Alkylation of Amines Tends To Be A Pretty Crappy Reaction

by James

in Amines

Trends in the basicity of amines provide a classic example of why chemistry is so interesting; it’s deciphering the delicate trade-offs between various general factors in specific cases that gives chemists a buzz. The miraculous starts looking obvious.
The Curious Wavefunction (Ash Jogalekar)

OK folks. Today, let’s talk about how to make amines. See if you can fill in the blank in the following skill-testing question:

 Let’s walk through the thought process. We need to form a C-N bond. The N of NH3 has a lone pair and is a good nucleophile. So, much like the Williamson ether synthesis, maybe we can use an alkyl halide like ethyl bromide (CH3CH2Br) and boom! done.

It doesn’t quite turn out that way. In the words of the scientist who investigated this thoroughly:

“The interaction of ethyl bromide and ammonia… conduces largely to the formation of triethylamine”.

[ref:  Werner,  J. Chem. Soc. 1918, p. 899]

Formation of a primary amine through alkylation of ammonia turns out to be a pretty crappy reaction. (For reasons of economy, we generally want to use reactions that form one dominant product in high yield to avoid tedious separations: this doesn’t qualify.)

This is part of the fun* of learning organic chemistry. Just when you think you might have things figured out, along comes a curveball. (*your definition of “fun” may vary)

Why does this reaction suck?

The Runaway Train That Is Amine Alkylation

Let’s look at the first reaction that would happen the moment that a solution of ammonia is combined with ethyl bromide: nucleophilic substitution (via the SN2 mechanism)

So far so good. We’ve formed our C-N bond. Since the reaction is in excess ammonia, an acid-base reaction can then lead to the neutralization of at least some of the ammonium salt, yielding us our neutral primary amine.

So now we have the primary amine. We’re done, right?

Not so fast. See, relative to substitution reactions, acid-base reactions are fast, so this deprotonation event will occur before the first substitution reaction has consumed all of the remaining ethyl bromide.

That means that ethylamine will be present in the same flask with ethyl bromide.


Well, ethylamine is itself a good nucleophile. In fact, it’s an even better nucleophile than ammonia itself, because of the electron-donating alkyl group [Link: 5 factors that affect the basicity of amines]

How much better? If we use pKaH as a proxy for nuclophilicity (reasonable,  since steric hindrance isn’t a factor here) we have a pKaH of 10.7 for ethylamine versus 9.2 for NH3. That’s about 101.5 = 30 times more nucleophilic.

The ethylamine created by the first SN2 will start gobbling up the ethyl bromide…  instead of NH3!

We’ve created a monster! 

Even if we try to stack the deck with (say) a 15-fold excess of NH3, the reaction of ethylamine will still be faster by a factor of (30/15) = 2. [That’s what Werner did, above; the yield of EtNH2 was 34% under those conditions, versus 11% for just 1 equiv of NH3.]

Now we have the diethylammonium product in the presence of excess base, which will lead to at least some formation of diethylamine.

You know what that means? Another nucleophile has been produced that will compete with NH3 for the remaining ethyl bromide.

How does the nucleophilicity compare to ammonia and ethylamine?

The pKaH of diethylamine is about 11.0, making it slightly more nucleophilic than ethylamine!

The runaway train 

Diethylamine, being a stronger nucleophile than both ammonia and ethylamine, will react faster with the remaining ethyl bromide than either of those two species.  This will form triethylammonium bromide, which can then be deprotonated (using any one of the amine bases in solution) to form neutral triethylamine.

Now we have three nucleophilic amine species all swimming around in solution at the same time, before all of our ethyl bromide has been consumed.

This soup containing multiple amine products is the kind of reaction that my friend Jeff would describe in his lab notebook as a “BFM”. (The “B” is for “big”, and the “M” is for “mess”….) 

The “runaway train” usually stops at the tertiary amine stage. In this specific case, the pKaH of triethylamine is 10.75, a little less than diethylamine. [Recall that pKaH refers to basicity, not nucleophilicity: the basicity of NEt3 is attenuated here due to the lowered solubility of the conjugate acid in water. [more here]. ] The nucleophilicity of triethylamine is less than that of diethyl amine largely because the triethylamine nitrogen is tertiary, which will increase steric hindrance and slow down the reaction rate.

For this reason, formation of tertiary amines from secondary amines via alkylation is thus, for the most part, exempt from our broad “amine alkylation is crap” statement.

Otherwise, workarounds must be used.  There are plenty. Azides are good nucleophiles for example, and they can be alkylated without incident and reduced to amines afterwards. The Gabriel Synthesis is another workaround. (We’ll cover these in due course). 

What about our skill-testing question from the top of the article?

Here’s a reaction that most organic chemists would consider to be the best general way to make amines: a reaction called reductive amination

Here’s a little sneak preview.

We’ll talk about how this reaction works next time.

Final Note: Exhaustive Methylation

Footnote. We saw that the alkylation of ammonia mostly stopped at the tertiary amine stage.

Although tertiary amines tend to be less nucleophilic than secondary amines, they are still nucleophiles, and when treated with excess alkyl halide under forcing conditions they can be converted to quaternary ammonium salts. 

The best example of this is with methyl iodide, in a reaction called exhaustive methylation. Recall that methyl halides are the fastest-reacting alkyl halides in SN2 reactions due to their low steric hindrance.

Treatment of amines with a large excess of methyl iodide leads to their quaternary ammonium salts. As we’ll see in a future post, these quaternary ammonium salts can behave as excellent leaving groups in elimination reactions (and, rarely, substitution reactions) to give alkenes, particularly in the Hofmann elimination.

Edit. Hm. Attempted to embed “Runaway Train” but it didn’t work.


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{ 1 comment… read it below or add one }

Nirdesh Singh

Dr.James greetings from my side.I have read lot of articles by you on different topics of organic chemistry and i really appreciate the way you make the students understand organic chemistry like a breeze.
I also refer your blogs to my students when they are stuck with some really difficult topics like complex mechanisms of the reactions and other topics.


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