[recon87]

http://i234.photobucket.com/albums/ee171/revinnova/RandS.jpg

Ok, so I get R and S for simple Fischer Projections; however, I'm confused which part gets the priority for more complex Fischer Projections. Like the second carbon, is bonded with CHO, OH, and H, what would the last one be, C? Can some one explain this to me? thanks!

[azmanam]

Ranking priorities is determined by examining what is bonded to the carbon atom in question and looking for the first point of differentiation.

The first step is to examine the 4 atoms directly attached to the atom in question.  For carbon 5 (I'll show you how, and you can try carbon 2 on your own), the four atoms are H, C, C, and O.  For this first step, it doesn't matter what is attached to those four atoms, we just want to know what those 4 atoms are. 

Now that we know, we can start to rank them in order from 1 to 4: #1 is O, #2(tie) are C,C, #4 is H.  We know priority 1 will be oxygen, we know priority 4 will be H, but we have a tie for priorities 2 and 3.

When we have a tie, we look at the three other atoms attached to the atoms that were tied in step 1.  Since we're looking at carbon 5, the two atoms that tied in step one were carbon 4 and carbon 6.  Carbon 4 is attached to H, O, and C.  Carbon 6 is attached to H, H, and O. 

At this point, it's often useful to condense these sets of atoms into a 3-letter unit, ranked in order of priority.  For example, the 3 atoms attached to carbon 4 are OCH, and the 3 atoms attached to carbon 6 are OHH.  When viewed this way, I find it's easier to see the first point of differentiation.  The first letter of each of the 3-letter unit is O - no differentiation.  The second letter of carbon 4 is C, while the second letter of carbon 6 is H. 

Finally, we have a point of differentiation.  C is higher priority than H.  Thus, carbon 4 is higher priority than carbon 6.

To sum up, the 4 groups attached to carbon 5, in priority order, are O, C4(OCH), C6(OHH), and H.  Stereocenter of carbon 5 is R.

Can you do carbon 2?

[recon87]

Thank you for the thorough explanation. Yes, I think I can do carbon 2. Here is how I did it, OH gets the priority because it has the highest atomic number and H will have the lowest priority. Therefore, I will have two carbons that are tied. The part bonded to carbon 1 is (OH) and the part bonded to carbon 3 is (OCH), but the O's are tied, so I go to the next bond. I see that C has a higher atomic number than H, so it'll have higher priority. So the carbon three bond will have a higher priority. Priorities are OH, C3(OCH), C1(OH), and H. Therefore, it'll go clockwise, so it's 2R.  ^_^. 

[azmanam]

Ah, double bonds... Forgot about those. 

The beginning of your analysis is correct.

Here's what you do for double bonds: when making your 3-letter units, carbons double bonded to things get counted twice.  Therefore, carbon 1 is bonded to O, O(again), and H, and carbon 3 is O, C, H.  3-letter units would be C1-OOH, C3-OCH.  Carbon 1 is actually higher priority (O vs C).  You will always be able to make 3-letter units in the event of a tie.  You should never get 2-letter units.

(if it were a triple bond, it would get counted three times)

Carbon 2 is R, though.  so you came to the right answer, but through the wrong path - check your procedure for assigning it R or S after you assign the 4 priorities.

[recon87]

Ah, I see missed the C=O lol. Thanks!