (a symbol that indicates that the reaction mixture is heated), the reaction will be an acid reaction.

Second, identify where you’re going in the mechanism. Converting an alcohol into an alkene usually involves dehydration (loss of water). But you may notice that the alkene in the product is in an unusual position, because dehydration usually puts the alkene adjacent to the starting alcohol. The product in this case has the double bond one carbon atom farther away from its expected position. So, you might expect something unusual in this mechanism that accounts for the unusual regiochemistry of the reaction.

Now that you know approximately where you’re going in this mechanism, you can begin to propose a route to get there. In an acid mechanism, the first thing you want to look for is a site on the molecule that can be protonated. The only basic site on the starting material in this case is the alcohol, so the first step would be protonation of the alcohol group by sulfuric acid.

The first step in acid mechanisms is almost always protonation.

As always, follow the conventions for arrow pushing, drawing full-headed arrows from electrons to where they’re going. (The only mechanisms that use single-headed arrows are free-radical mechanisms.)

Because many professors of organic chemistry are convinced that the innermost circle of Dante’s hell is reserved for those students who consistently draw protonation incorrectly, make sure to always draw the arrow from the electrons on the base to the acidic proton that the electrons are attacking. Never commit the unpardonable sin of drawing an arrow from an acid H+. Also, to make following the movement of atoms and charges easier, the atoms on the parts of the molecule that change in the reaction are drawn out explicitly with the full Lewis structure, as shown in the following figure.

The first step.
The first step.

Next, examine the consequences of the protonation. What can you do now that you couldn’t before? One thing you can do now that the alcohol is protonated is form a carbocation. Although hydroxide ion, OH, is a bad leaving group (strong bases are bad leaving groups), water is a good leaving group. So, as in a typical alcohol dehydration mechanism, the next step involves the loss of water to create a secondary carbocation, as shown in the following figure.

Making a carbocation.
Making a carbocation.

In alcohol dehydration, when the carbocation is formed, the next step is usually the formation of the double bond by eliminating a proton. Doing so at this step, however, would form the double bond in the wrong position. Somehow, the carbocation needs to move so that when hydrogen is eliminated, the double bond is formed in the correct position.

On closer inspection, you may notice that the carbocation is located next to a tertiary carbon center. The carbocation is currently a secondary carbocation, but it badly wants to become a tertiary carbocation. (Tertiary carbocations are more stable than secondary carbocations.) A more pragmatic thought process might be that this is where the cation must move to in order to get elimination in the correct place. To get the carbocation into the tertiary position, you perform a hydride shift, which moves the tertiary hydrogen atom and both of the electrons in the C-H bond to the cation center, leaving the tertiary carbon with the positive charge. This step is shown in the following figure.

Watch out for shifts in cation mechanisms.
Watch out for shifts in cation mechanisms.

Now, the elimination of a proton leads to the correct alkene. The deprotonation is shown in the following figure, using water to regenerate a molecule of acid. You could also use the conjugate base of sulfuric acid as the base here.

Forming the alkene.
Forming the alkene.
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