Van't Hoff on Ladenburg

From WikidChem

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Slide 12

This slide is simply a picture of Kekulé with his students in Bonn in 1872, one of whom was J. H. van't Hoff who studied with Kekulé for just a few months. Notice that he began studying with Kekulé three years after Landenburg proposed the three different formulas of benzene in his Remarks on the Theory of Aromaticity.

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Slide 13

At the young age of 24, van't Hoff published a response to Landenburg's benzene formula in 1876 (only four years after studying with Kekulé). J. H. van't Hoff disagreed with Landenburg and upheld Kekulé's theory because both have problems with the number of isomers and Kekulé's theory was simpler. Note that van't Hoff sees the models as they'd look in 3D, which he stresses through the dashed line. Most chemists of the time did not think that the geometric properties of these 3D models affected the nature of the molecules (and thought it was philosophically misleading to do so - because they thought there could not possibly be any experimental evidence that could decide such a question); young van't Hoff disagreed with them. 1) "The 1,2; 5,6; and 3,4 derivatives are completely similar, although differing from 4,5; 2,3; and 6,1."

Look at the prism. The first three derivatives have substituted positions connected by red lines and then next three by blue lines. Keeping in mind that this prism represents a 3D shape, van't Hoff is saying that the two pairs are simlar becuase they are both the diagonals of a square of the face of a prism. (It may be easier to think of it as a cube-take a box. The red/blue lines represent the distance from one corner to the opposite corner of one face of a box. No, it's different with a cube, because you can rotate a cube by 90° about the normal to a square face and get an indistinguishably oriented cube, but you can't do this with a triangular prism - JMM) They are even the same length as the sides are rectangular and therefore have equal diagonals. The red and blue lines only differ in that they are joining the opposite corners. As mirror images, they are not superimposable.

If you are still confused-I have a problem with this sort of thing which is why I'm speaking so in depth about it, take a piece of paper. Fold it twice (2 perpendicular folds of any length). Fold it into itself so you only see one of the three areas. Going around the rectangle, put the letters a, b, c, and d each at a corner of the perimeter. (In order, so that a-b, b-c, c-d, and a-d are the four lines of the perimeter.) Connect b-d with a line. That is "red." Connect a-c with a line. That's the "blue."

• Remarkably enough van't Hoff also gave instructions for paper folding to help chemists understand what he was talking about. Great minds… - JMM

2) Clearly in Kekulé's planar hexagon, the 1,3 product is different based on where A and B are located (which is double/single bonded to the carbon in between them). J. H. van't Hoff shows they are also different in Landenburg's theory because you cannot superimpose the two on each other.