I am told that this method yields ONE equivalent of water, but I seem to get two equivalents of water. Also does this mechanism actually work? I've never seen anything like it, but it seems to make sense. Would another Lewis acid such as boron trifluoride also work?

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    $\begingroup$ One of your two $\ce{H2O}$ molecules comes from the reactants, so it "doesn't count". $\endgroup$ – Philipp Sep 4 '14 at 16:37
  • $\begingroup$ What do you mean it doesn't count? $\endgroup$ – Dissenter Sep 4 '14 at 16:40
  • $\begingroup$ @Dissenter It means that the $\ce{H2O}$ isn't really generated from the reaction, but exists already. You added a $\ce{H2O}$ in the second step. $\endgroup$ – DHMO Jan 14 '17 at 12:53

In the reaction $\ce{{}^{$i$}Pr-OH + H2O ->[\ce{AlCl3}] C3H6 + 2H2O}$ the $\ce{H2O}$ on the reactant side and one of the $\ce{H2O}$s on the product side cancel, so you get $\ce{{}^{$i$}Pr-OH ->[\ce{AlCl3}] C3H6 + H2O}$.

I think the reaction would also work with other Lewis acids. But it will rather proceed through the $\mathrm{E}1$ mechanism in any case. And it is likely that you'll get $\mathrm{S}_\mathrm{N}1$ side reactions.

  • $\begingroup$ What do you mean by e1 and sn1 side reactions? That a hydrogen will be abstracted? But by what? Also what sn1 products? Alcohol is a poor leaving group. $\endgroup$ – Dissenter Sep 4 '14 at 16:48
  • $\begingroup$ @Dissenter The $\ce{OH}$ group is a bad leaving group, yes. But the Lewis acid turns it into a better leaving group, much as protonation would. As for $\mathrm{E}1$ and $\mathrm{S}_\mathrm{N}1$: You should know about them from your Organic chemistry courses. They work by first letting the leaving group leave, thus forming a carbocation, which is then either attacked by a weak nucleophile or eliminates a hydrogen. $\endgroup$ – Philipp Sep 4 '14 at 16:54
  • $\begingroup$ So you're saying the OH group is likely to leave by itself and then water attacks as a nucleophile and the electrophile/Lewis acid does nothing? $\endgroup$ – Dissenter Sep 4 '14 at 16:57
  • $\begingroup$ @Dissenter No, the OH group is coordinated by the Lewis acid (in a similar manner as a proton would bind to it - similar but definitely not the same) and will thus be a good leaving group. Then the water can act either as a nucleophile or as a base to effect either a $\mathrm{S}_\mathrm{N}1$ reaction or an $\mathrm{E}1$ elimination. $\endgroup$ – Philipp Sep 4 '14 at 17:01
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    $\begingroup$ @Dissenter It's the same situation as in $\ce{NaBH4}$. There you have $\ce{H3B^{\ominus}-H}$ and rather than $\ce{H3B^{\ominus}-H^{\oplus}}$. $\endgroup$ – Philipp Sep 4 '14 at 17:57

$\ce{BF3*OEt2}$ in $\ce{CH2Cl2}$ can be used too, at least for the dehydration of tertiary alcohols [DOI].

As for the moles of water yielded, please see Philipp's answer and use your common sense on whether the formation of a second molecule of water out of nothing is likely.

  • $\begingroup$ What do you mean out of nothing? Do you think there is no solvent? $\endgroup$ – Dissenter Sep 4 '14 at 16:51

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