Esterification

how does ethanoic acid react with propan-2-ol? ive seen on another thread the reaction will just occur on the 2nd carbon group but how would that be drawn

how does ethanoic acid react with propan-2-ol? ive seen on another thread the reaction will just occur on the 2nd carbon group but how would that be drawn

Have a look here: https://chemguide.co.uk/organicprops/alcohols/esterification.html (says pretty much the same as https://chemguide.co.uk/organicprops/acids/esterification.html)
And mechanism: https://chemguide.co.uk/physical/catalysis/esterify.html – might help with drawing it out/understanding, but it might just complicate things further

As it's a secondary alcohol, the carbon to which the -OH group is attached will be the one involved

i understand... but how would the overall product be drawn? im slightly confused

If you had propanoic acid (CH3CH2COOH) reacting with methanol (CH3OH), you'd get CH3-CH2-C(=O)-O-CH3.
If you had propanoic acid (CH3CH2COOH) reacting with ethanol (CH3CH2OH), you'd get CH3-CH2-C(=O)-O-CH2-CH3.

The difference between methanol and ethanol is, essentially, replacing one of the hydrogens (attached to the C attached to the -OH) with -CH3. If you replace another of these hydrogens with another -CH3 group, you have propan-2-ol. So When you're drawing the product of this esterification, you'd just do the same thing.


Does that help at all? I think it's probably better if you work the answer out properly for yourself so you understand it better

ah okay, so the name of the product formed (of propan-2-ol and ethanoic acid) would it be called 2-methyl-ethyl-ethanoate ??