Why does coupling of α-naphthol take place at δ-position (para to $\ce{-OH}$) rather than the β-position (ortho to $\ce{-OH}$)?

In the case of β-naphthol, we consider the aromaticity of the entire compound to decide the position.

Aromaticity is not disturbed even when the attack takes place at the ortho position. So I don't find any reason for the diazo compound to not attack there.

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    $\begingroup$ Does this answer your question? Regioselectivity in coupling reactions of α-naphthol $\endgroup$ – Safdar Faisal Aug 13 '20 at 11:49
  • $\begingroup$ In his picture, the OP says that ortho position is sterically hindered. I don't find any reason why. $\endgroup$ – DatBoi Aug 13 '20 at 11:55
  • $\begingroup$ OH can cause steric hinderance. Also, there is much greater -I effect on ortho compared to para.. $\endgroup$ – Safdar Faisal Aug 13 '20 at 13:02
  • $\begingroup$ We dont consider ortho effect in phenols because of the small size of OH. Why here then? $\endgroup$ – DatBoi Aug 13 '20 at 13:27
  • $\begingroup$ Take second case.. not first $\endgroup$ – Safdar Faisal Aug 13 '20 at 13:27