{"id":1018,"date":"2017-10-19T14:35:56","date_gmt":"2017-10-19T14:35:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1018"},"modified":"2018-10-03T19:52:27","modified_gmt":"2018-10-03T19:52:27","slug":"electrophilic-addition-reactions-of-alkenes","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/electrophilic-addition-reactions-of-alkenes\/","title":{"raw":"Electrophilic Addition Reactions of Alkenes","rendered":"Electrophilic Addition Reactions of Alkenes"},"content":{"raw":"<div class=\"elm-header\">\r\n<div class=\"elm-header-custom\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Objectives<\/h3>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div id=\"skills\">\r\n\r\n<span style=\"color: #000000\">After completing this section, you should be able to<\/span>\r\n<ol>\r\n \t<li><span style=\"color: #000000\">explain the term \u201celectrophilic addition reaction,\u201d using the reaction of a protic acid, HX, with an alkene as an example.<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">write the mechanism for the reaction of a protic acid, HX, with an alkene.<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">sketch a reaction energy diagram for the electrophilic addition of an acid, HX, to an alkene.<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key TERMS<\/h3>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div>\r\n\r\n<span style=\"color: #000000\">Make certain that you can define, and use in context, the key terms below.<\/span>\r\n<ul>\r\n \t<li><span style=\"color: #000000\">carbocation (carbonium ion)<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">electrophilic addition reaction<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div id=\"note\">\r\n<div class=\"textbox\">\r\n<h3 class=\"boxtitle\"><span style=\"color: #000000\">Study Notes<\/span><\/h3>\r\n<span style=\"color: #000000\">An <em>electrophilic addition reaction<\/em> is a reaction in which a substrate is initially attacked by an electrophile, and the overall result is the addition of one or more relatively simple molecules across a multiple bond.<\/span>\r\n\r\n<span style=\"color: #000000\">The mechanism for the addition of hydrogen halide to propene shown in the reading is quite detailed. Normally, an organic chemist would write this mechanism as follows:<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/85532\/7-7.png?origin=mt-web\" alt=\"hydrogen halide addition to propene\" \/><\/span>\r\n\r\n&nbsp;\r\n\r\n<span style=\"color: #000000\">However, the more detailed mechanism shown in the reading does allow you to see the exact fate of all the electrons involved in the reaction.<\/span>\r\n\r\n<span style=\"color: #000000\">In your previous chemistry course, you were probably taught the importance of balancing chemical equations. It may come as a surprise to you that organic chemists usually do not balance their equations, and often represent reactions using a format which is quite different from the carefully written, balanced equations encountered in general chemistry courses. In fact, organic chemists are rarely interested in the inorganic products of their reactions; furthermore, most organic reactions are non-quantitative in nature.<\/span>\r\n\r\n<span style=\"color: #000000\">In many of the reactions in this course, the percentage yield is indicated beneath the products: you are not expected to memorize these figures. The question of yield is very important in organic chemistry, where two, five, ten or even twenty reactions may be needed to synthesize a desired product. For example, if a chemist wishes to prepare compound D by the following reaction sequence:<\/span>\r\n\r\n<span style=\"color: #000000\">A \u2192\u00a0B \u2192\u00a0C \u2192\u00a0D<\/span>\r\n\r\n<span style=\"color: #000000\">and each of the individual steps gives only a 50% yield, one mole of A would give only<\/span>\r\n\r\n<span style=\"color: #000000\">1 mol \u00d7 50% 100% \u00d7 50% 100% \u00d7 50% 100% = 0.125 mol of D<\/span>\r\n\r\n<span style=\"color: #000000\">You will gain first-hand experience of such situations in the laboratory component of this course.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<span style=\"color: #000000\">This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide. Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.<\/span>\r\n<div id=\"section_1\">\r\n<h3 class=\"editable\"><span style=\"color: #000000\">Addition to symmetrical alkenes<\/span><\/h3>\r\n<span style=\"color: #000000\"><strong>What happens?<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\">All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.<\/span>\r\n\r\n<span style=\"color: #000000\">For example, with ethene and hydrogen chloride, you get chloroethane:<\/span>\r\n\r\n<span style=\"color: #000000\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140305\/padding.gif\" alt=\"image\" width=\"40\" height=\"15\" \/><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140306\/ethenehcl.gif\" alt=\"image\" width=\"248\" height=\"15\" \/><\/span>\r\n\r\n<span style=\"color: #000000\">With but-2-ene you get 2-chlorobutane:<\/span>\r\n\r\n<span style=\"color: #000000\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140305\/padding.gif\" alt=\"image\" width=\"40\" height=\"15\" \/><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140307\/but2enehcl.gif\" alt=\"image\" width=\"341\" height=\"37\" \/><\/span>\r\n\r\n<span style=\"color: #000000\">What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product.<\/span>\r\n\r\n<span style=\"color: #000000\">The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space.<\/span>\r\n\r\n<span style=\"color: #000000\">That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.<\/span>\r\n\r\n<\/div>\r\n<div id=\"section_2\">\r\n<h2 class=\"editable\"><span style=\"color: #000000\">Mechanism<\/span><\/h2>\r\n<span style=\"color: #000000\">The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of \u03c0 electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the\u00a0<a class=\"internal\" title=\"Organic Chemistry\/Reactions\/E1 Reaction\" href=\"\/Organic_Chemistry\/Reactions\/E1_Reaction\" rel=\"internal\">E1<\/a> reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds\u00a0to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good <a class=\"internal\" title=\"Organic Chemistry\/Reactions\/SN2 Reaction\/Nucleophile\" href=\"\/Organic_Chemistry\/Reactions\/SN2_Reaction\/Nucleophile\" rel=\"internal\">nucleophile <\/a>is used.<\/span>\r\n\r\n<span style=\"color: #000000\"><em>Mechanism of Electrophilic Addition of Hydrogen Halide to Ethene<\/em><\/span>\r\n\r\n<span style=\"color: #000000\"><em><img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140310\/general_rxn.gif\" alt=\"general rxn.gif\" width=\"690px\" height=\"155px\" \/><\/em><\/span>\r\n\r\n<span style=\"color: #000000\"><em>Mechanism of Electrophilic Addition of Hydrogen Halide to Propene<\/em><\/span>\r\n\r\n<span style=\"color: #000000\"><img class=\"default internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140312\/rxn_2.gif\" alt=\"File:Organic_Chemistry\/Hydrocarbons\/Alkenes\/Reactions_of_Alkenes\/Electrophilic_Addition_of_Hydrogen_Halides\/rxn_2.gif\" \/><\/span>\r\n\r\n<span style=\"color: #000000\">All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.<\/span>\r\n<div id=\"section_3\">\r\n<h3 class=\"editable\"><span style=\"color: #000000\">Reaction rates<\/span><\/h3>\r\n<span style=\"color: #000000\"><strong>Variation of rates when you change the halogen<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\">Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.<\/span>\r\n\r\n<span style=\"color: #000000\">When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.<\/span>\r\n\r\n<span style=\"color: #000000\"><strong>Variation of rates when you change the alkene<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\">This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.<\/span>\r\n\r\n<span style=\"color: #000000\">Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond.<\/span>\r\n\r\n<span style=\"color: #000000\">For example:<\/span>\r\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140314\/reactivity.gif\" alt=\"image\" width=\"382\" height=\"101\" \/><\/span><\/p>\r\n<span style=\"color: #000000\">There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.<\/span>\r\n\r\n<span style=\"color: #000000\">Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.<\/span>\r\n\r\n<span style=\"color: #000000\">Alkyl groups have a tendency to \"push\" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.<\/span>\r\n\r\n<span style=\"color: #000000\">The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.<\/span>\r\n\r\n<span style=\"color: #000000\">The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:<\/span>\r\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140317\/reactivity2.gif\" alt=\"image\" width=\"435\" height=\"284\" \/><\/span><\/p>\r\n<span style=\"color: #000000\">The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.<\/span>\r\n\r\n<span style=\"color: #000000\"><strong>Addition to unsymmetrical alkenes<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\"><strong>What happens?<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\">In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond.<\/span>\r\n\r\n<span style=\"color: #000000\"><strong>Orientation of addition<\/strong><\/span>\r\n\r\n<span style=\"color: #000000\">If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.<\/span>\r\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140319\/propenehcl.gif\" alt=\"image\" width=\"349\" height=\"197\" \/><\/span><\/p>\r\n<span style=\"color: #000000\">This is in line with Markovnikov's Rule which says:<\/span>\r\n<dl>\r\n \t<dt><\/dt>\r\n \t<dd><span style=\"color: #000000\">When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.<\/span><\/dd>\r\n<\/dl>\r\n<span style=\"color: #000000\">In this case, the hydrogen becomes attached to the CH<sub>2<\/sub> group, because the CH<sub>2<\/sub> group has more hydrogens than the CH group.<\/span>\r\n\r\n<span style=\"color: #000000\">Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH<sub>3<\/sub> group are totally irrelevant.<\/span>\r\n\r\n<\/div>\r\n<div id=\"section_4\">\r\n<h3 class=\"editable\"><span style=\"color: #000000\">Contributors<\/span><\/h3>\r\n<ul>\r\n \t<li><span style=\"color: #000000\"><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">William Reusch, Professor Emeritus (<a class=\"external\" title=\"http:\/\/www.msu.edu\/\" href=\"http:\/\/www.msu.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Michigan State U.<\/a>), <a class=\"external\" title=\"http:\/\/www.cem.msu.edu\/~reusch\/VirtualText\/intro1.htm\" href=\"http:\/\/www.cem.msu.edu\/%7Ereusch\/VirtualText\/intro1.htm\" target=\"_blank\" rel=\"external nofollow noopener\">Virtual Textbook of\u00a0Organic\u00a0Chemistry<\/a><\/span><\/li>\r\n \t<li><span style=\"color: #000000\"><a title=\"Organic_Chemistry_With_a_Biological_Emphasis\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\" rel=\"internal\">Organic Chemistry With a Biological Emphasis <\/a>by\u00a0<a class=\"external\" title=\"http:\/\/facultypages.morris.umn.edu\/~soderbt\/\" href=\"http:\/\/facultypages.morris.umn.edu\/%7Esoderbt\/\" target=\"_blank\" rel=\"external nofollow noopener\">Tim Soderberg<\/a>\u00a0(University of Minnesota, Morris)<\/span><\/li>\r\n \t<li><span style=\"color: #000000\">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>)<\/span><\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li><span style=\"color: #000000\"><span class=\"person_name\">John D. Robert <\/span>and <span class=\"person_name\">Marjorie C.<\/span> <span class=\"person_name\">Caserio <\/span>(1977) <em>Basic Principles of Organic Chemistry, second edition.<\/em> W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, \"You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format.\"<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"elm-header\">\n<div class=\"elm-header-custom\">\n<div class=\"textbox learning-objectives\">\n<h3>Objectives<\/h3>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div id=\"skills\">\n<p><span style=\"color: #000000\">After completing this section, you should be able to<\/span><\/p>\n<ol>\n<li><span style=\"color: #000000\">explain the term \u201celectrophilic addition reaction,\u201d using the reaction of a protic acid, HX, with an alkene as an example.<\/span><\/li>\n<li><span style=\"color: #000000\">write the mechanism for the reaction of a protic acid, HX, with an alkene.<\/span><\/li>\n<li><span style=\"color: #000000\">sketch a reaction energy diagram for the electrophilic addition of an acid, HX, to an alkene.<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key TERMS<\/h3>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div>\n<p><span style=\"color: #000000\">Make certain that you can define, and use in context, the key terms below.<\/span><\/p>\n<ul>\n<li><span style=\"color: #000000\">carbocation (carbonium ion)<\/span><\/li>\n<li><span style=\"color: #000000\">electrophilic addition reaction<\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div id=\"note\">\n<div class=\"textbox\">\n<h3 class=\"boxtitle\"><span style=\"color: #000000\">Study Notes<\/span><\/h3>\n<p><span style=\"color: #000000\">An <em>electrophilic addition reaction<\/em> is a reaction in which a substrate is initially attacked by an electrophile, and the overall result is the addition of one or more relatively simple molecules across a multiple bond.<\/span><\/p>\n<p><span style=\"color: #000000\">The mechanism for the addition of hydrogen halide to propene shown in the reading is quite detailed. Normally, an organic chemist would write this mechanism as follows:<img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/85532\/7-7.png?origin=mt-web\" alt=\"hydrogen halide addition to propene\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #000000\">However, the more detailed mechanism shown in the reading does allow you to see the exact fate of all the electrons involved in the reaction.<\/span><\/p>\n<p><span style=\"color: #000000\">In your previous chemistry course, you were probably taught the importance of balancing chemical equations. It may come as a surprise to you that organic chemists usually do not balance their equations, and often represent reactions using a format which is quite different from the carefully written, balanced equations encountered in general chemistry courses. In fact, organic chemists are rarely interested in the inorganic products of their reactions; furthermore, most organic reactions are non-quantitative in nature.<\/span><\/p>\n<p><span style=\"color: #000000\">In many of the reactions in this course, the percentage yield is indicated beneath the products: you are not expected to memorize these figures. The question of yield is very important in organic chemistry, where two, five, ten or even twenty reactions may be needed to synthesize a desired product. For example, if a chemist wishes to prepare compound D by the following reaction sequence:<\/span><\/p>\n<p><span style=\"color: #000000\">A \u2192\u00a0B \u2192\u00a0C \u2192\u00a0D<\/span><\/p>\n<p><span style=\"color: #000000\">and each of the individual steps gives only a 50% yield, one mole of A would give only<\/span><\/p>\n<p><span style=\"color: #000000\">1 mol \u00d7 50% 100% \u00d7 50% 100% \u00d7 50% 100% = 0.125 mol of D<\/span><\/p>\n<p><span style=\"color: #000000\">You will gain first-hand experience of such situations in the laboratory component of this course.<\/span><\/p>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide. Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.<\/span><\/p>\n<div id=\"section_1\">\n<h3 class=\"editable\"><span style=\"color: #000000\">Addition to symmetrical alkenes<\/span><\/h3>\n<p><span style=\"color: #000000\"><strong>What happens?<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.<\/span><\/p>\n<p><span style=\"color: #000000\">For example, with ethene and hydrogen chloride, you get chloroethane:<\/span><\/p>\n<p><span style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140305\/padding.gif\" alt=\"image\" width=\"40\" height=\"15\" \/><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140306\/ethenehcl.gif\" alt=\"image\" width=\"248\" height=\"15\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">With but-2-ene you get 2-chlorobutane:<\/span><\/p>\n<p><span style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140305\/padding.gif\" alt=\"image\" width=\"40\" height=\"15\" \/><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140307\/but2enehcl.gif\" alt=\"image\" width=\"341\" height=\"37\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product.<\/span><\/p>\n<p><span style=\"color: #000000\">The chlorine would be on a carbon atom next to the end of the chain &#8211; you would simply have drawn the molecule flipped over in space.<\/span><\/p>\n<p><span style=\"color: #000000\">That would be different of the alkene was unsymmetrical &#8211; that&#8217;s why we have to look at them separately.<\/span><\/p>\n<\/div>\n<div id=\"section_2\">\n<h2 class=\"editable\"><span style=\"color: #000000\">Mechanism<\/span><\/h2>\n<p><span style=\"color: #000000\">The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of \u03c0 electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the\u00a0<a class=\"internal\" title=\"Organic Chemistry\/Reactions\/E1 Reaction\" href=\"\/Organic_Chemistry\/Reactions\/E1_Reaction\" rel=\"internal\">E1<\/a> reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds\u00a0to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good <a class=\"internal\" title=\"Organic Chemistry\/Reactions\/SN2 Reaction\/Nucleophile\" href=\"\/Organic_Chemistry\/Reactions\/SN2_Reaction\/Nucleophile\" rel=\"internal\">nucleophile <\/a>is used.<\/span><\/p>\n<p><span style=\"color: #000000\"><em>Mechanism of Electrophilic Addition of Hydrogen Halide to Ethene<\/em><\/span><\/p>\n<p><span style=\"color: #000000\"><em><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140310\/general_rxn.gif\" alt=\"general rxn.gif\" width=\"690px\" height=\"155px\" \/><\/em><\/span><\/p>\n<p><span style=\"color: #000000\"><em>Mechanism of Electrophilic Addition of Hydrogen Halide to Propene<\/em><\/span><\/p>\n<p><span style=\"color: #000000\"><img decoding=\"async\" class=\"default internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140312\/rxn_2.gif\" alt=\"File:Organic_Chemistry\/Hydrocarbons\/Alkenes\/Reactions_of_Alkenes\/Electrophilic_Addition_of_Hydrogen_Halides\/rxn_2.gif\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.<\/span><\/p>\n<div id=\"section_3\">\n<h3 class=\"editable\"><span style=\"color: #000000\">Reaction rates<\/span><\/h3>\n<p><span style=\"color: #000000\"><strong>Variation of rates when you change the halogen<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">Reaction rates increase in the order HF &#8211; HCl &#8211; HBr &#8211; HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.<\/span><\/p>\n<p><span style=\"color: #000000\">When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>Variation of rates when you change the alkene<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don&#8217;t have to be.<\/span><\/p>\n<p><span style=\"color: #000000\">Reaction rates increase as the alkene gets more complicated &#8211; in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond.<\/span><\/p>\n<p><span style=\"color: #000000\">For example:<\/span><\/p>\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140314\/reactivity.gif\" alt=\"image\" width=\"382\" height=\"101\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">There are two ways of looking at the reasons for this &#8211; both of which need you to know about the mechanism for the reactions.<\/span><\/p>\n<p><span style=\"color: #000000\">Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.<\/span><\/p>\n<p><span style=\"color: #000000\">Alkyl groups have a tendency to &#8220;push&#8221; electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.<\/span><\/p>\n<p><span style=\"color: #000000\">The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.<\/span><\/p>\n<p><span style=\"color: #000000\">The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:<\/span><\/p>\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140317\/reactivity2.gif\" alt=\"image\" width=\"435\" height=\"284\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>Addition to unsymmetrical alkenes<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\"><strong>What happens?<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition &#8211; in other words, which way around the hydrogen and the halogen add across the double bond.<\/span><\/p>\n<p><span style=\"color: #000000\"><strong>Orientation of addition<\/strong><\/span><\/p>\n<p><span style=\"color: #000000\">If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.<\/span><\/p>\n<p style=\"text-align: center\"><span style=\"color: #000000\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05140319\/propenehcl.gif\" alt=\"image\" width=\"349\" height=\"197\" \/><\/span><\/p>\n<p><span style=\"color: #000000\">This is in line with Markovnikov&#8217;s Rule which says:<\/span><\/p>\n<dl>\n<dt><\/dt>\n<dd><span style=\"color: #000000\">When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.<\/span><\/dd>\n<\/dl>\n<p><span style=\"color: #000000\">In this case, the hydrogen becomes attached to the CH<sub>2<\/sub> group, because the CH<sub>2<\/sub> group has more hydrogens than the CH group.<\/span><\/p>\n<p><span style=\"color: #000000\">Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH<sub>3<\/sub> group are totally irrelevant.<\/span><\/p>\n<\/div>\n<div id=\"section_4\">\n<h3 class=\"editable\"><span style=\"color: #000000\">Contributors<\/span><\/h3>\n<ul>\n<li><span style=\"color: #000000\"><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/span><\/li>\n<li><span style=\"color: #000000\">Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/span><\/li>\n<li><span style=\"color: #000000\">William Reusch, Professor Emeritus (<a class=\"external\" title=\"http:\/\/www.msu.edu\/\" href=\"http:\/\/www.msu.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Michigan State U.<\/a>), <a class=\"external\" title=\"http:\/\/www.cem.msu.edu\/~reusch\/VirtualText\/intro1.htm\" href=\"http:\/\/www.cem.msu.edu\/%7Ereusch\/VirtualText\/intro1.htm\" target=\"_blank\" rel=\"external nofollow noopener\">Virtual Textbook of\u00a0Organic\u00a0Chemistry<\/a><\/span><\/li>\n<li><span style=\"color: #000000\"><a title=\"Organic_Chemistry_With_a_Biological_Emphasis\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry_Textbook_Maps\/Map%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\" rel=\"internal\">Organic Chemistry With a Biological Emphasis <\/a>by\u00a0<a class=\"external\" title=\"http:\/\/facultypages.morris.umn.edu\/~soderbt\/\" href=\"http:\/\/facultypages.morris.umn.edu\/%7Esoderbt\/\" target=\"_blank\" rel=\"external nofollow noopener\">Tim Soderberg<\/a>\u00a0(University of Minnesota, Morris)<\/span><\/li>\n<li><span style=\"color: #000000\">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>)<\/span><\/li>\n<\/ul>\n<ul>\n<li><span style=\"color: #000000\"><span class=\"person_name\">John D. Robert <\/span>and <span class=\"person_name\">Marjorie C.<\/span> <span class=\"person_name\">Caserio <\/span>(1977) <em>Basic Principles of Organic Chemistry, second edition.<\/em> W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, &#8220;You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format.&#8221;<\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":8,"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-1018","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1018","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":5,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1018\/revisions"}],"predecessor-version":[{"id":2286,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1018\/revisions\/2286"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1018\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1018"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1018"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1018"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1018"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}