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The book says there is no reaction.

However, I do not understand why.

The ROH group is resonance-stabilized with the benzene-like group. Its conjugate base is stable. There is an inductive effect of the cloud of electrons that make the oxygen more electronegative. However, resonance effects are always more important than inductive effects.

So, why is there no reaction?

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    $\begingroup$ If the OH group gets deprotonated it is not resonance stabilized by the naphthalin ring because it is not directly attached to it. What reaction are you expecting, exactly? Just deprotonation or something else? $\endgroup$ – Philipp Nov 16 '14 at 15:29
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The conjugated base of the depicted alcohol is not resonance-stabilized. The negative charge cannot be delocalized into the aromatic ring because the alcohol group is not a phenolic OH group, whose oxygen lone pairs are conjugated to the aromatic $\pi$ electrons, like shown below. Instead, it is attached to an $sp^3$-hybridized benzylic carbon which is not part of the conjugated system.

enter image description here

You might have mistaken the depicted compound with 2-naphtol, whose $\ce{pK_{a}}$ (9.51) is much lower than that of aliphatic alcohols and which can therefore be deprotonated by $\ce{NaOH}$. Aliphatic alcohols typically have $\ce{pK_{a}}$ values of about 16-18 (source), so there will be no deprotonation by sodium hydroxide.

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These would be the resonant structures of your deportonated compound:

enter image description here

As you can see, there will be no delocalization at al... If you try to move the electrons just a bit, you obtain a very unstable structure, which will not contribute to the resonance hybrid.

However, it is possible to deportonate it with a stronger base, like NaH, or LDA.

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