Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xiankai on July 27, 2006, 09:37:03 AM
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suppose i had 1-ispropyl-2,4-dimethylcyclohexane, of which to me there seems to be 3 chiral carbons located at 1st, 2nd, 4th positions.
CH3
|
|
|___
/ \
/ \___CH3
\ /
\___/
|
|
|
CH(CH3)2
how can i tell whether the other 2 alkyl groups connected are the same or not if these alkyl groups meet each other in the end (cyclic HC)?
???
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You want to go by what is on the carbon that immediately follows the stereo center you are scoping. OK so for the molecule you have there, let's look at the carbon with the methyl group. Now the carbon right above it, or to its immediate left is merely a CH2 group. But the carbon to its immediate right (or right below it) has an isoprpyl group on it and therefore takes priority over the other carbon. So the priority ranking is Isopropyl>Methyl>CH2>H. I'll leave it up to you to tell if it's R or S since I am not the greatest person to give you an answer (I can figure out priorities alright but suck at actually assigning R or S) ;D
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Well, I think that the structure of 1,4-dimethyl-propylocyclohexane is:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fcycloheksane_01.gif&hash=b7668847fd0440726adb6e3f01a1b477c539606e)
and the structure which post's author 'has drawn' is:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fcycloheksane_2.gif&hash=3f77f682b4e46cdf9e0480e39d01c84ee78e9d8b)
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thanks for pointing it out, im pretty new to this nomenclature :P
i've just started studying organic chemistry, so i have no idea what is R or S, or the priority of carbon groups :(
all i thought was that as long a carbon has 4 different groups attached to it, it becomes chiral. maybe u can explain this extension? :o
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ch-check it out
http://www.cem.msu.edu/~reusch/VirtTxtJml/sterism3.htm
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that seems very lengthy :-\, although my main goal was to find out to distinguish the chiral from the non-chiral carbons in cyclic HCs, since i do not know where the carbon chain terminates.
after reading part of the link though, i suppose we
2. If two substituents have the same immediate substituent atom,
evaluate atoms progressively further away from the chiral center until a difference is found.
..?
also i can somewhat confused by this example
3. If double or triple bonded groups are encountered as substituents, they are treated as an equivalent set of single-bonded atoms.
For example, C2H5– < CH2=CH– < HC?C–
to me the Mr of the groups is in decreasing order 29 < 27 < 25, which is not what i expected :o
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2. If two substituents have the same immediate substituent atom,
evaluate atoms progressively further away from the chiral center until a difference is found.
Well, it's very easy.
Look at the structure I have drawn below:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fstructure_03.gif&hash=af1e26e705878025ccef77b94117a5658fcdf795)
I hope you'll find chiral carbon atom. Now, look at substituents of this chiral carbon atom (compare atoms which are directly connected with stereogenic centre). As you can see chiral carbon is connected with hydrogen, bromine and two carbons. But - chiral carbon must have four different substituents and so that you must consider next pairs of atoms (in each substituent) to state if stereogenic centre has really four different substituents.
I just hope you've understood my english ;)
Bromine atom helps distinguish pair of similar substituents in the structure I have drawn.
3. If double or triple bonded groups are encountered as substituents, they are treated as an equivalent set of single-bonded atoms.
For example, C2H5– < CH2=CH– < HC?C–
It means that CH2=CH- we can show as:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fstructure_01.gif&hash=5786b0b0b117104ceb6a2a49447c8b3a8d860f9c)
and HC?C– as:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fstructure_02.gif&hash=0a1ddd5642e6f5ef41d9da3722d35c3d4ffffd83)
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so it can apply for cyclic HCs too?
thanks for the pictures, the wording was quite ambiguous :P
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thanks for the pictures, the wording was quite ambiguous
I can explain you everything but... in polish ;)
Nevermind.
Of course it can be apply also for cyclic hydrocarbons. Look at the picture:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fexample_01.gif&hash=82af95c6f866f68f9a0c5d0211f031b51e77c90d)
Stereogenic centre (on the 'top' of molecule) has four different substituents. Hydrogen and methyl group you can identify easily. And next two substituents are part of the ring. You should compare atoms in these substituents (which are part of the ring) to find difference between them. In the picture I've drawn arrows to show in which way you should compare particular atoms. Red arrows (and symbols) indicate where that difference is.
And next picture:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fexample_02.gif&hash=1fb7bfc60a602f9312647f59ac87f2f065b78059)
Now, tell if molecule above is chiral. That's your homework ;)
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hmm it doesnt seem chiral; even if i go in a loop and thus ending back where i started from originally, the atoms compared are still the same :D
now i have another question; when comparing atoms, what if one side branches? which of the branches do i pick from? :/
for example... take the molecule i was inspecting in the first place; 1-isopropyl-2,4-dimethylcyclohexane.
CH3
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|4__3
/ \
5/ \2__CH3
\ /
\6__/1
|
|
|
CH(CH3)2
in checking whether carbon #4 is chiral,
i start by comparing chains of carbon #5 and carbon #3.
next i move onto carbon #6 and carbon #2 respectively.
here i encounter a problem; should 4,3,2 carbon chain terminate at the methyl group or continue onto carbon #1?
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should 4,3,2 carbon chain terminate at the methyl group
Yes. Notice that carbon-2 is connected with methyl group and hydrogen and carbon-6 is connected with two hydrogens. It means that 4,3,2 carbon chain and 4,5,6 one are different.
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does that mean 4,3,2 has greater priority over 4,5,6?
when i try to visualise a tetrahedral structure for carbon #4, it gets quite impossible because the cyclic change is 'stuck' in a planar structure. does that it is achiral despite having 4 different groups?
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does that mean 4,3,2 has greater priority over 4,5,6?
Yes, it does.
Well, six-membered rings (and even five- and four-membered) aren't planar. Cyclohexane's ring usually occur in 'chair' conformation which can be drawn as below:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fcyclohexan_chair.gif&hash=5ad67776ce89bd56f66fb40cc204047e8d1d165c)
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hmm i see... how do manage to draw all these structures? im having a hard time with my ASCII art, lol
and next... when determining whether they are R or S, i have 2 questions:
let double lines signify going out of the plane of the paper while single lines signify going into the plane of paper
R1
/
/
/
R???C==\\
\ \\==R3
\
\
R2
I assume this is... (R)
------------------------
R1
/
/
/
R???C==\\
\ \\==R2
\
\
R3
and this is... (S)
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
is this the same as
R1
/
/
/
R???C==\\
\ \\==R2
\
\
R3
... (S) ...
------------------------
R1
\
\
\
//==C???R
R2=// /
/
/
R3
... (R) ...
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
are these 2 ways of expressing optical isomerism equivalent? because the second way is how my school goes about drawing it, and its obiviously different from the material online :o
lastly(hold your breath); is it that the shaded 'bonds' represent going out of the plane, while the striped 'bonds' represent going into the plane?
thanks for bearing with me :P
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Well, six-membered rings (and even five- and four-membered) aren't planar. Cyclohexane's ring usually occur in 'chair' conformation which can be drawn as below:
Something wrong with the picture - one of the axial hydrogen bonds have to be behind C-C bond.
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Something wrong with the picture - one of the axial hydrogen bonds have to be behind C-C bond.
Thanks. I've fixed it.
how do manage to draw all these structures?
Try to use some software - for example ChemSketch:
http://www.acdlabs.com/download/
then drawing molecules should be much more simpler ;)
R1
/
/
/
R???C==\\
\ \\==R3
\
\
R2
I assume this is... (R)
I'm afraid I don't understand your structure. You wrote that single lines signify bonds which lies in plane of paper and in your picture three bonds are signified single lines. This cannot be.
Stereogenic centre has tetrahedral geometry and structure of molecule (for example) should be:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Ftetrahedral_01.gif&hash=f95f13b769727dc2eca32ef8dd544f4f31ad1e2b)
So, it will be good idea to download some 'drawning molecules software'.
Shaded (bold) 'bonds' signify bonds which are closer to observator than other bonds. It means that shaded bonds are over plane of paper.
Striped bonds are under plane of paper.
Look at:
http://en.wikipedia.org/wiki/Fischer_projection
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everything fits now, i managed to distinguish the R and S enantiomers in the examples given in the link, yay!
now for the next step, what are diastereomers, meso-compounds and racemates? :)
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Diastereoisomers are stereoisomers which don't have an enantiomeric relationship. For example diastereoisomeric relationship have these two structure of 2-chloro-3-hydroksy-2,3-dimethylpentanoic acid:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fdiastereoisomers_01.gif&hash=cfd4331b97e438e835ba49a612386b993f28bd1b)
These molecules are of course stereoisomer but they aren't enantiomers because they don't have mirror-image relationship (enantiomer of first molecule would have S,R configuration on stereogenic centres, enantiomer of second molecule would have S,S configuration on stereogenic centres.
Diastereoisomerism can also occur in molecules which don't have chiral carbon atoms at all. For example - these compounds:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fdiastereoisomers_02.gif&hash=074a94e02c2a328bc142ebb36b2f9d3029e983b4)
are also diastereoisomers.
meso-compounds
Well, meso-compound is compound which has chiral carbon atoms but it itsef isn't chiral. For example:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fmeso_01.gif&hash=db3896eb95b6529f53f01e62ceea309f74e25926)
are meso-compounds. These two structures aren't enantiomers because it's possible to put first structure on second. You can easily recognise meso-compound because it has axis of symmetry:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fneecze.w.interia.pl%2Fmeso_02.gif&hash=3c807cd0fe6530652dc8b21e60d12d6e4196fccc)
racemates
Well, racemate is mixture of enantiomers of one compound in which there is 50% of first enantiomer an 50% of second enantiomer. Racemate has no optical activity.
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These two structures aren't enantiomers because it's possible to put first structure on second. You can easily recognise meso-compound because it has axis of symmetry:
Nice explanation by Neecze here. I just want to add that it is possible to have a meso compound that does not have a plane of symmetry, but instead has some other symmetry element that eliminates chirality. An example is an inversion center.