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Topic: R S Configuration  (Read 12944 times)

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Offline Nitin_Naudiyal

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R S Configuration
« on: May 11, 2012, 05:48:12 PM »
In my College Text Book its written in R S Configuration when a multiple bond is present both the atoms attached to the double or triple bond are considered to be duplicated or triplicated.
-C=C- -CH-CH-
            |    |
           (C)  (C)

  C       C
  |        |
-C=O   -C-O
            |  |
          (O) (C)
here they have given few Examples?
Please Explain what does this mean?  ???
What are Phantom Atoms?


Offline sodium.dioxid

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Re: R S Configuration
« Reply #1 on: May 11, 2012, 06:10:14 PM »
A double bond to something means two single bonds to that same thing. That is what they are doing. This is to help prioritize the different connecting groups.

-C=C- <--- Here, the left C is double bonded to the right C. Likewise, right C is double bonded to the left C.

Therefore, you can break the double bond and give each C a new C. This way, they each still have two bonds to C.

Again, this isn't literally true. We know that double bonds and single bonds are two different things. What they are showing here only applies to R S configuration - to help you prioritize the different attached groups.

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #2 on: May 12, 2012, 05:55:27 PM »
Thank u for u r help

Its written that in RS configuration the Priority sequence is determined by consideration of duplicated or triplicated structure in which there are Phantom atoms
But what are Phantom Atoms.

if u cud give me an example or link to an example .


Offline sodium.dioxid

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Re: R S Configuration
« Reply #3 on: May 12, 2012, 08:06:27 PM »
Thank u for u r help

Its written that in RS configuration the Priority sequence is determined by consideration of duplicated or triplicated structure in which there are Phantom atoms
But what are Phantom Atoms.

if u cud give me an example or link to an example .

Basically, all it means is that a double bond has higher priority than a single bond. That's it!
But I can give an example:
O=CH3
If you redraw it using "phantom atom", you get: O-CH3-O
(the double bond was broken into two single bonds). The extra Oxygen that was added is called the phantom atom.

But, why go through all this trouble? They are making things complicated when they should just tell you that a double bond has higher priority than a single bond.

Offline UG

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Re: R S Configuration
« Reply #4 on: May 12, 2012, 08:44:59 PM »
But I can give an example:
O=CH3
If you redraw it using "phantom atom", you get: O-CH3-O
:o

Offline sodium.dioxid

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Re: R S Configuration
« Reply #5 on: May 12, 2012, 09:54:39 PM »
But I can give an example:
O=CH3
If you redraw it using "phantom atom", you get: O-CH3-O
:o

Oops. It should be
O=CH2-(rest of molecule) :rarrow: O-CH2-(rest of molecule)
                                       |   |
                                     (C)  (O)

Oxygen is bonded to Carbon twice. Carbon is bonded to Oxygen twice.

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #6 on: May 15, 2012, 10:08:18 AM »
Thank You UG and Sodium Dioxide.

if i want to arrange the following sets of compounds in the Order or Priority, from Highest to Lowest ,what points do i need to look into.
as in the atom or group to which the carbon atom is attached to an so on.

here are the compounds
-CH=CH2,-CH(CH3)2,-CH2-OH




What decides whether the Atom attached to the Chiral Carbon atom is of Highest Priority or Lowest ?


Offline fledarmus

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Re: R S Configuration
« Reply #7 on: May 15, 2012, 10:27:13 AM »
Atomic number. Start with the atoms attached directly to the stereocenter. The one with the highest atomic number has the highest priority, the one with the lowest atomic number has the lowest priority. If two atoms attached to the stereocenter have the same atomic number, look at the next atom in the chain - the atoms attached to them. If those are identical, then look at the next atom in the chain... and so forth. Be careful to work your way through one atom at a time - priority has nothing to do with which is the largest substituent, it only has to do with whether that next atom has a higher atomic number or not.

So for the three substituents you list (I am assuming these are three different substituents attached to the same stereocenter?)

What is the first atom?

For group one, it is C. For group two, it is C. For group three, it is C. All three have the same atomic number. So far, all three groups are identical.

What is the second atom?

For group one, it is C, a "phantom" C for the double bond, and H. For group two, it is C, C, and H. For group three, it is H, H, and O. Group three has the highest atomic number (the O), so it is the highest priority. Groups one and two are still identical at this point.

And so on. Work your way through each group, one atom at a time. It really helps to have blinders on so you don't see any further than the next atom when you are working out these priorities.

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #8 on: May 16, 2012, 09:30:47 AM »
CH3- Br + NaOH  :rarrow: CH3 - OH + NaBr

Its written that after the reaction the tetrahedral structure of the reaction gets invert due to the Backside attack by the Nucleophile OH-

Y does this inversion happens?

I have attached an image of the reaction.
PS: sry for the bad image quality
I have drawn that in Paint .

Offline sodium.dioxid

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Re: R S Configuration
« Reply #9 on: May 17, 2012, 01:49:40 AM »


Inversion means bending the opposite direction to the current bend
In SN2, inversion happens because there is a backside attack. All SN2 reactions are backside attacks. When the OH is coming closer and closer, the groups have to bend opposite to the way OH is coming. Just visualize the process using the image. Put the image into motion using your imagination. Are you having trouble applying this to R and S stereochemistry? Or are you having trouble understanding what inversion means?
« Last Edit: May 17, 2012, 02:18:14 AM by sodium.dioxid »

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #10 on: May 17, 2012, 02:46:47 AM »
Having trouble with understanding inversion.

i visualized  the atoms at the vertices's of the tetrahedron and still the postions remain the same .
can u please elaborate about the Bends ?

Offline Dan

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Re: R S Configuration
« Reply #11 on: May 17, 2012, 03:03:09 AM »
This might help, the nucleophile is cyanide in this example:

My research: Google Scholar and Researchgate

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #12 on: May 17, 2012, 05:09:09 AM »
Oh that was really a Great Example.
Very Helpful
Thank you..

Great Community.
 

Offline Nitin_Naudiyal

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Re: R S Configuration
« Reply #13 on: May 28, 2012, 03:41:14 AM »
The Order of priority of few elements is given Below

OCOCH3 > NH2 > COOCH3 > COOH > COCH3 > C6H6

Can any one explain me how these priority were assigned.

If we consider C6H6 ....out of the 6 Carbon Atoms which one should i choose ... and if i choose any one of the Six then i should add their Atomic Numbers of the neighbouring elements and then find out the Largest and that would have the highest Priority. I tried this method with all the atoms but couldnt understand .What approach should i take.

Please Explain me atleast 3 of the above mentioned compounds so that i can get an idea of how to go about it.           


Offline sjb

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Re: R S Configuration
« Reply #14 on: May 28, 2012, 04:47:42 AM »
See e.g. http://www.google.co.uk/search?q=cahn+ingold+prelog ?

In your instance O > N > C for the atoms directly bonded, so the O-ester > amino > C-ester, for instance

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