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Why is the reaction between water and sodium more vigorous than that between ethanol and sodium? Also, why does the reaction between sodium and an alcohol becomes less vigorous the longer the hydrocarbon chain?

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The dominant issue explaining the less vigorous reaction of long-chain (i.e. greater than about 3 carbons) alcohols with metallic sodium is simply a function of the molecular-mass of the alcohol. I will address this aspect last.

At the lower end of the molecular-mass for this series (i.e. water, methanol, ethanol, isopropanol, etc.) the destabilization of the hydroxyl intermediates by the electron-rich hydrocarbon groups (of course absent in water) also slows the progress of the reaction to some degree.

The reaction of sodium metal with water/alcohols is:

1)$$\ce{2Na(s) + 2ROH(l) -> 2NaOR v + H2 ^}$$
where R=H for water, and is the hydrocarbon chain for alcohols.

The inductive destabilization effect mentioned above also largely controls the relative acidities of the alcohols, which in turn are directly related to the reactivity of the alcohol with sodium. The relative acidities of pure solutions of water and some selected $\ce{C1}$ to $\ce{C3}$ alcohols goes as:

$$\text{water > methanol > ethanol > isopropanol}$$

In the discussion above, we have considered the simple case in which the reactant of interest is also the solvent. Any comparison of relative concentrations of the different reactants of course requires that we prepare some concentration of the water or alcohol in an appropriate solvent. It is also desirable to dilute the water or alcohol because of the vigorous, highly exothermic nature of their reactions with sodium.

Because water is the most reactive of the compounds of interest, any sodium added to an aqueous solution of an alcohol would just preferentially react with the water solvent. And of course you could not measure different concentrations of water in itself! The solvent of choice in this case is DMSO. DMSO is not reactive toward metallic sodium, and is miscible with water, low molecular weight and even very high molecular weight alcohols. In this system, the relative acidities of neat solutions of the $\ce{C1}$ to $\ce{C3}$ alcohols shown above holds, but with one exception, water:

$$\text{methanol > ethanol > isopropanol > water}$$

Although the acidity of water is low in DMSO due to the poor solvation of the hydroxyl anion (the alkoxy anions are solvated by DMSO in order of increasing hydrocarbon chain length), its reactivity toward sodium remains high. This is because in the presence of sodium, the hydroxyl anion is the reactive species. It reacts quickly and does not need to remain solvated for the reaction to proceed.

Of course the effects discussed thus far are also applicable toward higher molecular weight alcohols. However, once you have a carbon chain of more than 3 or so, these effects diminish with increasing molecular weight of the alcohol. Here, the larger the molecular weight of the alcohol, the fewer molecules you have to react for a given weight or volume concentration of the alcohol. Even if you have a pure alcohol solution, think about the "concentration" of the reactive hydroxyl functional group. In the case of water, hydroxyl groups are everywhere. In the case of say a $\ce{C20}$ alcohol, the sodium is mainly surrounded by hydrocarbons with a few reactive hydroxyl groups mixed in and available for reaction.

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