Aromatic Compounds PDF

Aromatic Compounds Dr. Nirvana Ali Benzene It is a volatile liquid hydrocarbon Aromatic compounds were defined before as compounds having a “pleasant odor”. It is quite different from benzene used as car fuel (gasoline) which is a mixture of aliphatic hydrocarbons. The structure of Benz...

Aromatic Compounds Dr. Nirvana Ali Benzene It is a volatile liquid hydrocarbon Aromatic compounds were defined before as compounds having a “pleasant odor”. It is quite different from benzene used as car fuel (gasoline) which is a mixture of aliphatic hydrocarbons. The structure of Benzene and delocalization of electrons Benzene  Chemical formula= C6H6  Structure = Cyclic planner  Hybridization= SP2  Bond angles= 120 o  Bond Length= 1.4 Å Benzene  Benzene can be written as a six-membered ring with alternating single and double bonds (Kekulé structure).  Has six identical carbon–carbon bonds  The π electrons are delocalized  Benzene reacts by substitution reactions not with addition reactions.  benzene is remarkably stable; it doesn’t undergo addition and oxidation reactions characteristics of alkenes  In 1865, kekule proposed that benzene can be formulated as 1,3,5- cyclohexatriene. Kekule structure accounts for: 1-the exact molecular formula C6H6. 2- the presence of unsaturation in benzene.  It fails in : 1- stability of benzene, because it represents benzene as an alkene with localized double bonds. 2- benzene reacts by substitution and not by addition. 3- all C-C bonds are equal in length (1.4 Å) this value is intermediate in length between C-C single bond (1.48 Å) and C=C double bond (1.34 Å)  Resonance explanation of the benzene structure 1) According to resonance theory: benzene is a resonance hybrid of 2 kekule structures I & II. Each C-C bond is single in one structure & double in the other. so, C-C bond is intermediate in length between C-C & C=C bond (has partial π bond). Resonance explanation of the benzene structure The π electrons are delocalized (not in the same place). Benzene is a cyclic and conjugated system, this increases stability (lower energy) by resonance energy. Stability of benzene 1-hydrogenation of alkenes is exothermic reaction. The heat of hydrogenation (∆H) of cyclohexene = -28.6 kcal / mole. 2- calcd ∆H of benzene based on 1,3,5-cyclohexatriene: ∆H = 3 * -28.6 = -85.8 kcal / mole Stability of benzene 1- the actual heat of hydrogenation (∆H found) of benzene = -49.8 kcal / mole. This indicates that benzene has lower energy ( more stable) than imaginary 1,3,5-cyclohexatriene by 36 kcal / mole. The difference between ∆H calcd & ∆H found is called resonance energy. Stability of benzene resonance energy = ∆H calcd - ∆H found = -85.8 – (-49.8) = -36 kcal / mole This means that benzene is more stable & needs higher energy to lose aromaticity ( alkenes are hydrogenated at room temp. while benzene requires high temp. & pressure. Drawing resonance contributors Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons and of lone pairs that are adjacent to pi bonds. Never shift the location of electrons in sigma bonds. Rules for drawing resonance contributors: 1. In drawing resonance contributors, the electrons in one resonance contributor are moved to generate the next contributor. 2. Only electrons move. The nuclei of atoms NEVER MOVE. 3. The only electrons that can move are the π electrons and lone pairs of electrons. 13 4. The total number of electrons in the molecule does not change, and neither do the number of paired and unpaired electrons. 5. The arrow drawn between resonance contributors is a double headed arrow. 14 6. Electrons always move toward the more electronegative atom 7. Electrons move toward an sp2 carbon but never toward an sp3 carbon. 8. Radicals can also have delocalized electrons if the unpaired electron is on a carbon adjacent to an sp2 atom. 15 9. The electrons can move in one of three cases: a. Move π electrons toward a positive charge or toward a π bond (C=C-C+) or (C=C-C=) : b. Move lone-pair of electrons toward a π bond (C=C-C).. c. Move a single non-bonding electron toward a π bond (C=C-C). Case (a): moving π electrons toward a positive charge or toward a π bond. Case (b): moving lone pair of electrons toward a π bond (benzylic radicals). 17 Case (c): moving a single electron toward a π bond. 18 The Difference Between Delocalized and Localized Electrons 19 Drawing resonance contributors Aromaticity Aromaticity is used to describe certain properties of benzene and benzene like compounds: 1- increased stability: aromatic compounds are stable and possess large resonance energy. 2- special chemical reactivity: aromatic compounds react by substitution rather than addition. 3- C-C bond length: of aromatic rings is intermediate bet. C-C single bond & C=C double bond lengths. Requirements of Aromaticity: To be Aromatic, the structure must be: 1. cyclic. 2. planar. 3. fully conjugated, i.e. all ring carbons are sp2 hybridized 4. obey Huckel’s rule ( applied only when the molecule satisfies conditions 1-3): Requirements of Aromaticity: Huckel’s rule a. If the molecule contains 4n+2 π electrons (where n= 0,1,2…), it will be aromatic. b. If the molecule contains 4n π electrons, it will be antiaromatic. If the molecule doesn’t satisfy conditions 1-3, it is non aromatic 1- not Cyclic or 2- not Planar Non or 3- not Conjugated aromatic 1- Cyclic 2- Planar Aromatic 3- Conjugated 4- (4n+2 π electrons) 1- Cyclic 2- Planar Anti 3- Conjugated 4- (4n π electrons) aromatic Huckel’s rule must be applied to predict aromaticity or antiaromaticity of the molecule. Application of Huckel’s MO method: Huckel’s rule must be applied to predict aromaticity or antiaromaticity of the molecule. Benzene (C6H6): cyclic, planar, fully conjugated, has 6 π electrons 4n+2= 6 (4n+2 π electrons ) 4n= 4 AROMATIC n= 1 Application of Huckel’s MO method: Cyclobutadiene (C4H4): Cyclic, Conjugated, Planar, has 4 π electrons (4n) 4n+2= 4 (4n π electrons ) 4n= 2 ANTIAROMATIC n= 1/2 Application of Huckel’s MO method: Naphthalene: Cyclic, Conjugated, Planar, has 10 π electrons 4n+2= 10 (4n+2 π electrons ) 4n= 8 AROMATIC n= 2 Application of Huckel’s MO method: Cyclopentadiene: Cyclic, Planar, Not Fully Conjugated so it is not aromatic Non AROMATIC Application of Huckel’s MO method: Pyrrole:.. N H Cyclic, Planar, Fully Conjugated, has 6 π electrons 4n+2= 6 (4n+2 π electrons ) 4n= 4 AROMATIC n= 1 Application of Huckel’s MO method: Cyclopentadienyl cation: Cyclic, Planar, Fully Conjugated, has 4 π electrons 4n+2= 4 (4n π electrons ) 4n= 2 ANTIAROMATIC n= 1/2 Application of Huckel’s MO method: Tropylium anion: Cyclic, Planar, Fully Conjugated, has 8 π electrons 4n+2= 8 (4n π electrons ) 4n= 6 ANTIAROMATIC n= 3/2