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I've asked a question before and quickly realised that the resonance/mesometric effects played a large role in determining the C-13 NMR spectrum of phenol, shown below.

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This is the resonance structure of phenol:

enter image description here

But given this, I would have expected the carbons at $b$ to be more shifted than the ones at $d$. This is because although they both have partial negative charges, the inductive effect of the hydroxy group on $a$ should make $b$ more deshielded than $d$.

Why is this?

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    $\begingroup$ now it seems that you're accounting only for the mesomeric effect and not the induction. Consider the combined effects, especially with regard to how the positive charge distributes. $\endgroup$
    – Andrew
    Jun 11 at 14:54
  • $\begingroup$ @Andrew If I account for the inductive effect, shouldn't that would mean that d should be more shielded than b? This is because b is closer to the positive oxygen which will be withdrawing electron density. So why isn't this the case? $\endgroup$
    – John Hon
    Jun 12 at 0:11
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    $\begingroup$ I think the answer is gonna be more complicated than just inductive/resonance effects --- 13C shifts are not very straightforward, it's tempting to think of them like we do 1H shifts, but the underlying reasons for the chemical shifts are rather different $\endgroup$
    – orthocresol ♦
    Jun 12 at 0:26

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