Two Types of Elimination Reactions

by James

in Alkenes, Alkyl Halides, Organic Chemistry 1, Organic Reactions

Like I said in the introduction to substitution reactions, organic chemistry is an empirical, experimental science. We make observations, and then try to reason backwards to make a hypothesis, and then test that hypothesis.

The results of the experiments below weren’t predicted by those who did them in the first place! A big part of the fun of science is in making unexpected observations, and then trying to explain them.

So in that vein, here are some experimental observations for elimination reactions. As you’ll see, they seem to divide nicely into two classes.

How do we explain what is going on in each case? That’s for you to try to figure out.

Observation #1 : Byproducts

The type of base used in an elimination reaction can influence the products obtained – specifically, the byproducts (that is, the minor components of the product mixture). In the first example, we take a sample of 2-bromobutane as a single enantiomer. Treatment with the strong base sodium ethoxide (NaOEt) gives two alkenes (trans and cis) which follow Zaitsev’s rule. The trans product dominates over the cis product (due to less steric crowding), but what’s really interesting is the byproduct obtained: 2-ethoxy butane, obtained with inversion of stereochemistry. This is the product of a substitution reaction  – specifically, an SN2 reaction.

Now, if the same starting material is treated with water (a weaker base) and heated, we also obtain elimination products. However, the substitution product that is formed (2-butanol) is obtained as a mixture of enantiomers. In other words, we have a mixture of retention and inversion of the stereocenter. This is recognizable as an SN1 process.

So one type of elimination (with strong bases) tends to compete with SN2 reactions, while the other (with weak bases) tends to compete with the SN1 pathway.

Observation #2: Stereochemistry

Here’s a quirky example of an elimination reaction. If you treat this substituted cyclohexane with the strong base NaOEt, you might expect to get the more substituted (tetrasubstituted) alkene with  double bond between C1 and C2. But that’s not what we actually get! In this case we only get the trisubstituted alkene shown below.

However, if you dissolve this same alkyl bromide in water, and heat it, you do obtain the tetrasubstituted alkene. The trisubstituted alkene is formed too, albeit only to a minor extent.

So clearly the stereochemistry of the starting material has some influence on the product of the first reaction, but seemingly not the second.

Observation #3: Rate laws

Finally, we can measure the rates of each of these reactions and determine their dependence on the concentration of substrate and base.

Looking at the first reaction below, the rate is dependent on the concentration of both the substrate and also of the base. Note how if we double the concentration of either the substrate or the base, the rate also doubles.

Contrast that with the second reaction, where the reaction rate is dependent *only* on the concentration of the substrate, but not on the base.

So these are the facts. How can we use these facts to come up with some hypotheses to explain how these reactions work? That will be the subject of the next posts.

Next Post: The E1 Mechanism

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