[Sis290025]

Select any molecule that has more than 2 ionizable groups. Write its structures, showing every atom, at pH levels 1 and 7.


The molecule that I chose is geranyl pyrophosphate (http://en.wikipedia.org/wiki/Geranyl_pyrophosphate) :


____________________________________O_____O
____________________________________||____||
CH3C(CH3)=CHCH2CH2C(CH3)=CHCH2--O—P--O--P--O-
____________________________________|_____|
____________________________________O-____O-

This one does have 3 ionizable groups, right?

At pH 1, the molecule is under acidic conditions, so will all the ionizable groups be protonated?

____________________________________O_____O
____________________________________||____||
CH3C(CH3)=CHCH2CH2C(CH3)=CHCH2--O--P--O--P--OH
____________________________________|_____|
____________________________________OH____OH


At pH = 7, the presence of water can be presumed(?), so does this arise?


____________________________________O
____________________________________||
CH3C(CH3)=CHCH2CH2C(CH3)=CHCH2--O—P--OH and
____________________________________|
____________________________________O-

____O
____||
HO--P--O-
____|
____O-


Thank you.
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[Yggdrasil]

The hydrolysis can occur at pH 7 (although, hydrolysis is catalyzed by acid or base, so I would say the hydrolysis would occur faster at pH 1).  However, I think the question isn't asking about the hydrolysis of geranyl pyrophoshpate, but which of the ionizible groups will be protonated and unprotonated at pH 1 and 7.

[Sis290025]

So not all the ionizible groups will be protonated or unprotonated?

For example, at pH 1, protonation will occur? If so, then are all three groups going to be protonated? Or will a certain number from the three ionizable groups be protonated if protonation occurs?

For pH 7, will all three groups remain unprotonated if unprotonation is promoted, or will only a certain number from the three ionizable groups be unprotonated if unprotonation is promoted?

[Borek]

It all depends on pH and pKa of particular group. Each acid will dissociate according to equation:

HA = H⁺ + A^-

(note: for now it doesn't matter whether the molecule HA was already dissociated earlier, it changes only charge of A^- product)

This reaction has its own dissociation equilibrium:

Ka = [H⁺][A^-]/[HA]

taking logs and rearranging gives us:

log([HA]/[A^-]) = pKa - pH

(unless I have messed signs). For every ionizable group ratio of concentrations of unionized and ionized forms is a function of pH.

[Sis290025]

I'm still a bit confused over determining which of the three ionizable groups become protonated and remain unprotonated. How do I apply the Henderson-Hasselbach equation to this problem? I thought it was more of a qualitative problem than a quantitative problem?
Thanks again.

[Borek]

What I wrote is just a version of HH equation. Note that there is no situation like "group is protonated" or "group is not protonated". We are talking about huge amount of molecules, some of them have their groups protonated, some don't. All you can say is what is a ratio of those protonated to those not - and this is where HH (or the equation I wrote) comes into play. You just have to know which pKa refers to which group.

If the pH is far enough from pKa you may assume that all molecules in the solution have particular group protonated/not protonated - but it all depends on the situation. Sometimes even 0.01% of protonated groups can be too much, sometimes 10% doesn't matter.

[Sis290025]

If there are three ionizable groups, then a different pka value corresponds with each pair of protonated/unprotonated groups?

As another example before I do pyrophosphate, take phosphatidylserine.
___________________________+
R1C(=O)OCHCH2OP(=O)OCH2CH(NH3)COO-
________|______|
________|______O-
________|
________CHOC(=O)R2

http://en.wikipedia.org/wiki/Phosphatidylserine

The three ionizable groups are +NH3, COO-, and (P-)O-?

For each group:

pH = pKa + log([A-]/[HA])

pH will equal 1 or 7.

Ratios for each ionizable group of the molecule are as follows?

1) NH2/NH3
2) COO-/COOH
3) (P-)O-/OH

The pKa value corresponds to the protonated group on the bottom? As for the pKa value, the number that I use is the pKa of phosphatidylserine with the protonated group on it?

Thanks again. 

[Yggdrasil]

From the HH, you can see that if the pH of the solution exceeds a group's pKa by 1 pH unit then (i.e. pH - pKa = 1), then

log ([A-]/[HA]) = 1
[A-]/[HA] = 10

So, in these cases the concentration of the deprotonated form is ten times the concentration of the protonated form.  For simple, deprotonated/protonated cases, you can say that this corresponds to deprotonated since the majority of the molecules are deprotonated.

Similarly, if the pH is below the groups pKa by 1 pH unit (i.e. if pH - pKa = -1), then

log ([A-]/[HA]) = -1
[A-]/[HA] = 1/10

And the protonated form is ten times as prevalent as the deprotonated form.


Now, lets say the pKa of the acidic/basic groups of phosphatidylserine are as follows (note: these are just made up values):
carboxylate - 2
phosphate - 6
amine - 8
By the criteria established above:
The carboxylate will be protonated at pH < 1 and deprotonated at pH > 3
The phosphate will be protonated at pH < 5 and deprotonated at pH > 7
The amine will be protonated at pH < 7 and deprotonated at pH > 9

This means that if:
pH < 1, all of the groups will be protonated
2 < pH < 5, only the carboxylate will be deprotonated
pH 7, only the amine will protonated
pH > 9, all groups will be deprotonated

[Sis290025]

Thanks for making it clearer. I suppose my real question is:

Since you used fabricated numbers for the pKa values of the amine, carboxylate, and phosphate groups, where can I find the bona fide pKa values for these ionizable groups? (Are the fabricated values anywhere near the true values?)