Elimination Reactions
The E1 Reaction
Last updated: January 15th, 2020 |
The E1 Reaction – Three Key Pieces of Evidence, and a Mechanism
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.
Table of Contents
- How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)
- First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)
- The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates
- The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the SN1 Reaction
- Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation
1. How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)
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.
2. First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)
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.
3. The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates
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?
4. The Third Clue About The Mechanism Of The E1 Elimination Reaction: It 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.
5. Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation
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!
Thanks for posting these blogs!! :) :) This is really helping me really learn and enjoy organic chemistry!! Keep doing this, and more people will enjoy the beauty of this subject. Thanks again!!
Sir,
1-chlorobutane on elimination will give 1-butene or 2-butene?
Thank you
Well, elimination will remove a proton from the carbon directly adjacent to the leaving group, so C2. The initial product will thus be 1-butene.
sir if we use OH- instead of H2O in E1 reaction then what might happen
E2, most likely.
Since -OH is a strong base, right?
Yes, since HO- is a strong base, it will go down the E2 pathway instead of the E1 pathway.
Great article, it is really helpful!
One question, why is water able to remove the H from the cycle?
Thanks
Water is a weak base and the lone pair from the oxygen forms a new bond with H+. The lone pair from the C-H bond then forms a new pi bond.
Thank you so much for taking the time to create these straightforward, easy to follow, posts that concisely outline the subject at hand. Truly, you are helping me out a lot more than my current professor. These posts have been invaluable to acing my organic chem midterms. Thanks again, you wonderful human!
i really didn’t understand why substitution product also came as one of the by-product. what if we take a molecule with no chiral centres ? Will the same happen ? Please do resolve my doubt with an illustration. And thanks for explaining the mechanism so well.You are truly wonderful !
There’s an intermediate carbocation which can undergo several fates. One of those fates is elimination. Another possibility is substitution. When heated, some substitution products will revert back to the carbocation intermediate and undergo elimination, resulting in the conversion of these substitution products to elimination products.
Thank you so much for a very comprehensive explanation of what I thought to be a complex mechanism. But you made it so much easier for me. Been using your site as reference for most of my Organic Chemistry exams. :) THANK YOU!!!
among chlorocyclopropane,chlorocyclobutane &chlorocyclopentane which one undergoes fastest Sn1 reaction and why?
That’s a good exam question! I invite you to look into the special bonding in cyclopropane and cyclobutane and reason out an answer as to whether carbocations would be more or less stable within a 3-membered ring or a 5-membered ring.
Hi James,
Great work.
You have the presented the work, the way it has to be. Most of the books do it in reverse way. They deal with mechanism first and then support it with the reactions.
One thing that bothers me is the name of these reactions. Is the E1 reaction called as elimination uni molecular or elimination first order?
I like presenting the evidence first and the mechanism last, kind of like the solution to a puzzle. Re: your question, I don’t think it matters. It’s just the E1 Mechanism.
Awesome explanation, so far the best notes on organic chemistry i have ever reffered to. I am glad that I found your website. Thanks a lot sir.
would it be possible to make it so that an E2 reaction results in only one product (vs major and minor products)? if so, how would this be done?
An E2 – certainly, because it’s controlled by stereochemistry. A good example would be in some substituted cyclohexanes where there is only one hydrogen anti (trans) to the leaving group.
An E1 (the topic of this post) – the best situation would be one where only one alkene is possible as a product. For example the E1 reaction of C6H5-CH(OH)-CH3 would only give one alkene.
Otherwise, the selectivity is going to be determined by the difference in stability between the alkenes, and since that’s usually quite small, 70:30 or 80:20 ratios are common.