Lecture 10 Halogen substituents in electrophilic aromatic substitution Halogens are deactivator and o p directors Why To understand that we need to understand how they interact with a cations For methoxy substituent we said that there is an inductive effect which takes electron density away from the rest of the molecule and this effect destabilizes the cation The second effect is a resonance effect and this effect stabilizes the cation With O N S and P substituents resonance effect is strionger and these substituents are activators and o p directors Overall the cation is more stable with Ome than with a secondary cation without this substituent With halogenes there is the same inductive effect and the resonance effect However with all halogens the inductive effect is stronger than the resonance effect so the overall result is that halogens are mildly destabilizing for the cation Important to note that the resonance effect is still stabilizing but cannot overcome the inductive effect In elecrtophilic aromatic substitution in ortho and para position to the halogen substituent halogen can exert both the inductive and resonance effect the result will be that ortho and para positions are mildly deactivated relative to benzene This shows the reaction in ortho position Same is for the para position ends up right next to Br Substitution in the meta position never places the charge next to bromine so bromine can only exert the inductive effect The net effect is that meta position is deactivated inductive effect without activating resonance effect As a result the meta position is most deactivated one and the halogens are ortho para directors For export Page 1 Lecture 10 Monday January 26 2015 2 35 PM When there are two or more substituents in the molecule there are 3 general situations Work together Selectivities reinforce each other both substituent prefer the substitution at the same position in the molecule Stronger one wins When the substituent do not agree where the substitution should go then the stronger effect wins That s why it is good to know from the table who is strong activator week activator and so on Also if you understand inductive and resonance effects you can figure it out on your own Sterics If electronic effects of the substituents is roughly the same and there is no strong preference for one product over the other than steric effects come in to play In this case substitution exclusively away from the bigger t Bu group This allows us to plan and execute a synthesis of molecules like this one In this case only route b will give us the two substituents in the desired orientation So starting from benzene we have to introduce acyl group first and then do the bromination The other way would give us orto para substituted product For export Page 2 Lecture 10 Monday January 26 2015 2 35 PM Order of reactions in synthesis of aromatic compounds Because Br and Acyl direct electrophilic aromatic substitution the order in which we introduce the two substituents matters We can get the meta product only by first introducing the acyl group and then use its presence to direct the bromination to the meta position We also have to worry about the reactivity we may have to introduce the substituents in specific order to avoid the problems with reactivity Both groups are meta directing so we should be fine either way The problem is that Friedel Crafts alkylation and acylation do not happen with strongly deactivated arene Halo arenes are ok So nitrobenzene would not work As a result only path B works First we introduce the acyl group than the nitro group For export Page 3 Lecture 10 Monday January 26 2015 2 35 PM Conversion of one functional group to the another with different selectivity It s good to know that we can convert some functional groups into another and in the process change the way the substituent effects electrophilic aromatic substitution Examples In both cases we go from meta directors to o p directors Think of an example where we go in the other direction For export Page 4 Lecture 10 Wednesday January 28 2015 4 28 PM Nucleophilic aromatic substitution Aromatic halides can also undergo nucleophilic aromatic substitution But the mechanism is completely different than for Sn2 and Sn1 reactions In this case there is no Sn1 because Aryl cation would be very unstable SN2 is not posible because it is not possible to have a back side attack That s where the aromatic ring is In general there is no SN2 on sp2 hybridized carbons Despite all this the Nucleophilic substitution can still work However the mechanism is different and involves addition followed by eleimination Addition is the slow step Remember arene does not want more electron Addition elimination mechanism For export Page 5 Lecture 10 Wednesday January 28 2015 5 07 PM Consistent with this mechanism is the fact that this reaction requires the presence of a strong electron withdrawing group Nitro cyano acyl And more such groups there are faster the reaction Also consistent with the mechanism is the fact that you want electron withdrawing group ortho or para to the leaving group That way the electron withdrawing group can have the full inductive and resonance effect on the stability of the anion In ortho and para position the anion can actually makes it s way to the carbon directly attached to the electron withdrawing group Substituent in a meta position is not good enough because it cannot effectively stabilize the negative charge in the ring In this case only inductive effect is operative and that is not strong enough to support this mechanism Also consistent with the mechanism is the order of reactivity of aryl halides Fluoride is the best at stabilizing the negative charge that develops in the first step And that is why it is most reactive in these types of reactions Fluoride is the worst leaving group but that step is fast and does not matter for the overall rate of the reaction Leaving group reactivity Finally these reactions have a second order kinetics which is also consistent with the addition elimination mechanism For export Page 6