Chemistry - Difference in acid strength of oxalic acid and malonic acid

Solution 1:

Linear saturated dicarboxylic acids

$n$ = number of $\ce{-CH_2}-$ groups in chain

 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724    
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0 
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid

Aromatic carboxylic acids

                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ---- 
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid 
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids

$n$ = number of $\ce{-CH=CH}-$ trans groups in chain

n                            pka1      pka2 
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87              
3                   

Straight-chain, saturated carboxylic acids

$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).

n                     pKa   
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8 

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the carboxylate ion to solvate, which increases its acidity.

In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization , the second ionization

Solution 2:

First, notice that a carboxylic carbon ($\ce{COOH}$) has an oxidation state of $\mathrm{+III}$ (except in formic acid where it is $\mathrm{+II}$). This means, that the carbon is very electron deficient and wishes to draw electrons towards it from sources that are not the two oxygen atoms (because they are stronger). We term this an inductive effect, as we cannot easily draw mesomeric structures to explain it.

Since the inductive effect is mainly based on electrostatic interactions, it gets exponentially weaker with increasing distance. In oxalic acid, there is one one bond separating the two carboxylic groups, so they can exert a strong inductive effect. In malonic acid the effect is weaker due to the separating $\ce{CH2}$ group. And in succinic and subsequent diacids it is almost unmeasurable due to the distance between the groups.

But how do we rationalise each group exerting a $+I$ effect on the corresponding other group? Well — we cannot deprotonate twice immediately (I am introducing this as an axiom, but it does have reasoning that I feel too lazy to explain right now). So we have to pretend one side is going to be inert while the other is deprotonated. And if we do that, we see that the inductive effect of one side is strong in drawing away electron density from the $\ce{COOH}$ group we are deprotonating. Since this means there is a lower effective negative charge, deprotonation happens more easily, i.e. at lower $\ce{pH}$.

We can use this simple picture since the groups are isotopic, i.e. they can be transformed into each other by a $C_2$ rotation of the molecule’s inherent symmetry.