# Why is the alkene a more stable species in this case?

How can we account for $$\mathrm{S_N1}$$ substitution in this case versus elimination in others?

When we dehydrate alcohols, we usually carry out the reaction in concentrated sulfuric acid and at high temperature. The hydrogen sulfate ($$\ce{HSO4-}$$) present after protonation of the acohol is a weak nucleophile, and at high temperature the highly reactive carbocation forms a more stable species by losing a proton and becoming an alkene. (Emphasis mine) Furthermore, the alkene is usually volatile and distills from the reaction mixture as it is formed, thus drawing the equilibrium toward alkene formation. The net result is an $$\mathrm{E1}$$ reaction.

(Solomons, 11th edition page 511)

The author seems to imply that the formation of an alkene from a carbocation is energetically more favorable than the formation of an $$\mathrm{S_N1}$$ product. Why?

Once the carbocation is formed, the step leading to the $$\mathrm{S_N1}$$ product requires the formation of a sigma bond whereas the step leading to the $$\mathrm{E1}$$ product requires the formation of one pi bond.

It seems to me that the $$\mathrm{S_N1}$$ product should be more favorable (not by a lot though, since the rate limiting step for both reactions is the same).

• – Mithoron Aug 28 '17 at 22:09

I don’t think that your interpretation is what the author meant. I think that they are comparing the reactivity (or inverse stability) of the alkene to that of the carbocation. I think we can both agree that a carbocation is a much more reactive species than an alkene.

When it comes to comparing the products of a nucleophilic substitution to those of an elimination, I don’t think there’s an easy call that can be made. It depends far too much on variable factors such as the steric congestion around the reactivity site, the substitution pattern of the double bond, the size and shape of the nucleophile etc.

In the case of this specific reaction, however, we can make that call. The only mildly nucleophilic species is the hydrogen sulphate anion $\ce{HSO4-}$ deriving from deprotonated sulphuric acid. Hydrogen sulphate has three equivalent oxygens that each carry one third of the negative charge, and they are also practically symmetry-equivalent (not ideally, though, since the proton will angle off). The high oxidation state of the central sulphur atom reduces the electron density on the oxygens. Each of these oxygens is therefore a very bad nucleophile — compare to typical sulphonic acid residues such as tosylate or mesylate which are equally terrible nucleophiles. This should make a nucleophilic attack highly unlikely.

A second thing to remember is that sigma bond does not equal sigma bond. Usually when discussing sigma bonds, $\ce{C-C}$ or $\ce{C-H}$ bonds are meant which are very unpolar, have a bond dissociation energy of something around $\pu{400 kJ/mol}$ and are thus very hard to dissociate. Here, we would be making a $\ce{C-O}$ bond. But not any kind of $\ce{C-O}$ bond: the oxygen belongs to a sulphate residue. This makes the $\ce{C-O}$ bond very weak and very easily cleaved, especially since sulphate is a very good leaving group.

Therefore, in this specific case we can actually make the call and say that the elimination product is more stable than a product of a nucleophilic substitution.

One more thing: you do write cleavage of a sigma bond. But what is actually happening is akin to a deprotonation: a proton leaves leaving its electrons behind in a type of acid-base reaction. As you may know, these are typically very rapid as protons have an easy time ‘jumping’ from one molecule to another if the overall conditions for acid-base reactions are favourable.

• Okay, I got it. Just one last question: you wrote in your edit that HSO4- is a poor nucleophile. Why wouldn't it be a poor base also? – xasthor Aug 28 '17 at 13:23
• @xasthor I can’t follow. $\ce{HSO4-}$ is a (very) poor base as it is the conjugate base of a strong acid. – Jan Aug 28 '17 at 13:26
• You wrote: HSO4- is a very poor nucleophile. 'This should make a nucleophilic attack highly unlikely.' Why would an attack on a hydrogen atom be any less likely? – xasthor Aug 28 '17 at 13:30
• @xasthor Because we have a very reactive carbocation and the full reaction is $\ce{R2CH-\overset{+}{C}R2 + HSO4- <=>> R2C=CR2 + H2SO4}$. The overall reaction is exergonic because the alkene is an even weaker base than hydrogen sulphate. – Jan Aug 28 '17 at 13:35