According to my teacher, the inductive effect in $\ce{CH3COOH}$ exists because of the electron deficiency on $\ce{C}$ (partial positive charge $\delta +$) as a result of the electronegativity of the double bonded $\ce{O}$. Although the single-bonded $\ce{O}$ is more electronegative than $\ce{C}$, the electron density of the single bonded $O$ moves to $C$ in order to “neutralize” the $\delta+,$ leaving $\ce{H}$ less attached to the molecule:
Why do we consider that double-bonded $\ce{O}$ establishes that $\delta+$ on our $\ce{C}$ (and creates an electron withdrawing inductive effect −I) and not the single-bonded $\ce{O}$? Is this because of the double π-bond? Why?
Now supose we have the following carboxylic acids:
$$ \begin{array}{rr} \hline \text{Acid} & \mathrm{p}K_\mathrm{a} \\ \hline \ce{CH3COOH} & 4.80 \\ \ce{CH2FCOOH} & 2.66 \\ \ce{^-OOCCH2COOH} & 5.69 \\ \ce{^-OOC(CH2)4COOH} & 5.41 \\ \hline \end{array} $$
For me it makes sense that $\ce{CH2FCOOH}$ is more acidic than $\ce{CH3COOH}$ because $\ce{F}$ is an electronegative element that creates a partial negative charge attracting the electron density and leaving $\ce{H}$ in the carboxyl group less attached and easier to remove.
On the other hand, the fact $\ce{^-OOCCH2COOH}$ is less acidic than $\ce{^-OOC(CH2)4COOH}$ seems less intuitive. My teacher's presentation states that $\ce{O2^-}$ linked to an alkyl chain will create an electron donating inductive effect +I. Why? Isn't there a withdrawing resonance effect −R?
If $\ce{O}$ is an electronegative element, then why doesn't it create an −I effect instead of +I? Because if that was the case adding carbons $\ce{(CH2)4}$ in the chain would reduce the the −I effect and the $\ce{^-OOC(CH2)4COOH}$ would be less acidic than $\ce{^-OOCCH2COOH}$, and that isn't the case.