What would be the order of $\mathrm{p}K_\mathrm{a}$ values of the following:
- $\ce{O2N-CH2-CO2H},$
- $\ce{Me3\overset{+}{N}-CH2-CO2H},$
- $\ce{NC-CH2-CO2H},$
- $\ce{HO-CH2-CO2H}?$
Since only the groups attached to $\ce{-CH2-CO2H}$ are changing, we must look at the point of difference in each group and I was able to figure out that since there is no conjugation, we cannot compare these groups on the basis of the mesomeric effect.
So, we must compare them on the basis of negative inductive effect as the more stable the conjugate base will be, the more acidic our compound is and hence the lesser is its $\mathrm{p}K_\mathrm{a}$ value. But I am not able to come to a final conclusion which matches with what's given in my book.
Is there some other factor playing role in this as well? Because if we just look at it on the basis of inductive effect, the order for increasing values of $pK_a$ should be
$$\ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{O2N-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
which seems to be wrong.
The order for the increasing $pK_a$ values given in my book is:
$$\ce{O2N-CH2-CO2H} < \ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
Note that the order given in my book may be subject to errors.
EDIT:
After pondering over this for a lot of time and with no convincing answers or explanations, I finally came to a potential explanation for this. Could Hydrogen bonding be the reason for the compound with $-NO_2$ to be more acidic than the one with $-Me_3\overset{+}{N}$ group ?
EDIT-2:
Thanks to Buck, it's now clear that my book's answer is right. Could someone please give a possible explanation to this from organic point of view?