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Chem 263 Oct Chem 263 Oct. 4, 2016 How to determine position and reactivity: Examples The strongest donating group wins: NO2 NO2 HNO3 + HNO3 + H SO 2 4 NO2 H2SO4 NO O2N NO2 OH OH OH 2 OH OH HNO3 H2SO4 NO2 O2N NO2 OH 2,4,6-trinitrophenol picric acid This reactivity can be explained by the following resonance forms: OH O O O H H H Meta-director example: O N O HNO3 H2SO4 N N O O O O This reactivity can be explained by the following resonance forms: H H H H H H H H H H H H H H H H H H H H N N N N O O O O O O O O The electrophile is positive and reacts at sites that are not positive (ortho para) but at those that are more negative (meta) as shown below O N O O O O N N N O O O N N N N O O O O O O O O A Different Example : NO2 HNO 3 + H SO 2 4 NO2 Cl Cl Cl In this reaction, chlorobenzene has a chloro (-Cl) group attached on the benzene ring. When we perform nitration, only ortho and para substitution occurs. The chloro group is an electron withdrawing group inductively, however, the lone pairs of electrons are conjugated to the benzene ring through resonance as an electron donating group. As a result, resonance beats inductive effect, which gives ortho and para substitution products. However, chlorbenze is less reactive than benzene Example: Acetophenone is meta directing O O HNO3 H2SO4 NO 2 O O Cl + HCl AlCl3 When the electrophile attacks, it avoids the partial positive charges on the ortho and para positions: O H O H O H O H H H H H H Aromatic compounds with an alkyl chain are considered inductively donating (weak) and will direct substitution to the ortho or para positions. R Example: Br AlBr3 AlCl4 This cation is stabilized by resonance as well as the donation of electrons from the t-Bu group via induction effects (shown by the red arrow) H The reaction occurs at the para position instead of the ortho position due to sterics. Notice how the order that reactions are done affect the following two examples: Cl HNO3 H2SO4 AlCl 3 O N 2 Cl HNO3 H2SO4 AlCl3 O2N O2N The products of the above two reactions are structural (or constitutional) isomers. Aromatic compounds that have an electron-withdrawing group attached but that do not fall into the above categories (e.g. CF3) are inductively withdrawing and direct substitution to the meta position. F F F Example: NO2 If ortho: HNO3 H2SO4 F F F F F F NO2 F F F disfavored (not observed) H -CF3 is electron-withdrawing NO 2 NO2 and does not want to withdraw electrons from a carbon that is already positively-charged. H F F F F F F Multiple Substitution If there is more than one group on an aromatic ring, electrophilic aromatic substitution is controlled by the strongest donating group. An example is the dialkylation of methoxybenzene with methyl iodide by Friedel-Crafts alkylation. Since the methoxy group is an electron-donating group (resonance donator), it will direct the first alkylation to the ortho and para positions. The second alkylation is also directed to the ortho and para positions relative to the methoxy group because it is a stronger director (resonance donator) than the methyl group (inductive donator). The second alkylation will go ortho to the methoxy rather than ortho to the methyl. resonance from methoxy inductive inductive from methyl from methyl CH3I CH3I AlCl3 AlCl3 resonance from methoxy O O O O Similarly, nitration occurs at the ortho position for the following example: OCH3 OCH3 NO2 HNO3 H2SO4 CH 3 CH3 In the following case the nitro group is an electron-withdrawing group, so the methyl group is the strongest donating group and takes precedence in directing where the acetyl group will go. O CH3 CH3 O CH3 H3C Cl NO2 NO2 BCl3 NO2 O In this last example, the position between the two meta substituents is too sterically hindered. Therefore, substitution in that position is not observed. Follow-up: How do you synthesize the starting material in the previous reaction? Answer: CH3 HNO3 CH3Cl AlCl H2SO4 NO 3 2 NO2 If alkylation was done first, followed by nitration, the meta-substituted aromatic compound could not be obtained, since the methyl group is an ortho/para-directing group. How to determine position and reactivity: 6 & 5 membered rings are favoured O O SOCl2 HO thionyl chloride O + (this reaction will be covered later in the course) O phthalic anhydride O O O Cl AlCl3 O O anthraquinone After the first two reactions, the substituent (a ketone) would normally direct the second acylation to the meta position as a resonance withdrawer. This does not occur, as this would produce a very strained structure. In this example, the second acylation goes to the ortho position to relieve the strain to produce anthraquinone. 6-rings are favoured, as are intramolecular reactions. Synthesis Example: The sequence of steps is critical How to make versalide, shown below, from benzene? O The answer is below. Be able to rationalize this sequence. Cl O O O Cl Cl Cl AlCl3 AlCl AlCl3 3 O If you start to add ethyl group first, polyalkylation happens: CH3CH2Cl AlCl3 Alkylation must also be last because the acetyl group will direct to the meta position. Reaction of Side Chains of Aromatic Compounds: 1) Alkyl Side Chain Oxidation 2) Clemmensen Reduction 3) Nitro Reduction 4) Amino Diazotization 5) Diazo Replacement + Coupling Alkyl Side Chain Oxidation Reaction If the carbon directly attached to the aromatic ring has > 1 hydrogen attached to it, it can be oxidized to the corresponding carboxylic acid using hot aqueous potassium permanganate (KMnO4) as shown below. Carbon dioxide (CO2) is one of the side products H O R' KMnO C 4 R OH H2O, heat O KMnO4 OH heat ! O KMnO OH KMnO 4 4 no rxn + CO heat 2 heat (no adjacent H's to the aromatic ring) propyl benzene O KMnO 4 OH H2O, heat More examples of alkyl side chain oxidation: KMnO4 heat HO O O O KMnO4 HO heat O The above answer is accepted as it fits the rule given, but in reality, the methyl group of the ketone is also oxidized to a carboxylic acid (special case) O O KMnO4 HO heat HO O Another example CO2H KMnO4 heat (phthalic acid) CO2H tetralin.
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