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A2 Chem Help: Deductive Chemistry - Acylation watch

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    Compunds P and Q are isomers of ethyl ethanoate.

    P reacts with sodium carbonate (aq) to produce a gas which turns limewater milky (ie produces Carbon Dioxide).

    Q reacts with sodium hydroxide (aq) to produce propan-2-ol.

    On reduction, P forms 2-methylpropan-1-ol.

    Identify P and Q

    Help!?
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    acid + carbonate ---> salt + water + carbon dioxide

    carboxylic acid + alcohol <---> ester + water
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    (Original post by 2strong)
    Compunds P and Q are isomers of ethyl ethanoate.

    P reacts with sodium carbonate (aq) to produce a gas which turns limewater milky (ie produces Carbon Dioxide).

    Q reacts with sodium hydroxide (aq) to produce propan-2-ol.

    On reduction, P forms 2-methylpropan-1-ol.

    Identify P and Q

    Help!?
    Like EVS hinted out, P must contain a -CO2H group, that is how it liberated the CO2 gas when reacting with the carbonate.
    so P is something-CO2H

    What reacts with NaOH to form alcohol back? Actually ester could react with strong base to form the alcohol and an acid back(but with sodium in place of hydrogen meaning you have -CO2Na instead)

    So, since you produce propan-2-ol, your ester should be formed from propan-2-ol and methanoic acid in order to get the overall 4 carbon in the ester(hence isomer of ethyl ethanoate).

    You also deduce that P get reduced to alcohol, it is a primary alcohol, hence when oxidized, it turns to acid. Hence from the structure of the alcohol, and the fact that a carboxylic acid is present, and your molecular formula is C4H8O2, you should be able to work out structure of P.

    As for Q, it should be an ester as it reacts with the base to form the alcohol back. From the alcohol structure which you know now, you also know it must react with methanoic acid(but Na in place of the H at the -CO2H end), to form the ester and NaOH back. Q structure is therefore that of the ester.

    Hope that helps.
 
 
 
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