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In chemistry; chirality is generally defined in 2 ways.

  1. Lord Kelvin's definition: "I call any geometrical figure, or group of points, chiral, and say it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself" (source: 1. chirality.org , 2. Biochemistry book by Lubert Stryer.)

  2. "If a tetrahedral (sp3) carbon atom contains a different chemical group on each of its 4 hands, the central carbon atom is called chiral. Though there are some exceptions where we do not found any chiral carbon still some compounds can be chiral. Chiral molecules cannot be superposed with their 3-dimensional mirror image" (common classroom-definition for chirality).

But it seems the first definition is the basic definition; and the second definition seems to be a corollary or application of the first definition.

Now I want to know; is it possible to 'prove' or 'derive' the second definition (better to say second statement or second description) from the first (Lord Kelvin's) definition mathematically or geometrically or logically? and if possible then how to prove the second definition from the first definition?

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    $\begingroup$ Build two enantiomeric models out of matchsticks and chewing gum, and try to superimpose them. If that isn't enough of a proof to you, then what is? $\endgroup$ – Ivan Neretin Sep 1 at 12:44
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    $\begingroup$ The two definitions are not trying to define the same thing, to begin with. (1) refers to the chirality of a molecule, and (2) refers to chiral carbons, and is not directly related to the chirality of the entire molecule (that's why there is a disclaimer saying that there are some exceptions). So you need to be clear about what you're asking: you cannot prove the definition of A using the definition of B. $\endgroup$ – orthocresol Sep 1 at 12:45
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    $\begingroup$ If you want mathematic proof, take a seat and ingest two semesters of linear algebra. The keyword is point groups. Unfortunately most chemistry faculties dont care. ;) $\endgroup$ – Karl Sep 1 at 14:20
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    $\begingroup$ @AlwaysConfused — Geometrically speaking, the two rules are equivalent. It is demonstrable that if a point (i.e. the (a)chiral center) two identical groups, there always is a type of symetry (point, plane or rotation symetry). As a consequence, any chiral tetrahedral point MUST have 4 different groups. $\endgroup$ – SteffX Sep 1 at 15:34
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    $\begingroup$ Exhaustive trial and testing is a totally valid mathematical method to prove sth. The possibilities are very limited, after all. $\endgroup$ – Karl Sep 1 at 16:15
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Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.

So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.

Chemistry

In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.

Geometry

Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.

Showing that the carbon is chiral

If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.

3D illustration

see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')

Is this trivial?

No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.

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  • $\begingroup$ Another way to show this is to first define improper torsion angles, and then show that they switch signs upon reflection in a mirror. $\endgroup$ – Karsten Theis Sep 4 at 19:59
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I don't think there is any mathematical way to prove it.It's only experimental. See Chirality is defined for the compounds whose mirror image doesn't superimpose on each other. enter image description here

enter image description here

-source : LG Wade Orgainc chemistry. if you want more explanation then you can take help of LG Wade Or Solomons Organic Chemistry(both are available free on net)

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    $\begingroup$ Yes empirical observations demonstrate it, but the proof is nothing to do with that: it is pure geometry. A rigid tetrahedron with 4 differently labelled vertices cannot be the same as its mirror image. $\endgroup$ – matt_black Sep 2 at 15:45

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