{"id":5988,"date":"2012-09-19T07:16:21","date_gmt":"2012-09-19T11:16:21","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=5988"},"modified":"2026-04-18T06:21:20","modified_gmt":"2026-04-18T11:21:20","slug":"the-e1-reaction","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2012\/09\/19\/the-e1-reaction\/","title":{"rendered":"The E1 Reaction"},"content":{"rendered":"

The E1 Reaction – Three Key Pieces of Evidence, and a Mechanism<\/strong><\/p>\n

Last time in this walkthrough on elimination reactions, we talked about two types of elimination reactions.<\/a>\u00a0In 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.<\/p>\n

\"summary-E1<\/a><\/p>\n

Table of Contents<\/strong><\/p>\n

    \n
  1. How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)<\/a><\/li>\n
  2. First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)<\/a><\/li>\n
  3. The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates<\/a><\/li>\n
  4. The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the SN<\/sub>1 Reaction<\/a><\/li>\n
  5. Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation<\/a><\/li>\n
  6. Notes<\/a><\/li>\n
  7. Quiz Yourself!<\/a><\/li>\n
  8. (Advanced) References and Further Reading<\/a><\/li>\n<\/ol>\n
    \n

    <\/a>1. How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)<\/h2>\n

    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\u2013C(\u03c0) bond, and breaking a C\u2013H and C\u2013leaving group (which is Br in this case)<\/em><\/span> bond.<\/p>\n

    \"elimination<\/p>\n

    But now we want to know more than just “what happens”. We want to understand\u00a0how<\/strong> it happens. What’s the sequence of bond-forming and bond breaking?\u00a0To understand HOW it happens, we need to look at what the data<\/strong> tells us.<\/p>\n

    That’s because chemistry is an empirical science; we look at the evidence, and then work backwards.<\/p>\n

    <\/a>2. First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)<\/strong><\/h2>\n

    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.<\/p>\n

    However, if we vary the concentration of the base (here, H2<\/sub>O) the rate of the reaction doesn’t change at all.<\/p>\n

    \"e1<\/p>\n

    What information can we deduce from this? \u00a0The rate determining step for this reaction (whatever it is) therefore\u00a0does\u00a0not\u00a0<\/strong>involve the base. Whatever mechanism we draw will have to account for this fact.<\/p>\n

    The reaction is\u00a0first-order\u00a0<\/strong>in the\u00a0substrate\u00a0<\/strong>(alkyl halide) and\u00a0zero order\u00a0<\/strong>in base (i.e. doubling concentration does not double the rate).<\/p>\n

    <\/a>3. The Second Clue About The Mechanism Of The E1 Elimination Reaction – <\/strong>Rate Is Fastest For Tertiary Substrates<\/strong><\/h2>\n

    Another interesting line of evidence we can obtain from this reaction is through varying the\u00a0type<\/strong> 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\u00a0tertiary<\/strong> alkyl halide), the rate is faster than for the middle alkyl halide (a\u00a0secondary<\/strong> alkyl halide) which is itself faster than a\u00a0primary<\/strong> alkyl halide (attached to only one carbon in addition to Br).\u00a0 (See post: Primary, Secondary, Tertiary<\/a><\/em>)<\/p>\n

    So the rate proceeds in the order\u00a0tertiary (fastest) > secondary >> primary (slowest)<\/strong><\/p>\n

    \"reaction<\/p>\n

    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?<\/p>\n

    <\/a>4. The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the SN<\/sub>1 Reaction<\/strong><\/h2>\n

    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<\/strong> obtain the expected elimination product.<\/p>\n

    However<\/em>, we also get substitution<\/strong> reactions in the product mix as well. Remember – substitution reactions involve breakage of C-(leaving group) and formation of C-(nucleophile).<\/p>\n

    What makes this particular starting alkyl halide particularly interesting is that the carbon attached to Br is a\u00a0stereocenter.\u00a0<\/strong>And if we start with a single enantiomer of starting material here, we note that the substitution product formed is a\u00a0mixture of stereoisomers.\u00a0<\/strong>Note that\u00a0both inversion\u00a0and<\/strong> retention of stereochemistry at the stereocenter has occurred.<\/p>\n

    We’ve seen this pattern before – it’s an SN<\/sub>1 reaction! (See article:<\/span> The SN1 Reaction<\/a><\/em>)<\/p>\n

    \"competing<\/p>\n

    This last part is a very important clue. If an SN<\/sub>1 reaction is occurring in the reaction mixture, looking back at the mechanism of the SN<\/sub>1<\/a> could help us think about what type of mechanism might be going on in this case to give us the elimination product.<\/p>\n

    <\/a>5. Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation<\/h2>\n

    Taking all of these clues into account, what’s the best way to explain what happens? This:<\/p>\n

    \"mechanism<\/p>\n

    The reaction is proposed to occur in two steps:\u00a0first, the leaving group leaves,\u00a0<\/strong>forming a carbocation.\u00a0Second,\u00a0<\/strong>base removes a proton<\/strong>, 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<\/em><\/span>].<\/p>\n

    Similar to the SN<\/sub>1 mechanism, this is referred to as the E1 mechanism<\/strong> (elimination, unimolecular).<\/p>\n

    So what’s going on in the\u00a0other type of elimination reaction? That’s the topic for the next post!<\/p>\n

    Next Post: The E2 Mechanism<\/strong><\/a><\/p>\n

    \n

    <\/a>Notes<\/h2>\n

    Related Articles<\/strong><\/p>