alanan84

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Kaplan's got toluene reacting with HNO3 in H2SO4. Then it says that 63% of the product will be with NO2 at the ortho position and 34% at the para position. Shouldn't it be the other way around?
 

jefff

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Kaplan's got toluene reacting with HNO3 in H2SO4. Then it says that 63% of the product will be with NO2 at the ortho position and 34% at the para position. Shouldn't it be the other way around?
I believe it should be the other way around also in order for it to be further from the carbonyl
 

jefff

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Yes, I agree that its gotta be the other way around... but,

Where is the carbonyl? Isn't it due to the steric hindrance from CH3?
you're right i apologize :\
i was thinking of the wrong structure
 
May 22, 2009
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Kaplan's got toluene reacting with HNO3 in H2SO4. Then it says that 63% of the product will be with NO2 at the ortho position and 34% at the para position. Shouldn't it be the other way around?
No it's not gonna be the other way around, CH3 doesnt really have steric hindrance, so you get equal substitution at the 2 ortho and one para positions, all equally, but you gotta realized u have TWO ortho positions and ONE para. They're all getting equal (33%), but since there's two ortho, you have ~66% ortho and ~33% para. Close enough ;p
 
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alanan84

alanan84

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No it's not gonna be the other way around, CH3 doesnt really have steric hindrance, so you get equal substitution at the 2 ortho and one para positions, all equally, but you gotta realized u have TWO ortho positions and ONE para. They're all getting equal (33%), but since there's two ortho, you have ~66% ortho and ~33% para. Close enough ;p
Ah ha! I get it, thanks!
 

jefff

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No it's not gonna be the other way around, CH3 doesnt really have steric hindrance, so you get equal substitution at the 2 ortho and one para positions, all equally, but you gotta realized u have TWO ortho positions and ONE para. They're all getting equal (33%), but since there's two ortho, you have ~66% ortho and ~33% para. Close enough ;p
nice explanation, thanks
 
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alanan84

alanan84

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No it's not gonna be the other way around, CH3 doesnt really have steric hindrance, so you get equal substitution at the 2 ortho and one para positions, all equally, but you gotta realized u have TWO ortho positions and ONE para. They're all getting equal (33%), but since there's two ortho, you have ~66% ortho and ~33% para. Close enough ;p
I guess I just assumed you'd have more of the para product than the ortho b/c para is more stable...

So you're saying if that question said "Pick the major product" it would be both products in equal proportions?
 
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I guess I just assumed you'd have more of the para product than the ortho b/c para is more stable...

So you're saying if that question said "Pick the major product" it would be both products in equal proportions?
I guess you'd determine major product by % of which one u get right? I guess here it would be ortho since you're getting 66% ortho, and the two substitutions are the same.

I'm not too sure about para being more stable. The reason you get more para sometimes is due to the steric hindrance of the activator/inactivator. For example if you have a t-butyl group, it's o/p directing, but you're not gonna get much ortho due to the hindrance of the t-butyl group, then the para is likely to be the major product.
 
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alanan84

alanan84

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I guess you'd determine major product by % of which one u get right? I guess here it would be ortho since you're getting 66% ortho, and the two substitutions are the same.

I'm not too sure about para being more stable. The reason you get more para sometimes is due to the steric hindrance of the activator/inactivator. For example if you have a t-butyl group, it's o/p directing, but you're not gonna get much ortho due to the hindrance of the t-butyl group, then the para is likely to be the major product.
Got it, makes perfect sense.