[co_life]

Hi all
This is a simple question about a mechanism...
I want to transform 2-methylbutan-2-ol and H3PO4 into 2-methylbutenes.
I know the products will be  2-methylbut-2-ene (primary) and  2-methylbut-1-ene (less qty).
First of all, the H from a OH (coming from the H3PO4) gets attacked by a lone pair from the  2-methylbutan-2-ol. Then in the organic molecule, the C-O bond goes to the O and a H2O is liberated. We now have a sec carbocation.
In order to get the   2-methylbut-2-ene, we send a lone pair (from H2PO4-) to a H from the 3rd carbon (main chain) ----> this is the ZAITSEV rule
And in order to get the  2-methylbut-1-ene, we should send the lone pair (still from H2PO4-) to a H from the end of the chain (from C#1 or C from the methyl). -------> this is the HOFFMAN rule
Finally, we send the C-H bond toward the C, creating a c=c bond.

Now, is that right?? are the names (Zaitsev Hoffman) and the rules used right?
Thie reaction does not imply the Markovnikov or the anti-Markov rule, does it?

Thanks to all 

[azmanam]

You have the right overall concept... let's tighten up the language a bit.

The first step is good.  A lone pair on the O atom of the alcohol attacks a proton on the acid, and the alcohol becomes protonated, with a full positive charge on oxygen.

A better way to phrase the second step:  The C-O bond heterolytically cleaves, and both electrons in the former C-O sigma bond move onto the oxygen atom.  This liberates a molecule of water (H₂O) and forms the tertiary carbocation (not secondary as you noted).

In the third step, a lone pair from a base does attack a proton on a carbon next to the carbocation (the beta position).  However, as the dehydration of alcohols is usually catalytic in acid, it is usually a lone pair on the oxygen atom of a molecule of water which effects the transformation.

The Zaitsev's product arises, as you said, from attack of base on the proton of carbon 3.  The electrons in the former C-H sigma bond collapse down to form a pi bond with the empty p orbital on the carbocation to form the alkene.

The minor product indeed arises from the same mechanism - however the base attacks a proton on carbon 1.  I would not call this the Hoffman product.  In my mind, when you change the reagent such that the less-substituted double bond is the major product, that is the Hoffman rule.  for example, if this were the dehydrohalogenation of an alkyl halide, use of a small base (CH₃O^-) would favor the more-substituted double bond (Zaitsev's rule).  Use of a big, bulky base (like (CH₃)₃CO^-) would favor the less-substituted double bond (Hoffman rule).  In this case, the less-substituted double bond is simply the minor product, imho.

This mechanism has nothing to do with Markovnikov's rule.

[co_life]

Thanks azmanam! you answer is very clear and complete - it helped.
And the picture as well was an amazing idea.

Though I still have a few precise questions:
    - Agreeing that the minor product is not formed by Hoffman's rule, can we say that it is just formed without respecting Zaitsev's rule?
    - I do not fully comprehend why you use the water in the third step. In fact, as we stated with H3PO4 and, during the reaction, made it "change" to H2PO4-, why don't we regenerate the catalyst (H3PO4) instead of using water (thus generating H3O+)  (H2PO4-'s got a lone pair as well).

Thanks a lot!!

[azmanam]

can we say that it is just formed without respecting Zaitsev's rule?

yes.  we can say the minor product is formed in spite of Zaitsev's rule, or in contradiction to Zaitsev's rule.

as we started with H3PO4 and, during the reaction, made it "change" to H2PO4-, why don't we regenerate the catalyst (H3PO4) instead of using water (thus generating H3O+)  (H2PO4-'s got a lone pair as well).

With respect, you're thinking about it too literally.  There are a couple of answers to your question, though.  First think about the different acidic equilibria that are actually going on.  What probably actually happens first is acid base reaction between phosphoric acid and water: 

H₃PO₄ + H₂O  --><--  H₂PO₄^- + H₃O⁺

The equilibrium is fast.  The implication of this is that the actual active catalyst is probably H₃O⁺, not really H₃PO₄.  Thus, while it's acceptable to have the alcohol pick up the proton from phosphoric acid, in all actuality, it probably takes the proton from the hydronium ion.

Second, it's not as if the alcohol thinks to itself, "Ok, I'm under acidic conditions... Phosphoric acid is the catalyst...  I'll swim around until I find phosphoric acid and then pick up a proton."  Rather, the acidic proton is probably shuttling around between several different molecules very quickly.  The acidic proton starts on phosphoric acid.  Water picks up the proton to form H₃O⁺.  Another water molecule takes the proton from the first water molecule.  A molecule of the alcohol takes the proton from the protonated water molecule, but doesn't heterolytically cleave to the carbocation (this is plausible.  It requires a non-trivial amount of energy to heterolytically cleave a protonated alcohol.  It is spontaneous, but it is the rate limiting step of the reaction).  A different alcohol molecule might take the acidic proton from the first alcohol molecule.  Then a third molecule of water might pick up the proton for a while..  So what I'm saying is that it's not so cut and dry as 'add phosphoric acid.  alcohol takes proton from phosphoric acid...'  There are a number of acid/base equilibria occuring: phosphoric acid and water, hydronium ion and water, hydronium ion and alcohol, phosphoric acid and alcohol, all are fast, and all are reversible.  Where is the actual acidic proton at the time the alcohol takes it and heterolyitcally cleaves?  Who knows.

To rectify this in our minds, organic chemists don't usually think about it in such literal, quantitative terms.  We know there's acid, we know it's a catalyst, but it really doesn't matter which exact molecule has the proton that the alcohol takes.  It only matters that it picks up a proton.

What organic chemists usually do when drawing a mechanism under acid catalysis is to just say the proton comes from some acidic molecule HB (Where B^- is any conjugate base: H₂PO₄^-, H₂O, R-OH (an un-protonated alcohol).  The first step of the mechanism would be the alcohol taking a proton from this generic acid, HB, to generate protonated alcohol, R-OH₂⁺, and a conjugate base, B^-.  After the C-O bond heterolytically cleaves to form the carbocation, we usually just say some generic base, B^-, takes the beta proton and forms the alkene.

It is not usually taught this way, though.  Ambiguity doesn't go over very well with students in a lecture class where exams and grading are involved .  If you'd like to think about it that the phosphoric acid donates the proton to the alcohol, then the hydrogen phosphate anion removes the proton to form the alkene, that's fine and you wouldn't be marked wrong (if I were grading it... you may want to check with your instructor for the actual grading policy...)  All I was trying to say is think about the statistics:  very small amount of acid catalyst.  Many more water molecules in solution (you even generate more water as the reaction progresses when the C-O bond heterolytically cleaves to liberate a molecule of water).  It's just statistically more probable that water collides with the carbocation to form the alkene than hydrogen phosphate anion.

But, I wouldn't get too hung up on it (even if I was a bit wordy in my explanation...).  You have the right idea about the dehydration.  You know what's going on.  You seem to have a good grasp on the mechanism.

[co_life]

Well, thanks a lot again for you answer and your patience.
It's true that we don't always learn the truth behind mechanism in class and when I rethink about, it's true that it can't work as perfectly as we think. It's logical that H3O+ is the catalyst, since the the H3PO4 (or other strong acids) is often diluted in water.

Your answers really do help in order to build confidence. THANKS!
Your explanations make me somehow be interested in chemistry.

Thanks again

[azmanam]

Well... since I've piqued your interest... care to step it up a notch?

Consider:  in the alkene chapter you probably learned about the acid-catalyzed hydration of alkenes to form alcohols.  The general scheme of this reaction is:

  acid
R₂C=CR₂ + H2O   ------->  R₂CH-CR₂-OH

Note that this is the reverse reaction of the acid-catalyzed dehydration of alcohols we just talked about:

acid
  R₂CH-CR₂-OH   ------->  R₂C=CR₂ + H2O

Interestingly, the mechanisms for the two reaction are exactly the same... just in reverse order.  For the dehydration mechanism, alcohol picks up a proton to form a protonated alcohol, water leaves to form a carbocation, base removes a proton to form an alkene.  for the hydration, alkene picks up a proton to form a carbocation, water adds to form a protonated alcohol, base removes the proton to form the neutral alcohol.

(this is known as the priniciple of microscopic reversibility, http://www.cem.msu.edu/~reusch/VirtTxtJml/alcohol1.htm#alcrx3)

In reality, these two reactions are in equilibrium with each other, and both reactions are happening in both directions at the same time.  On paper, if you set up an acid-catalyzed dehydration of an alcohol, there's nothing keeping the newly formed alkene from reacting with acid and water to form an alcohol.  In theory, you should not be able to control this reaction, and you should end up with a statistical mixture of alkene and alcohol...

acid
  R₂CH-CR₂-OH   ---><---  R₂C=CR₂ + H2O

In practice, one can actually do these reactions to isolate either the alcohol or the alkene you desire.  It is possible to control this equilibrium and achieve high yield of the product you want.

My question to you, then, is how?  How can you control this reaction so that you can preferentially, and purposefully, force the equilibrium in one direction or the other?  Do you remember a concept from gen chem that might help control the direction of this equilibrium?

[co_life]

Well, it must be the concentrations.

Now, to HYDRATE, we need water, so the concentration of, lets say, H3PO4 must have a small value (diluted).
On the other hand, in order to DEHYDRATE, we must eliminate water... therefore, the concentration should be higher.
Is that right?

(sorry if I didn't see your answer sooner)

[azmanam]

Yes.  concentration will definitely have something to do with it. 

You talk about modulating the concentration of acid.  This is true, but you talk as if you're keeping the amount of water constant and changing the amount of acid.  Remember... both reactions are catalytic in acid, so the gross amount of acid will probably be the same.

It turns out that even though the amount of acid is kept constant, it's the amount of water that is modulated.  How will changing the amount of water help determine the outcome of the reaction?  This is related to a principle from Gen Chem.  Do you know which one I'm talking about?