[polly]

Hi,

In lab, we used cyclohexanol with H3PO4 as a catalyst to prepare cyclohexene.  My question is why does the equilibrium strongly favor the reverse reaction, hydration of the alkene?  Does it have to do with a double bond which is less stable?  Also, would you expect the rate of the acid catalyzed dehydration of 1-methylcyclohexanol to be slower, faster or about the same as for cyclohexanol?  I think the answer is slower but I am not sure I understand why.  Thank you!

Edit: edited title for better indexing. Mitch

[GCT]

The equilibrium is a result of pi bond electron density availability combined with the presence of oxidizing agents.  It does not have anything is particular to do with the stability of one compound over another.

[polly]

so your saying the equilibrim strongly favors hydration of an alkene because the pi bond electron density is readily available??  confused...

[GCT]

I should have been more specific.  You should study the mechanism of acid catalyzed dehydration.  For instance if there was a base present, the reaction would have favored the products I presume through an E2 reaction.  However, since no strong base is present (as required by E2 reactions) one needs to supply heat for the reaction to favor the products.  Remember that dehydration also requires the removal of an hydrogen atom and involves the formation of acarbocation.

[polly]

i am really trying to understand this... hydration of the alkene is favored bc it is easier to do as opposed to dehydration which requires heat??

also, the rate of acid catalyzed dehydration of 1-methylcyclohexanol is slower then cyclohexanol bc there is a tertiary carbocation?

[GCT]

I apologize, it has been a while since I have covered this concept.

The dehydration goes through an E1 mechanism since a strong base is not present.  And yes, the formation of the carbocation is unfavorable...which raise the activation energy for the reaction (and thus decreasing the rate).  

Which is thermodynamically more stable?  The way to determine this is to sum up all of the bond energies for each compound.  I would imagine that in this case the alcohol would be more stable.  Since an C-O bond is stronger than C-C bond.

But notice that in entropical terms the product is more stable, since the cleavage of the hydroxyl group occurs, more molecules in the product.  Part of the reason the reaction favors the product (alkene) when heated is due to dTS in the free energy equation.  The overall reaction is unfavorable in dH terms, but as the temperature is raised dTS increases making dG more positive the forward reaction unfavorable.

 

[polly]

ok- starting to make more sense- but what is an E1 mechanism, dts and dg??  Are you talking about the reaction pathway (delta g?)?

[GCT]

You should review the E1 mechanism, it stands for elimination and the rate of the reaction depends only on 1 molecule, thus E...1.

The important element in a E1 reaction as pertinent to this topic is that of the formation of a carbocation.  

Although the hydronium ion is a good leaving good, it still results in the formation of a carbocation, plus we don't have a very good base...thus the next activation energy (abstraction of a hydrogen by water) is unlike relative to if there were a strong base present.

[polly]

thanks for getting back to me and for all your help.  we did not discuss E1,E2, Sn1 or Sn2 yet in class.  the answer was faster bc of the tertiary carbocation.

[polly]

thanks for getting back to me and for all your help.  we did not discuss E1,E2, Sn1 or Sn2 yet in class.  the answer was faster bc of the tertiary carbocation.