0
$\begingroup$

If we consider the experimentally determined pka values, it looks like hydroxide ion should be the right answer but in many books it has been mentioned that all alcohols are less acidic than water implying that all alkoxide ions are stronger bases than hydroxide ion. This has created a lot of confusion.

$\endgroup$
  • $\begingroup$ Could you mention where you saw the $\mathrm pK_\mathrm{a}$ values and what the values were? $\endgroup$ – Safdar Aug 5 at 5:52
  • $\begingroup$ Yes sir. I got the values from Solomon and Fryhle organic chemistry book. pka of methanol is given as 15.5 and that of water as 15.7 $\endgroup$ – Anagha Aug 5 at 6:28
  • $\begingroup$ chemistry.stackexchange.com/questions/35068/…. $\endgroup$ – Safdar Aug 5 at 6:49
  • $\begingroup$ If methanol is more acidic than water does that not mean that it's conjugate base (methoxide) is less basic than the conjugate base of water (hydroxide)? $\endgroup$ – Anagha Aug 5 at 6:59
  • $\begingroup$ @Safdar Alright I do not know how much I am correct. I shall get back to this after some more reading. Apparently, NMR spectroscopy has proved that CH3- is electron widrawing in methanol. I am not sure. chemistry.stackexchange.com/questions/42535/… Otherwise the reason of basicity is same. CH3O- has I+ effect of the CH3- group causing it to be more basic than OH-. $\endgroup$ – Aditya Roychowdhury Aug 5 at 7:45
2
$\begingroup$

We have $\ce{MeO-}$ and $\ce{HO-}$

$\ce{-CH3}$ has a +I effect. Therefore it destabilizes the methoxide ion. $\ce{-H}$ has a comparatively lesser +I

This leads to the conclusion that $\ce{OH-}$ is more acidic than $\ce{CH3O-}$.

This seems like a valid argument, but the order varies in gaseous/aqueous phases, because we consider polarizabilty and solvation effect, rather than inductive effect there

| improve this answer | |
$\endgroup$
  • $\begingroup$ Question in the title.. :). That has to be fixed in the given answer though. $\endgroup$ – Safdar Aug 5 at 7:56

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