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To dehydrate ethanol one uses concentrated sulfuric acid:

$$\ce{C2H5OH ->[\text{conc.} H2SO4] C2H4 + H2O}$$

but to go in the reverse direction, dilute sulfuric acid is used:

$$\ce{C2H4 + H2O ->[\text{dil.} H2SO4] C2H5OH}$$

Why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place?

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    $\begingroup$ Dilute sulfuric acid implies the presence of a lot of water ... and dehydration conditions call for the removal of water. Having water around when one is trying to remove water doesn't help with the removal of water. $\endgroup$
    – Dissenter
    Nov 23, 2014 at 9:41
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    $\begingroup$ "For that matter, how does the sulfuric acid even help in the first place?" Dehydration again implies the loss of water; note that in your dehydration reaction there is an hydroxyl functional group (-OH) which resembles water in that the -OH is one hydrogen away from H2O (water). An acid catalyst helps tack a H+ onto the -OH and this makes a good, stable leaving group. $\endgroup$
    – Dissenter
    Nov 23, 2014 at 10:04

1 Answer 1

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Think of the reaction as an equilibrium equation:

$$\ce{CH3CH2OH <=>[H2SO4] CH2=CH2 + H2O}$$

Now, using Le Châtelier's Principle, we can deduce Dissenter's comments.

In concentrated sulfuric acid there is little to no water (note that we also start off with little to no ethylene). The equilibrium shifts to produce products.

In dilute sulfuric acid there is lots of water (and little to no ethanol). The equilibrium shifts to produce reactants.

As for the role of sulfuric acid, it acts as a proton source to enable the loss of leaving group in the forward direction and the electrophilic addition in the reverse direction.

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  • $\begingroup$ Does the hydration of ethylene happen in a significant amount? I'm just curious, but it is rather different from adding $\ce{HBr}$ across the double bond since the bromide can immediately attack the carbocation (in the transition state), whereas $\ce{HSO4-}$ is such a bad nucleophile that it will not add. But this would yield a primary substituted, very unstable and unfavorable, carbocation intermediate, right? Or does $\ce{HSO4-}$ add and is later attacked by $\ce{H2O}$ in an $\ce{S_{N}2}$ type mechanism? $\endgroup$
    – Jori
    Nov 23, 2014 at 12:40
  • $\begingroup$ In the hydration in dilute $\ce{H2SO4}$, water molecules far outnumber sulfuric acid molecules. Water is also a far better nucleophile than $\ce{HSO4-}$. $\endgroup$
    – Ben Norris
    Nov 23, 2014 at 17:04
  • $\begingroup$ Yes, but what is the mechanism precisely? $\endgroup$
    – Jori
    Nov 23, 2014 at 18:05
  • $\begingroup$ @Jori $\ce{H2SO4}$ protonates the double bond, and the carbocation is captured by $\ce{H2O}$. Subsequent deprotonation of the $\ce{OH2^{+}}$ group yields the alcohol. $\endgroup$
    – Jannis Andreska
    Dec 2, 2014 at 18:13
  • $\begingroup$ @JannisAndreska Thank-you. I seem to have completely forgotten that $\ce{H2O}$ is the solvent and therefore is very available (that confused me since primary carbocations normally do not form). $\endgroup$
    – Jori
    Dec 2, 2014 at 20:00

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