Arrange the following organic compounds in decreasing order of acidity.

P: 4‐methylphenol; Q: 4‐ethylphenol; R: 4‐(propan‐2‐yl)phenol; S: 4‐tert‐butylphenol


S, R, Q, P.


According to my knowledge the answer should be P, Q, R, S. As we proceed from P to S we are increasing the number of methyl groups, and so the number of electron donating groups increases, decreasing the polarity of of the C−O bond, which is the opposite to the answer. This obviously means that whatever I'm thinking is the right reason is the wrong logic, but I'm not sure how.

  • $\begingroup$ The major factor is hyperconjugation and not inductive effect. $\endgroup$ – Nisarg Bhavsar Apr 30 at 12:09

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions.

A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S.

This means that the number of destabilizing structures that are produced via this resonance decreases as you go down the list. This leads to greater stability in the deprotonated form compared to each other.

Therefore the correct order becomes S, R, Q, P.


For alkyl groups attached to the phenol ring, there are two main points that your reasoning does not take account of -

  1. Inductive effect is distance dependent. As the distance increases, its effect becomes very small.
  2. Even if the group were near, the general trend is that hyperconjugation has stronger effect than inductive effect.

You can refer the following links for an explanation on hyperconjugation - https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hyperconjugation https://en.wikipedia.org/wiki/Hyperconjugation

More the number of hydrogen atoms attached to the benzyl carbon, more is the number of hyperconjugative structures possible and more is the withdrawal of positive charge (or equivalently, donation of negative charge) into the benzene ring.

Phenol forms phenoxide ion on losing H+ and the phenoxide tries to delocalize its negative charge into the ring. If hyperconjugation already donates electrons into the ring, the extent of delocalization decreases, making phenoxide less stable and hence phenol less acidic.

As a result, less the number of benzylic hydrogen, more is the stability of phenoxide and more the acidity. That explains the order S > R > Q > P.

NOTE : Hyperconjugation, like resonance (mesomeric effect), operates only at the ortho and para positions. In case the same groups were attached at meta, then hyperconjugation would have no role and your order based on inductive effect would be correct.

  • $\begingroup$ Are you implicitly saying hyperconjugation dominates the delocalization of electron density of O? Why should hyperconjugation even take place when a phenoxide ion is required to be formed? Your answer makes sense, but I have a doubt about this now. $\endgroup$ – ljm Apr 30 at 10:01
  • $\begingroup$ I meant to say that due to hyperconjugation the electron density in the benzene ring increases. As a result the extent of delocalization of electron density of the O atom is decreased. As to why hyperconjugation takes place, it can't be controlled (as far as I know) and takes place regardless of any other substituents. $\endgroup$ – TRC Apr 30 at 11:12

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