Before I can even begin to contemplate the products of a given SN1 reaction, I need to be able to determine what my leaving group will be, which should be a species with a strong conjugate acid, and as I understand it the species should be relatively stable once it has left the substrate. Please correct me if I am making any mistakes in my understanding.

A hydroxide ion will be a terrible leaving group, since its conjugate acid is water, and hydroxide is not very stable by itself.

So when I am asked this question "Draw the major organic product(s) for this SN1 reaction:

Enter image description here

How do I know what will be the best leaving group? Hydroxide is not very good, so perhaps $\ce{CH_3OH}$ will work better? How do I know it is not $\ce{CH_3CH_2OH}$ instead?

Is this only one decent option, and I am over thinking this?

  • $\begingroup$ this example, to be honest, is bad since OH is bad leaving group. also, H-Br is used for hydrogenation of alkenes. on the other hand, this Sn1 reaction can be done by first toscilating it with TsCl to make a very good leaving group! $\endgroup$
    – user10075
    Nov 30, 2014 at 3:01

2 Answers 2


This case (and similar cases) you need to think about the following:

  1. When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.
  2. What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?
  • 1
    $\begingroup$ I guess I was thinking the hydroxide would be near impossible to leave on its own, but I suppose it reacts with the hydrogen to make water and the leaving group could be replaced by left over bromide. I started giving myself the idea that the 2-3 carbon bond would break and separate the entire thing but that is bad thinking considering the obvious affects of the hydrogen bromide. $\endgroup$
    – Leonardo
    Dec 2, 2012 at 22:38
  • $\begingroup$ Its just the opposite. The only feasible leaving group on this molecule is -OH group. In general, a carbon group will never be a leaving group, because it would leave as a carbanion, and that is like the worst leaving group ever. C-C bond breaks are rare in organic chemistry, and they will definitely never take place as a consequence of a R group leaving the molecule. $\endgroup$ Nov 30, 2014 at 3:17

To answer this question, you must first inspect the species that are reacting first. What are these species? What is acid? what is base?...not necessarily focusing on the leaving group. In general, acid-base reactions have extremely high Keq (Ka) and in the system will occur first. It is so rapid that the exchange of a proton is close to the diffusion-controlled rate limit. That happens first. The alcohol is protonated and oxygen becomes formally +. It is now the leaving group, water. As it leaves a stable tertiary carbenium ion is formed. Bromide, the conjugate base of the acid used then attacks to form the Sn1-type product.

Someone mentioned TsCl...tosylation will not be a good choice. True it will activate the alcohol to be a weaker base. But since this is a tertiary alcohol, aa back-sided displacement is not going to happen to give you an alkyl bromide, you will get the elimination instead of substitution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.