Can you help me understand the difference between absolute and relative configuration?


1 Answer 1

  1. Let's say that we have synthesized a new compound, C(1)(2)(3)(4), where (1)-(4) represent 4 different groups attached to the central carbon, so the molecule is chiral.
  2. Let's say that we find a method to separate the two enantiomers present in the new compound and that we placed the enantiomers into separate test tubes.
  3. Next we might take each enantiomer by itself and place it in a polarimeter. We would find that the material in one test tube rotates light in the (+) direction and the material in the other test tube rotates light in an equal, but opposite direction (-).
  4. At this point, all we know is that we have two enantiomers that, as expected, rotate light in equal and opposite directions. We don't know which test tube contains the "R" enantiomer or which contains the "S" enantiomer. We just know their relative configuration with respect to each other, they are mirror images of each other.
  5. To find out which is "R" and which is "S" we would either need to 1) put a crystal of one of enantiomers in an X-ray diffractometer and get a three-dimensional picture of the structure of the molecule from which we could see whether it is the "R" or "S" enantiomer or 2) carry out a series of chemical transformations where the stereochemical consequences of each step are known (e.g. does each step proceed with inversion or retention of configuration at the chiral center) and convert it to a compound of known absolute configuration
  6. Let's say we pick one of our test tubes and are able to convert that enantiomer to a chiral compound C(A)(B)(C)(D) where we know that R-C(A)(B)(C)(D) rotates light in the (+) direction.
  7. We find that the C(A)(B)(C)(D) we created from our C(1)(2)(3)(4) enantiomer rotates light in the (-) direction; therefore we have produced the "S" enantiomer of C(A)(B)(C)(D).
  8. For simplicity, let's say that all of the reactions we used in our conversion of C(1)(2)(3)(4) into C(A)(B)(C)(D) are known to proceed with retention of configuration. Then we know that the test tube of C(1)(2)(3)(4) used in our reaction contains the S-C(1)(2)(3)(4) enantiomer. Now we know the absolute configuration of our sample.

EDIT: d, l and D,L

The lower case "d" and "l" can be used interchangeably with "+" and "-" respectively. The letters stand for dextrorotatory and levorotatory. They simply indicate how the plane of polarized light will be rotated when a sample of the material is placed in a polarimeter. So l-2-butanol is the same as (-)-2-butanol. Use of "d" and "l" is discouraged because it can create confusion with "D" and "L".

Upper case "D" and "L" are somewhat archaic. Way back when, the two enantiomers of glyceraldehyde were separated, purified and placed in a polarimeter. One enantiomer rotated the light in a dextrorotatory manner; this enantiomer was called D-glyceraldehyde. The other enantiomer rotated light in the levorotatory direction and was referred to as L-glyceraldehyde. If you take what ever chiral molecule you are studying and convert the chiral carbon to glyceraldehyde, and it rotates light in the "+" direction, then you've made D-glyceraldehyde, and if all of your reaction steps involved retention of configuration (see above) then you would have the "D" enantiomer of your material. The terms "D" and "L" relate whatever chiral center you are studying to the chiral carbon in glyceraldehyde.

  • $\begingroup$ You should mention that the (+) isomer can be labeled D (for dextrorotatory). $\endgroup$
    – LDC3
    Oct 25, 2014 at 15:05
  • $\begingroup$ @LDC3 I just wanted to keep it focused on the OP's specific question. $\endgroup$
    – ron
    Oct 25, 2014 at 15:41
  • $\begingroup$ But it does point out that the D and L labels are applied to relative enantiomers. In fact, they can sometimes be applied to a series base on one compound (all natural amino acids are labeled L even though some have dextrorotatory rotation). $\endgroup$
    – LDC3
    Oct 25, 2014 at 15:47
  • $\begingroup$ Thank you very much, that was a very useful answer. Just a question regarding the comments. Does D- (capital D) and capital L- mean dextrorotary and leverorotary? I thought it was small d and l. Would you be able to explain the difference? $\endgroup$
    – RobChem
    Oct 25, 2014 at 18:10

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