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Thread: Alcohol

  1. #1 Alcohol 
    Forum Masters Degree thyristor's Avatar
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    Feb 2008
    What is the reason a carbon atom cannot bind more than one hydroxyl group? It seems strange to me. The only thing I can think of is that the hydroxyl groups repel each other too much, but it seems unlikely.

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  3. #2  
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    Sep 2008
    They can as far as I know, or at least I don't know any rules that preclude this. That said I can't remember ever seeing such a molecule, so can only assume it must be extremely unlikely rather than impossible. I suppose the 2nd -OH would rapidly steal a H on the carbon or another carbon, or failing that on the other alcohol to give a water and a double bond, in the last case giving a ketone.

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  4. #3  
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    Feb 2009
    Farm boy, I think your second answer is most likely correct.

    When making a hemiacetal or hemiketal, the rxns are acid catalysed. Traditionally, you would synthesise the acetal or ketal from the carbonyl species in the presence of an alcohol and an acid catalyst to give R2C(OH)(OR). The reaction is reversable, existing in equilibrium between the carbonyl and the ketal/acetal product. My point is that if the R2C(OH)(OR) can decompose to regenerate the carbonyl, the 1,1-bishydroxy species most likely would aswell. There is an entropic argument which supports this idea. Basically 2 species is more entropically favourable than 1, and water is an entropically viable leaving group.

    I suspect then, that you could actually synthesise a 1,1-bishydroxy species, but the conditions for this would be very difficult to determine. The species decmposition could be catalysed by acid or base. The species would most likely be water intollerable, meaning it would have to be kept under an inert nitrogen environment, and most probably at reduced temperature.
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  5. #4  
    gc is offline
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    Sep 2009
    It can, but as farmboy mentioned the ketone is (usually) more stable because of steric hindrance.
    If you dissolve acetone in water, the ketone will be in equilibrium with a small amount (less than 1%) of the corresponding diol.
    For formaldehyde, the diol form is favoured. This is because it does not have methyl groups to repel eachother like acetone.
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