3
$\begingroup$

In my teacher's power-point states that we must consider that inductive effect in $CH_{3}COOH$ exists because of the electron deficiency in $C$ (partial positive charge $\delta^{+}$) as a result of the electronegativity of the double bonded $O$. As you may notice in the figure 1 although the single bonded $O$ is more electronegative than $C$, the electron density of the single bonded $O$ moves to $C$ in order to "neutralize" the $\delta^{+}$, leaving $H$ less attached to the molecule.

Why do we consider that double bonded $O$ establishes that $\delta^{+}$ on our $C$ (and creates an electron withdrawing inductive effect $-I$) and not the single bonded $O$? Is this because of the $\pi$ double bond? Can you tell me why?

asd

Figure 1: Inductive effect created by the electron deficiency of $C$. As we can see the $H$ is now less attached to our molecule

Now supose we have the following carboxylic acids:

$CH_{3}COOH$, pka = $4.80$

$FCH_{2}COOH$, pka = $2.66$

$O_{2}^{-}CCH_{2}COOH$, pka = $5.69$

$O_{2}^{-}C(CH_{2})_{4}COOH$, pka = $5.41$

For me it makes sense that $FCH_{2}COOH$ is more acidic than $CH_{3}COOH$, because $F$ is an electronegative element and he creates a partial negative charge moving de electron density to him, leaving our $H$ in the carboxyl group less attached and easier to remove.

On the other hand the fact $O_{2}^{-}CCH_{2}COOH$ is less acidic than $O_{2}^{-}C(CH_{2})_{4}COOH$ for me it's less intuitive. In my teacher's power-points states that $O_{2}^{-}$ linked to an alkyl chain will create an electron donating inductive effect $+I$. Can you explain to me why? Isn't there a withdrawing resonance effect $-R$? If $O$ is an electronegative element why doesn't it create an $-I$ effect instead of $+I$? Beacause if that was the case adding carbons $(CH_{2})_{4}$ in the chain would reduce the the $-I$ effect and the $O_{2}^{-}C(CH_{2})_{4}COOH$ would be less acidic than $O_{2}^{-}CCH_{2}COOH$, and that isn't the case.

$\endgroup$
3
$\begingroup$

Why do we consider that double bonded O establishes that δ+ on our C (and creates an electron withdrawing inductive effect −I) and not the single bonded O?

Look at the resonance structures for acetic acid in the top line of the following drawing. We can see that the middle resonance structure takes the pi electrons involved in the $\ce{C=O}$ double bond and moves them to oxygen, creating a positive charge at the carbonyl carbon. In the third, or left-most resonance structure we see that the "single-bonded oxygen" is also involved in resonance delocalization, but it does not create any positive charge at the carbonyl carbon. Aside from resonance factors, both oxygens will create partial positive charge at the carbonyl carbon through inductive effects.

enter image description here

To answer the question about the relative $\ce{pK_{a}s}$ of these acetic acid derivatives let's proceed as follows. First, let's consider resonance effects of the substituent ($\ce{X}$) to be negligible. This is because 1) the substituent is not directly attached to the carbonyl group or the resulting carboxylate anion, 2) it turns out there is a node at $\ce{C2}$ in the carboxylate anion (zeroth order approximation), so the $\ce{X}$ group is disconnected in a resonance sense. So we need only consider the inductive effect of $\ce{X}$ upon both the reactant and the product and this should allow us to estimate how the equilibrium involving acid ionization will shift. $\ce{X=H}$ is our baseline and we will compare the other two compounds against it. When $\ce{X=F}$ the starting acid ($\ce{A}$) will be inductively destabilized due to the electron withdrawing inductive effect of fluorine. On the other hand, the inductive effect of fluorine will serve to stabilize the resultant carboxylate anion ($\ce{B}$). These two effects (destabilization of the reactant and stabilization of the product) will act in concert to push our equilibrium to the right, making fluoroacetic acid a stronger acid than acetic acid. In the case of $\ce{X=CO2^-}$, the inductive electron donating effect of the $\ce{-CH2CO2^-}$ group should be stronger than the $\ce{CH2-H}$ group in acetic acid, so our reactant should be more stabilized. On the other hand, the inductively electron donating $\ce{-CH2CO2^-}$ should destabilize the already electron rich carboxylate anion ($\ce{B}$). Here both inductive effects again work in concert, but now to stabilize the reactant and destabilize the product making this derivative less acidic than acetic acid. In addition to this inductive effect there is also an electrostatic effect. When $\ce{X=CO2^-}$, we will have two negative charges in $\ce{B}$. This electrostatic effect further destabilizes $\ce{B}$. The combination of inductive and electrostatic effects in this last case will make this acid much less acidic than acetic acid.

$\endgroup$
  • $\begingroup$ I still have one last doubt. I edited my question and introduced $O_{2}^{-}C(CH_{2})_{4}COOH$. It's about the electron donor effect of $CO_{2}^{-}$. Can you check again my question and explain why this group acts like an electron donor instead of electron withdrawing like $F$? $\endgroup$ – 21Brunoh Jun 4 '14 at 0:03
  • $\begingroup$ In this case the $\ce{CO2^-}$ group is too far away to have an inductive effect. The inductive effect of the $\ce{(CH2)_{4}}$ group should be just a bit more electron donating than a methyl group (acetic acid) which should make this acid slightly less acidic than acetic acid. But we that this compound is much less acidic than acetic acid. Again the destabilizing electrostatic effect of having two negative charges in the same molecule must be the major factor making this acid so much less acidic than acetic acid. $\endgroup$ – ron Jun 4 '14 at 0:13
1
$\begingroup$

The last pka is for the negative ion? Isn't it obvious, that in that case leaving hydrogen cation has to overcome attraction from two negative charges instead of one and have troubles with it?

$\endgroup$
  • $\begingroup$ You're right, I was thinking in another compound, but the next quetions still remains, just forget the sentence "On the other hand the fact O−2CCH2COOH is less acidic than CH3COOH for me it's less intuitive". I'm going to edit the question and show you the other compound $\endgroup$ – 21Brunoh Jun 3 '14 at 22:55
  • $\begingroup$ I would be very careful with making any conclusions using the second compound. It has chain long enough to form internal cycle, requiring much more accurate estimation of effects other them +/- M +/- I. And the difference is low. $\endgroup$ – permeakra Jun 4 '14 at 6:11
1
$\begingroup$

I think both oxygens will have an inductive effect on the bonded carbon, but since the single bonded oxygen is also attached to a hydrogen, its inductive effect is decreased relative to that of the double-bonded oxygen. You can think of both oxygens as having an inductive effect, but the net inductive effect is definitely away from the carbon and in the direction of the double-bonded oxygen.

Also did you mean to write negatively charged diatomic oxygen? I think you meant to write $\ce{O^{2-}}$.

$\endgroup$
  • $\begingroup$ But still why the electron density from the single bonded $O$ tends to move towards the $C$? Because of eletrostatic interactions? Yes I meant to write the negatively charged diatomic oxygen, where you have one single bonded $O$ and a double bonded $O$ therefore existing withdrawing Resonance effect $-R$ I think $\endgroup$ – 21Brunoh Jun 3 '14 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.