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chemical formula of an ester (1 Viewer)

smegthehead

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hey i was just wondering how do you work out the symbolic equation of an ester? I know how to draw them from their names and name them from the drawings, but when you write out the symbolic equation do you just write it from left to right like its any other compund (eg ethanol CH3CH2OH).

for example, would:

Methanol + Ethanoic acid <---> Methyl ethanoate + water


CH3OH + CH3COOH <---> CH3OCOCH3 + H2O

is this right?
 

BlackJack

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Hmm... when writing those condensed structureal formulas, all the atoms attached to C come after it, according to functional groups.

Don't be confused by this. It means that methyl ethanoate should actually be written CH3COOCH3, with the methyl off the right hand side.

This groups the two Os together with their carbon, which shows the actual chemistry as they act more like a group.
This is also more consistent with the alkanoic acids, COOH..
 

smegthehead

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so anoic acid written first, alcohol written second? and the oxygen link between the two is put in the middle?
 
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mojako

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Yes. alternatively, the -yl is written after COO...
-oic before COO.
 

Li0n

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[C(n-1) H(2n-1) COO] [CnH(2n+1)]
the [] indicate the 'first' and 'second' part of the name of the ester
 

CM_Tutor

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Originally posted by Li0n
[C(n-1) H(2n-1) COO] [CnH(2n+1)]
the [] indicate the 'first' and 'second' part of the name of the ester
Be careful writing general formulae, as this is a general formula for an alkyl alkanoate where the alkanol and alkanoic acid starting materials have the same number of carbons, which is a specific type of ester... The general alkyl alkanoate would be C<sub>n</sub>H<sub>2n+1</sub>C(=O)-O-C<sub>k</sub>H<sub>2k+1</sub>, where both n and k are integers, n => 0 and k => 1, and even this is a fairly restricted subset of the esters.

Esters only require the presence of the -C(=O)-O-C moiety, where the final C is not involved with certain other functional groups. As an example, aspirin is an ester, with the formula CH<sub>3</sub>COOC<sub>6</sub>H<sub>4</sub>COOH, but it is clearly not an alkyl alkanoate ester.
 
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