Last time in this walkthrough on elimination reactions, we talked about two types of elimination reactions. In this post, we’re going to dig a little bit deeper on one type of elimination reaction, and based on what experiments tell us, come up with a hypothesis for how it works.
Here’s the reaction. First, look at the bonds that are being formed and broken. This is a classic elimination reaction – we’re forming a new C–C(π) bond, and breaking a C–H and C–leaving group (Br here) bond.
But now we want to know more than just “what happens”. We want to understand how it happens. What’s the sequence of bond-forming and bond breaking? To understand HOW it happens, we need to look at what the data tells us. That’s because chemistry is an empirical science; we look at the evidence, and then work backwards.
First Clue – The Rate Laws
Let’s look at the first important clue for this reaction. We can measure reaction rates quite readily. When we vary the concentration of the substrate, the reaction rate increases accordingly. In other words, there is a “first-order” dependence of rate on the concentration of substrate.
However, if we vary the concentration of the base (here, H2O) the rate of the reaction doesn’t change at all.
What information can we deduce from this? The rate determining step for this reaction (whatever it is) therefore does not involve the base. Whatever mechanism we draw will have to account for this fact.
Second Clue – Dependence of Rate on Substrate
Another interesting line of evidence we can obtain from this reaction is through varying the type of substrate, and measure the rate constant that results. So if we take the simple alkyl halide on the far left (below) where the carbon attached to Br is also attached to 3 carbons (this is called a tertiary alkyl halide), the rate is faster than for the middle alkyl halide (a secondary alkyl halide) which is itself faster than a primary alkyl halide (attached to only one carbon in addition to Br).
So the rate proceeds in the order tertiary (fastest) > secondary >> primary (slowest)
Any mechanism we draw for this reaction would likewise have to account for this fact. What could be going on such that tertiary substrates are faster than primary?
Third Clue – This Elimination Reaction Competes with the SN1 Reaction
A final interesting clue about the mechanism of this reaction concerns the by-products that are often obtained. For instance, when the alkyl halide below is subjected to these reaction conditions, we do obtain the expected elimination product. However, we also get substitution reactions in the product mix as well. Remember – substitution reactions involve breakage of C-(leaving group) and formation of C-(nucleophile). What makes this particular starting alkyl halide particularly interesting is that the carbon attached to Br is a stereocenter. And if we start with a single enantiomer of starting material here, we note that the substitution product formed is a mixture of stereoisomers. Note that both inversion and retention of stereochemistry at the stereocenter has occurred.
We’ve seen this pattern before – it’s an SN1 reaction!
This last part is a very important clue. If an SN1 reaction is occurring in the reaction mixture, looking back at the mechanism of the SN1 could help us think about what type of mechanism might be going on in this case to give us the elimination product.
Taking all of these clues into account, what’s the best way to explain what happens? This:
The reaction is proposed to occur in two steps: first, the leaving group leaves, forming a carbocation. Second, base removes a proton, forming the alkene. This nicely fits in with the three clues mentioned above. [Also note that the more substituted alkene is formed here, following Zaitsev’s rule].
Similar to the SN1 mechanism, this is referred to as the E1 mechanism (elimination, unimolecular).
So what’s going on in the other type of elimination reaction? That’s the topic for the next post!