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    Half equation to show what happens to the oxidising agent in oxidation reactions. For example, pentan-3-ol into pentanoic acid would it be:

    4e- + C5H11OH + H2O ----> C5H10O2 + 4H+
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    (Original post by kiiten)
    Half equation to show what happens to the oxidising agent in oxidation reactions. For example, pentan-3-ol into pentanoic acid would it be:

    4e- + C5H11OH + H2O ----> C5H10O2 + 4H+
    First things first. Pentan-3-ol would oxidise to pentan-3-one not pentanoic acid. So what's the question now?
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    (Original post by TeachChemistry)
    First things first. Pentan-3-ol would oxidise to pentan-3-one not pentanoic acid. So what's the question now?
    In my textbook alcohols oxidise to aldehydes then carboxylic acid. In this example, the product wasn't distilled so wouldn't it be a carboxylic acid?
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    A primary alcohol can oxidise to an aldehyde and then further oxidise to a carboxylic acid.

    A secondary alcohol, such as pentan-3-ol, can oxidise to a ketone. When oxidising a secondary alcohol, one might as well just use reflux, to maximise the rate of the reaction, since there is no need to try to be selective over which product forms (as you would worry about with a primary alcohol).
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    Amazingly coherent, considering how drunk I was last night.

    I checked this morning, saw that I'd posted and my heart sank.
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    (Original post by Pigster)
    Amazingly coherent, considering how drunk I was last night.

    I checked this morning, saw that I'd posted and my heart sank.
    A drunken TSR post :rofl:

    Is it correct though :curious:
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    It seems correct, I know that a secondary alcohol can only oxidise once and a tertiary cannot be oxidised at all. Ill check if it does oxidise to a ketone
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    No need to check, I was quite correct.
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    (Original post by TeachChemistry)
    First things first. Pentan-3-ol would oxidise to pentan-3-one not pentanoic acid. So what's the question now?
    would it oxidise to C5H10O and is the structural formula CH3CH2OCH2CH3
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    For the equation I got:

    C5H11OH + 2e- ---------> C5H10O + 2H+
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    (Original post by kiiten)
    would it oxidise to C5H10O and is the structural formula CH3CH2OCH2CH3
    Yes. It would look like this (5 Carbons, 10 Hydrogens, 1 Oxygen at the 3rd carbon):
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    (Original post by Es0phagus)
    Yes. It would look like this (5 Carbons, 10 Hydrogens, 1 Oxygen at the 3rd carbon):
    Thanks - is the half equation correct - see earlier post
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    You lose Hydrogens when a molecule is oxidize, so I think so. I am not sure where those 2e- are from though, but the product side (on the right) looks right.
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    (Original post by Es0phagus)
    You lose Hydrogens when a molecule is oxidize, so I think so. I am not sure where those 2e- are from though, but the product side (on the right) looks right.
    The 2e- balance the 2+ charge from the H ions

    Posted from TSR Mobile
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    (Original post by kiiten)
    would it oxidise to C5H10O and is the structural formula CH3CH2OCH2CH3
    You wouldn't form diethylether!

    I suspect you meant CH3CH2COCH2CH3.
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    (Original post by kiiten)
    For the equation I got:

    C5H11OH + 2e- ---------> C5H10O + 2H+
    C5H11OH ---> C5H10O + 2H+ + 2e-
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    (Original post by TeachChemistry)
    C5H11OH ---> C5H10O + 2H+ + 2e-
    ??? - that doesn't make sense
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    (Original post by Pigster)
    You wouldn't form diethylether!

    I suspect you meant CH3CH2COCH2CH3.

    Yup, it was a typo
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    (Original post by kiiten)
    ??? - that doesn't make sense

    In what way does it not make sense? Both sides balanced for atoms AND charge.
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    (Original post by TeachChemistry)
    In what way does it not make sense? Both sides balanced for atoms AND charge.
    Sorry ive confused myself...
 
 
 
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