R&S Help. (Mistake in TBR) Orgo 1, Chapter 3, Ex. 3.3

[drechie]

Can't figure out how TBR "prioritized" the groups off of "carbon 3" in example 3.3 (Orgo 1, Chapter 3, TBR).
Carbon 3 is the second carbon that is considered in the attached image. Please either inform me, or tell me their mistake.

On how I think the groups should be prioritized on carbon 3, I think they got #1 right (has a carbonyl down the branch), but i think 2 and 3 should be switched.

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[JMMTB]

Remember with prioritizing you look at the immediate atom to which the carbon is bound not groups.
So for carbon 3 we have the hydrogen going back getting priority 4. Then the other three atoms are all carbons (equal priority). To distinguish these carbons we need to see what they are bound to. Carbon 2 is bound to 1 Oxygen, 2 Carbons, and 1 hydrogen. Carbon 4 is bound to 2 carbons, 1 oxygen, and a hydrogen. And the ethyoxy carbon is bound to 1 oxygen, 1 carbon, and 2 hydrogens. This ethyoxy carbon is now our priority group number 3 because it is bound to only one carbon whereas the other 2 carbons are bound to 2 (1 oxygen equals out). We need to then walk out one more carbon to pick our priorities 1 and 2. So moving off of carbon 2 to carbon 1 we get a carbon bound to 2 oxygens, 1 carbon, and 1 hydrogen. On carbon 5 we have a carbon bound to 2 carbons, and 2 hydrogens. So carbon 1 wins out over carbon 5. Now walking back to carbon 3 we have our priorities 1 to the left, 2 to the right, 3 coming out of the plane, and 4 going back. This gives us a ccw circle and S.

An important note is that when we went to carbon 1 and 5 we didn't choose to go out to the alcohol groups because they were equivalent. If they were not equivalent we would need to take both branches into effect.

For carbon 4 remember that the hydrogen is coming out of the page and thus the stereochemistry that you assign needs to be inverted.

Good luck.

 

[BerkReviewTeach]

Hey drechie, as pointed out by JMMTB, you need to move a couple carbons away from the stereogenic center to get your priorities aligned. Think of the third carbon of an aldohexose. The carbon connected to the aldehyde carbon will be top priority in all linear monosaccharides, as long as all of the other groups are alcohols. There is no typo here.

Excellent explanation JMMTB! This is a common example in sugars, but for some reason when drawn in the stick-and-wedge format causes troubles for people.

 

[drechie]

Thank you so much... i follow your generously detailed description and used the same thought process as you mention. However, it seems to me that TBR didnt use the same prioritization as you list --- the exothy group (which they prioritize as #2) and carbon 4 which they prioritize as number 3. Just want to make sure I am on the same page. Thank you for both of your help!