MagicJigPipe - 27-1-2010 at 16:42
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*
Thanks in advance.
[Edited on 1-28-2010 by MagicJigPipe]
Paddywhacker - 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]
Bikemaster - 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.
JohnWW - 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?
psychokinetic - 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]
Bikemaster - 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
MagicJigPipe - 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]
Bikemaster - 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]
VestriDeus - 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.