# Behavior of amines as acids

I came across this question which asked us to find in which alternative does the amine behave as an acid the most:

In which of the following reactions does the amine behave as an acid
(a) $$\ce{(C2H5)2NH}$$ + $$\ce{H2PtCl6}$$
(b) $$\ce{CH3NH2}$$ + $$\ce{H2O}$$
(c) $$\ce{((Me)2CH)2NH}$$ + $$\ce{n-C4H9Li}$$
(d) $$\ce{(C2H5)3N}$$ + $$\ce{BF3}$$

(c)

Now coming to my thought process:
So without considering the reagents (because I don't know any of these reactions!), I considered the acidic and basic nature of amines.
We know that secondary amines are the most basic while primary amines are less basic due to lesser $$+I$$ effect. So, my reasoning says that the answer should be (b), which is evidently wrong.

Any ideas about the reactions and the solution?

• It is a relative term. For instance, water also acts as an acid or base based on which reagent it is reacting with. Basically, anything with $\mathrm{p}K_\mathrm{a} \lt 7$, water acts as an base (e.g., acetic acid). Anything with $\mathrm{p}K_\mathrm{a} \gt 7$, water acts as an acid (e.g., ammonia). – Mathew Mahindaratne Jun 2 at 4:15
• Question does not ask which amine is most acidic, but in which reaction amine acts as acid! – Mithoron Jun 2 at 12:19

Acid/base nature of any reagent is a relative term. For instance, water $$(\mathrm{p}K_\mathrm{w} = 14)$$ also acts as an acid or base based on which reagent it is reacting with. Basically, any reagent with $$\mathrm{p}K_\mathrm{a} \lt 7$$, water acts as an base (e.g., acetic acid; $$\mathrm{p}K_\mathrm{a} = 4.75$$): $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$ Then again, if the conjugate acid of any reagent has $$\mathrm{p}K_\mathrm{a} \gt 7$$, water acts with that reagent as an acid (e.g., the conjugate acid of ammonia with $$\mathrm{p}K_\mathrm{a} = 9.25$$): $$\ce{NH3 + H2O <=> NH4+ + OH-}$$

Now, let's look at the given reactions:

(a) $$\ce{(C2H5)2NH}$$ + $$\ce{H2PtCl6}$$:
This is the though one among all four choices. The name of $$\ce{H2PtCl6}$$ is hexachloroplatinic acid, and thus, it is safe to assume that this acts like an acid in water meaning $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{H2PtCl6}$$ is $$\lt 7$$. Ammonia with water acts as a acid $$(\mathrm{p}K_\mathrm{a} \gt 7)$$ and hence the amine in this case must acts as a base.

(b) $$\ce{CH3NH2}$$ + $$\ce{H2O}$$:
$$\ce{CH3NH2}$$ acts like ammonia with water because $$\mathrm{p}K_\mathrm{a}$$ of its conjugate acid is about $$10.5$$. Thus, $$\ce{CH3NH2}$$ acts like a base here.

(c) $$\ce{((Me)2CH)2NH + n-C4H9Li}$$:
$$\ce{n-C4H9Li}$$ is a strong base with $$\mathrm{p}K_\mathrm{a} \approx 50$$ for conjugate acid of $$\ce{n-C4H9-}$$, $$\ce{n-C4H10}$$. Therefore, the amine acts here as an acid, giving away its lone proton.

(d) $$\ce{(C2H5)3N + BF3}$$:
Here, $$\ce{BF3}$$ is a Lewis acid, $$\ce{B}$$ in which does not fulfil the octet. Thus, it is craving for a electron pair, $$\ce{N}$$ in $$\ce{(C2H5)3N}$$ would gladly donate. Therefore, amine acts as a base here.

Therefore, no doubt, the choice C is the answer.