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I'm trying to figure out the major product. At the moment I'm leaning towards an $\mathrm{S_N2}$ reaction, because there's a polar aprotic solvent, and sort of unhindered electrophile. However to me it could just as easily be an $\mathrm{S_N1}$ reaction (electrophile does have some hinderence and there's a weak nucleophile and weak base). $\mathrm{S_N1}$ would attack the tertiary carbocation and $\mathrm{S_N2}$ would go for the primary.

So how do I know if it's $\mathrm{S_N1}$, $\mathrm{S_N2}$, or even $\mathrm{E1}$/$\mathrm{E2}$?

I guess the main question is I don't know if the electrophile is hindered "enough" so we can say that an sn2 reaction is definitely not happening.

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    $\begingroup$ E2 requires strong base, SN2 requires good nucleophile. Water is a lousy base and a lousy nucleophile. Even if you have a polar aprotic solvent you probably wouldn't get any SN2 product imo $\endgroup$ Commented Nov 12, 2015 at 8:56

2 Answers 2


Generally, your substance can react in one of the following four ways that you have already identified:

  • $\mathrm{S_N1}$ solvolysis on the tertiary chlorine resulting in 4-chloro-2-methylbutan-2-ol;[1]
  • $\mathrm{S_N2}$ solvolysis on the primary chlorine resulting in 3-chloro-3-methylbutan-1-ol;
  • $\mathrm{E1}$ elimination from the tertiary chlorine resulting in 4-chloro-2-methylbut-1-ene or 1-chloro-3-methylbut-2-ene;[1]
  • $\mathrm{E2}$ elimination from the primary chlorine resulting in 3-chloro-3-methylbut-1-ene.

It is highly unlikely that one of the $\mathrm{X1}$ mechanisms happen on the primary chlorine due to the very badly stabilised intermediate primary carbocation we would need, highly unlikely for an $\mathrm{S_N2}$ reaction to proceed on the tertiary chlorine due to steric hindrance (especially since a primary one is available) and moderately unlikely (but not impossible) for $\mathrm{E2}$ to happen on the tertiary chlorine.

We thus need to consider:

  • How good is the leaving group (influences everything strongly except for $\mathrm{E2}$)?[2]
  • How strong is a base possibly present in solution (influences which elimination reaction can happen and how well)?[3]
  • How good is our nucleophile (influences the probability of $\mathrm{S_N2}$ most)?[4]

Starting at the base. This is a case of rule 6:[6] There is no base present in the mixture! (Yes, technically water can act as a base, but you need a strong acid for it to happen. Water’s $\mathrm{p}K_\mathrm{b}$ is $15.7$ if I recall correctly.) This definitely rules out $\mathrm{E2}$ which requires a strong base, and mildly rules out $\mathrm{E1}$ which requires a weak base.

Let’s check out potential nucleophiles next. Well, one of our nucleophiles is water, also known as very lousy, and another possible one is acetone, also known as even lousier. There is hardly any chance that either of them will nucleophilicly attack the primary chlorine site ruling out $\mathrm{S_N2}$.

Next, what about our leaving group? Well, we have chlorine which is also a pretty lousy leaving group. It means that $\mathrm{S_N2}$ will be discouraged even more and the generated chloride will probably hang around close enough to the carbocation required for the $\mathrm{X1}$ reaction paths to disallow $\mathrm{E1}$.

Finally, there are solvent effects, too. Here, the solvent is polar protic; a combination generally preferring the $\mathrm{X1}$ pathways.[5]

All things considered, this leaves $\mathrm{S_N1}$ as the only mildly probable pathway remaining. And thus I would strongly assume the main product to be 4-chloro-2-methylbutan-2-ol.

Of course, one should always include reaction times and temperatures (and yields) to any reaction-predicting exercise. Raising the temperature should increase the likelyhood of $\mathrm{E1}$ over $\mathrm{S_N1}$ because of the entropic factor (two particles become three in $\mathrm{E1}$ resulting in a rise in entropy).[7] However, in exercises, this is usually indicated by the presence or absence of a Δ above or below the reaction arrow. Since we have no Δ, we can assume room temperature.

[1] C. A. Grob, A. Waldner, Helv. Chim. Acta 1979, 62, 1854.
[2] https://en.wikipedia.org/wiki/SN2_reaction#Leaving_group
[3] https://en.wikipedia.org/wiki/Elimination_reaction#E2_mechanism
[4] https://en.wikipedia.org/wiki/SN2_reaction#Nucleophile
[5] https://en.wikipedia.org/wiki/SN1_reaction#Solvent_effects
[6] Rule 6: There is no rule six!
[7] https://en.wikipedia.org/wiki/Elimination_reaction#E2_and_E1_elimination_final_notes

  • $\begingroup$ I assumed that acetone was the solvent and H2O a reagent, but I agree, SN1 is most likely. $\endgroup$ Commented Nov 12, 2015 at 12:44
  • $\begingroup$ @orthocresol How does that change anything? Acetone is protic (albeit slightly). $\endgroup$
    – Jan
    Commented Nov 12, 2015 at 12:47
  • $\begingroup$ if acetone is protic, then it certainly doesn't change anything. I always thought it was considered aprotic. $\endgroup$ Commented Nov 12, 2015 at 14:52

This is a poorly written exercise. There is no temperature and no reaction time. Students have to be taught that these 'dimensions' are essential to describe synthetic reactions. With heating, there shouldn’t be any substitution of the primary chloride (SN2) but you would see the SN1 solvolysis product i.e. the tertiary alcohol. The carbocation intermediate can also yield the E1 alkene products (the more substituted alkene being the dominant one). This is a reference you can check: Helvetica Chimica Acta, 62(6), 1854-65; 1979. I only have access to an abstract where dioxane is the solvent in the reaction schemes but they report that acetone was also used. Since by using acetone you would achieve lower temperatures than dioxane, it is likely that you will see less of the E1 alkene products.


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