I've been studying acids and bases in organic chemistry. My textbook says that alcohols are very weak acids, therefore they cannot react with strong bases. the conclusion i formed from this was: very low acid / base strength leads to not being able to react with stronger bases / acids. But then, while reading the chapter about Amines, i read that aromatic amines (such as aniline) ONLY react with strong acids. Therefore my question is: why is it that very weak acids cannot interact with strong bases, however very weak bases can ONLY interact with strong acids?
5 Answers
Qualitative attributes like strong, weak, very weak, very very weak (acid/base) have no widely accepted agreement/definition what it exactly means quantitatively. As consequence, different sources may give different boundaries between them.
That is the reason why you have ended with different evaluation for acids and bases.
In context of water solutions, acids have the characteristic acidity parameter $\mathrm{p}K_\mathrm{a}$. It is equal to $\mathrm{pH}$ when $\pu{50 \%}$ of the acid is dissociated.
\begin{array} {|l|c|} \hline \mathrm{pH} & \text{% of acid dissociation} \\ \hline \mathrm{p}K_\mathrm{a} - 3 & \approx \pu{0.1 \%} (\frac{1}{1001}) \\ \hline \mathrm{p}K_\mathrm{a} - 2 & \approx \pu{1 \%} (\frac{1}{101}) \\ \hline \mathrm{p}K_\mathrm{a} - 1 & \approx \pu{9 \%} (\frac{1}{11}) \\ \hline \mathrm{p}K_\mathrm{a} & \pu{50 \%} (\frac{1}{2}) \\ \hline \mathrm{p}K_\mathrm{a} + 1 & \approx \pu{91 \%} (\frac{10}{11}) \\ \hline \mathrm{p}K_\mathrm{a} + 2 & \approx \pu{99 \%} (\frac{100}{101}) \\ \hline \mathrm{p}K_\mathrm{a} + 3 & \approx \pu{99.1 \%} (\frac{1000}{1001}) \\ \hline \end{array}
Mutual reactions of acids with bases have to be assessed in sense what $\mathrm{pH}$ wrt the acid $\mathrm{p}K_\mathrm{a}$ is set by the base excess.
In context of Broensted-Lawry acid-base theory, bases accept H+ ion:
$$\ce{B(aq) + H+(aq) <=> BH+(aq)\text{(conjugate acid)}}$$
where $\ce{B}$ may be a molecule or a ion.
If the base has the basicity constant $\mathrm{p}K_\mathrm{b}$ then its conjugate acid has the acidity constant (at $\pu{25 ^\circ C}$)
$\mathrm{p}K_\mathrm{a, conj}$ = 14 - $\mathrm{p}K_\mathrm{b}$
We can then compare the relative strength of the given acid and the conjugate acid of the given base. (That applies on a solvent too, if it can act as a base like water)
For the stoichiometric ratio of the given acid and base, we can estimate the degree of neutralization.
\begin{array} {|c|r|} \hline \mathrm{p}K_\mathrm{a,conj} - \mathrm{p}K_\mathrm{a} & \text{Neutralization degree} \\ \hline 8 & \pu{0.01 \%} \\ \hline 6 & \pu{0.1 \%} \\ \hline 4 & \pu{1 \%} \\ \hline 2 & \pu{9 \%} \\ \hline 0 & \pu{50 \%} \\ \hline -2 & \pu{91 \%} \\ \hline -4 & \pu{99 \%} \\ \hline -6 & \pu{99.9 \%} \\ \hline -8 & \pu{99.99 \%} \\ \hline \end{array}
There exist several sorts of weak acids. They can be weak, very weak and very very weak acids.
Weak acids react with weak and strong bases.
Very weak acids do not react with weak bases. They do react with strong bases.
Very very weak acids do not with weak and with ordinary weak bases.
Ethanol is such a very very weak acid. It does not react significantly with strong bases. I only reacts with sodium metal in a redox reaction.
Aniline is a weak base, but not a very very weak base. It reacts with strong acids.
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1$\begingroup$ I think I understand now. Alcohols are very very weak acids so they do not react with weak bases at all, and their reactivity to strong bases is negligible. Since my textbook said that alcohols do not react with strong bases, I thought that they could interact with weaker ones (if A is not B then it is C). Is it fair to conclude that if something does not have significant reactivity to strong bases, then as a consequence it will NOT neutralize a weak base? (sorry if the questions seem trivial, I am a high school student barely scratching the surface of chemistry) $\endgroup$– mralCommented Mar 3, 2023 at 7:27
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$\begingroup$ @mral The essential condition is what "does not react" mean quantitatively. The threshold is a subjective, deliberate parameter. If fraction $\pu{e-3}$ reacted, it reacts. What about $\pu{e-4}$? $\pu{e-6}$? $\pu{e-10}$? $\pu{e-15}$? Any molecule? $\endgroup$– PoutnikCommented Mar 3, 2023 at 7:41
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$\begingroup$ The weak base $\ce{NH3}$ with amount concentration $\pu{1 M}$ has $\mathrm{pH} \approx 14 - 4.75/2 \approx 11.63$. If I added a small amount (wrt $\ce{NH3}$) of a very weak acid $\ce{HA}$ with $\mathrm{p}K_\mathrm{a} = 12.63$, it would have $\pu{\approx 9 \%}$ in dissociated form $\ce{A-(aq)}$, complemented by $\ce{NH4+(aq)}$. So very weak acids do react with weak bases. // Similarly, Aniline would react with acetic acid too. $\ce{Ph-NH3+}$ has pKa 4.63, acetic acid 4.75. In solution with equimolar concentration, both pairs would have roughly equal concentration of respective acid/base. $\endgroup$– PoutnikCommented Mar 3, 2023 at 11:32
Well, weak acids are even worse with weak bases. So try logic from 2 or 3 points rather than one.
The alcohol proton is very weakly acidic, pKa 17, but much more so than hydrogen covalently bonded to alkane, around pKa 50.
A weaker base, such as ammonia, pKa 9.3, will not touch it, but a very strong base, such as lithium diisopropyl amide, pKa 35, will remove it.
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1$\begingroup$ pKa 9.3 for ammonia $\ce{NH3}$, without mentioning it is the acidity constant of ammonium $\ce{NH4+}$ , may be confusing for the OP. $\endgroup$– PoutnikCommented Mar 3, 2023 at 6:59
To add to the existing answers, you can make the statement in your textbook more correct by adding "even":
Alcohols are very weak acids, therefore they cannot react with even strong bases.
Without that word, it was somewhat misleading, naturally leading to your question.
Your question employs the Bronsted theory, where acids are proton donors, and bases are proton acceptors. Similar to e.g., a farmers' market with sellers and buyers, acid and base can action only in presence of the other. pKa values are a convened scale to sort them according to the likelihood they exchange goods (here, the theory deals with protons) in a set of parameters (temperature, solvent, ...).
Mayr's scales of nucleophilicity and electrophilicty numbers of reagents (link to the project / project's database) illustrate «trading criteria» can equally be applied elsewhere.