Help me Assign R/S for Galactose!

[justadream]

Can someone assign the R/S for Carbons 2-5?

TBR says that "all sugars drawn on the right in a Fischer projection have R stereochemistry".

Let's assume we oxidize galactose and convert both terminal carbons (C1 and C6) to COOH groups. Now we have a meso compound.

In meso compounds, there should be OPPOSITE R/S configurations as you move away from the plane of symmetry.

Does this occur here?

I believe it is:
C2: R
C3: S
C4: S
C5: R

But meso compounds should have opposite R/S:

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[mspeedwagon]

Sure, your diagram:
For C2 = OH is 1, C double bond O is 2, the chain is 3 and H is 4. But H is on the outward bow. So invert and it's R.
For C5 = OH is 1, CH2OH is 2, long chain is 3, H is 4. H s outward bow. So it's also R.

PS - For some reason, only the top half of your post was displayed when I typed my response. I agree with your designations and can't quite answer the latter part of your question, so will defer to someone else.

 

[techfan]

Galactose isn't meso C1 and C6 don't have the same things connected. If you added HNO3 then C1 would have COOH and C6 would have COOH and then C2 would be R, C3 would be S, C4 would be R (the =O on C6 gets higher priority than the -OH on C2) and C5 would be S
C2: R
C3: S
C4: R
C5: S IF you add HNO3 to make C1 and C6 equivalent

 

[justadream]

@techfan

Okay I see how you got that. You would agree that this oxidized galactose (changing C1 and C6 to COOH) IS meso, correct?

If so, this example illustrates what I'm trying to say about #47 in the other thread.

The line of symmetry in the oxidized galactose it between C3 and C4. Do you notice how the stereochemistry of the chiral centers are OPPOSITE of each other as you move away from the line of symmetry?

In #47 you have something that is NOT meso, yet the sterochemistry is opposite.

 

[techfan]

Yeah, but in the other thread the line of symmetry is through C4 not between 2 carbons so you don't know which way the -OH group is facing.