[spirochete]

Why doesn't dehydration of an alcohol with H2SO4 and other similar acids give any SN1 product.  The conjugate base is certainly a weak nucleophile, but plenty of weak nucleophiles can add to carbocations.  What makes this situation different?

[azmanam]

Short answer - perhaps it does.  If it does (and I've never heard of this happening) it will be a reversible addition.  Because it is reversible, the bisulfate anion will leave and reform the carbocation. 

ok - assume for a second that the 'nucleophilic addition' is favored.  Even if it is favored, it will be reversible.  But let's even say for sake of argument that the equilibrium favors the C-O-S bond 99:1.  (that is, for every 100 molecules, 99 will have the conjugate base participate in the 'nucleophilic addition' and 1 will have the bisulfate anion leave and the organic compound will exist as the carbocation).  That one molecule will compete between two possible reaction pathways.  It could potentially reform the 'nucleophilic addition' compound or E1 eliminate to form the pi-bond and complete the alkene.  That alkene will be significantly more stable than both the carbocation and the 'nucleophilic addition' compound.  When the carbocation eliminates to the alkene, the 99:1 equilibrium is disrupted, and Le Chatlier's principle will force another molecule of the 'nucleophilic addition' compound to convert to the carbocation... which will eliminate to the alkene which will disru....  That elimination/le chatlier's principle will act as a siphon driving the reaction to the expected alkene product.

But that probably doesn't happen.  The lone pair of electrons on the bisulfate anion (the ones left over from loss of proton and the ones that would presumable participate in the 'nucleophilic addition') will be shared over four atoms (3 oxygen atoms and a sulfur atom) through resonance (draw structures to convince yourself).  Those electrons will be so delocalized they will not be organized enough to participate in the 'nucleophilic addition.'

interesting question, though.  Good to see you're thinking beyond the book and really trying to understand the chemistry.

[spirochete]

Makes sense so far.  I was taught that your average SN1/E1 rxn gives roughly 80% SN1, but I never learned if that was for kinetic or thermodynamic reasons. 

Looking at a prototypical SN1/E1 with water as the nucleophile it appears that the substution product is more stable, since an RO-H bond formed is much stronger than the C-C pi bond.  On the other hand, looking at the substitution product with the sulfate ion it's clear that the bond formed is going to be significantly weaker since homolytic cleavage will produce a highly delocalized radical.  I don't have the exact bond dissociation energy for the new bond formed, but I'd guess, based on my understanding of your explanation, that it should be lower than 65 kcal/mole (vale for C-C pi bond formed in elimination).

If not, I'm still a little bit confused.

Or are BDE's not a good way to think about this?

[azmanam]

You  might be able to use BDE's, but that's not where I was coming from.

I'm thinking more along the lines of energy of activation and principle of microscopic reversibility.  More of a qualitative analysis; although, a quantitative analysis can probably be derived by the same principles.

I've redrawn the energy diagram (inspired by this page: http://www.cem.msu.edu/~reusch/VirtTxtJml/alcohol1.htm#alcrx3) with the possible alternative pathway.  I haven't done any qualitative analysis, but it would seem to me (assuming I've drawn the energy diagram accurately) that the 'nucleophilic addition' pathway would be fast (low [delta]E₂[double dagger]) and reversible (small [delta]E).  But the E1 reaction would be of intermediate speed ([delta]E₁[double dagger] and essentially irreversible - or the reverse reaction is significantly slower than forward reaction (large [delta]E and larger [delta]E₃[double dagger]).  The reverse reaction (depending on the various substituents on the new alkene) could be essentially irreversible under the reaction conditions.

with water as the nucleophile it appears that the substution product is more stable, since an RO-H bond formed is much stronger

Careful.  You're not forming a strong R-O-H bond, you're initially forming a weak R-O⁺H₂ bond when water is the nucleophile.

homolytic cleavage will produce a highly delocalized radical

probably true, but under the reaction conditions, any bond cleavage will be heterolytic.  No radicals in this mechanism.

[azmanam]

here's a good overview chart.

http://www.cem.msu.edu/~reusch/VirtTxtJml/alhalrx3.htm#hal9

[spirochete]

The diagrams were helpful for thinking things through.  In general it seems my thinking was leaning more toward thermodynamics, yours more toward kinetics.  For example, in comparing substitution by water vs. substitution (hypothetically) by the bisulfate anion.  Kinetically, the bisulfate anion might be favored, because the final product is formed immediately.  The final product of the substitution by water is certainly more stable, but there's a kinetic barrier to get past in forming the intermediate with a positively charged oxygen.

   I looked at my textbook again and realized they sort of explain this very question, just not directly.  Even with a strong acid catalyst dehydration won't go to completion, because the hydrate is more stable and the rxn is easily reversable.  The solution is to distill away the alkene: alcohols have a higher boiling point than alkenes so they stay in solution.   I'd guess the bisulfate substitution product would have the same or higher boiling point than alcohol.

And just to clarify what I was saying before about homolytic cleavage:

probably true, but under the reaction conditions, any bond cleavage will be heterolytic.  No radicals in this mechanism.

I was just using radical stability as a way to estimate bond dissociation energy for an unusual bond (R-Bisulfate), not saying the mechanism actually contained a radical.  For example, allylic hydrogens have BDE's around 10 KCals/Mol lower than regular primary hydrogens.  This is due to the stablized radical that's formed from cleavage of the allylic C-H bond.  This is reflected in the thermodynamics of any reaction involving a molecules with allylic hydrogens, regardless of the mechanism.

Radical stability is just one factor in BDE's though.  For example bonds between hydrogen and SP2/SP carbons are stronger because of the higher S character in the carbons' hybrid orbitals.

Sorry if you knew all that already, I just learned about this stuff and think it's pretty cool.

[Yggdrasil]

I would agree that the reaction is under thermodynamic control and the reason that the major product of R-OH + H₂SO₄ is the alkene because the substitution products are not as stable as the alkene. 

Think of which substitution products are possible.  Water could act as a nucleophile and attack the carbocation to reform the alcohol, but this is not favored for two reasons.  First, the reaction occurs under acidic conditions, H₂SO₄ will protonate most of the newly formed water molecules forming a very non-nucleophilic species (H₃O⁺).  Second, if water does successfully add, the acid in the solution will often protonate the alcohol, creating R-OH₂⁺ which has a good leaving group.

The other substitution reaction that could occur is the addition of a bisulfate anion, but a bisulfate group is a good leaving group and can easily dissociate.  So, yes, I think your BDE argument makes sense.  The newly formed C=C bond will be thermodynamically favored over forming a C-OSO₃H bond, and protonation of the water molecules prevents the carbocation from re-forming the alcohol.