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I don't follow this reaction. This is after all an acid catalyzed dehydration reaction; why then, is the final product the alkene and water? Water wasn't the original acid. Hydronium ion is the original acid. Shouldn't the final product be not water but hydronium ion (produced through deprotonation which in turn forms the alkene)?

In addition, this is the book's reaction mechanism, which clearly shows the formation of hydronium ion:

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But this isn't a one-off sort of thing with the water as a "product" of acid-catalyzed dehydration.

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So, is there a reason water is listed as the product in acid-catalyzed dehydration? Am I missing something?

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    $\begingroup$ Hi Dissenter: acid-catalyzed means hydronium is the catalyst, not a product. Dehydration means water is the product. Not clear what's tripping you. $\endgroup$ – Silvio Levy Aug 4 '14 at 1:40
  • $\begingroup$ Okay, I think I see it now. I was expecting the hydronium ion to be regenerated as it is a catalyst, but on the other hand, the problems don't show the hydronium ion as being consumed. $\endgroup$ – Dissenter Aug 4 '14 at 1:41
  • $\begingroup$ it is regenerated on the last step when $\ce{H2O}$ pulls a proton off the carbon. $\endgroup$ – Silvio Levy Aug 4 '14 at 1:45
  • $\begingroup$ I know, that's why I was expecting hydronium ion as a product, not water, but I see where the water is coming from now and since hydronium ion isn't being explicitly consumed, it isn't shown either as a product (explicitly). $\endgroup$ – Dissenter Aug 4 '14 at 1:46
  • $\begingroup$ Yes, I see the difficulty now. The product water leaves at step 2. $\endgroup$ – Silvio Levy Aug 4 '14 at 1:47
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Consider systematically:

The mechanism is an $\bf E_1$ reaction in which the substrate is a protonated alcohol.

Step I

In this step, an acid–base reaction, a proton is rapidly transferred from the acid to one of the unshared electron pairs of the alcohol. In dilute sulfuric acid the acid is a hydronium ion; in concentrated sulfuric acid the initial proton donor is sulfuric acid itself. The leaving group is a molecule of water.

Step II

The carbon–oxygen bond breaks heterolytically. The bonding electrons depart with the water molecule and leave behind a carbocation. The carbocation is, of course, highly reactive because the central carbon atom has only six electrons in its valence level, not eight.

Step III

Finally, in step 3, a water molecule removes a proton from the $\beta$ carbon of the carbocation. The result is the formation of a hydronium ion and an alkene. In step 3, also an acid–base reaction, any one of the nine protons available at the three methyl groups can be transferred to a molecule of water. The electron pair left behind when a proton is removed becomes the second bond of the double bond of the alkene. Notice that this step restores an octet of electrons to the central carbon atom.


Though water gives proton in step one and other molecule of water takes proton in last step, it is usually written as a produce of alkene and water until a mechanism is specifically mentioned.This is also reminding of the catalytic nature of water in this reaction.

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