$$\ce{2Na(l) +2NH3 (g) ->2NaNH2(s) + H2(g)}$$
$$\ce{CH3Cl (alc) +NH3 (alc) -> CH3NH2 (alc) +HCl(alc)}$$
Doesn't ammonia donate a proton in both cases?
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5The first reaction is a redox reaction where you can't give $\ce{NH3}$ the role of a proton donor in the same sense as an acid is a proton donor. In the second reaction, you have to look at the reaction mechanism: it's a $\mathrm{S}_{\mathrm{N}}2$ reaction. Here, the $\ce{NH3}$ does not donate a proton directly but it first substitutes the $\ce{Cl}$ on methylchloride, thus forming an (much more acidic) ammonium salt, and then the ammonium salt donates the proton acting as an acid. – Philipp Sep 25 '13 at 06:36
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@Philipp Thanks for the comment. Unfortunately, I don't understand how it forms an ammonium salt. Doesn't ammonium have to be $\ce{NH4+}$? Where does the extra proton come from? Also, why is it that in a redox reaction $\ce{NH3}$ cannot act as a Bronsted-Lowry proton donor? – user2246 Sep 25 '13 at 06:55
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3Ammonium doesn't mean $\ce{NH4^{+}}$ in general. There can be any organic group attached to $\ce{N}$ in an ammonium salt. In this case you have $\ce{CH3NH3^{+}}$, which is formed when $\ce{NH3}$ kicks out the $\ce{Cl^{-}}$ group in the first step of the $\mathrm{S}_{\mathrm{N}}2$ reaction thus leading to a positively charged ammonium cation. As for your second question: In the first reaction $\ce{NH3}$ doesn't dissociate like a normal acid, i.e. there is no reaction like $\ce{NH3 <=> NH2^{-} + H^{+}}$, instead the $\ce{H}$ that is lost ends up with the oxidation number $0$, i.e. not a proton. – Philipp Sep 25 '13 at 07:18
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1@Philipp: That's probably enough to form a decent proper answer rather than a comment. – Aesin Sep 25 '13 at 11:02
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Yes. I agree. Thanks @Philipp! It sure answered my question. – user2246 Sep 25 '13 at 11:43
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@Aesin Ok, I added some arguments involving $\mathrm{p}K_{\mathrm{a}}$ values and hope that the answer is coherent. – Philipp Sep 25 '13 at 16:29
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Ammonia has a $\mathrm{p}K_{\mathrm{a}}$ value (a description what that means can be found in this answer of mine) of 33 and is thus not acidic at all - you'd need a $\mathrm{pH}$ value around 30 to deprotonate $\ce{NH3}$ to any appreciable amount! This means that $\ce{NH3}$ very rarely acts as a proton donor itself.
So, what happens in your first reaction is the following: It is widely known that when you dissolve $\ce{Na}$ in liquid ammonia you get solvated electrons (in the form of electride complexes). But electrons have of course very strong reducing power, so after a while they attack $\ce{NH3}$ abstracting one of its $\ce{H}$ atoms under the liberation of hydrogen gas. The important point here is that there is no dissociation of $\ce{NH3}$ here, i.e. no reaction $\ce{NH3 <=> NH2^{-} + H^{+}}$, so $\ce{NH3}$ does not act as a proton donor in the same sense a Bronsted acid would.
Your second reaction is an $\mathrm{S}_{\mathrm{N}}2$ reaction where $\ce{NH3}$ acts as a nucleophile. So, it doesn't act as a proton donor in this reaction neither. What it does is that it replaces (substitutes) the $\ce{Cl}$ in $\ce{CH3Cl}$, thus forming a methylammonium salt, $\ce{CH3NH3^{+}Cl^{-}}$.
Now, the $\mathrm{p}K_{\mathrm{a}}$ of an alkylammonium cation lies somewhere around 10 (the $\mathrm{p}K_{\mathrm{a}}$ of $\ce{NH4^{+}}$ is 9.24). So, it is much more acidic than $\ce{NH3}$ and also more acidic than the surrounding alcohol (I assume that "(alc)" means that the reaction is done in alcohol), which has a $\mathrm{p}K_{\mathrm{a}}$ between 15 and 17, and thus it (the ammonium salt) acts as a proton donor.
