{"id":357,"date":"2018-11-26T15:17:25","date_gmt":"2018-11-26T15:17:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry2\/?post_type=chapter&p=357"},"modified":"2022-02-19T15:43:06","modified_gmt":"2022-02-19T15:43:06","slug":"14-1-electrophilic-aromatic-substitution-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry2\/chapter\/14-1-electrophilic-aromatic-substitution-reactions\/","title":{"raw":"14.1. Overview","rendered":"14.1. Overview"},"content":{"raw":"
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Electrophilic aromatic substitution (EAS) reactions - the general picture<\/h3>\r\nAlthough the delocalized pi electrons in aromatic rings are much less reactive than those in isolated or conjugated alkenes, they can undergo electrophilic reactions given a powerful enough electrophile.\u00a0 Overall<\/em> electrophilic addition<\/em> to aromatic double bonds, however, is not generally observed - this would be energetically unfavorable because it would result in a loss of aromaticity in the product.\r\n\r\nInstead, aromatic double bonds undergo electrophilic aromatic substitution reactions, (abbreviated EAS, sometimes called SE<\/sub>Ar), which have an electrophilic addition step followed immediately by an electrophile elimination step.\u00a0 In many cases, these are preceded by the formation of the electrophile\r\n\r\nThe general mechanism usually consists of three steps:\r\n

1. Electrophile formation (various mechanisms):<\/h4>\r\n\"\"\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>
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2. Formation of the Wheland Intermediate (WI) (Electrophilic addition):<\/h4>\r\n\"Electrophile\r\n

3. Loss of H+ of the WI to form the final product (Electrophile elimination):<\/h4>\r\n\"Loss\r\n

Free energy reaction coordinate diagram for electrophilic substitution of benzene using a generic electrophile E+<\/sup><\/h3>\r\nBelow is a free energy reaction coordinate diagram<\/a> showing the reaction of benzene with an electrophile E+<\/sup>.\u00a0 The intermediate formed is sometimes referred to as a Wheland intermediate.\u00a0 Remember that this intermediate is not aromatic<\/a><\/strong>, <\/em>even though it is somewhat stabilized by resonance, so it is much less stable than the aromatic starting material and aromatic product.\r\n\r\n\"\"\r\n\r\nAn aromatic ring is much more stable than a simple alkene.\u00a0 This means that if we compare an electrophilic addition step of an aromatic ring with a simple alkene, the aromatic ring starts from a much lower energy level than the alkene.\u00a0 This in turn means that the aromatic has a higher energy barrier to climb over than the alkene has.\u00a0 Even though the EAS Wheland Intermediate has some stabilization from resonance, it has still lost the aromaticity and is much less stable than the starting aromatic.\u00a0 The outcome is that alkenes are much more reactive towards electrophiles and electrophilic addition than are aromatics.\u00a0 The following diagram shows the reaction diagrams for the rate-limiting electrophilic addition step in each:\r\n\r\n\"Reaction\r\n\r\nBecause of this, simple aromatics only react with highly reactive electrophiles, whereas alkenes can react with less reactive ones.\u00a0 There are two things that can promote EAS reactions with weaker electrophiles - either a catalyst (typically a Lewis acids such as AlCl3<\/sub>, which activates the electrophile) or an electron-donating substituent (EDG) on the aromatic ring (which activates that ring). The reaction coordinate diagram below shows the effect of such an activating substituent on the reactivity of an aromatic ring:\r\n\r\n\"Rxn\r\n\r\nIn this case, the electron-withdrawing group stabilizes the Wheland intermediate and also lowers the energy of the nearby transition state.\u00a0 This reduces the activation energy, thereby speeding up the reaction when EDGs are present (recall Hammond's postulate, section 5.5.<\/a>) .\u00a0 The opposite effect can also be seen when electron-withdrawing<\/em> groups are present - the EAS reaction is slowed down<\/em> because the Wheland intermediate is less<\/em> stable.\u00a0 These effects will be discussed in more detail in section 14.3<\/a>.\r\n\r\n<\/div>\r\n<\/div>\r\n
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References<\/h3>\r\n
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  1. Klein, David R.\u00a0Organic Chemistry II as a Second Language.\u00a0Hoboken, NJ: John Wiley & Sons, 2006<\/li>\r\n \t
  2. Parsons, A.F.\u00a0Keynotes in Organic Chemistry.\u00a0Oxford; Malden, MA: Blackwell Science, 2003<\/li>\r\n \t
  3. Taylor, R.\u00a0Electrophilic Aromatic Substitution.\u00a0Chichester, West Sussex, England; New York: J. Wiley, 1990<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    Exercises<\/h3>\r\n
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    1. Label the hybridization on all the carbons in a) reacting benzene ring, b) intermediate (including resonance forms), and c) product (monosubstituted benzene ring)<\/li>\r\n \t
    2. Is the energy of activation higher in the first step or second step of the mechanism? Explain your reasoning.<\/li>\r\n \t
    3. If you wanted to halogenate benzene, what sort of reagent and catalyst (if needed) would you use?<\/li>\r\n \t
    4. Which hydrogren is used in order to regain aromaticity after the electrophile has added to the ring?<\/li>\r\n \t
    5. Critical Thinking Question: Mentioned above was the fact that electrophilic aromatic substitution can and does happen when there are substituents already present on the ring. Already present substituents will determine where something adds onto the ring in relation to itself (ortho, meta, or para position). What sort of factors do you think influence where addition will occur?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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      Answers<\/h3>\r\n[reveal-answer q=\"770270\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"770270\"]\r\n
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      1. a) All are\u00a0sp<\/em>2<\/sup><\/strong>\u00a0b) Five carbons are\u00a0sp<\/em>2<\/sup><\/strong>, the carbon with the attached E is\u00a0sp3<\/sup><\/strong>\u00a0hybridized c) All are\u00a0sp<\/em>2<\/sup><\/strong><\/li>\r\n \t
      2. The activation energy is higher for the first step- aromatic compounds are lower in energy (\"happier\") than their non-aromatic forms. Therefore it takes more energy to change a compound from being aromatic to non-aromatic, than it does in the reverse direction.<\/li>\r\n \t
      3. $$X_2 (X=Cl, Br) + FeX_3$$<\/li>\r\n \t
      4. The hydrogen used is the one attached to the same carbon that E added to<\/li>\r\n \t
      5. Critical Thinking: If you came up with a) Inductive Effects (correlates to electronegativity) b) Resonance effects c) Electrostatics d) Steric Hindrance then good job! All of these influence where a substituent will add to the ring. Go to\u00a0Activating and Deactivating Benzene Rings<\/a>\u00a0to find out more about how position is determined.[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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        Contributors<\/h3>\r\n