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Chemistry - Organic Chemistry.

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    Propanol reacts with hot mixture of sulphuric acid and potassium dichromate (VII). Write the molecular formula of the organic compound that you would expect to be formed. What type of organic compound is this product?

    So I know that, if I'm correct, when propanol react sulphuric acid the product is ethene and water (as sulphuric acid is a good dehydrating agent).

    And when propanol react with potassium dichromate (VII), the oxidizing agent, the product is carboxylic acid and water.

    But I have no clue on what the product will be if propanol were to react with the hot MIXTURE of sulphuric acid potassium dichromate (VII)

    Please help me understand this. Thanks in advance.
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    (Original post by Golden Boy)
    Propanol reacts with hot mixture of sulphuric acid and potassium dichromate (VII). Write the molecular formula of the organic compound that you would expect to be formed. What type of organic compound is this product?

    So I know that, if I'm correct, when propanol react sulphuric acid the product is ethene and water (as sulphuric acid is a good dehydrating agent).

    And when propanol react with potassium dichromate (VII), the oxidizing agent, the product is carboxylic acid and water.

    But I have no clue on what the product will be if propanol were to react with the hot MIXTURE of sulphuric acid potassium dichromate (VII)

    Please help me understand this. Thanks in advance.
    Potassium dichromate (VI) - (potassium dichromate (VII) doesn't exist ) - is a very strong oxidising agent. Hence, you'd look for the reaction of either forming an aldehyde or a carboxylic acid if you mean propan-1-ol, or forming a ketone if it's propan-2-ol.

    I presume you mean propan-1-ol, so I'd say forming the aldehyde (propanal), as there is no reflux involved, which is what you need to get the carboxylic acid (propanoic acid).

    Also, you would only get a dehydration reaction with concentrated H2SO4 (around 85%), whereas for the oxidation you need it to be dilute, as you only want the K2Cr2O7 to be acidified a bit.

    Lastly, dehydrating propan-1-ol (or propan-2-ol) would produce propene as the organic product; not ethene .
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    oops my mistake there. Yes it should be propene

    Anyway, my note stated clearly that it is Potassium Dichromate (VII). I believe you though, and I think this note is a bit incorrect. Thank you for your correction .

    So, since the there's no mention of concentrated on the Sulphuric acid, I guess no dehydration reaction take place.

    But what about the potassium dichromate? Isn't it supposed to be acidified? If the potassium dichromate is not acidified, will the oxidation of alcohol becomse successful?

    Why does it need to be acidified? Sorry to add up another question :/
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    (Original post by Golden Boy)
    oops my mistake there. Yes it should be propene

    Anyway, my note stated clearly that it is Potassium Dichromate (VII). I believe you though, and I think this note is a bit incorrect. Thank you for your correction .

    So, since the there's no mention of concentrated on the Sulphuric acid, I guess no dehydration reaction take place.

    But what about the potassium dichromate? Isn't it supposed to be acidified? If the potassium dichromate is not acidified, will the oxidation of alcohol becomse successful?

    Why does it need to be acidified? Sorry to add up another question :/
    Well, simply having the fact that you have Potassium dichromate (VI) there implies that oxidation takes place, as it is one of the strongest oxidising agents known to man.

    If the sulfuric acid is dilute, then you'll get the oxidation taking place. If the sulfuric acid is concentrated, you'll get a bit of the dehydration reaction taking place, but the oxidation reaction will still occur (albeit at a slower rate, I think), simply due to the strength of the oxidising agent present.

    Basically, there's one more step between aldehyde --> carboxylic acid that you don't need to know about at A level, which involves nucleophilic addition, for which you need H+ ions (protons). The dilute sulfuric acid provides these protons for the oxidation of primary alcohol --> carboxylic acid.

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