Comparing acidic strength between 5 alcohols and carboxylic acids

The correct answer is E

But I do not understand why? Shouldn't (v) have the lowest $\mathrm{p}K_\mathrm{a}$ since it does not have a carboxyl group making it a strong acid? How do we determine the strength of acids?


First of all, let's make something crystal clear: The more acidic is a compound, they smaller will be its $pK_a$.

To evaluate the acidity of these compounds you need to analyze the stability of the corresponding conjugate base of each one, in effect, the stability of the same compound but without a proton and a negative charge on an oxygen atom.

First of all, you have two kinds of compounds:

  • Two alcohols: V and II
  • Two carboxylic acids: I, III and IV

From those compounds, alcohols will always be less acidic than carboxylic acid. Why? Because the negative charge left by a proton leaving the molecule can be delocalizaed by resonance troughout the carbonyl, but in the case of the alcohols it will be localized on the only oxygen atom, which makes the conjugate base less stable, and therefore, the compound less acidic.

That being said, from the two alcohols, V will be more acidic, because the presence of the chlorine atoms substract charge from the carbon they are bond to (chlorine is more electronegative than carbon), and helps stabilizing the negative charge that is left in the molecule after the proton is substracted.

In regard to the carboxylic acids, you can use a similar reasoning for the carboxylic acid. The less acidic is the one in which R is just a hydrocarbon chain, which has no effect in contrast to I, which has 1 chlorine atom that stabilizes the negative charge, and to III, which has 3 chlorine atoms that stabilize the negative charge of the conjugate base even more.

Therefore, the order presented in option E is justified.

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