enter image description here

This is how I went about it, although I got a wrong answer:

  1. 3,5-Dimethyl-4-nitrophenol (2) shows the nitro group to be sterically hindered, due to which it should go out of plane and this will be an example of the steric inhibition of resonance. Hence, there will only be a weak withdrawing inductive effect, and no resonance. Thus, this should be the least acidic.

  2. 4,5-Dimethyl-2-nitrophenol (3) shows a hydrogen bonding between the nitro and the hydroxyl group, due to which there will be a lower tendency to release the proton. Hence, even this should be less acidic.

  3. Between 2,6-dimethyl-3-nitrophenol (1) and 2,4-dimethyl-5-nitrophenol (4), (1) should be weakly acidic due to the hyperconjugation destabilisation caused by the two methyl groups, and hence the answer should be (4), with lower hyperconjugation, lower induction from the methyl groups, and higher induction from the nitro group (no resonance at the meta position).

However, the answer is given to be (2), which I ruled out first.

Can someone please explain what is happening, and where I've gone wrong?


2 Answers 2


The right answer should be 3 as the $\ce{NO2}$ group has a negative mesomeric effect, which is only possible with 3. In 2, due to steric inhibition of resonance, it exhibits only -I effect, and although 3 contains H-bonding, it slightly affects acidity.

  • 1
    $\begingroup$ So hydrogen bonding is less effective than the -M effect, you say? Can we say that although the hydrogen bonding decreased the acidity slightly, it is still greater than the -I effect groups due to the -M effect? However, this answer doesn't match with the given answer either... $\endgroup$
    – Abhigyan
    May 31, 2018 at 3:28
  • $\begingroup$ Turns out that 2 is slightly more acidic than 3. $\ce{pKa(2) = 8.25 @ 25 ºC}$ source:10.1021/ja01187a054 $\ce{pKa(3) = 8.55 @ 28 ºC}$ source:10.1021/ja01161a043 I could not find data for compounds 1 and 4. $\endgroup$
    – Dennis Cao
    Jun 8, 2018 at 19:22

$\ce{-NO2}$ is an electron-withdrawing group and $\ce{-CH3}$ is an electron-donating group and the effect of these groups can be observed at ortho- and para- position of the ring with respect to the functional group attached to the ring.

enter image description here

enter image description here

in the above resonating structure of toluene you can see -ve charge in o-, p- position, similarly in nitrobenzene you can see positive charge in o-, p- positions, and $\ce{-OH}$ group attached to electron deficient carbon or positively charged carbon release $\ce{H+}$ more easily. Considering all this condition second option is the only option where carbon to which $\ce{-OH}$ is attached is getting positive charge and not getting a negative charge, so $\ce{-OH}$ is ready to loose $\ce{H+}$.

  • $\begingroup$ Are you familiar with Steric Inhibition of Resonance ? $\endgroup$
    – Soumik Das
    May 30, 2018 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.