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Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):

$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$ $$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$

Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?

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    $\begingroup$ They react, but not "in the same way". The reaction with sodium is a redox reaction (note the changing oxidation state of hydrogen from +1 to 0), whereas that with NaOH is merely acid-base. As for the reaction with NaOH, it will certainly proceed, but in general will probably not go to completion (see also vapid's comment on the answer). $\endgroup$ – orthocresol♦ Sep 17 '16 at 6:51
  • $\begingroup$ @orthocresol If an electrophile (like CS2) is added to the solution of Methanol in NaOH, do sodium methoxide will attack to CS2 to form sodium salt of xanthate? $\endgroup$ – Khan Sep 17 '16 at 10:57
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The proposed reaction:

$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$

According to wikipedia:

The solid hydrolyzes in water to give methanol and sodium hydroxide

The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:

In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.

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    $\begingroup$ This answer would benefit from quoting $\mathrm{p}K_\mathrm{a}$ values and how they influence the equilibrium. $\endgroup$ – Jan Sep 17 '16 at 16:23

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