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Question regarding one organic chemistry question. Help please!

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ColumbiaOrtho

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Hi, I've been working on one orgo problem and I just could not figure out the reasoning behind the textbook answer. The answers above are for I) meta-Nitrotoluene, and as you can see I thought the bromine halogenation would occur at the meta position of nitro group since nitro group is a strong deactivator that directs meta position whereas methyl is a weak activator that directs orto-para position. However, the textbook shows that the halogenation would still occur at the ortho position of methyl. Why is this happening?
 

Syxx

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Activating groups will typically have a greater directing effect over deactivating groups. Activating groups donate electrons into the ring which makes them more reactive towards electrophiles (which Br2 in FeBr3 is) at the ortho and para positions. And that position is least sterically hindered of the available ortho/para positions to the methyl so the bromine ends up there.

And a little more detail. If you remember aromatic substitution reactions go through an arenium ion intermediate. The major product will be the substituted position that gives the most stable intermediate. If you were to draw the intermediates possible for this reaction you would see that the 2-bromo-5-nitrotoluene and 2-bromo-4-nitrotoluene products give an intermediate with the positive charge placed next to the electron donating alkyl group. These will be your most stable intermediate and the route the reaction will go through. And due to steric effects 2-bromo-5-nitrotoluene would be your most favorable product. The 3-bromo-5-nitrotoluene product will give you no such favorable intermediate structures.
 
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ColumbiaOrtho

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Activating groups will typically have a greater directing effect over deactivating groups. Activating groups donate electrons into the ring which makes them more reactive towards electrophiles (which Br2 in FeBr3 is) at the ortho and para positions. And that position is least sterically hindered of the available ortho/para positions to the methyl so the bromine ends up there.

And a little more detail. If you remember aromatic substitution reactions go through an arenium ion intermediate. The major product will be the substituted position that gives the most stable intermediate. If you were to draw the intermediates possible for this reaction you would see that the 2-bromo-5-nitrotoluene and 2-bromo-4-nitrotoluene products give an intermediate with the positive charge placed next to the electron donating alkyl group. These will be your most stable intermediate and the route the reaction will go through. And due to steric effects 2-bromo-5-nitrotoluene would be your most favorable product. The 3-bromo-5-nitrotoluene product will give you no such favorable intermediate structures.
Thank you very much!
 

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Activating groups will typically have a greater directing effect over deactivating groups. Activating groups donate electrons into the ring which makes them more reactive towards electrophiles (which Br2 in FeBr3 is) at the ortho and para positions. And that position is least sterically hindered of the available ortho/para positions to the methyl so the bromine ends up there.

And a little more detail. If you remember aromatic substitution reactions go through an arenium ion intermediate. The major product will be the substituted position that gives the most stable intermediate. If you were to draw the intermediates possible for this reaction you would see that the 2-bromo-5-nitrotoluene and 2-bromo-4-nitrotoluene products give an intermediate with the positive charge placed next to the electron donating alkyl group. These will be your most stable intermediate and the route the reaction will go through. And due to steric effects 2-bromo-5-nitrotoluene would be your most favorable product. The 3-bromo-5-nitrotoluene product will give you no such favorable intermediate structures.
I have carefully drawn out all the intermediates to possible products. The position ortho to the methyl as well as para to the methyl will give intermediates that are similar in energy. Clearly the intermediate argument is invalid. One can try to argue steric. However, the methyl group has a greater steric demand than a nitro group. Referencing the Advanced Organic Chemistry text by Jerry March and using values of CH3 ...1,7 kcal/mole vs NO2 of 1.1 Kcal/mole...even though on a cyclohexane,,,,,,it gives us an idea that Methyl is more sterically demanding. Alas.....both our explanations fail. Welcome to the world of Advanced Organic Chemistry. This is an example that most orgo teachers do NOT want to give because it is an anomaly. Usually we can predict product ratio by examining the stability of the arenium ion intermediate, but not here. The answer would clearly lie in orbital theory where we would have to examine what is called the coefficients .As you can see, the " usual " organic chemistry by resonance theory fails here. Proper Huckel theory is not particularly good when there are strongly electronegative atoms present such as in the case you have shown. The product ratio depends on the ENERGIES of the higher occupied orbitals. One would require fairly sophisticated Quantum Mechanic calculations to show the fact that the position ORTHO to the methyl group is electronically favored. The bottom line is this......This is not a DAT question, but a PhD question that is still being argued today. I hope this gives you some idea to the fact that Quantum Mechanics often is invoked when anomalies arise. However, for most purposes, The Huckel and Resonance theory works quite well. If you would like to go deeper into this, the book by Ian Fleming , Director of Organic Chemistry at Cambridge discusses this in great detail in the book Frontier Orbitals and Organic Chemical Reactions.

Hope this helps

Dr. Romano
 
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