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    so the equation is:

    CH3C0CH2 + 3[O] --------> CH3COOHCH2 + H20

    please could someone explain what is happening here?
    Thanks
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    You should try drawing it using full displayed formulae making sure that carbons have four bonds, oxygens two and hydrogens one.
    Propanal is CH3CH2CHO - what you drew kind of looked like propanone.
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    (Original post by Pigster)
    You should try drawing it using full displayed formulae making sure that carbons have four bonds, oxygens two and hydrogens one.
    Propanal is CH3CH2CHO - what you drew kind of looked like propanone.
    Ok ive drawn out the displayed formula, where do the 3 [o] go?

    Is the product CH3CH2COOH ?

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    (Original post by kiiten)
    Ok ive drawn out the displayed formula, where do the 3 [o] go?

    Is the product CH3CH2COOH ?

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    It should be, yes, propanoic acid. Although your equation looks kinda odd
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    (Original post by AngryRedhead)
    It should be, yes, propanoic acid. Although your equation looks kinda odd
    Yeah, the first post is wrong - do you mind if i ask you a few questions about the oxidation of alcohols?
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    (Original post by kiiten)
    Yeah, the first post is wrong - do you mind if i ask you a few questions about the oxidation of alcohols?
    Fire away
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    ANYONE CAN ANSWER (if you know).

    In a secondary alcohol like propan-2-ol you remove one H from the OH and another H from the same carbon. Is this the same for all primary and secondary alcohols?

    What do you do for primary to aldehyde. Then aldehyde to carboxylic. Or primary to carboxylic.

    What about secondary to ketone.

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    (Original post by kiiten)
    ANYONE CAN ANSWER (if you know).

    In a secondary alcohol like propan-2-ol you remove one H from the OH and another H from the same carbon. Is this the same for all primary and secondary alcohols?

    What do you do for primary to aldehyde. Then aldehyde to carboxylic. Or primary to carboxylic.

    What about secondary to ketone.

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    Primary and secondary alcohols are different things.

    Primary alcohol --> Aldehyde --> Carboxylic acid
    Secondary alcohol --> Ketone

    To oxidise a primary or secondary alcohol you use acidified potassium dichromate and the conditions are heat and reflux

    If you want to reduce a ketone or an aldehyde you use NaBH4 (sodium tetraborohydride)
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    (Original post by AngryRedhead)
    Primary and secondary alcohols are different things.

    Primary alcohol --> Aldehyde --> Carboxylic acid
    Secondary alcohol --> Ketone

    To oxidise a primary or secondary alcohol you use acidified potassium dichromate and the conditions are heat and reflux

    If you want to reduce a ketone or an aldehyde you use NaBH4 (sodium tetraborohydride)
    Sorry i mean in terms of atoms if that makes sense e.g. like do you remove a hydrogen etc.

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    (Original post by kiiten)
    Sorry i mean in terms of atoms if that makes sense e.g. like do you remove a hydrogen etc.

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    Yes I think that is correct
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    Pottasium dichromate reacts with sulfuric acid to form chromic acid and potassium sulphate.

    It is the chromic acid that reacts with the alcohol.

    An H from the alcohol is lost as well as a hydroxy group from the chromic acid (condensation) and an chromate eseter is formed.
    The lone pair of electrons on the Oxygen of the water moleule gives the O a negative polarity and its electrons are transferred to one of the delta positive Hydrogen bonded to the alpha Carbon.
    The bonding pair of electrons in the C-H bond move to the C-O causing a C=O to form and the O-Cr bonding electrons return to the Cr.

    Your oxidised compound is formed as well as protonated water and a hydrogen chromium trioxide ion.
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    Thanks guys - im still a little unsure though.

    Im going to look at my textbook again and ill post a specific example if i still dont understand it.

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