I not agree with this solution, for me the red bond formed must be a dotted line in the drawing, and the hydrogen atom musn't be represent with a dot. Am I right or not?

enter image description here

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    $\begingroup$ Assuming that the "red dot" on I-2 is used in the conventional sense and indicates that the hydrogen is projecting above the plane of the drawing and on the same face of the molecule as the ester group in I-2, then you are correct, I-2 is not the correct product for the disrotatory motion shown by the arrows. The disrotatory motion indicated by the arrows would place both the ester group and hydrogen below the plane of the screen. The other possible disrotatory motion (move both arrowheads to the other end of the curved lines) would produce I-2. $\endgroup$ – ron Nov 14 '15 at 17:08
  • $\begingroup$ @ron that exactly what i was thinking thank you :) $\endgroup$ – ParaH2 Nov 14 '15 at 17:12
  • $\begingroup$ @ron I’m curious, how are arrows ‘defined’ to point above or below? What would the other arrow look like? Could you add that as an answer? $\endgroup$ – Jan Nov 15 '15 at 13:59
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    $\begingroup$ @Jan To me, the arrows do not indicate "above" or "below", but rather indicate that, in the present case, the tops of the two terminal p-orbitals are rotating inwards. If the arrowheads were both moved to the other end of the curved lines, they would indicate the other possible disrotatory motion where the tops of both terminal p-orbitals are rotating outwards. The dotted lines indicate the pi bond axis, and usually the arrows are drawn closer to the dotted lines to better indicate what is really going on. $\endgroup$ – ron Nov 15 '15 at 14:44

You are both correct and incorrect. (And this entirely without superimposition of quantum states!)

You are taking an achiral molecule and performing a reaction in achiral media to form a chiral product. Symmetry dictates that the reaction must be racemic, so there must be equal amounts of upward-pointing and downward-pointing carboxylates.

However, the drawing is also incomplete. It does not make sense to discuss the reaction without labelling the entire stereochemistry at the newly formed bond. It is important to realise that it generates only two of the four possible stereoisomers: the two syn ones ((S,R) and (R,S)). You should add an upwards-pointing hydrogen to the right-hand carbon of the new bond.

Judging by your comments, you are only interested in the arrow direction and resulting movement. Well, the arrows are not clear. They are 2D on a 2D plane and trying to signify a 3D movement. You could grab the curved bit of the arrow like a handle and:

  • either pull it upwards — then you have a rotation which would move the carboxylate below your screen’s plane and create the downward-pointing product ((R) on the α-carboxylate carbon),

  • or push it downwards — then you have the rotation which would move the carboxylate above the screen’s plane and create the upward-pointing product ((S) on the α-carboxylate carbon).

So the arrows’ directions are arbitrary, we do not know if the are shaped below or above the plane.

  • $\begingroup$ I hope I am correct with R and S. If anyone could supply me with the IUPAC numbering of atoms, that would be great; I don’t have access to ChemDraw again until Monday ^^' $\endgroup$ – Jan Nov 14 '15 at 16:40
  • $\begingroup$ I didn't said that the solution precise the final solution is racemic what i only ask is about the movement of the arrows and the product we have :) $\endgroup$ – ParaH2 Nov 14 '15 at 16:40
  • $\begingroup$ @Shadock Well, racemic implies that both up and down are generated, so … ;) $\endgroup$ – Jan Nov 14 '15 at 16:41
  • $\begingroup$ Yes but to generate the other one the arrows have to go in the other way $\endgroup$ – ParaH2 Nov 14 '15 at 16:42
  • $\begingroup$ For me if the blue arrows go in the other way i agree we have the product mention in the picture. $\endgroup$ – ParaH2 Nov 14 '15 at 16:43

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