# How to determine the order of acidity of the following dimethyl nitrophenols

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?

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.
• 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. – Dennis Cao Jun 8 '18 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.
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 $$\ce{H+}$$.