[refid]

How many of the following aldaric acids are not optically active?



1-5 left to right

The answer is #3...when I checked the R,S configuration :



I Got both R's... isn't is it suppose to be R & S which cancels eachother out

&

but isnt molecule #1 achiral since there is two carboxylic acid groups?

am I misunderstanding something?

[Yggdrasil]

I get that #3 is R,S.  I don't know how you are figuring out the R/S designation, but the way I've been taught is to position the lowest priority group (the hydrogen) away from you, then going from highest priority to lowest priority, trace a circle.  A clockwise circle means R and a counterclockwise circle means S.  Using this system #3 is R,S.

I agree that #1 is achiral. #4 is the mirror image of #3 and therefore should also be meso (achiral).

[edit: I'm an idiot.  See other posts]

[AWK]

Think about meso-tartaric acid

[Dolphinsiu]

From wikipedia,

http://en.wikipedia.org/wiki/Meso_compound

2 and 5 are meso compound!

3 and 4 are enantiomers. Don't be misled!

I know why you will get 2R's wrongly in Fischer projection

You may regard it as

    l
-------
    l
-------
    l

But in actual case,

     l
-------- (C2)
     l
--- l --- (C1)
     l

Do u see the difference? Maybe Newman projection can make you know that the attached groups in C1 and C2 are antiperiplanar so that they can exert less steric effect, more chemically stable!

[refid]

still a little confused why i get two R's.... I see Dolphinsiu shows  fischer projection and Newman projection, still dont know

[Dolphinsiu]

Please refer to D-glucose in p.5 of the following pdf. document

http://www.sdsc.edu/~jss/Draw_Rep_Proj.pdf

The three attached groups (-H,-OH,-COOH) of C1 must be drawn as being in antiperiplanar position of that of C2. (Dihedral angle = 180 Degree)
 

[refid]

so bottom line my R,R is correct for #3

[Dolphinsiu]

Yes, you are right! Sorry! Actually my concept give to you is wrong! So trust yourself!

As you can refer to:
http://atom.chem.wwu.edu/pavia/chap5d.ppt

[Dan]

Hurrah for sugars! I find the easiest way to tell if a compound represented by a fisher projection is this:

Draw the mirror image fischer projection.

If the mirror image is related to the original by a 180 degree rotation in the plane of the page then it is the same compound, and thus achiral.

eg. 2 and 5 are mirror images, but can be related by a 180 degree rotation in the plane of the page => 2 and 5 are the same compound, and the substance is achiral.

The same is true for 1 (but it may be easier to see because the sp3 centre only has 3 different substituents, H, OH and 2 x COOH)

3 and 4 are mirror images, but cannot be related by a 180 degree rotation in the plane of the page => 3 and 4 are enantiomers, and thus chiral.

note: although not an issue here, beware when using this method when you have a terminal CH₂OH - this substituent can point either left or right in the Fischer projection and not change the identity of the sugar because the terminal C is not a chiral centre - it is drawn pointing right by convention.

[refid]

thanks you guys for clearing that up !