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Author: Subject: Help With Organic Nomenclature
MagicJigPipe
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[*] posted on 27-1-2010 at 16:42
Help With Organic Nomenclature


I am trying to help someone with their organic chemistry. I am only in my first semester of organic (and it is the beginning at that). They need the structure of a compound:

3-(2-methylcyclobutyl)-4,6-diethyl-7-(l, I-dimethylethyl)decane

I don't know what the "capitol i" means, though. Also, how would I make this the levo- isomer?

I have this so far: *Damn, I messed up. Shift all substituents up one carbon*
Methcyclane.jpg - 7kB

Thanks in advance.

[Edited on 1-28-2010 by MagicJigPipe]




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Paddywhacker
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[*] posted on 27-1-2010 at 19:31


I would call the compound drawn 2-(2-methylcyclobutyl)-3,5-diethyl-6-t-butyldecane and expect to be perfectly understood, even if it is not the latest fashion in nomenclature.

So try drawing again. 1,1-dimethylethyl is tha same as t-butyl, but looks uglier, to me.

[Edited on 28-1-2010 by Paddywhacker]
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[*] posted on 27-1-2010 at 20:23


The name of your compound can vary between some names, all will depend on the convention that you use. For exemple, at my school we are oblige to use the IUPAC name convention.

This is the step to name a molecule by this convention.

1. Find the longest chain of carbone.
2. Find the side to start giving the number of the ramification.
(ex: Smaller number for OH group, for the double and triple liasion or try the have the smallest number in total) ( they have also other things, but you can start with this)
3.Right the molecule name in alphabetical order for the ramification and not in the order that they come on the molecule.
(exception: For the alphabetical order, ''di'' and ''tri'' don't play a role, but we have to consider ''cyclo'' and ''tert-''. I don't know why, but someone have decide this...)


The bad thing is that I never name a ramification with a ramification (2-methylcyclobutyl). But let say that it really have to be place in alphabetical order, it will give.

3,5-diethyl-2-(2-methylcyclobutyl)-6-tert-butyldecane

The thing that I'm not sure is that complexe ramification go always on the begining of the name...

Did you seem the possible isomer???

isomer : Cis)trans or E)Z
optical isomer: R)S or l)d

Because your drawn have the posibility to be trans or cis (2-methylcyclobutyl), but because the drawn is not in three dimentions, you can't tell this.
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JohnWW
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[*] posted on 27-1-2010 at 23:15


That compound appears to have 6 optically active (stereogenic) carbons. Have any of the 64 possible stereoisomers been actually synthesized and characterized?
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psychokinetic
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[*] posted on 28-1-2010 at 00:31


It's starting to look like one of those theoretical molecules that some folk make because they want to call it penisane as it looks like a penis.

EDIT: I am the champion of run-on sentences.

[Edited on 28-1-2010 by psychokinetic]




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[*] posted on 28-1-2010 at 07:28


As JohnWW said, they have 6 carbon with 4 different ramification. We call this chirality or ''asymmetric carbon atom''.

The probleme is that you can't tell the final form of the molecule because it is in two dimentions, so you can't tell more on this compound.

On the other drawn, know you can tell the optical isomer (maybe you will only see this at the end of the semester...)

(S)-1-bromopropan-1-ol




Sans titre.bmp - 440kBSans titre1.bmp - 246kB
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MagicJigPipe
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[*] posted on 28-1-2010 at 17:26


I think you guys may not fully understand what I am saying. I was given the name of the compound and I needed to draw the structure. That picture was my primary attempt. And I don't know what the capitol i means in the nomenclature. Was that a typo and the L and i were meant as 1s? Maybe that's why it confused me.

Thanks for all the help. I really appreciate it.

P.S. In short, the information given to me was the name of the compound and I was trying to come up with the bond-line structure.

[Edited on 1-29-2010 by MagicJigPipe]




"There must be no barriers to freedom of inquiry ... There is no place for dogma in science. The scientist is free, and must be free to ask any question, to doubt any assertion, to seek for any evidence, to correct any errors. ... We know that the only way to avoid error is to detect it and that the only way to detect it is to be free to inquire. And we know that as long as men are free to ask what they must, free to say what they think, free to think what they will, freedom can never be lost, and science can never regress." -J. Robert Oppenheimer
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[*] posted on 29-1-2010 at 20:39


Sorry, I read your post too fast.
I talk with my teacher and the ''I'' in the name is a mistake. I am pready sure that it should be the number one. She also told me that the ''complexe ramification'' always go at the beginning of the molecule.

Your other question was how can this molecule be levo- isomere?
The bad thing is that l/d is not really in relation with R/S so even if you drawn the molecule in 3 dimention, you can't say if is levo or dextro. The only way to know is by making the light deviation test with differente isomer to see the light deviation. Levo = left deviation of the light.

I espect that it help.




[Edited on 30-1-2010 by Bikemaster]



[Edited on 30-1-2010 by Bikemaster]

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[*] posted on 4-2-2010 at 20:26


I got:

3,5-diethyl-2-(2-methylcyclobutyl)-6-tert-butyldecane

Your teacher's a [insert derogatory name here]

And if he/she doesnt take the "tert" part, then he/she should get out of teaching.
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