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Hydration of an alkene

  1. Oct 1, 2015 #1
    One of the first addition reactions we learn in a basic organic chemistry class is the hydration of an alkene, that is breaking a double bond by adding water (the H and the OH specifically) across the double bond.

    The only requirement is that the solution be acidic.

    I'm wondering if the reaction still would take place even in pure water but just much slower.

    obviously, adding Sulfuric acid increases the presence of hydronium in the solution, and the hydronium will then act as the electrophile, for the reaction. But even in pure water, hydronium is still present, so the pi bond should still break.... it just happens slower....

    Am I right or is there a completely different reason that we need acid to be present.
     
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  3. Oct 1, 2015 #2

    Bystander

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    Yup.
     
  4. Oct 1, 2015 #3
    Excellent, I wish my book would just say that instead of "lying" for my benefit.

    I guess we would call the acid a catalyst in this case.

    Now I'm wondering where the hydroxide is coming from. The H+ is readily available to leave the hydronium, now we have a section of a carbocation.... is that going to "force" lone water to split up to donate a hydroxide to the site?
     
  5. Oct 1, 2015 #4

    epenguin

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    I have seen books say specifically weak acids so maybe it is general acid catalysis as illustrated in link below rather than hydronium ion - I wouldn't like to rule. To the carbocation a bond is then formed from oxygen's lone pair of a water molecule and the proton given back to the solution.

    You can find many illustrations googling but e.g. http://www.chemguide.co.uk/physical/catalysis/hydrate.html#top

    If you find a certain amount of repetition helpful :oldsmile:, it is done in detail here https://www.khanacademy.org/science.../v/addition-of-water-acid-catalyzed-mechanism
     
    Last edited: Oct 2, 2015
  6. Oct 2, 2015 #5

    epenguin

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    Didactic detail plus OPEN QUESTION OF CHEMICAL PRINCIPLE

    Further relevant points to know:

    • This hydration is one of the biggest industrial chemical processes making most of the industrial (around 100%) ethanol. (I thought therefore it must be also for other alcohols like propanol, but looking into that - it gets complicated.)
    • For real chemistry here you need to understand not just bond rearrangements and mechanism but the physicochemical influences, of which here http://www.chemguide.co.uk/physical/equilibria/ethanol.html#top . (I notice it says that without catalyst the reaction is far too slow).
    • A favourite exam and test question is about what way round do OH and H add, when there is a choice - Markonikov's rule. Favourite I guest because the chemists are glad to find at least one example of predictivity that works. (Except when it doesn't or they don't want it to - then there is the free radical reaction.)
    • You need to know about the reverse reaction, dehydration of ethanol, to produce ethylene and also the halfway dehydration that produces diethyl ether (requiring milder conditions). Both are favoured by the physics, product being gas in the first, and very easily distillable volatile liquid (b.p. about 35°) in the second. The first of these would be pointless industrially but the second is important and largescale. Anyway, it's a reversible reaction, but I have not been able to find what the equilibrium constant is, would be glad to hear. Here more on etherification
      http://www.masterorganicchemistry.com/2014/11/14/ether-synthesis-via-alcohols-and-acid/
      http://www.wikipremed.com/03_organicmechanisms.php?mch_code=030207_070
    • I went to town on these things here for two reasons. One is the dehydration is the reaction I remembered most from school: I am 75% convinced I actually made some ether this way, but it is generally not considered for schools because of its unsafety.
    • It occurred to me that just as ether is made by halfway dehydration of ethanol, so it should be possible to make it from halfway hydration of ethylene. 2C2H2 + H2O → (H3C)2O . Looking it up, this too is done industrially. I didn't know that, maybe not a lot of people know that. :oldsmile: https://en.m.wikipedia.org/wiki/Diethyl_ether
    • The second reason interesting me is theoretical. School, and I see they are still teaching it this way, fixed in my mind sulphuric acid had these three big types of reaction: oxidation which I leave aside, and then acidity and then dehydration. The dehydration had nice lab. illustrations, such as turning sugar into carbon, and if I remember it got so hot in the process it caught fire. As for acidity, it is the most acid substance in the ordinary laboratory. Actually only for its first dissociation, but the pK is given as around -3. But that tells you about 55M water, and I reckon if the water is about 0.5 M practically all the water molecules should be protonated. It even protonates itself to form H3SO4+ !
    So we were taught, and I still am seeing now looking at the educational things on the web that sulphuric acid has these two properties - acidity and water abstraction.

    But I ask - aren't these two just the same thing?!

    If you consider e.g. ether formation you have the equilibria

    2,EtOH ⇔ Et-O-Et + H2O

    H2O. +. H2SO4. ⇔ H3+O + HSO4-

    So the sulfuric acid is (as well as catalyzing) dragging the equilibrium towards ether formation, removing the H2O co-product by the protonation reaction?

    Is teaching missing a point here? Or am I?
     
    Last edited: Oct 3, 2015
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