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?
- A methyl-group is not t-butyl, so it isn't all that crowded.
- There are two ortho-positions.
- 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.
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.