[Nescafe]

Hi,

I always tend to get lost in translation when people throw Pka values around like peanuts. What I remember from my undergraduate studies and now reading into it as a graduate student, if you want to deprotonate lets say a phenol which has a Pka of 10, you will need to use a base with a higher basicity. But, most of the time when I am looking into finding how strong of a base something is, I often find the Pka of their conjugate acid (The conjugate acid Pka = XX). For instance, LDA, I was looking into it and all I can find is the LDA-H pka which is 36, so what is its pka when it is deprotonated (LDA), I imagine it would be really high.Does anyone know why it is most often this way?

Let's say I have an aniline, a simple aniline, the Pka of that aniline is 30, the Pka of the conjugate acid of this aniline is 4.6 (so when its the anilinium ion), could I use a strong base such as LDA to deprotonate it, hypothetically speaking.

Any advice would be appreciated,

Thanks,

Nescafe.

[Dan]

At a basic level:

AH + B^- A^- + BH

If pK_a(AH) < pK_a(BH) then the equilibrium lies on the right, and if pK_a(AH) > pK_a(BH) then it lies to the left.

So yes, LDA will deprotonate aniline (pK_a ~30) because aniline is 10⁶ times more acidic than ⁱPr₂NH [the conjugate acid of LDA] (pK_a ~36).

I don't understand why you want the pK_a of LDA, I don't think it's practically possible to deprotonate it, the conjugate base of LDA will be a bis-anion, probably:

^-CH₂CH(Me)N^-CHMe₂

Do you mean the pK_b of LDA?

[fledarmus]

You're right, this can be very confusing and some of the people that create pKa charts aren't particularly careful about pointing out what reaction they are considering. In particular, primary and secondary amines tend to be confusing, because depending on the reaction you may be concerned with the pKa for the protonated amine going to the free base (when you are reacting the amine with an acid) or with the deprotonation of the amine by an even stronger base. Fortunately there are some much more complete charts out there that show both the starting material and the product for the reaction they are reporting a pKa for.

This is one of the more complete sources that I've found. It at least lists the ammonium compounds (protonated) separately from the amine compounds. In each case, the pKa is listed for the removal of a proton from the species shown. http://www.chem.wisc.edu/areas/reich/pkatable/index.htm

[Nescafe]

At a basic level:

AH + B^- A^- + BH

If pK_a(AH) < pK_a(BH) then the equilibrium lies on the right, and if pK_a(AH) > pK_a(BH) then it lies to the left.

So yes, LDA will deprotonate aniline (pK_a ~30) because aniline is 10⁶ times more acidic than ⁱPr₂NH [the conjugate acid of LDA] (pK_a ~36).

I don't understand why you want the pK_a of LDA, I don't think it's practically possible to deprotonate it, the conjugate base of LDA will be a bis-anion, probably:

^-CH₂CH(Me)N^-CHMe₂

Do you mean the pK_b of LDA?

I see. When I asked what's the pka of LDA I guess what I should really be asking is the pkb of it. I didn't mean the pka of conjugate base of lda, I don't even think that's possible. So when you read pka of a base is let's say 10 does that mean the pka of the conjugate acid or the unprotonated form of it. Let's say dipea, before it picks up a proton it only has an electron pair. When they say the pka of dipea is 10 doe they mean pka of the conjugate base?

Cheers,

Nescafé.

[Dan]

So when you read pka of a base is let's say 10 does that mean the pka of the conjugate acid or the unprotonated form of it. Let's say dipea, before it picks up a proton it only has an electron pair. When they say the pka of dipea is 10 doe they mean pka of the conjugate base?

Yes, this is a convention that annoys me a bit. Sources will often quote the pK_a of an amine to be in the 9-11 range, when what they mean is the pK_a of the protonated amine (the conjugate acid of the amine). The pK_a of amines themselves (if they have an NH) are around the 35 mark. Tertiary amines like DIPEA would be (theoretically) deprotonated at C, since there is no NH, and would have a pK_a probably 50+. It is the conjugate adic of DIPEA, diisopropylethylammonium, that has a pK_a ~10.

[Nescafe]

So when you read pka of a base is let's say 10 does that mean the pka of the conjugate acid or the unprotonated form of it. Let's say dipea, before it picks up a proton it only has an electron pair. When they say the pka of dipea is 10 doe they mean pka of the conjugate base?

Yes, this is a convention that annoys me a bit. Sources will often quote the pK_a of an amine to be in the 9-11 range, when what they mean is the pK_a of the protonated amine (the conjugate acid of the amine). The pK_a of amines themselves (if they have an NH) are around the 35 mark. Tertiary amines like DIPEA would be (theoretically) deprotonated at C, since there is no NH, and would have a pK_a probably 50+. It is the conjugate adic of DIPEA, diisopropylethylammonium, that has a pK_a ~10.

I'm sorry but what do you mean by at C?

Can you really assign something that doesn't have an acidic proton a pka? Like you know how you said Dipea without the H has a pka of 50+, wouldn't it be more correct to give something that doesn't have a proton a pkb instead?

Ok so pka of aniline ( neutral nh2) is 30, pka of its conjugate acid is 4.6ish so NH3(+)

Cheers,

Nescafé.

[orgopete]

Can you really assign something that doesn't have an acidic proton a pka? Like you know how you said Dipea without the H has a pka of 50+, wouldn't it be more correct to give something that doesn't have a proton a pkb instead?

Ok so pka of aniline ( neutral nh2) is 30, pka of its conjugate acid is 4.6ish so NH3(+)

pKa and pKb are related to each other. For water, pKa is the negative log of [H⁺]. pKb is the neg log of [HO^-]. If the pH is low (very acidic) the hydroxide concentration will be low.

pKa is the general convention for measuring the acidity and because they are related, the basicity. The conjugate base of a weak acid can deprotonate a stronger acid. LDA can deprotonate an ester or ketone, but not an alkane. We know that from the pKa's of the acids, diisopropylamine 36, ester 26, and ethane 50.

Theoretically, any proton can have an acidity, that is, how strong is the proton-electron pair bond? The further the electron pair is pulled from the proton, the weaker will be the bond and the more acidic the acid. If you increase the nuclear charge, it will increase the nuclear field to pull electrons toward it and it will also increase the nuclear-proton force repelling the proton. In my opinion, this simple physics, the inverse square law. This will explain why methane has a long bond and is a weak acid and HI has a long bond and is a strong acid. It isn't the nucleus-proton distance that matter (as much) as the proton-electron pair distance.

When reading pKa tables, one must use caution as to how they maybe written. The structure of an amine may be given, ammonia, but the pKa may state or have a footnote explaining it is actually for the conjugate acid of ammonia, NH4⁺. So a compound like aniline can have two pKa values, one for aniline (~30) and one for anilinium ion (~4). Typographically, it is easier and less confusing (for some) to list the pKa of the conjugate acid while showing the structure or formula of the base. Presumably, the electrons of the ammonium nitrogen are unaffected by the counter ion. Therefore, separate entries are not required for ammonium chloride, bromide, iodide, bisulfate, triflate, etc. 

[Borek]

pKa is the negative log of [H⁺]. pKb is the neg log of [HO^-].

Really? 

[orgopete]

pKa is the negative log of [H⁺]. pKb is the neg log of [HO^-].

Really? 

Oops, pH

[Dan]

I'm sorry but what do you mean by at C?

Abstraction of a proton from a C-H bond.

Can you really assign something that doesn't have an acidic proton a pka? Like you know how you said Dipea without the H has a pka of 50+, wouldn't it be more correct to give something that doesn't have a proton a pkb instead?

DIPEA has 19 hydrogen atoms. All H atoms have some degree of acidity, even alkanes.

[Nescafe]

At a basic level:

AH + B^- A^- + BH

If pK_a(AH) < pK_a(BH) then the equilibrium lies on the right, and if pK_a(AH) > pK_a(BH) then it lies to the left.

Your helping me figure out things i used to think to be too complicated to understand. But there is still one thing I don't understand. The equation above, why would the equilibrium shift to the right if pka(BH) > pka (AH). Memorizing is one thing, I want to understand why it is so,

Cheers,

Nescafé

[Dan]

Qualitatively, the system will shift to an energy minimum where the stronger base is protonated and the stronger acid is deprotonated.

So if pK_a(BH) > pK_a(AH), this means that AH is a stronger acid than BH, and it follows that B^- is a stronger base than A^-. So at equilibrium B^- will be mostly protonated (because it is the strongest base in the system) and AH (the stronger acid) will be mostly deprotonated.

Mathematically, if you write an expression for the equilibrium constant K_eq, and the dissociation constants K_a(BH) and K_a(AH):

K_eq = [A^-][BH]/[AH][B^-] (1)
K_a(BH) = [B^-][H⁺]/[BH]  (2)
K_a(AH) = [A^-][H⁺]/[AH]  (3)

if you substitute (2) and (3) into (1), you get:

K_eq = K_a(AH)/K_a(BH)

you can then take logs and rearrange to:

K_eq = 10^x, where x = pK_a(BH) - pK_a(AH)

Therefore: if pK_a(BH) > pK_a(AH), then x > 0, and K > 1 which means the equilibrium lies on the right.

[Nescafe]

Qualitatively, the system will shift to an energy minimum where the stronger base is protonated and the stronger acid is deprotonated.

So if pK_a(BH) > pK_a(AH), this means that AH is a stronger acid than BH, and it follows that B^- is a stronger base than A^-. So at equilibrium B^- will be mostly protonated (because it is the strongest base in the system) and AH (the stronger acid) will be mostly deprotonated.

Mathematically, if you write an expression for the equilibrium constant K_eq, and the dissociation constants K_a(BH) and K_a(AH):

K_eq = [A^-][BH]/[AH][B^-] (1)
K_a(BH) = [B^-][H⁺]/[BH]  (2)
K_a(AH) = [A^-][H⁺]/[AH]  (3)

if you substitute (2) and (3) into (1), you get:

K_eq = K_a(AH)/K_a(BH)

you can then take logs and rearrange to:

K_eq = 10^x, where x = pK_a(BH) - pK_a(AH)

Therefore: if pK_a(BH) > pK_a(AH), then x > 0, and K > 1 which means the equilibrium lies on the right.

This has been most useful.

Thank you,

Cheers,

Nescafe.