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November 01, 2025, 12:15:27 pm

Author Topic: Naming Esters  (Read 1212 times)  Share 

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chemkid_23

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Naming Esters
« on: May 09, 2011, 07:41:58 pm »
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Hi all,
I'm revising over esters and how to name them, got a bit stuck. Looking at one resource, it says to split the molecule between the carbon and single bond to oxygen; another resource says to split the bond after the oxygen with the single bond to R1. Hope you understand that description.

vea

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Re: Naming Esters
« Reply #1 on: May 09, 2011, 08:03:33 pm »
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I guess both descriptions are kind of right. You can use either method and then see which side the carbon-oxygen double bond is, this being the -oate part while the other side is the -yl part. Another way of looking at it is by the names of the reactants used to form each ester. The alkanol will get the -yl part while the alkanoic acid will get the -oate ending.

Hope I made sense.
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chemkid_23

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Re: Naming Esters
« Reply #2 on: May 09, 2011, 08:19:34 pm »
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Yep thanks for that. I know which part is called and all that, but was just worried if the actual splitting affected anything, but I see what you mean.

vea

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Re: Naming Esters
« Reply #3 on: May 09, 2011, 08:36:32 pm »
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The actual splitting does kind of matter if you wanted to get the two reactants. In this case, the first description is correct because in esterification, the alkanoic acid loses OH while the alkanol loses H. From this, you can see that you should separate it so that the two Os in the ester are not together.
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chemkid_23

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Re: Naming Esters
« Reply #4 on: May 09, 2011, 09:14:35 pm »
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So we DO split it at the C-O bond and not after that oxygen?
It should look like this O=C/-O rather than O=C-O/-

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Re: Naming Esters
« Reply #5 on: May 10, 2011, 02:27:53 am »
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I don't think it matters.

Considering the name follows 'alkyl carboxylate', so long as the carbons belong to the right group, you will get the right name.

As for the actual esterification reaction, it is true that the -COOH loses OH in the process, but this is not VCE knowledge.
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