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Topic: Alkane Nomenclature  (Read 2384 times)

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jacksonpeebles

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Alkane Nomenclature
« on: May 14, 2012, 01:19:41 PM »
Can anyone please kindly help me on how to go about naming this molecule (step-by-step, please - if I just wanted the name I could probably Google the condensed structure!)?



So far, I know that the longest chain is octane and that we have a 3-methyl branch and some sort of branching on 4 and 5, I know these are alphabetized, but I do not know how to name these branches (would the two branches on 5 become 5-diethyl?  I have no clue for the 4 branch).

Offline Kate

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Re: Alkane Nomenclature
« Reply #1 on: May 14, 2012, 06:55:10 PM »
You have to give the substituents the smallest number possible.

First you have to identify the longest chain. And then you start numbering the carbons of that chain.

You have 1 methyl group on C3, on C4 you have a (1-methylpropyl) group and on C5 you have an isopropyl group.

That weird group on C4, since it's has a branch, which is a methyl group, you have to give its position in relation to the propyl group. That position is 1.

I think IUPAC accepts the isopropyl name; propyl because there's 3 carbons and iso because the group is symmetrical.

The final name has to be in alphabetical order.
« Last Edit: May 14, 2012, 07:07:05 PM by Kate »

Offline discodermolide

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Re: Alkane Nomenclature
« Reply #2 on: May 14, 2012, 08:57:20 PM »
Can anyone please kindly help me on how to go about naming this molecule (step-by-step, please - if I just wanted the name I could probably Google the condensed structure!)?



So far, I know that the longest chain is octane and that we have a 3-methyl branch and some sort of branching on 4 and 5, I know these are alphabetized, but I do not know how to name these branches (would the two branches on 5 become 5-diethyl?  I have no clue for the 4 branch).


Here is a picture and the name courtesy of ChemDraw. The longest chain is shown in red.  Perhaps it will help you figure it out.
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jacksonpeebles

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Re: Alkane Nomenclature
« Reply #3 on: May 14, 2012, 09:06:30 PM »
Thank you!

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