Why is ortho more stable than para

The most stable carbocations in each case will look like those below. Make sure you know how to obtain these using curved arrows! I will spare you from putting a third big diagram in here to describe halogens like F, Cl, Br, and I, but they are often a source of confusion.

Sometimes your ortho product is more than your para. But for this reaction the para product is the major product. And once again steric hindrance is one factor to think about when you're doing these kinds of reactions here. So there's of course another position on your aromatic ring. So if we installed the nitro group on this position, we would call this the meta product. And the meta product is not observed in high yield for this reaction. So we say that the methoxy group is an ortho para director.

And we could also label this as being the ortho position on this side. And we could say this is the meta position because of symmetry. But the meta product is not seen in a large yield, in a high yield.

And let's go ahead and look at why by drawing a bunch of resonance structures and thinking about the mechanism for electrophilic aromatic substitution. And so let's go ahead and start with an ortho attack. And so we know that when you're doing a nitration the sulfuric acid acts as a catalyst to generate the nitronium ion from nitric acid. And that functions as the electrophile in your mechanism here. So if we're going to do an ortho attack, we need to show the nitro group adding onto the ortho position.

So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here.

I'm showing the nitro group adding onto the ortho position. And remember there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. We can show some resonance structures. So we can show some resonance stabilization of this cation here.

So I could show these pi electrons moving over to here. And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here.

So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro group in the ortho position. And we have these pi electrons here. And now we show the movement of those pi electrons over to here. So let me go ahead and highlight those. Already Have an Account? Why is Ortho xylene more stable than para xylene? Class 12 Question. Answer to Question. Prakhar Maheshwari Sep 11, Methyl is an ortho-para directing group.

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