Usually when an ortho-para directing substituent is present on the benzene ring for an electrophilic aromatic substitution reaction, the para product is the major product (exceptions can be there when hydrogen bonding or ortho effect of COOH group makes the ortho product a major one.) But I don't understand why the ortho product in nitration of toluene is the major product (and the para product isn't). The ortho position is supposed to have more steric crowding, right?

  • $\begingroup$ Now I know nothing about this particular nitration but one thing that tends to make ortho products favorable is the inductive effect, due to which ortho position of toulene is slightly more activated than para. $\endgroup$
    – Sawarnik
    Apr 9, 2017 at 18:49
  • $\begingroup$ i read the answers and now I'm confused . If this is the reason then what if there is an ethyl group instead of methyl . Ortho position should be major because of +I effect but this is not the case as para is major .why is that $\endgroup$
    – utkarsh
    Apr 10, 2017 at 12:19
  • $\begingroup$ @utkarsh Then steric hindrance increases and the nitro group will go to the para position. $\endgroup$
    – dr.drizzy
    Mar 2, 2018 at 20:29
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/88752/… $\endgroup$ Mar 3, 2018 at 3:43

4 Answers 4

  1. A methyl-group is not t-butyl, so it isn't all that crowded.
  2. There are two ortho-positions.
  3. There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.

Given all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.


There are three effects: sterics, electronics and statistics.

Yes, there is more steric crowding in the ortho position, but a methyl group isn't that big. In addition, electronic effects are stronger in ortho position than in para, since it is much closer to the +I substituent. And last but not least: there are two ortho positions which can react, but only one para-position, which can have a rather big influence if reactivity towards ortho and para position is similar.


One other factor favoring the ortho isomer: hydrogen bonding. Yes, hydrogen bonding to the methyl group.

Following the molecular orbital description of hydrogen bonding, an electron pair from the nitro group oxygen atoms can overlap the antibonding orbitals in the methyl groups, delocalizing what are formally the carbon-hydrogen bonds in the methyl group. We can represent this interaction with an additional contributing structure that no octet-deficient atoms:

enter image description here

A para, or for that matter meta attack would not allow this delocalization to take place due to the remoteness of the nitro and methyl functions.


Ortho nitrotoluene is a major product while para nitrotoluene is a minor product . Reasons: In case of reactivity: Ortho product is more reactive comparing with para product because electron density is more enriched at ortho position compared to the para position. In case of stability: Para product is more stable because is away from substituent group so it does not affected by steric hinderence as the ortho position I'd more steric hindered.


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