Aromatic Compounds - 4-Pages Summary

Aromatic Compounds Naming Aromatic Compounds Page 1 Using “benzene” as the parent can also be encountered, even though it is not what you’d commonly see: Benzene is the “first and simplest” aromatic compound and many monosubstituted derivatives of benzene are named systematically by adding the name of the substituent to “benzene” which is the parent: NO2 Br Cl OH Cl Cl 3-chlorophenol nitrobenzene bromobenzene chlorobenzene OH synonyms More comonly used 1-chlo-3-hydroxynbenzene ethylbenzene To name other polysubstituted aromatic compounds, you can follow these steps: When treated as a substituent, the benzene ring is 1) Identify and name the parent. If it is not one of the common names, then use benzene. 1-Phenyloctane called a phenyl group: 2) Identify and name the substituents. There are also monosubstituted benzene rings that have common names. And when naming an 3) Number the ring to give the substituents the smallest possible number. aromatic compound with one of these rings, you need to use the common name as the parent and 4) Put the substituents alphabetically followed by the parent name. not the “benzene”. Below is the list of these common names: CH3 NH2 OH O O C C OH H 5 4 Toluene Phenol Benzoic acid Aniline Benzaldehyde C SO3H Br 1 Br 1 Br 2 2 3 4 5 Br Br 6 6 5 3 1,2-dibromobenzene or o-dibromobenzene O O Br Br 1 6 2 2 3 4 3 1 6 Br 5 4 Br Br NOT NOT 1,2,4-tribromobenzene 1,6-dibromobenzene 1,4,6-tribromobenzene CH3 If the numbering does not make a difference, then give a smaller locant based on alphabetical priority: Anisole Benzenesulfonic acid Acetophenone Styrene NO2 2 1 Br 1 Disubstituted Derivatives of Benzene First, know the relative positions: ortho meta o-Xylene m-Xylene p-Xylene CH3 CH3 CH3 Br 2 NOT 1-bromo-2-nitrobenzene para NO2 1-nitro-2-bromobenzene If there are more than two groups, and numbering doesn’t make a difference, start from the alphabetical priority and number the ring toward the next alphabetical priority: CH3 Br Br CH3 1 CH3 O2N ortho OH meta para OH OH o-Cresol m-Cresol OH OH p-Cresol NOT O2N 3 Cl 4 1 2 3 2 6 5 1-bromo-3-chloro-5-nitrobenzene 4 6 5 Cl 1-bromo-3-nitro-5-chlorobenzene OH OH CH3 If two common names are possible, for example phenol and aniline, we need to go based on the priority OH Catechol CH3 OH Hydroquinone Resorcinol 2-methylphenol 3-methylphenol CH3 4-methylphenol of the groups. In this case, we are comparing the amine and alcohol, and since OH stands above NH2 in the priority chart, the parent chain should be an alcohol (phenol) rather than aniline. For the other rings with common names, start numbering the ring from the substituent that is part 1 4 6 1 OH Cl 3 2 3 2 4 OH > NH2 1 NH 2 6 Br Cl 5 3-chlorophenol 4-bromo-3-chloro aniline H2N 5 3 4 5-amino-2-bromophenol 3 2 2 Br 6 5 OH OH of the common ring such that the other groups get the smallest possible locants: Synonym H2N Br 4 5 1 6 4-bromo-3-hydroxyaniline IUPAC-preferred gevorg@chemistrysteps.com Copyright © 2019 Chemistry Steps Aromatic Compounds Page 2 Identifying Aromatic Compounds Common Examples of Aromatic Compounds Benzene, and all aromatic compounds are associated with a characteristic high stability due to a Many heterocyclic compounds and ions, including the nucleobases, are aromatic because of a lone cyclic, planar, fully conjugated structure. pair being part of the conjugated system. For example, unlike other compounds with double The Structure of Benzene N 2 sp carbons C=C bond when reacted with bromine. trigonal planar geometry s om at H sp2 orbital Pyrrole + an eo ft Br pl p orbitals - 90o to the plane e N Br Br2 he H H not part of p orbital 4 e- - double bonds 2 e- - lone pair N H Br H H - Pyridine Br H H 6 e- - Aromatic sp2 orbital N bonds, no electrophilic addition to benzene occurs Br2 6 e- - Aromatic N No Reaction 4 e- - double bonds 2 e- - lone pair N H Aromatic, Antiaromatic, or Nonaromatic Compounds p orbital p orbital Imidazole N 6 e- - Aromatic N H sp2 orbital 6 e- - double bonds p orbital closes the loop Tropylium ion Not all the compounds that are cyclic, planar, and fully conjugated are aromatic. According to the Hückel’s rule, aromatic compounds contain 4n+2 p electrons, where n is a whole number starting There is a ton of examples of nonaromatic compounds as they have to be either not cyclic, not planar, from 0. If the first three criteria are met, but the system has 4n p electrons, the molecule is or not fully conjugated. Two interesting examples that may seem to be antiaromatic are shown below: antiaromatic. If any one of these criteria is not met, then it is a nonaromatic compound - nothing Cyclooctatetraene is not planar - Nonaromatic to do with aromaticity. Br2 Br o Br 100% Cyclic Planar Conjugated (Huckel) 4n+2 e- 4n -50 C Any of these not met Cyclooctatetraene Nonaromatic p orbitals of adjacent double bonds cannot overlap. Reacts readily with Br2 even at lower temperatures. [10] - annulene is seemingly aromatic as it resembles naphthalene, however, it is not planar due to the e- (Mobius) sterics of two internal hydrogens, and therefore, it is nonaromatic. H H Both classes meet these criteria They only differ in the number of e- not fully conjugated H H not planar Naphthalene Aromatic Nonaromatic Benzene Pyridine Naphtaline Antiaromatic Ions Pentalene 6e 6e [10]-annulene planar Nonaromatic H2O Base - Aromatic - 10 e 4 e- 4 e- The molecule is not planar Unusual Reactivity Explained by Aromaticity N - The molecule is 8 eH Antiaromatic H pKa 44 Contrary to aromatic compounds, antiaromatic compounds are generally H H pKa = 15 very acidic Br Aromatic Structure and Aromaticity of Cyclopentadienyl Anion pentalene Note: Whenever possible the molecule will “count or ignore” some electrons to be aromatic and avoid being antiaromatic as these are generally quite unstable. 6 e- - Aromatic 4 e- - double bonds 2 e- - lone pair Antiaromatic Cyclopentadienyl Cation o very unstable. For example, pentalene is unstable above -100 C. OH Very slow p orbital p orbital 4 e- - Antiaromatic e- 4 - double bonds p orbital closes the loop H sp2 orbital H sp2 orbital gevorg@chemistrysteps.com Copyright © 2019 Chemistry Steps Electrophilic Aromatic Substitutions Electrophilic Aromatic Substitution How do you determine the position of EA substitution if there is more than one group on the ring? E E+ F2 reacts too vigorously. X X2 1) I I2 AlCl 3 or FeBr3 SO3H 4) O R Cl R O O NO2 Br NO2 Br2 FeBr3 O O NO2 + CH3O- and NO2 direct same way major Cl benzene, it is classified as an activator or a deactivator. Activators donate electron density to the O O O Cl Cl2 Depending on whether a group makes the electrophilic substitution faster or slower compared to R AlCl 3 Friedel-Crafts Alkylation Resonance stabilization sterics O Step 2. Proton transferRestoring Aromaticity Activating and Deactivating Groups 6) AlCl 3 Sigma Complex Scenario 2: The groups point to different positions. H2SO4 R Scenario 1: Both groups point to the same position. Br Nitration 5) The rule here is that the stronger activator wins. Let's consider a couple of specific cases. NO2 HNO3 Sulfonation E Bfast Steps1. Addition of the electrophileMaking the C-E bond HNO3 H2SO4 E H slow Halogenation 3) E + BH 2) SO3 Ortho-, Para- and Meta- in Disubstituted Rings The Mechanism of Electrophilic Aromatic Substitution Below are the most common electrophilic aromatic substitution reactions. X = Cl or Br Page 3 NO2 AlCl 3 Cl CH3O- is an Cl activator and + + NO2 NO2 NO2 it wins. too sterically hindered ring either by inductive or resonance effect thus making the ring more reactive. Scenario 3: If both groups are deactivators - no selectivity, a mixture is obtained. Friedel-Crafts Acylation Electron withdrawing groups (EWG), on the other hand, are deactivators since they reduce the electron density of the ring. Limitations of Electrophilic Aromatic Substitution Reactions CH3 NO2 Changing the Common Substituents NH2 * Friedel-Crafts alkylation does not work for aryl and vinyl halides. H R H COOH 1) KMnO4 or Na2Cr2O7 Br CH3 excess NBS will tribrominate Br2, h 2) H3O Donating-inductive EWG NBS or Cl Donating-resonance AlCl3 sp2 Carbocation- Cl AlCl3 too unstable to form Increasing the rate of Electrophilic Substitution * Friedel-Crafts alkylations and acylations are the slowest in EA reactions, so the reaction does not work when a strongly deactivating group is present. O R R NO2 Zn (Hg) NH2 Zn / HCl HCl or Fe / HCl, Sn / HCl, Pd-C / H2 1 O O R' OR OOR" O OH 2 1 6 ortho (1,2) meta (1,3) para (1,4) 3 undergo a rearrangement to form more stable carbocations: 4 Br aldehydes and ketones to carboxylic or Cl acids and esters respectively. Cl AlCl 3 if R = H No Reaction * Direct alkylation works for methyl and ethyl halides only. Longer hydrocarbon chains 2 6 5 Br 4 4 1 2 3 5 3 5 The Baeyer-Villiger oxidation converts Br AlCl3 Br Br Br 6 R Cl + Disubstituted benzenes are classified based on the relative position of the substituents. Clemmensen reduction or Wolff–Kishner by H2NNH2, HO- O NO2 What are Ortho-, Para- and Meta- Positions Ortho-, Para- and Meta- Directors H AlCl3 H Cl AlCl3 2o carbocation 1, 2-hydride shift Formed When a functional group is already present on the aromatic ring, the substitutions occur at specific positions depending on whether the group is an ortho-, meta-, or a para- director. Reactions of Arenediazonium Salt Cl N N NH2 NaNO2 N2 is an excellent leaving group making all the reactions possible. HCl H3PO2 CuCl CuBr KI NaOH SnCl 2 H Cl Br I OH NH2NH2 Deactivating * Cl-, Br- halogenations are the fastest among EA reactions, so when a strongly activating group is NHR OH OR NHR2 N H Cl, Br, I O R- Alkyl groups O R R R = H, OH, Cl, OR NO2 CN NH3 NH2R SO3H + FeBr3 Br2 Br NH2 Br No catalyst needed No selectivity for monohalogenation major Br O O * Amines are strong Lewis bases and coordinate to the Lewis acid catalyst and become deactivators. Br2 FeBr3 NHR2 NH2 Br para ortho Strongly Deactivating O R F, present, there is no need for a Lewis acid. Br Br2 NH2 O NaOH CN Deactivators Moderately activating CuCN HCl Activators Strongly activating * Acylations proceed without rearrangement so you get what you plan. Br They need to be protected/deactivated first. O meta - major NH2 All the activators are ortho-, para- director. All the deactivators are meta-directors, except the H2N AlCl3 AlCl3 NH2 O R Cl HN R Moderately activating can do Friedel- halogens which are deactivators and yet, ortho-, para-directors. gevorg@chemistrysteps.com Copyright © 2018 Chemistry Steps Deactivated ring No Friedel-Crafts possible Doing practice problems is the only way to learn - https://www.chemistrysteps.com/category/organic-chemistry/electrophilic-aromatic-substitution/ Crafts Aromatic Compounds Nucleophilic Aromatic Substitutions Notice from the left column that three criteria must be satisfied for SNAr: Some aryl halides, generally with a strong electron-withdrawing substituent(s) on the ring can undergo nucleophilic substitution (SNAr) instead of electrophilic substitution: X EWG X = F, Cl, Br, I EWG = Electron Withdrawing Group 1. The ring must contain a strong electron-withdrawing group (typically a nitro group). 2. The ring must contain a leaving group (rate F >> Cl Br I). 3. The leaving group must be either ortho or para to the EWG. Nu Nu Page 4 X + EWG The Elimination-Addition Mechanism X NH2 NH2 - The reaction with OH requires extreme conditions, however, works at lower temperatures with NaNH2 X + because it is a very strong bases and the reaction goes through a triple-bonded benzyne intermediate. There are two main mechanisms: addition-elimination or elimination-addition, also known as the benzyne benzyne mechanism which is sometimes considered as a separate reaction. The benzyne mechanism works H NH2 mainly with a very strong base and nucleophile, NH2- and does not require an EWG group, although its presence is not a hurdle. SNAr by Addition-Elimination Cl 2) H3O+ 14C Cl OH g) -30 2) H3O+ O2N OH NO2 O2N 1) NaOH 1) NaNH2/NH3, 2) H3O+ NO2 NH2 + o + For the traditional SNAr, it is easy as the nucleophile appears where the leaving group was. -30 C ring, a mixture of isomers is obtained. If the ring contains an inductively EWG , the nucleophile ends up on the carbon that is farther away from the EWG group. Cl Cl 1) NaOH 1) NaNH2/NH3, No Reaction d) i) 70 oC NO2 For meta, the H closer to EWG is removed - more acidic. No Reaction meta F O CH3O = CF3! CH3 Cl X o 1) NaNH2/NH3, j) 67 oC NO2 NO2 NH2 + -30 oC - EWG EWG - EWG m NH2 NH2 EWG does not work EWG NH2 + - NH2 NH2 m NH2 EWG EWG - NH2 EWG - EWG EWG NH2 p plane of the F OH where the orbitals are, which is H2N X HO benzyne has to occur in the benzene ring, + F the nucleophilic attack on a m X stabilized carbanion called a Meisenheimer complex: F hard to form since + EWG H The reaction starts with the addition of the nucleophile to the aromatic ring generating a resonance- HO Ortho is generally from EWG NH2 OCH3 The SNAr by Addition-Elimination F NH2 farther the EWG works major HO (-) closer to NH2 NH2 OCH3 CF3 CF3 EWG - NH2 EWG 2) H3O+ CF3 EWG EWG NH2 CH3ONa e) O For the benzyne mechanism, remember this: When no electron-withdrawing group is present on the 2) H3O+ NO2 35 oC O NH2 The Regiochemistry of Nucleophilic Aromatic Substitution NH2 h) NH2 1) NH22) H3O+ Cl NO2 Cl oC 2) H3O+ NO2 Cl c) NH2 1) NaNH2/NH3, 1) NaOH, 100 oC NO2 protonation Nu attack halide (see reaction "h") or trapping the intermediate by a Diels-Alder reaction. f) No Reaction b) H sp2 orbital loss of LG The mechanism (formation of benzyne) can be confirmed using 14C isotope labeling, substituted aryl OH 1) NaOH, 350 oC 1) NaOH, heat Cl NH2 SNAr by Elimination-Addition (Benzyne) Cl a) NH2 H O NH2 deprotonation slow - RDS p orbitals benzyne Cl Cl p NH2 NH2 sterically hindered. Preferred ion -OH O2N Only inductively EWGs are relevant here as the electrons of the anion are at 90o to the ring and -FNO2 Nucleophilic addition-slow NO2 resonance-stabilized carbanion Meisenheimer complex NO2 NO2 resonance effects cannot be communicated through the aromatic system. Therefore, OCH3, for example, is electron-withdrawing equivalent to CF3! Electron donating groups do not enforce regioselectivity as Loss of LG gevorg@chemistrysteps.com Copyright © 2019 Chemistry Steps much, however, opposite trends will be expected.