I tried writing an analogy to explain the difference between enantiomers and diastereomers. However, I'm not entirely sure if it's actually a correct representation of the differences. It might also be more confusing than it has to be:

To use an everyday object as an illustration of the difference, imagine two forks with four tines each. If fork 1 has its first and third tine longer than the other ones, and fork 2 has its second and fourth tine longer than the other ones, then the two forks would not match completely if you tried to stack them on top of each other. However, if you held fork 2 up to a mirror, then you would see it now has both its elongated tines in the same place as fork 1. These forks would be enantiomers of each other because they are the same thing, but mirror images of each other

If you changed fork 2 to have its first and second tine longer than the other ones, instead of the second and the fourth, then the two forks would instead be diastereomers. This is because now they both cannot match completely when stacked on top of each other, and holding either one up to a mirror will not make them look any more similar. This is exactly how diastereomers work too, because it only partially mirrors its counterpart. As you can see, fork 2 now has its first elongated tine actually match the first elongated tine of fork 1, while also retaining the effect that holding it up to a mirror will have its second elongated tine match that of fork 1. That is, in the mirror fork 2 will have its third tine elongated, just like fork 1 does when not viewed through a mirror. In other words, fork 2 has its first elongated tine match fork 1’s, but its second elongated tine is the mirror image of fork 1’s. This makes them diastereomers because they are the same thing, but neither structurally identical nor complete mirror images of each other.

So, is there anything off about my analogy, or does it correctly and comprehensibly explain the difference between the two?


1 Answer 1


Enantiomers are non-superposable mirror images (all the signs of one spatial coordinate are reversed). Diastereomers are formed by inverting some but not all stereocenters.

pp. 11-13. Excellent overall.

Have some fun with your prof! An acyclic, aliphatic, unstrained, tetrahedral carbon atom is $\ce{sp^3}$-hybridized. It is an uncharged, valence-saturated singlet. It bears four freely rotating, rigorously identical substituents, e.g., $\ce{CR4}$, with identical bond lengths to that central carbon. Said carbon is a chiral center. It has no ${S_n}$ improper axes of symmetry.

1) Show how this can be true.
2) Use any system of nomenclature to name left-handed and right-handed stereoisomers.

Acyclic, unstrained, etc., trigonal planar $\ce{sp^2}$-hybridized carbons ${R2>C=C<R2}$ are each identical stereocenters. Each carbon and the olefin overall have no ${S_n}$ improper axes of symmetry. The olefin is resolved chiral. (1) and (2) as above.

  • $\begingroup$ C'mon, folks. Show me how isolated CR_4 can be a chiral center. The answer is trivial. End of the week I'll show you how you already know the answer simply by reading the question. $\endgroup$
    – Uncle Al
    Commented Mar 2, 2014 at 16:11
  • $\begingroup$ The enantiomer of CR_4 is CS_4 for four idenntical substituents, Cahn-Ingold-Prelog notation. There is no way to name chirality. If you suggest by default, consider CR_aR_bR_cR_d in which the "R" priorities are arranged in S order. The olefin uses the same loophole. A mirror plane may not pass through a chiral center. The mirror image of R is S, and the reverse. $\endgroup$
    – Uncle Al
    Commented Mar 6, 2014 at 23:22
  • $\begingroup$ Go into HyperChem or whatever, make dodecahedrane. Put a tetrahedral sp^3 carbon at its center, make four internal bonds to the shell (remove those four external hydrogens). Minimize. Make the inversion. Superpose the two central carbons, and rotate to exactly align all four bonds. Enantiomers! Name the central carbons' chiralities. Good luck on that, for all four "groups" in each molecule are identical, point group T (not T_d or T_h). $\endgroup$
    – Uncle Al
    Commented Mar 6, 2014 at 23:28

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