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The carboxylation in Kolbe's reaction occurs predominantly at the **ortho** position to the phenoxide because

Aphenoxide oxygen chelates $Na^+$ and delivers $CO_2$ ortho
Bpara gives an unstable carbocation
Cpara is sterically blocked by the OH
Dortho is the only kinetic site available
Answer & Solution
Correct answer: A. phenoxide oxygen chelates $Na^+$ and delivers $CO_2$ ortho
The sodium of the phenoxide coordinates with $CO_2$, holding it in proximity to the ortho position. The intramolecular delivery makes ortho substitution overwhelmingly favoured over para, even though both are activated.
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