Orgo Q

[pbehzad]

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Ok well my first organic test is thursday, and i am ferveshily tryin to understand this useless crap. Could someone explain R/S priority in stereochemistry. If you have the lower priority at the wedge or dash lines do you reverse the order (make it the opposite direction)? How do you determine if it goes clockwise or counterclockwise based on the position of the atom w/the lowest priority? Also when you have enantiomers and you are looking at the mirror image, the things that are at the wedge and dash line swap places when trying to superimpose on the original structure, correct? thanks

 

[Doctora Foxy]

whoa, don't you have a textbook that would explain this more clearly than a response on here?

I think you have to prioritize the substituents and then deal with the rest.

 

[Dr. Dodger Dog]

Originally posted by pbehzad
Ok well my first organic test is thursday, and i am ferveshily tryin to understand this useless crap. Could someone explain R/S priority in stereochemistry. If you have the lower priority at the wedge or dash lines do you reverse the order (make it the opposite direction)? How do you determine if it goes clockwise or counterclockwise based on the position of the atom w/the lowest priority? Also when you have enantiomers and you are looking at the mirror image, the things that are at the wedge and dash line swap places when trying to superimpose on the original structure, correct? thanks

So if you are given a question, you first prioritize all the groups. Next look at the position of the lowest priority atom. If it is at the dashed wedge, then you are good to go. If it is at the solid wedge, just remember to change your designation at the end to the opposite. If your lowest priority group is at a solid line (in the plane of the page), swap it with the group having the dashed wedge and remember to change the designation at the end to the opposite. Remember, swapping the position of the two groups will change R to S and S to R.

Then, around the central carbon stereocenter, draw a curved arrow, either clockwise or counterclockwise, starting with the highest priority, going to the second highest priority, and ending ur arrow at the third highest priority. This will give you the "as is" designation of R or S. If the molecule has the lowest priority at the dashed wedge, the "as is" designation is the correct one. Now, if the lowest priority group is at the solid wedge, reverse the "as is" to the opposite. If you were forced to swap two groups to get the lowest priority at the dashed wedge, reverse the "as is" to the opposite as well.

I hope this helps... it's really hard to explain through a post. I would recommend using models. It really helped me a lot.

 

[Tweetie_bird]

Here is a sure-fire way that worked for me. Easiest thing to do is----

1. First, make the molecule flat. So it should look like a + sign. This is your "ORIGINAL MOLECULE."

2. Now, you look at each of the groups on the + sign. Number all groups according to their priority. Now, here's the catch-- The lowest group SHOULD be on the vertical axis. If it's not, you swap one of the groups with the lowest to put it on the Vertical axis. This is your "CURRENT MOLECULE."

3. Now, with the lowest priority group on the vertical axis, you follow each sequential group. I.e. You look at 1, then at 2, then at 3. Does it make a clockwise circle or counterclockwise circle?

3. If it's clockwise, then you label the CURRENT molecule as "R."

4. And this is the most crucial part. Since you had to swap two groups to put the lowest priority on the vertical axis, your ORIGINAL MOLECULE is actually the opposite of the CURRENT MOLECULE. Therefore, if your CURRENT MOLECULE was an "R"....then your "ORIGINAL MOLECULE" was actually an "S."

Why does this happen? because you are actually changing the chirality/positioning of the molecule's substituents by swapping two of the groups. Therefore, after swapping, whatever answwer you got . . . . the original molecule is actually the OPPOSITE of the answer you got. Make sense?

Now, if your ORIGINAL MOLECULE already had the lowest priority on the vertical axis, you needn't swap anything. Just number the groups, and go from 1 to 2 to 3 and if it's clockwise, it's R. if it's coutner-clockwise, it's S.

This always worked for me. Hope it does for you too. 🙂

 

[pbehzad]

thanks for all your help. your responses made it alot clearer.

 

[CD]

The way I keep R and S strait is by locating the lowest priority and ignoring it. Then using the other three, note which direction they proceed. Treating the three like a steering wheel if I turn to the (R)ight then it's an R and if it's to the left it's an S.

If the lowest priority (for example H) is away from you (dashed line) and you determine the configuration to be R then you are good to go and R it is. If however, the lowest priority is toward you (wedged line) and you determine the configurationto be R then the designation is actually S. (is this what you were asking?)

 

[pbehzad]

Originally posted by Dr. Dodger Dog
If your lowest priority group is at a solid line (in the plane of the page), swap it with the group having the dashed wedge and remember to change the designation at the end to the opposite.

So you dont rotate the atoms so that the lowest priority ends up on the wedge, you just swap it even if the lowest priority is in the plane and has to skip over the wedged atom? For example, In the original molecule you put the substituent that was in the dashes on the plane where the lowest priority atom was and put the lowest priority atom on the dashes and leave the other 2 substituents in place regardless if you have to skip over them to swap the lowest substituent to get to the dashes? i think its starting to click i just want to make sure i understand it fully. thanks.

 

[Dr. Dodger Dog]

Originally posted by pbehzad
So you dont rotate the atoms so that the lowest priority ends up on the wedge, you just swap it even if the lowest priority is in the plane and has to skip over the wedged atom? For example, In the original molecule you put the substituent that was in the dashes on the plane where the lowest priority atom was and put the lowest priority atom on the dashes and leave the other 2 substituents in place regardless if you have to skip over them to swap the lowest substituent to get to the dashes? i think its starting to click i just want to make sure i understand it fully. thanks.

yeah... that's correct.. do lots of examples and make sure it works every time

 

[Sonya]

Higher molecular weight --> higher priority.

two groups have equal MW --> look at MW of thing bondend to the atoms of tied MW.

Now, imagine the molecule in 3D. Put the H (or gro up of smallest priority) in the back. Now, go from highest to lowest priorty among the three with highest priority.

If you are spinning right --> R

Spinng Left -->S

Often it is hard to imagine the lowest priority molecule in the back, if it is drawn in the front. Thus, if the lowest priority is in the front then:

Spinng right (highest to lowest priority) --> S
Spinning Left --> R

How do I remember R = right, and S equals left?

Well, it's my own way, might confuse you more.
Imagine the Letter capital "R" now start at the top of the letter, and follow the curvature. You go to the right.

Imagine the capital letter "S". STart at the top and follow curvature, you go the the left.
(yes, i made up this memorization technique on my own...so it's strange)



"Also when you have enantiomers and you are looking at the mirror image, the things that are at the wedge and dash line swap places when trying to superimpose on the original structure, correct? "

yes, that's c orrect.

Sonya