Why is $\ce{H-OH}$ more acidic than $\ce{CH3CH2-OH}$? A common answer to this question is because of the $+I$-effect of $\ce{CH3}$ the negative charge at $\ce{O}$ increases making it less acidic. But if we compare electronegativity difference, the difference between $\ce{O}$ and $\ce{H}$ is more than that between $\ce{O}$ and $\ce{C}$ so there should be even more negative charge at $\ce{O}$ in case of $\ce{H2O}$. In this way $\ce{H2O}$ should be less acidic than $\ce{CH3CH2-OH}$. This is also the reason that $+M$ (electron donating ability) of $\ce{-OH}$ (which is having more negative charge at $\ce{O}$) is greater than the $+M$ of $\ce{-OR}$.
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asked Mar 19, 2015 at 19:06
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$\begingroup$ What data are you using to support this? According to Evans' pKa table methanol is slightly more acidic than water in water and notably more acidic in DMSO evans.harvard.edu/pdf/evans_pKa_table.pdf $\endgroup$– bonMar 19, 2015 at 19:15
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$\begingroup$ Yes, that's an exception. Edited the question. $\endgroup$– Ayush PateriaMar 19, 2015 at 19:20
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\begin{array} {|l|r|r|} \hline &\mathrm{pK_a~in~water} &\mathrm{pK_a~in~DMSO}\\ \hline \ce{H2O} &15.74 &31.4\\ \hline \ce{CH3CH2OH} &16 &29.8 \\ \hline \end{array}
As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in.
However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.
http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf http://www.chem.wisc.edu/areas/reich/pkatable/index.htm
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1$\begingroup$ Is solvation the only reason? How does inductive effect plays it role? And how will you explain ethanol being more acidic in DMSO? $\endgroup$ Mar 19, 2015 at 22:53