{"id":1252,"date":"2017-10-12T15:25:07","date_gmt":"2017-10-12T15:25:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1252"},"modified":"2017-10-12T15:25:07","modified_gmt":"2017-10-12T15:25:07","slug":"delocalized-electrons-explain-benzenes-structure","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/delocalized-electrons-explain-benzenes-structure\/","title":{"raw":"Delocalized Electrons Explain Benzene\u2019s Structure","rendered":"Delocalized Electrons Explain Benzene\u2019s Structure"},"content":{"raw":"<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n\r\nKekul\u00e9 was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.\r\n<div id=\"s3639\">\r\n\r\n<img class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145006\/kekuleatoms.gif\" alt=\"\" width=\"97px\" height=\"111px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <img class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145007\/kekule.gif\" alt=\"\" width=\"63px\" height=\"72px\" \/>\r\n\r\nThis diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing - and that must be attached to a hydrogen atom.\r\n<div id=\"section_1\">\r\n\r\n\r\n<h3 id=\"Problems_with_the_Kekul.C3.A9_structure-3639\">Problems with the Kekul\u00e9 structure<\/h3>\r\nAlthough the Kekul\u00e9 structure was a good attempt in its time, there are serious problems with it regarding chemistry, structure and stability.\r\n\r\n<strong>The Kekul\u00e9 structure has problems with the chemistry.<\/strong> Because of the three double bonds, you might expect benzene to have reactions like ethene - only more so! Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms. Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new.\r\n\r\n<strong>The Kekul\u00e9 structure has problems with the shape.<\/strong> Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekul\u00e9 structure. The problem is that C-C single and double bonds are different lengths.\r\n<ul>\r\n \t<li>C-C (0.154 nm)<\/li>\r\n \t<li>C=C (0.134 nm)<\/li>\r\n<\/ul>\r\nThat would mean that the hexagon would be irregular if it had the Kekul\u00e9 structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.\r\n\r\n<strong>The Kekul\u00e9 structure has problems with the stability of benzene.<\/strong> Real benzene is a lot more stable than the Kekul\u00e9 structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekul\u00e9 structure, you get an answer which is wrong by about 150 kJ mol<sup>-1<\/sup>. This is most easily shown using <a title=\"Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Robert's_%22Basic_Principles_of_Organic_Chemistry%22\/11%3A_Alkenes_and_Alkynes_II._Oxidation_and_Reduction_Reactions._Acidity_or_Alkynes\/11.3%3A_Heats_of_Hydrogenation\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)\/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes\/11.3%3A_Heats_of_Hydrogenation\" rel=\"internal\">enthalpy changes of hydrogenation<\/a>. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:\r\n\r\n\\[\\ce{CH_2=CH_2 \u00a0+ \u00a0H_2 \\rightarrow CH_3CH_3} \\tag{1}\\]\r\n\r\nCH_2=CH_2+H_2\u2192CH_3CH_33\r\n\r\nIn order to do a fair comparison with benzene (a ring structure) we're going to compare it with cyclohexene. Cyclohexene, C<sub>6<\/sub>H<sub>10<\/sub>, is a ring of six carbon atoms containing just one C=C.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145008\/chenefull.gif\" alt=\"\" width=\"80px\" height=\"90px\" \/>\r\n\r\nWhen hydrogen is added to this, cyclohexane, C<sub>6<\/sub>H<sub>12<\/sub>, is formed. The \"CH\" groups become CH<sub>2<\/sub> and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekul\u00e9 structure for benzene is simplified - by leaving out all the carbons and hydrogens.\r\n<p style=\"text-align: center\"><img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145009\/cyclos.gif\" alt=\"\" width=\"198px\" height=\"96px\" \/><\/p>\r\nIn the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH<sub>2<\/sub>. The hydrogenation equation could be written:\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145010\/hydrogeq1.gif\" alt=\"\" width=\"201px\" height=\"34px\" \/>\r\n\r\nThe enthalpy change during this reaction is -120 kJ mol<sup>-1<\/sup>. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.\r\n\r\nIf the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene - that is, -240 kJ mol<sup>-1<\/sup>.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145011\/hydrogeq2.gif\" alt=\"\" width=\"203px\" height=\"34px\" \/>\r\n\r\nIn fact, the enthalpy change is -232 kJ mol<sup>-1<\/sup> - which isn't far off what we are predicting. Applying the same argument to the Kekul\u00e9 structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol<sup>-1<\/sup>, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.\r\n\r\n<img class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145013\/hydrogeq3.gif\" alt=\"\" width=\"202px\" height=\"31px\" \/>\r\n\r\nIn fact what you get is -208 kJ mol<sup>-1<\/sup> - not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions \"fall down\" to the same end point.\r\n<p style=\"text-align: center\"><img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145014\/energetics.gif\" alt=\"\" width=\"337px\" height=\"283px\" \/><\/p>\r\nHeavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekul\u00e9 form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol<sup>-1<\/sup> more stable than the Kekul\u00e9 structure gives it credit for. This increase in stability of benzene is known as the <strong>delocalization energy<\/strong> or <strong>resonance energy<\/strong> of benzene.\r\n\r\n<\/div>\r\n<div id=\"section_2\">\r\n\r\n\r\n<h3 id=\"Contributors-3639\">Contributors<\/h3>\r\nJim Clark (<a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\" href=\"http:\/\/www.chemguide.co.uk\" target=\"_blank\" rel=\"external nofollow noopener\">Chemguide.co.uk<\/a>)\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<p>Kekul\u00e9 was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.<\/p>\n<div id=\"s3639\">\n<p><img decoding=\"async\" class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145006\/kekuleatoms.gif\" alt=\"\" width=\"97px\" height=\"111px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <img decoding=\"async\" class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145007\/kekule.gif\" alt=\"\" width=\"63px\" height=\"72px\" \/><\/p>\n<p>This diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing &#8211; and that must be attached to a hydrogen atom.<\/p>\n<div id=\"section_1\">\n<h3 id=\"Problems_with_the_Kekul.C3.A9_structure-3639\">Problems with the Kekul\u00e9 structure<\/h3>\n<p>Although the Kekul\u00e9 structure was a good attempt in its time, there are serious problems with it regarding chemistry, structure and stability.<\/p>\n<p><strong>The Kekul\u00e9 structure has problems with the chemistry.<\/strong> Because of the three double bonds, you might expect benzene to have reactions like ethene &#8211; only more so! Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms. Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new.<\/p>\n<p><strong>The Kekul\u00e9 structure has problems with the shape.<\/strong> Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekul\u00e9 structure. The problem is that C-C single and double bonds are different lengths.<\/p>\n<ul>\n<li>C-C (0.154 nm)<\/li>\n<li>C=C (0.134 nm)<\/li>\n<\/ul>\n<p>That would mean that the hexagon would be irregular if it had the Kekul\u00e9 structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same &#8211; intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.<\/p>\n<p><strong>The Kekul\u00e9 structure has problems with the stability of benzene.<\/strong> Real benzene is a lot more stable than the Kekul\u00e9 structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekul\u00e9 structure, you get an answer which is wrong by about 150 kJ mol<sup>-1<\/sup>. This is most easily shown using <a title=\"Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Robert's_%22Basic_Principles_of_Organic_Chemistry%22\/11%3A_Alkenes_and_Alkynes_II._Oxidation_and_Reduction_Reactions._Acidity_or_Alkynes\/11.3%3A_Heats_of_Hydrogenation\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)\/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes\/11.3%3A_Heats_of_Hydrogenation\" rel=\"internal\">enthalpy changes of hydrogenation<\/a>. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:<\/p>\n<p>\\[\\ce{CH_2=CH_2 \u00a0+ \u00a0H_2 \\rightarrow CH_3CH_3} \\tag{1}\\]<\/p>\n<p>CH_2=CH_2+H_2\u2192CH_3CH_33<\/p>\n<p>In order to do a fair comparison with benzene (a ring structure) we&#8217;re going to compare it with cyclohexene. Cyclohexene, C<sub>6<\/sub>H<sub>10<\/sub>, is a ring of six carbon atoms containing just one C=C.<\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145008\/chenefull.gif\" alt=\"\" width=\"80px\" height=\"90px\" \/><\/p>\n<p>When hydrogen is added to this, cyclohexane, C<sub>6<\/sub>H<sub>12<\/sub>, is formed. The &#8220;CH&#8221; groups become CH<sub>2<\/sub> and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekul\u00e9 structure for benzene is simplified &#8211; by leaving out all the carbons and hydrogens.<\/p>\n<p style=\"text-align: center\"><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145009\/cyclos.gif\" alt=\"\" width=\"198px\" height=\"96px\" \/><\/p>\n<p>In the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH<sub>2<\/sub>. The hydrogenation equation could be written:<\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145010\/hydrogeq1.gif\" alt=\"\" width=\"201px\" height=\"34px\" \/><\/p>\n<p>The enthalpy change during this reaction is -120 kJ mol<sup>-1<\/sup>. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.<\/p>\n<p>If the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene &#8211; that is, -240 kJ mol<sup>-1<\/sup>.<\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145011\/hydrogeq2.gif\" alt=\"\" width=\"203px\" height=\"34px\" \/><\/p>\n<p>In fact, the enthalpy change is -232 kJ mol<sup>-1<\/sup> &#8211; which isn&#8217;t far off what we are predicting. Applying the same argument to the Kekul\u00e9 structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol<sup>-1<\/sup>, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.<\/p>\n<p><img decoding=\"async\" class=\"internal\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145013\/hydrogeq3.gif\" alt=\"\" width=\"202px\" height=\"31px\" \/><\/p>\n<p>In fact what you get is -208 kJ mol<sup>-1<\/sup> &#8211; not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions &#8220;fall down&#8221; to the same end point.<\/p>\n<p style=\"text-align: center\"><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05145014\/energetics.gif\" alt=\"\" width=\"337px\" height=\"283px\" \/><\/p>\n<p>Heavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekul\u00e9 form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol<sup>-1<\/sup> more stable than the Kekul\u00e9 structure gives it credit for. This increase in stability of benzene is known as the <strong>delocalization energy<\/strong> or <strong>resonance energy<\/strong> of benzene.<\/p>\n<\/div>\n<div id=\"section_2\">\n<h3 id=\"Contributors-3639\">Contributors<\/h3>\n<p>Jim Clark (<a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\" href=\"http:\/\/www.chemguide.co.uk\" target=\"_blank\" rel=\"external nofollow noopener\">Chemguide.co.uk<\/a>)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":1,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1252","chapter","type-chapter","status-publish","hentry"],"part":26,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1252","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1252\/revisions"}],"predecessor-version":[{"id":1986,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1252\/revisions\/1986"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/26"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1252\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1252"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1252"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1252"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1252"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}