# What happens to a quaternary ammonium cation at a pH above its pKa?

I am working with a lipid and I have firm reasons to believe that its pKa is around 8. This lipid is a cationic lipid where the nitrogen is a quaternary amine with two lipid chains and two methyl groups.

Now, my question is that at a pH above the pKa of this lipid, the quaternary ammonium ion should get deprotonated, or basically act as an acid. But the quaternary ammonium cation doesn't have any protons to lose, so what would happen to it? My guess would be it accepts a lone pair and becomes negatively charged.

• You seem to think that the pKa of a quaternary salt is the same as the pKa of any other ammonium salt (ie around 9), but that’s not true. The pKa of something is a direct measure of how easily it’s deprotonated. Ammonium salts like R3NH+ are relatively easy to deprotonate, so their pKas are relatively small. A quaternary ammonium salt having no N-H bonds is not easy to deprotonate, so its pKa is large. Things can certainly happen at high pH values, eg elimination or substitution, but a pH of 14 (for example) is not larger than the pKa of the quaternary salt, and it won’t behave as an acid. – orthocresol Apr 30 '18 at 0:03
• If you have "firm reasons" that the $\ce{pK_a}$ is around 8, then acid group must be located at the lipid chains. So what lipid groups are you talking of? – aventurin Apr 30 '18 at 6:13
• there is no acid chain. the long chains are long aliphatic chains. with no double bonds or other groups present... my basic question is that in case of a quaternary ammonium ion with no H+ to deprotonate what really happens when pH exceeds its pKa? – ansari Apr 30 '18 at 14:26
• You keep saying pKa. What is the pKa of a quaternay ammonium ion? What exactly is the deprotonation you're looking at? – Zhe May 1 '18 at 13:21
• hi! its me again. I am trying to find more data on quaternary ammonium ion exchanging their counter ions, like at what pH it is triggered, or in what conditions. or is it possible that above a certain pH the counterions can shield the quaternary ammonium ion from interacting with other species in the solvent.. can someone help with these questions or guide me to some literature regarding this? – ansari Aug 20 '18 at 9:00

• There is no exchangeable $\ce{H}$ on Quaternary $\ce{N}$, thus, theoretically, it doesn't have $K_\mathrm{a}$. – Mathew Mahindaratne Apr 30 '18 at 15:40