6

I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some ...


5

This would be the mechanism I propose for the reaction: I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.


3

While the user seems to have vanished, its worth pointing out the misconception they are having. Its not that the (E) and (Z) enantiomers of cyclooctene are atropisomers. Rather, (E)-cyclooctene has two atropisomeric conformations, which can be interconverted with out breaking the double bond. Rotation is hindered around this bond, but it can still occur due ...


2

For reaction with $\ce{KMnO4}$, the given compound must have a carbon atom attached directly to the ring and that carbon atom must have at least one hydrogen atom. But in case of tert-butylbenzene this condition isn't followed and hence is a special case. The final product obtained on reaction of tert-butylbenzene with $\ce{KMnO4}$ is Pivalic acid $\ce{(...


2

Given are keto form of keto-enol tautomerism of reference compounds. Answers are given according to the percentages of enol forms. Your duty is to find which enol form is dominant. In cyclohexa-2,5-dienone, the enol form (phenol) dominate 100% in solid or in solutions, regardless of the solvent because of the aromaticity (100% B). In the case of pyridones, ...


1

I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group.


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