SN1/SN2/E1/E2 Decision

By James Ashenhurst

Deciding SN1/SN2/E1/E2 – The Solvent

Last updated: April 17th, 2024 |

Secondary Alkyl Halides With Strongly Basic Nucleophiles. The “Ask Your Instructor” Edition

  • In the previous four articles in this series, we covered how to identify where an SN1/SN2/E1/E2 reaction could take place, and then discussed the various roles of the substrate (primary, secondary, tertiary), the nucleophile/base, and temperature.
  • We’ve seen that secondary alkyl halides with poorly basic nucleophiles (such as RS(-), N3)(-), (-)CN and halides) tend to give SN2 products, particularly in polar aprotic solvents such as dimethyl sulfoxide (DMSO), N,N-dimethylformamide (DMF), acetone, and acetonitrile.
  • Following the consequences of each of these variables only leaves us with one tricky situation we need to deal with:
    What about secondary alkyl halides with strongly basic nucleophiles like hydroxide (HO ) and alkoxide (RO).

with a secondary alkyl halide and strongly basic nucleophile which pathway dominates sn2 or e2

  • The chemical literature is pretty clear that these reactions result in mostly elimination (E2). Additionally, most textbooks state that elimination (E2) is favored – even in polar aprotic solvents, which tend to favor substitution reactions with poorly basic nucleophiles.
  • However, that doesn’t mean that this topic is taught consistently among all schools or instructors. There are certainly examples of instructors giving exam questions where the expected answer is that this reaction gives substitution. (SN2).
  • For that reason, I strongly suggest asking your instructor their opinion on what happens when a strongly basic nucleophile reacts with a secondary alkyl halide (and whether or not a polar protic or polar aprotic solvent will affect this!)  and prepare accordingly.

summary-sn2 reactions on secondary alkyl halides with strongly basic nucleophiles give mostly E2

Table of Contents

    1. The Question Many Instructors Avoid: Secondary Alkyl Halides With A Strongly Basic Nucleophile
    2. What Do Experiments Say? (E2)
    3. What Do Textbooks Say? (E2… mostly)
    4. Some Potentially Ambiguous Exam Questions
    5. Bottom Line: Ask Your Instructor
    6. Notes
    7. Quiz Yourself!
    8. (Advanced) References and Further Reading

1. The Question Many Instructors Avoid: Secondary Alkyl Halides With A Strongly Basic Nucleophile

You’re in an exam. You’re given the following question.

a test question with secondary alkyl halide without any other cues

What do you draw as the answer?

Let’s start by walking through the process for narrowing down whether a reaction is SN1/SN2/E1/E2:

  • LG – Identify a good leaving group (Cl)
  • sp3 – Ensure that it’s on an sp3-hybridized carbon (it is)
  • 123– Identify that carbon as primary, secondary or tertiary (secondary). This doesn’t rule out anything.
  • N – the nucleophile is NaOCH3 (potassium methoxide).  It’s a strong nucleophile due to the negative charge on oxygen, but also a strong base. (For our purposes,  anything equally basic to HO(-) or RO(-) qualifies as a strong base.  That rules out SN1/E1
  • T – the temperature is not indicated. If heat were indicated, elimination would be likely.

Since SN1 and E1 have been ruled out, we’re left with SN2 and E2 here. We have no further clues.

What’s the product? 

  • If the question is, “what would happen in an actual reaction flask”, the answer is elimination (E2). [Ref] How would you know this? You would have to be told explicitly either by your instructor or by your textbook, because this is an experimental observation, not something that could be predicted from first principles.
  • If the question is, “what answer would you give on an exam?” then I would say, “ask your instructor, because the answer to this specific type question is not always taught consistently. It could be SN2 or E2.”

As far as the way that organic chemistry is taught in North America, there is no greater question I know of that has as much variance as the question of “secondary alkyl halide with strongly basic nucleophile”.

Quiz of instructors on whether secondary alkyl halide plus alkoxide gives sn2 or e2

Small sample size, maybe, but there’s a lot of variation in the answers.  And many instructors just flat-out avoid asking this question.

Welcome to the “ask your instructor” edition of the SN1/SN2/E1/E2 decision!

2. What Do Experiments Say? (E2!)

When I was just starting grad school, I was trying to do a substitution reaction (SN2) on a secondary alkyl halide as part of my research project with a fairly basic nucleophile (a Gilman reagent, to be precise).

My product was almost exclusively elimination (E2). Very little of my desired SN2 product was formed.

When I showed the results to my boss, his reaction was “yeah, the SN2 is a pretty crappy reaction on secondary alkyl halides”.

This initially came as a shock. After all, I’d spent so much time in sophomore organic chemistry learning the various factors that determined whether a reaction was SN1/SN2/E1/E2. Surely they wouldn’t spend all this time teaching us the SN2 if it wasn’t useful in the real world, would they?

Well… yes. In the real world, the SN2 can be a great reaction for primary alkyl halides, but for secondary alkyl halides, it only tends to work well if the nucleophile is poorly basic, like RS(-), N3(-), (-)CN, and halide ions.  We covered this in the post on nucleophiles – see article.  In addition, a polar aprotic solvent like DMF (N,N-dimethylformamide), DMSO (dimethyl sulfoxide), acetonitrile or acetone will assist the rate of SN2 reactions. 

sn2 reactions on secondary alkyl halides are good with weakly basic nucleophiles

However, as soon as the nucleophile becomes strongly basic (such as with alkoxides, hydroxide, and acetylides, or Gilman reagents) elimination (E2) becomes the dominant pathway.

Here is the results of an early study on 2-bromopentane. [Ref]

example of sn2 vs e2 on with alkoxide on 2-bromopentane gives 82 per cent elimination products

Even at room temperature it’s 82% alkene (elimination). As the temperature is increased, the proportion of elimination products also increases. (See article – Elimination Reactions Are Favored By Heat).

The absolute best case I could find was isopropyl bromide with sodium ethoxide in ethanol / H2O, which gives a 47% yield of SN2 product (and only 21% in pure ethanol). 

This is as good as it gets! 

the best case scenario for favoring sn2 over e2 using isopropyl halides

Even changing out one of the methyl groups for ethyl (i.e. to make 2-bromobutane) makes the yield drop considerably to the 20% range.

Cyclic alkyl halides aren’t any better. Cyclohexyl bromide, for example, gives about a 1% yield of SN2 product:

with more hindered alkyl halides like cyclohexyl bromide sn2 is almost negligible

You might think you can improve the yield of SN2 by using a polar aprotic solvent like dimethyl sulfoxide (DMSO). But in that case the yield of substitution product drops from 47% (in ethanol/H2O)  to about 3%.

using polar aprotic solvent with secondary alkyl halides and alkoxides actually gives more elimination products

The bottom line here is that yes, indeed,  reactions of secondary alkyl halides with hydroxide and alkoxides give predominantly E2 products. [Note 1]

3. What Do The Textbooks Say? (E2… mostly)

Given to the amount of angst that SN1/SN2/E1/E2 questions cause students in introductory organic chemistry, one might reasonably expect that the issue of competing SN2/E2 reactions with secondary alkyl halides would get significant coverage.

From what I see, textbooks do not discuss this issue in any great depth. It gets a sentence, or perhaps a short paragraph.

Here’s what some of the more popular organic chemistry textbooks have to say on the topic of “secondary alkyl halide + strongly basic nucleophile”.

  • Klein (1st, p. 378) – E2 (major), SN2 (minor). “When the substrate is both a strong nucleophile and a strong base, bimolecular mechanisms will dominate (SN2 and E2).”
  • McMurry (OpenStax): For secondary alkyl halides… E2 elimination predominates if a strong base is used.  Reference
  • Clayden – shows a chart where strongly basic, unhindered nucleophiles (e.g. RO(-) ) give E2. Weakly basic nucleophiles (e.g. RS(-) give SN2).
  • Solomons (12th) – E2 major.  “With secondary halides, however, a strong base favors elimination because steric hindrance in the substrate makes substitution more difficult.”  Gives the example of isopropyl bromide with NaOH. 79% E2, 21% SN2.
  • Wade  (8th, p. 272) – “with a strong base, either the SN2 or E2 reaction are possible”.  Nothing more specific than that.
  • Streitweiser (4th ed. p. 232) – gives an example (secondary alkyl halide with alkoxide) that gives an 85:5 ratio of E2 to SN2.

So from this relatively small sample, textbooks generally say that secondary + strongly basic nucleophile favors E2, but there is a considerable amount of hedging.

4. Clear Up This Question With Your Instructor Ahead of Time

Instructors want to see that you understand that SN2 reactions result in  inversion of configuration.

For that reason, expect to see examples of SN2 reactions on secondary alkyl halides containing a chiral center.

Hopefully, they’ll do this using questions involving a poorly basic nucleophile like RS(-), N3(-), (-)CN or a halide.

But they might not!  Be prepared for a situation where the use hydroxide or an alkoxide to demonstrate this point:

from roberts stewart caserio organic chemistry methane to macromolecules

This is from “Organic Chemistry” by John D. Roberts et. al., 1971. I love John Roberts (the father of NMR), but question the choice of nucleophile to demonstrate SN2 here. 

Despite the fact that the literature is pretty unambiguous that E2 is dominant, and textbooks are generally on the side of E2, sometimes instructors don’t explicitly tell you whether or not E2 or SN2 is favored for cases with a secondary alkyl halide and a strongly basic nucleophile.

That means you might have to guess the product on your own.

You should ask your instructor ahead of time whether or not reactions like the ones below give predominantly E2 or SN2 products.

Here’s the first example. Substitution or elimination?

is the product sn2 or e2

The bulky base (t-butoxide) tends to only give SN2 with methyl halides, but you should get a definitive answer from your instructor on the expected product with ethoxide.  (The answer key gave both answers as elimination). 

Another example. Substitution or elimination as major product?

-is the product sn2 or e2

In this case the stereochemistry of the leaving group is specified, and the solvent is polar aprotic (acetone). If you weren’t told ahead of time that these reactions favor elimination, there’s a good case to be made that this could be SN2. (The answer key said E2). Check this with your instructor! 

Below is another example where the stereochemistry is specified. Is this substitution or elimination?

is the product sn2 or e2

In this case the answer key also said elimination.

5. Summary: All I Can Really Say Is, “Ask Your Instructor”.

At some point before your big midterm on substitution vs elimination reactions, you should try to pin down your instructor on this question.

“Is SN2 ever the major product with a strongly basic nucleophile like hydroxide or alkoxide?”.

A simple, one-sentence question. Feel free to use any of the examples in this section to make your point.

That way, at least you’ll be told explicitly what to expect.


In previous versions of this article, I have said that secondary alkyl halides with strongly basic nucleophiles tend to give SN2 products with polar aprotic solvents. After doing a deep dive into the old chemical literature, I find that there is really zero evidence for this view. Therefore, I am now advising that the major products of these reactions should be E2 in all circumstances. For not having fixed this earlier, I apologize.

Note 1. One example that does work for the formation of ethers on secondary alkyl halides, but kind of a special case, is here.

It works because  RO(-) is the conjugate base of phenol (pKa = 10) which is roughly as acidic as a thiol. So elimination isn’t as much of a concern as it would be with the more basic hydroxide and alkoxide ions.

Note 2. Table of results from a 1948 study by Ingold et. al. The results are not presented in a way that is easy to digest, but the key value is top left, where isopropyl bromide at 45°C in 60% EtOH/40% H2O gives a mixture that is 53% alkene (olefin) and presumably 47% substitution product. All other results with isopropyl halides are worse.

SN2 vs E2 Ingold 1948

Quiz Yourself!


(Advanced) References and Further Reading

  1. Mechanism of elimination reactions. Part VII. Solvent effects on rates and product-proportions in uni- and bi-molecular substitution and elimination reactions of alkyl halides and sulphonium salts in hydroxylic solvents
    K. A. Cooper, M. L. Dhar, E. D. Hughes, C. K. Ingold, B. J. MacNulty and L. I. Woolf
    J. Chem. Soc. 1948, 2043-2049
    DOI: 10.1039/JR9480002043
    For the reaction of isopropyl bromide with hydroxide ion in ethanol at 55°C, the reaction gives 29% SN2 products and 71% E2 products (see table V).  (The amount of SN2 can be increased to 46% by adding 40% H2O by volume).
  2. Mechanism of elimination reactions. Part X. Kinetics of olefin elimination from isopropyl, sec.-butyl, 2-n-amyl, and 3-n-amyl bromides in acidic and alkaline alcoholic media
    M. L. Dhar, E. D. Hughes and C. K. Ingold
    J. Chem. Soc. 1948, 2058-2065
    DOI: 10.1039/JR9480002058
    This article studies the SN2/E2 ratios of various secondary alkyl halides such as 2-bromobutane and 2-bromopentane and finds that they almost exclusively (>80%) give E2 products.
  3. The Reaction of Primary and Secondary Alkylaryl and Alkyl Sulfonates with Potassium t-Butoxide in Dimethyl Sulfoxide
    Carl H. Snyder and Aida R. Soto
    The Journal of Organic Chemistry 1964 29 (3), 742-745
    DOI: 10.1021/jo01026a055
    Study on SN2/E2 ratios in primary and secondary alkyl sulfonates. In this article the authors state that essentially no SN2 products are isolated from the reaction of secondary alkyl sulfonates (e.g. ROTs) with KOtBu in dimethyl sulfoxide.
  4. Hydroxide-Promoted Elimination Reactions: Alkyl Halides as Substrates†
    Manfred Schlosser, Claudio Tarchini
    Helvetica Chim. Acta. 1977, 60, 3060-3068
    DOI: 10.1002/hlca.19770600858.
    Secondary alkyl halides are shown as giving almost exclusively elimination products here.
  5. Solvation of ions. XIV. Protic-dipolar aprotic solvent effects on rates of bimolecular reactions. Solvent activity coefficients of reactants and transition states at 25°.
    R. Alexander, E. C. F. Ko, A. J. Parker, and T. J. Broxton
    Journal of the American Chemical Society 1968 90 (19), 5049-5069
    DOI: 10.1021/ja01021a002
    Rates of 78 SN2 and E2 reactions for primary and secondary alkyl halides in various solvents. In this article the yield of the SN2 reaction of isopropyl bromide with NaOCH3 in DMSO is given as 3% as measured by vapor phase chromatography (97% elimination).
  6. How Alkyl Halide Structure Affects E2 and SN2 Reaction Barriers: E2 Reactions Are as Sensitive as SN2 Reactions
    Paul R. Rablen, Brett D. McLarney, Brandon J. Karlow, and Jean E. Schneider
    The Journal of Organic Chemistry 2014 79 (3), 867-879
    DOI: 10.1021/jo4026644
    A more recent article about competition between SN2 and E2 reactions.


Comment section

45 thoughts on “Deciding SN1/SN2/E1/E2 – The Solvent

  1. Relating to the last sentence I said… Alcohol Is a polar protic solvent… (but it’s less polar than water… and the textbook kept saying that a less polar solvent favours elimination…)
    My last sentence more fittingly refers to the Williamson synthesis.
    Sorry for the several comments, I am trawling this site for my chemistry final in 2 weeks.

  2. Oh a couple of other people have asked about the alcoholic and aqueous KOH already-
    I’ve found the source with the explanation:
    The book says that alkoxide ion prefers to be a base rather than a nucleophile for steric reasons, as opposed to OH- which can go in for an SN2 easier.
    The textbook is Pradeep’s New Course Chemistry for Class 12
    So yeah, exception to polar protic solvents favouring E2 over SN2.

  3. The second reaction tripped me up because ethoxide ion in ethanol is the usual reagent for the Williamson ether synthesis, which is an SN2 reaction.
    (Not on topic, but my textbook also had something confusing about whether it’s better to have a tertiary alkyl halide substrate or a tertiary alkoxide nucleophile for that SN2 reaction…I may ask about it on the comments section of that page)
    But as you say it’s not that the SN2 reaction won’t occur at all, it’s that the E2 reaction will occur in greater proportion, I guess. The Williamson page said how we get alkoxide ion from alcohol in situ with Na or NaH, and if the alkoxide ion is given beforehand, we could use a polar aprotic solvent to prevent elimination from competing.
    Also, in my course we hear that aqueous KOH leads to an SN2 reaction while alcoholic KOH leads to an E2 reaction. I heard the explanation is that OH- is a good nucleophile for an SN2 reaction, and when there’s alcohol and KOH the OH- can deprotonate the alcohol to get alkoxide ion, which is a stronger base than a nucleophile, so takes a proton for an E2 reaction. Does this hold up?

  4. F is a very weak nucleophile because it would rather take a H+ and leave than attack an electrophilic center. This article is incorrect in this regard. This is a special exception.

  5. Can one claim that the polarity of the solvent affects also E1/SN1 ratio?
    Both pathways form the same carbocation as the product of the rate-determining step. But in the second step, the solvent can interact with the (usually not charged) base-nucleophile. The higher the solvent’s polarity, the stronger the interaction (that lowers the nucleophilicity more than basicity).
    Are these interactions negligible here? Does it have empirical support?
    Generally, what effect the solvent has in E1/SN1 reactions?

  6. My question is if polar protic solvents favour E2 over SN2 then why does a alkyl halide when reacted with aq KOH give alcohol as a major product , even though H20 is a more stronger protic solvent than alcoholic media.Even if it isn’t strong , it is still protic so it shouldn’t give SN2 , but it does !!🤔

    1. Primary or secondary alkyl halide? Because if it’s primary, it should have been clear from the start that it’s an SN2. The real question is what happens with secondary alkyl halides. There is vast inconsistency in how instructors handle this, so I strongly suggest SN2 for polar aprotic solvents, and E2 for polar protic.

  7. Thanks a lot James for giving us concepts for free what our Professors themselves are confused about. Will you please do us a favour by creating a post on ( though not related to this topic) electrophilic aromatic substitution in disubstituted benzene rings…. I am confused over the exceptional cases.

  8. My prof told me that polar protic solvent will favour sn1 reaction because it solvates the carbocation thus increasing the stability. Any comment?

  9. than why does water, even though being a polar protic solvent favour sn2 mechanism.
    CH3ChHBrCHH3 +OH-(AQ.) = CH3CHOHCH3 (SN2)

  10. is it true that polar aprotic solvents are not used in SN1 reactions because some of them can react with the carbocation intermediate and give an unwanted product. Rather, polar protic solvents are preferred ? how eh ?

    1. Polar aprotic solvents cannot react with the carbocation intermediate. Rather, polar protic solvents are used because they are generally more polar than polar aprotic solvents.

  11. I am confused. I thought protic solvents favored sn1/e1 over sn2/e2 vs aprotic solvents favoring e2/sn2 rxns over e1/sn1. Why is that you have compared sn2/e2 here for “protic” solvents than sn1/e1?

    1. I encourage you to go back and read the two previous posts. This is the third in a series of posts on how to deduce SN1/SN2/E1/E2 by asking 4 questions. In the first post we looked at the substrate, and said that tertiary substrates were most likely to undergo sn1/e1 and primary substrates the least likely to undergo sn1/e1. In the second post, we examined the nucleophile, and said that charged nucleophiles/bases would favor sn2/e2 and uncharged nucleophiles would favour sn1/e1. Just by answering those two questions, you can determine whether the reaction will be sn1/e1 or sn2/e2 for most cases. In this third post we show how sn2/e2 can be distinguished in the case of secondary alkyl halides: polar protic solvents tend to favor e2 moreso than sn2.

  12. This post was very helpful, but an answer from my lecturer about this confused me. The reaction was a primary haloalkane with NaOH, with water as the solvent. I thought that since water is obviously a polar protic solvent, an E2 reaction would be favoured, but my lecturer told me that the reaction is Sn2. Now I’m really confused. Any explanation?

    1. If it’s a primary alkyl halide, it’s almost certainly SN2. the only exception would be if you have an extremely bulky base/nucleophile like t-BuO- , and then, possibly, you might get elimination (e2)

  13. Hi! I’m a sophomore chem major who is currently in organic chemistry, and I just wanted to say thanks for this resource! I studied this stuff for ages but couldn’t get it to all line up correctly in my head. This gave me a great frame of reference and an excellent systematic approach with which to approach these problems, which was just what I needed. ^^ Thanks again! You may have earned me ten points on my final!

  14. Dear James,

    I’ve been getting conflicting information from different sources, or maybe I’m just not interpreting it correctly. From what you say here, when the situation is set up for a bimolecular reaction, a polar aprotic solvent will favor SN2 and protic will favor E2. However I have also seen sources that list polar aprotic as a “good” solvent for E2, and this makes sense to me, because with a protic solvent wouldn’t you run the risk of protonating the base before it can abstract the beta-H?

    1. I think the same question! But I guess the effect of polar protic solvents on the nucleophiles will be much more severe than that on the bases. Is that right?

      1. Yes! The nucleophile is attacking carbon here (specifically orbitals on carbon) which are much more difficult to access (and thus sensitive to steric hindrance) than the 1s orbital on a hydrogen [ie when something acts as a Bronsted acid]

  15. Thank you very much for this. Honestly, I chose this over Khan Academy (mainly because I’m in quite a rush to learn this before an exam and you gave a wonderful yet concise explanation). I honestly preferred this over Khan Academy and even the 2nd language book for organic chemistry. I definitely plan on recommending this to classmates and even to my professor so that he may recommend it to future students. Cheers!

    1. Yes!!!!! I completely agree!!! Love this site and I have already told my classmates about it! Thanks so much!!! So helpful and I love how you use funny pictures and explanations, stuff sticks better that way! Orgo final tomorrow!

  16. great blog, I have learned a lot, thanks for all of your hard work, if I may ask a question though, I am sure I am just missing this but WHY are polar protic solvents preferred for E2? I would think that there are a bunch of H+ floating around and for E2 we need a strong base, that this would bind to the protons before ever reaching the nucleophile?

    1. It’s not so much that polar protic solvents are preferred for E2, it’s that polar protic solvents are less favoured for SN2. Polar protic solvents lead to hydrogen bonding of the nucleophile, which you can think of as making the nucleophile more “sterically hindered”. Since nucleophilic substitution is more sensitive to steric hindrance than acid-base reactions (i.e. attacking the H of the C-H bond) a polar protic solvent will tend to decrease the rate of substitution (relative to elimination) .
      In a situation where both reactions compete (e.g. secondary alkyl halide with a strong base like NaOH) this can be the difference between favouring substitution (SN2) vs favouring elimination (E2).

      1. My Organic Chem Prof said this is wrong. Polar protic solvents slow down second order reactions because Protic solvents weaken bases and Nucleophiles.

    2. E2 and Sn2 are favoured by polar aprotic i think that’s a mistake question 2 can’t be E2 it is E 1

  17. 2 comments :)

    -I think it may be important to say that in the truly ‘toss up’ cases (like #2 above), if we ran the reaction and analyzed the results by, say, GC/MS (something we do in our ugrad OChem labs… tho not for this reaction), we’d probably see BOTH E2 and SN2 products… but with protic solvents the E2 is the major organic product, and with aprotic solvents the SN2 is the major organic product. It’s not as if one solvent gives 100% of elimination, and another solvent gives 100% of substitution.

    -I also have my students analyze the reactions in this way. What does the nucleophile tell us? What does the electrophile tell us? What does the solvent tell us? I use these to generate ‘evidence for’ a mechanism, or ‘evidence against’ a mechanism.

    That’s my comment. I would hesitate to use a phrase like ‘rule out’ since psychologically that’s exclusionary thinking. I could conjure up an example where the evidence from the nucleophile ‘rules out’ an SN2 mechanism… but based on the other evidence SN2 is exactly the pathway (off the top of my head, KOtBu + MeBr). But if I tell myself a Big Bulky Base ‘rules out’ SN2, I exclude that possibility mentally. ‘Evidence against’ mentally tells me this is not my first choice… but it’s not impossible.

    (Also, I tell my students solvent can sometimes lean us in one direction or another, but solvent should never be used as ‘evidence against’ a particular mechanism… especially because our SN2 lab uses ethanol as the solvent!)

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