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This is in reference to Kolbe's Reaction, actually I think that Sodium Phenoxide is the most activating because, in Kolbe's Reaction we don't make react $\ce{CO2}$ directly with phenol for the electrophilic substitution(Maybe because of it's comparatively less activating effect), but we do make Sodium Phenoxide react with $\ce{CO2}$ (Maybe because of it's comparatively more activating effect) so that it will get attach to the Ortho position of Benzene Ring. Along with this, please also tell me why in Kolbe's Reaction, the main product is Ortho directing & in most of other reactions, the main product is para directing?

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Sodium Phenoxide is much more activating than phenol itself. Because here directly one negative charge comes and bond with the C below and the pi bond shifts to the ortho position as negative charge. And it also answers your doubt that why ortho product is major.enter image description here You can refer to this mechanism. After reacting with OH- then the tautomerization takes place but before it no effect where as in case of phenoxide it can perform the same reaction even without OH- So activation order Anisole

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  • $\begingroup$ What about anisole? In the title of the question, anisole was also mentioned... $\endgroup$ – Tan Yong Boon Dec 12 '18 at 3:58
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Out of $\ce{O-}$, $\ce{-OH}$ and $\ce{-NH2}$, I would say that O- is the hardest nucleophilic centre,so by HSAB, phenoxide should be more activated.

For Kolbe's reaction, since it is carried out in basic medium, so ortho position gets more activated as $\ce{O-}$ has a positive mesomeric as well as a positive inductive effect on the molecule

Also, it may depend on the stabilization of the transition state by the base. Take a look at this (During electrophilic substitution by carbon dioxide)-

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