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Topic: Aldol Condensation product  (Read 11694 times)

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Offline MK23

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Aldol Condensation product
« on: April 26, 2007, 12:12:21 PM »
I accidently posted this in the general forum.... sorry.

Greetings, I was performorming an aldol condensation in Organic Lab, through IR and HNMR I have determined that my starting materials were p-tolualdehyde and Acetone.  I am fairly certain I have the proper structure but I am having some difficulty with the nomenclature, in other words I can't seem to name it(IUPAC rules). Any help would be greatly appreciated. Thanx.

Offline english

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Re: Aldol Condensation product
« Reply #1 on: April 26, 2007, 12:24:32 PM »
Draw your structure, then use the program's built-in tools menu to generate the compound's name.

http://www.acdlabs.com/download/

Offline MK23

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Re: Aldol Condensation product
« Reply #2 on: April 26, 2007, 12:54:38 PM »
My laptop is punking out on me...

It is essentially a dibenzylacetone with an extra methyl group in the para position of each ring.

Offline english

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Re: Aldol Condensation product
« Reply #3 on: April 26, 2007, 01:21:17 PM »
I don't know if I got this right...but this is what I'm thinking.

Offline MK23

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Re: Aldol Condensation product
« Reply #4 on: April 26, 2007, 02:11:12 PM »
bingo

Offline english

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Re: Aldol Condensation product
« Reply #5 on: April 26, 2007, 04:04:26 PM »
bingo

Well, my intuition tells me that you, nor I, no how the hell to name this thing given our current chemical knowledge.   :P

So when your laptop begins to work properly again, download the freeware here, draw the compound, go to

Tools ----> Generate Name for Structure.

The program should be able to name any compound so long as it's 50 atoms or less.  That is, it will give you the name for this compound.  It involves a bit of nomenclature I'm not currently familiar with.

Offline Custos

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Re: Aldol Condensation product
« Reply #6 on: April 30, 2007, 02:10:54 AM »
The longest alkyl chain is 5 so it's a pentane.
It has a ketone at carbon 3 so it's a pentan-3-one.
It has double bonds starting at carbon 1 and 4 so it's a pentan-1,4-diene-3-one.
It has 4-methylphenyl substituents at carbons 1 and 5 so it's:
1,5-bis(4-methylphenyl)pentan-1,4-diene-3-one.
If you want to get fancy, and you're sure the stereochemisty of the Double bonds are trans the name would be:
(1E,4E)-1,5-bis(4-methylphenyl)pentan-1,4-diene-3-one

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