joonkimdds

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how do I know whether it's faster or slower?
 

WOAHHI

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Hmm.. since nitration is an electrophilic aromatic substitution reaction it is affected by the other substituents on the benzene ring. In this case, you have a monosubstituted benzene ring with a CO2H. This is an electron withdrawing deactivating group. Therefore, it would decrease the rate of the nitration reaction. On the other hand, the nitration of Benzene would be faster because there are no deactivating groups and thus nothing to slow the rate of the nitration reaction. Thus, nitration of Benzene is faster than the first molecule.

By the way, what study material is this from? Thanks.
 

gentile1225

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nice reply woahhi! perfect explanation.
 
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doc3232

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Repeat the question but this time, instead of the COOH group, put a halogen (for example, bromine). Now, it is an interesting scenario.
bromine can actually donate its electrons in some resonance structures even though it deactivates the ring by being electronegative.
hence, it will direct ortho/para.
 

WOAHHI

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Repeat the question but this time, instead of the COOH group, put a halogen (for example, bromine). Now, it is an interesting scenario.
bromine can actually donate its electrons in some resonance structures even though it deactivates the ring by being electronegative.
hence, it will direct ortho/para.

o_O.. so in this case what would be the correct answer?

If I were to guess I would have to go off the fact that we have learned halogens to be deactivators, therefore the nitration of a benzene ring with a Bromine substituent will still be slower than the nitration of an unsubstituted benzene ring? Is this a possible test question? Or would we need experimental/modeling data to make a real decision here?
 

doc3232

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Repeat the question but this time, instead of the COOH group, put a halogen (for example, bromine). Now, it is an interesting scenario.
bromine can actually donate its electrons in some resonance structures even though it deactivates the ring by being electronegative.
hence, it will direct ortho/para.
o_O.. so in this case what would be the correct answer?

If I were to guess I would have to go off the fact that we have learned halogens to be deactivators, therefore the nitration of a benzene ring with a Bromine substituent will still be slower than the nitration of an unsubstituted benzene ring? Is this a possible test question? Or would we need experimental/modeling data to make a real decision here?
This is definately a fair question. bromine deactivates the ring (which is good experimentally so you can end up with a product, activators do NOT experimentally work). But it can donate its electrons by resonance, just like Oxygen would. The structure will look a bit weird, but it is correct (carbon double bonded to a Bromine). so ortho para directing.
 

WOAHHI

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Errr.. now I am sorta confused - I know that halogens are ortho / para directors but how does that tell us which reaction would occur faster (I thought that was the question)?

I also do not really know what you mean when you say that bromine deactivating the ring is good experimentally. I was under the impression that activating groups enhance substitution reactions and deactivating groups inhibit. Therefore, wouldn't activating the ring be better experimentally because the reaction is helped along a bit?
 

WOAHHI

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Also, I flipped through my Kaplan book and found that it classifies F, Cl, Br, and I as:

Deactivating, ortho/para-directing substituents (weakly electron withdrawing)

Does this mean that the electron withdrawing effects of these highly electronegative atoms outweighs the stabilization gained by the resonance with the lone pairs?
 

doc3232

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Also, I flipped through my Kaplan book and found that it classifies F, Cl, Br, and I as:

Deactivating, ortho/para-directing substituents (weakly electron withdrawing)

Does this mean that the electron withdrawing effects of these highly electronegative atoms outweighs the stabilization gained by the resonance with the lone pairs?
They are two separate concepts.
First, if a group is activating then the reaction will NOT work experimentally.
Because the benzene will pick up 1 group, but now it is more nucleophilic and hence will attack another molecule.
So if the group actually is eletron withdrawing (such as Bromine), now we will see if it is ortho/para or meta directing. Now we look at resonance structures. And they show ortho/para.
Makes sense?
 

joonkimdds

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I am guessing that halogens will slow the nitration because it's still deactivator, electron withdrawing regardless of the fact that it causes ortho and para.

WOAHHI//I got this question from Kaplan.
 

sciencegod

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I am guessing that halogens will slow the nitration because it's still deactivator, electron withdrawing regardless of the fact that it causes ortho and para.

WOAHHI//I got this question from Kaplan.
its very simople. anything that has a deactivator will react slower thant benzen and activator faster.
 

joonkimdds

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its very simople. anything that has a deactivator will react slower thant benzen and activator faster.
I guess that's where they came with the name activator and deactivator :D

Funny thing is that I found the similar question from destroyer #70. I guess it's good to use more than 1 book to study :)
 
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INH

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I believe electron withdrawing groups, which direct metta (except for halogens which direct ortho/para) tend to be deactivating towards future reactions in EAS reactions. In contrast, e- donating groups woud be activating causing the reaction to occur faster than in would have for a deactivator
 

supergenius310

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I believe electron withdrawing groups, which direct metta (except for halogens which direct ortho/para) tend to be deactivating towards future reactions in EAS reactions. In contrast, e- donating groups woud be activating causing the reaction to occur faster than in would have for a deactivator
Dude this is 5 years old rofl
 
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