{"id":1426,"date":"2017-10-12T14:25:00","date_gmt":"2017-10-12T14:25:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&p=1426"},"modified":"2018-10-05T19:00:26","modified_gmt":"2018-10-05T19:00:26","slug":"the-criteria-for-aromaticity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/the-criteria-for-aromaticity\/","title":{"raw":"The Criteria for Aromaticity","rendered":"The Criteria for Aromaticity"},"content":{"raw":"
\u00a0In 1931, German chemist and physicist Erich H\u00fcckel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has 4n+2 \u03c0 electrons, it is considered aromatic<\/a>. This rule would come to be known as H\u00fcckel's Rule.<\/div>\r\n
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Four Criteria for Aromaticity<\/h3>\r\nWhen deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.\r\n
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  1. The molecule is cyclic (a ring of atoms)<\/li>\r\n \t
  2. The molecule is planar (all atoms in the molecule lie in the same plane)<\/li>\r\n \t
  3. The molecule is fully conjugated<\/span> <\/a>(p orbitals at every atom in the ring)<\/li>\r\n \t
  4. The molecule has 4n+2 \u03c0 electrons (n=0 or any positive integer)<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    Why 4n+2 \u03c0 Electrons?<\/h3>\r\nAccording to H\u00fcckel's Molecular Orbital Theory<\/span><\/a>, a compound is particularly stable if all of its bonding molecular orbitals<\/a> are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by n<\/em>), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of 4n+2 \u03c0 electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 \u03c0 electrons. Its first 2 \u03c0 electrons fill the lowest energy orbital, and it has 4 \u03c0 electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.\r\n\r\n\"benzene\r\n\r\n<\/div>\r\n
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    How does the 4n+2 Rule Work?<\/h3>\r\nTo apply the 4n+2 rule, first count the number of \u03c0 electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six \u03c0 electrons:\r\n\r\n4n + 2 = 6\r\n4n = 4\r\nn = 1\r\n\r\nFor benzene, we find that n=1, which is a positive integer, so the rule is met.\r\n\r\n<\/div>\r\n
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    How Can You Tell Which Electrons are \u03c0\u00a0 Electrons?<\/h3>\r\nPerhaps the toughest part of H\u00fcckel's Rule is figuring out which electrons in the compound are actually \u03c0 electrons. Once this is figured out, the rule is quite straightforward. \u03c0 electrons lie in p orbitals<\/span><\/a>. Sp2<\/sup>hybridized<\/a> atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2<\/sup> hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the \u03c0 electrons. A simple way to know if an atom is sp2<\/sup> hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This video<\/a> provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is sp2<\/sup> hybridized and has a p orbital. Let's look at our previous example, benzene:\r\n\r\n\"benzene<\/a>\r\n\r\nEach double bond (\u03c0 bond) always contributes 2 \u03c0 electrons. Benzene has 3 double bonds, so it has 6 \u03c0 electrons.\r\n\r\n<\/div>\r\n
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    Aromatic Ions<\/h3>\r\nH\u00fcckel's Rule also applies to ions. As long as a compound has 4n+2 \u03c0 electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp2<\/sup> hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp2<\/sup> hybridized is if an atom has 1 or more lone pairs and is attached to an sp2<\/sup> hybridized atom, then that atom is sp2<\/sup> hybridized also. This video<\/a> explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 \u03c0 electrons and fulfills the 4n+2 rule.\r\n\r\n\"aromatic<\/a>\r\n\r\n<\/div>\r\n
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    Heterocyclic Aromatic Compounds<\/h3>\r\nSo far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp2<\/sup> hybridized. But is the oxygen atom sp2<\/sup> hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp2<\/sup> hybridized atom, so it is sp2<\/sup> hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are \u03c0 electrons? An sp2<\/sup> hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp2<\/sup> orbital. So, only 1 of oxygen's 2 lone electron pairs are \u03c0 electrons. Furan has 6 \u03c0 electrons and fulfills the 4n+2 rule.\r\n\r\n\"heterocyclic<\/a>\r\n\r\n<\/div>\r\n
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    Exercises<\/h3>\r\n
    \r\n\r\nUsing the criteria for aromaticity, determine if the following molecules are aromatic:\r\n\r\n\"problems<\/a>\r\n\r\n<\/div>\r\n
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    Answers<\/h3>\r\n[reveal-answer q=\"70524\"]Show Answers[\/reveal-answer]\r\n[hidden-answer a=\"70524\"]\r\n\r\n1. Aromatic - only 1 of S's lone pairs counts as \u03c0 electrons, so there are 6 \u03c0 electrons, n=1\r\n\r\n2. Not aromatic - not fully conjugated, top C is sp3 hybridized\r\n\r\n3. Not aromatic - top C is sp2 hybridized, but there are 4 \u03c0 electrons, n=1\/2\r\n\r\n4. Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not \u03c0 electrons, there are 6 \u03c0 electrons, n=1\r\n\r\n5. Aromatic - there are 6 \u03c0 electrons, n=1\r\n\r\n6. Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as \u03c0 electrons, so there 8 \u03c0 electrons, n=1.5\r\n\r\n7. Not aromatic - there are 4 \u03c0 electrons, n=1\/2\r\n\r\n8. Aromatic - only 1 of N's lone pairs counts as \u03c0 electrons, so there are 6 \u03c0 electrons, n=1\r\n\r\n9. Not aromatic - not fully conjugated, top C is sp3 hybridized\r\n\r\n10. Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not \u03c0 electrons, there are 6 \u03c0 electrons, n=1[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    References<\/h3>\r\n
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    1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function<\/u>. 5th ed. New York: W. H. Freeman & Company, 2007.<\/li>\r\n \t
    2. Berson, Jerome. Chemical Creativity: <\/span>Ideas from the Work of Woodward, H\u00fcckel, Meerwein, and Others<\/u>. New York: Wiley-VCH, 1999.<\/li>\r\n \t
    3. Badger, G.M. Aromatic Character and Aromaticity<\/span><\/u>. London, England: Cambridge University Press, 1969.<\/li>\r\n \t
    4. Lewis, David and David Peters. Facts and Theories of Aromaticity<\/span><\/u>. London, England: Macmillan Press, 1975.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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      Internal Links<\/h3>\r\n