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SN1/SN2/E1/E2 Decision

By James Ashenhurst

Deciding SN1/SN2/E1/E2 (4) – The Temperature

Last updated: March 26th, 2019

The Quick N’ Dirty Guide To Determining SN1/SN2/E1/E2, Part 4

In previous installations of the Quick N’ Dirty Guide, we’ve examined the substrate, the base/nucleophile, and the solvent. Today, we’ll address the final variable to consider: the temperature.

If you’ve been following so far, you may have noticed that by this point we should be able to differentiate all cases where SN2 is favored over E2 (and vice versa) but are still left with this dilemma: when a carbocation is formed, how do we determine whether SN1 or E1 products are favored?

First of all, note that the first step of the SN1 and E1 reactions is the same:  loss of a leaving group to give a carbocation. Since both of these reactions proceed via the same intermediate, in practice a mixture of both SN1 and E1 products will be found whenever the reaction proceeds through a  carbocation [where possible].  Given that, however, we would still like to have a rule of thumb that tells us what type of product should be the major product in most cases.

Generally speaking, SN1 products tend to predominate over E1 products at lower temperatures.  However, recall that elimination reactions are favored by heat.  In cases where substitution reactions and elimination reactions are in competition with each other, increasing the temperature tends to increase the amount of elimination products produced. 

Here is a representative example:

Loss of bromide ion from the substrate leads to the formation of a tertiary carbocation [stable, hence no rearrangement]. At low temperatures, the SN1 pathway (above) will dominate: attack at the carbocation by CH3OH, followed by loss of proton to give the ether. The bottom pathway – removal of hydrogen from the carbon adjacent to the carbocation – will be minor at low temperature [note the formation of the more substituted alkene here – Zaitsev’s rule in action]. As temperature is increased, the amount of elimination relative to substitution should gradually increase.

This leads to the following Quick N’ Dirty rule of thumb.

Quick N’ Dirty Rule #6:  When carbocations are formed, at low temperatures, the SN1 pathway will dominate over the E1 pathway. At higher temperatures, more E1 products will be formed.

(Note: before applying these reaction patterns to the substrate, make sure to examine the carbocation that is formed. If a more stable carbocation can be formed through a hydride or alkyl shift, do this rearrangement first!)

Let’s go back to the examples we’ve been working on.

The third case – addition of H2SO4 to a tertiary alcohol – is a case where a carbocation is formed in the absence of a good nucleophile [the negatively charged oxygen on the conjugate base, [HSO4(-)] is stabilized through resonance, reducing its reactivity]. The fact that heat is being applied helps to tip the balance even further toward E1 being dominant over SN1.

In the fourth example we have a tertiary halide [which will form a stable carbocation] in a polar protic solvent [will help to stabilize the intermediate carbocation] and heat is not indicated. Therefore using Quick N’ Dirty Rule #6, we can say that SN1 products will dominate. [E1 products will form as well, but they will not be the major products].

This is truly a Quick N’ Dirty rule. It is not applied evenly and there are plenty of exceptions. Your mileage will vary widely. I wish I had a hard, concrete example to show you that clearly demonstrates the relationship between increased heat and a greater proportion of elimination (E1) versus substitution (SN1) products. Sadly, I cannot find a good example at this time. So a hand-wavy “elimination increases as heat is applied” will have to do for now.*

In the next post we’ll summarize all the Quick N’ Dirty Rules for determining whether a reaction goes SN1/SN2/E1/E2.

Next Post: Wrapup of the Quick N’ Dirty Guide



* I am annoyed by the lack of hard data available when making pronouncements about the SN1/SN2/E1/E2 decision. Wohler’s quote about the  “monstrous and boundless thicket… into which one may well dread to enter” seems appropriate here.  I would love to see the control experiments with various substrates run under identical sets of conditions that clearly delineate the impact of each variable. I have not seen this. Any undergraduate labs out there with a desire for performing this valuable public service?

A final note/editorial.

One Difference Between “Real ” Chemistry and “Exam Question” Chemistry

The post above is not so much about understanding some facet of organic chemistry as it is about how to answer some arbitrary question from a textbook or exam. For the purpose of understanding organic chemistry, it’s enough to know that heat favors elimination reactions. For the purpose of knowing how to answer a particular kind of question on an exam, it will of course depend on the examiner. There are wide variations. However, I will share with you one common observation that I’ve seen in 2.5 years of seeing exams from all over the country.

First, background.  In the laboratory, it is extremely common to heat reactions to get them to go at a reasonable speed, such as in this example [from March’s Advanced Organic Chemistry 5th ed; Cooper, K.A. et. al. J. Chem. Soc. 1948, 2038:]

However, on an exam, instructors – for various reasons, including a well-intentioned desire not to overwhelm the student – will often omit some of the data. For exam purposes, if the above reaction were written as a  question it will often look like this:

 Note how “heat” has been omitted, which is in accord with the principle of least effort. The expected answer in this instance would be t-BuOH, the product of an SN1 reaction. Depending on the question wording, some instructors will also insist that the E1 product be drawn as well.

Here’s the observation I see in many (but certainly not all) courses.  If the word “heat” is written in the exam question, it is often a clue from the instructor that an elimination is to take place. In the following reaction, for example, the question would point to elimination (2-methyl propene) being the major product.

If you are a student and your goal is to answer a particular type of question on an exam correctly, I advise you to double check this issue with your instructor and get their answer on it. There is tremendous inconsistency in this practice nationwide.



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Comment section

21 thoughts on “Deciding SN1/SN2/E1/E2 (4) – The Temperature

  1. If we have tertiary alkyl halide,water or alcohol and 22 degree celcius what will be reaction?is it E1 or E2?

    1. Likely SN1 – tertiary alkyl halide (not SN2) and weak nucleophile (water or alcohol) [ no E2] in the absence of heat.

  2. I just wanted to thank you for the awesome website, and method of explanation of concepts you have here. I was struggling with this chapter for days, and reading your instructions, and rules, I understood the concept in an hour! and actually am answering the questions correctly. THANK YOU.

  3. I think it should be “In the fourth example we have a tertiary halide” instead of “In the fourth example we have a tertiary alcohol”.

    1. If your question is related to an exam question, the answer is likely “yes”. Applying Quick N’ Dirty rules, tertiary rules out SN2, ethanol (neutral) rules out E2, and room temperature rules out E1. That leaves SN1.

      If your question is related to a real-life reaction, my advice would be to go to March’s Advanced Organic Chemistry and look up “solvolysis rates” to get an idea of how quick it will be.

  4. Well, a good example of this temperature thing is the addition of H2SO4 to primary alcohol. At 413K, elimination takes place to form alkene, but at 443K, ether is formed by SN1. Thank you so much for these thumb rules. Very helpful!

  5. Dear Dr James

    You discussed primary secondary tertiary carbons. What about alilic and benzilic carbons. Will primary benzilic carbon behaves like primary carbon?

    1. Allylic and benzylic carbons will be more prone to going through carbocation pathways (SN1/E1) than typical primary carbons. If a charged nucleophile is present, using the Quick N’ Dirty rules I would generally expect substitution via SN2, and if the nucleophile is neutral (or using the solvent as nucleophile) the SN1 and E1 pathways start to become feasible, particularly with heat.

    1. It is vague. But “heat” would generally refer to heating a solution to its boiling point (“reflux”) and the boiling point of most typical solvents for our purposes is 60-100 °C .

    1. In the second example the “nucleophile” is KOEt which is charged and therefore “strong” (according to our “Quick N’ Dirty” rules). In the fourth example the “nucleophile” is CH3OH which is neutral (and therefore “weak” according to these rules).
      The species containing Cl (in the second example) and Br (in the fourth example) are called the “substrate”.

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