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.
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$\begingroup$ Could you mention where you saw the $\mathrm pK_\mathrm{a}$ values and what the values were? $\endgroup$– Safdar FaisalCommented Aug 5, 2020 at 5:52
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$\begingroup$ 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 $\endgroup$– AnaghaCommented Aug 5, 2020 at 6:28
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$\begingroup$ chemistry.stackexchange.com/questions/35068/…. $\endgroup$– Safdar FaisalCommented Aug 5, 2020 at 6:49
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$\begingroup$ 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)? $\endgroup$– AnaghaCommented Aug 5, 2020 at 6:59
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$\begingroup$ @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-. $\endgroup$– Solid - NMRCommented Aug 5, 2020 at 7:45
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1 Answer
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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
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$\begingroup$ Question in the title.. :). That has to be fixed in the given answer though. $\endgroup$ Commented Aug 5, 2020 at 7:56