{"id":1700,"date":"2017-10-10T16:12:54","date_gmt":"2017-10-10T16:12:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1700"},"modified":"2018-10-05T19:28:54","modified_gmt":"2018-10-05T19:28:54","slug":"nmr-spectroscopy-and-proton-equivalence","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/nmr-spectroscopy-and-proton-equivalence\/","title":{"raw":"NMR Spectroscopy and Proton Equivalence","rendered":"NMR Spectroscopy and Proton Equivalence"},"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\nAfter completing this section, you should be able to\r\n<ol>\r\n \t<li>identify those protons which are equivalent in a given chemical structure.<\/li>\r\n \t<li>use the <sup>1<\/sup>H NMR spectrum of a simple organic compound to determine the number of equivalent sets of protons present.<\/li>\r\n<\/ol>\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 key-takeaways\">\r\n<h3>Key Term<\/h3>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div>\r\n\r\nMake certain that you can define, and use in context, the key terms below.\r\n<ul>\r\n \t<li>diastereotopic<\/li>\r\n \t<li>enantiotopic<\/li>\r\n \t<li>homotopic<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3 class=\"boxtitle\">Study Notes<\/h3>\r\nIt is important at this stage to be able to identify equivalent protons in any organic compound given the structure of that compound. Once you know the number of different groups of equivalent protons in a compound, you can predict the number (before coupling) and relative strength of signals. Look at the following examples and make sure you understand how the number and intensity ratio of signals are derived from the structure shown.\r\n<table style=\"height: 309px\">\r\n<tbody>\r\n<tr style=\"height: 30px\">\r\n<th style=\"height: 30px\" scope=\"col\">Structure<\/th>\r\n<th style=\"height: 30px\" scope=\"col\">Number of Signals<\/th>\r\n<th style=\"height: 30px\" colspan=\"2\" scope=\"col\">Ratio of Signals<\/th>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\">$\\ce{\\sf{CH3OCH2CH2Br}}$<\/td>\r\n<td style=\"height: 15px\">3<\/td>\r\n<td style=\"height: 15px\">A : B : C<\/td>\r\n<td style=\"height: 15px\">3 : 2 : 2<\/td>\r\n<\/tr>\r\n<tr style=\"height: 56px\">\r\n<td style=\"height: 56px\"><img src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87099\/13-8a.png?origin=mt-web\" alt=\"cyclopentane\" \/><\/td>\r\n<td style=\"height: 56px\">1<\/td>\r\n<td style=\"height: 56px\"><\/td>\r\n<td style=\"height: 56px\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 71px\">\r\n<td style=\"height: 71px\"><img src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87100\/13-8b.png?origin=mt-web\" alt=\"2,2-dimethyloxetane\" \/><\/td>\r\n<td style=\"height: 71px\">3<\/td>\r\n<td style=\"height: 71px\">A : B : C<\/td>\r\n<td style=\"height: 71px\">2 : 2 : 6 (or 1 : 1 : 3)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 77px\">\r\n<td style=\"height: 77px\"><img src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87101\/13-8c.png?origin=mt-web\" alt=\"1,3-cyclohexadiene\" \/><\/td>\r\n<td style=\"height: 77px\">3<\/td>\r\n<td style=\"height: 77px\">A : B : C<\/td>\r\n<td style=\"height: 77px\">2 : 4 : 2 (or 1 : 2 : 1)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"height: 30px\"><img src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87102\/13-8d.png?origin=mt-web\" alt=\"1-methoxy-2-butanone\" \/><\/td>\r\n<td style=\"height: 30px\">4<\/td>\r\n<td style=\"height: 30px\">A : B : C : D<\/td>\r\n<td style=\"height: 30px\">3 : 2 : 2 : 3<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"height: 30px\"><img src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87103\/13-8e.png?origin=mt-web\" alt=\"3-methylpyridine\" \/><\/td>\r\n<td style=\"height: 30px\">5<\/td>\r\n<td style=\"height: 30px\">A : B : C : D : E<\/td>\r\n<td style=\"height: 30px\">3 : 1 : 1 : 1 : 1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\nIf all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, the information in the previous paragraph would be interesting from a theoretical standpoint, but would not be terribly useful to organic chemists. Fortunately for us, however, resonance frequencies are not uniform for all protons in a molecule. <em>In an external magnetic field of a given strength, protons in different locations in a molecule have different resonance frequencies, because they are in non-identical electronic environments.<\/em>\u00a0 In methyl acetate, for example, there are two \u2018sets\u2019 of protons. The three protons labeled H<sub>a<\/sub> have a different - and easily distinguishable \u2013 resonance frequency than the three H<sub>b <\/sub>protons, because the two sets of protons are in non-identical environments: they are, in other words, chemically nonequivalent.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154125\/image006.png\" alt=\"image006.png\" width=\"173\" height=\"79\" \/>\r\n\r\nOn the other hand, the three H<sub>a<\/sub> protons are all in the same electronic environment, and are chemically equivalent to one another.\u00a0 They have identical resonance frequencies. The same can be said for the three H<sub>b<\/sub> protons.\r\n\r\nThe ability to recognize chemical equivalancy and nonequivalency among atoms in a molecule will be central to understanding NMR.\u00a0 In each of the molecules below, all protons are chemically equivalent, and therefore will have the same resonance frequency in an NMR experiment.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154127\/image008.png\" alt=\"image008.png\" width=\"612\" height=\"164\" \/>\r\n\r\nYou might expect that the equitorial and axial hydrogens in cyclohexane would be non-equivalent, and would have different resonance frequencies.\u00a0 In fact, an axial hydrogen <em>is<\/em> in a different electronic environment than an equitorial hydrogen.\u00a0 Remember, though, that the molecule rotates rapidly between its two chair conformations, meaning that any given hydrogen is rapidly moving back and forth between equitorial and axial positions.\u00a0 It turns out that, except at extremely low temperatures, this rotational motion occurs on a time scale that is much faster than the time scale of an NMR experiment.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154129\/image010.png\" alt=\"image010.png\" width=\"382\" height=\"110\" \/>\r\n\r\nIn this sense, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed - the result is a blurred image.\u00a0 In NMR terms, this means that all 12 protons in cyclohexane are equivalent.\r\n\r\nEach the molecules in the next figure contains <em>two<\/em> sets of protons, just like our previous example of methyl acetate, and again in each case the resonance frequency of the H<sub>a<\/sub> protons will be different from that of the H<sub>b<\/sub> protons.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154132\/image012.png\" alt=\"image012.png\" width=\"568\" height=\"155\" \/>\r\n\r\nNotice how the symmetry of <em>para<\/em>-xylene results in there being only two different sets of protons.\r\n\r\nMost organic molecules have several sets of protons in different chemical environments, and each set, in theory, will have a different resonance frequency in <sup>1<\/sup>H-NMR spectroscopy.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154135\/image014.png\" alt=\"image014.png\" width=\"528\" height=\"192\" \/>\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154139\/image016.png\" alt=\"image016.png\" width=\"676\" height=\"265\" \/>\r\n\r\nWhen stereochemistry is taken into account, the issue of\u00a0 equivalence vs nonequivalence in NMR starts to get a little more complicated.\u00a0 It should be fairly intuitive that hydrogens on different sides of asymmetric ring structures and double bonds are in different electronic environments, and thus are non-equivalent and have different resonance frequencies. In the alkene and cyclohexene structures below, for example, H<sub>a<\/sub> is <em>trans<\/em> to the chlorine substituent, while H<sub>b<\/sub> is <em>cis<\/em> to chlorine.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154141\/image018.png\" alt=\"image018.png\" width=\"303\" height=\"86\" \/>\r\n\r\nWhat is not so intuitive is that diastereotopic hydrogens (section 3.10) on chiral molecules are also non-equivalent:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154145\/image020.png\" alt=\"image020.png\" width=\"256\" height=\"154\" \/>\r\n\r\nHowever, enantiotopic and homotopic hydrogens are chemically equivalent.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154147\/image022.png\" alt=\"image022.png\" width=\"487\" height=\"151\" \/>\r\n<div>\r\n<div id=\"example\">\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<div>\r\n<div id=\"example\">\r\n\r\nHow many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154150\/image024.png\" alt=\"image024.png\" width=\"503\" height=\"396\" \/>\r\n<h3>Solution<\/h3>\r\n[reveal-answer q=\"980947\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"980947\"]a) 8 b) 8 c) 5 d) 18 [\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_1\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<div id=\"section_1\">\r\n<div id=\"s61718\">\r\n<div id=\"section_20\">\r\n\r\nHow many non-equivalent hydrogen are in the following molecules; how many different signals will you see in a H<sup>1<\/sup> NMR spectrum.\r\n\r\nA. CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>Br\r\n\r\nB. CH<sub>3<\/sub>OCH<sub>2<\/sub>C(CH<sub>3<\/sub>)<sub>3<\/sub>\r\n\r\nC. Ethyl Benzene\r\n<h3>Solutions<\/h3>\r\n[reveal-answer q=\"14204\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"14204\"]\r\n\r\nA. 3; B. 3; C. 5; D. 7\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_2\">\r\n<h3 class=\"editable\">Contributors<\/h3>\r\n<ul>\r\n \t<li><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>)<\/li>\r\n \t<li>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>)<\/li>\r\n \t<li><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)<\/li>\r\n<\/ul>\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>After completing this section, you should be able to<\/p>\n<ol>\n<li>identify those protons which are equivalent in a given chemical structure.<\/li>\n<li>use the <sup>1<\/sup>H NMR spectrum of a simple organic compound to determine the number of equivalent sets of protons present.<\/li>\n<\/ol>\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 key-takeaways\">\n<h3>Key Term<\/h3>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div>\n<p>Make certain that you can define, and use in context, the key terms below.<\/p>\n<ul>\n<li>diastereotopic<\/li>\n<li>enantiotopic<\/li>\n<li>homotopic<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3 class=\"boxtitle\">Study Notes<\/h3>\n<p>It is important at this stage to be able to identify equivalent protons in any organic compound given the structure of that compound. Once you know the number of different groups of equivalent protons in a compound, you can predict the number (before coupling) and relative strength of signals. Look at the following examples and make sure you understand how the number and intensity ratio of signals are derived from the structure shown.<\/p>\n<table style=\"height: 309px\">\n<tbody>\n<tr style=\"height: 30px\">\n<th style=\"height: 30px\" scope=\"col\">Structure<\/th>\n<th style=\"height: 30px\" scope=\"col\">Number of Signals<\/th>\n<th style=\"height: 30px\" colspan=\"2\" scope=\"col\">Ratio of Signals<\/th>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\">$\\ce{\\sf{CH3OCH2CH2Br}}$<\/td>\n<td style=\"height: 15px\">3<\/td>\n<td style=\"height: 15px\">A : B : C<\/td>\n<td style=\"height: 15px\">3 : 2 : 2<\/td>\n<\/tr>\n<tr style=\"height: 56px\">\n<td style=\"height: 56px\"><img decoding=\"async\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87099\/13-8a.png?origin=mt-web\" alt=\"cyclopentane\" \/><\/td>\n<td style=\"height: 56px\">1<\/td>\n<td style=\"height: 56px\"><\/td>\n<td style=\"height: 56px\"><\/td>\n<\/tr>\n<tr style=\"height: 71px\">\n<td style=\"height: 71px\"><img decoding=\"async\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87100\/13-8b.png?origin=mt-web\" alt=\"2,2-dimethyloxetane\" \/><\/td>\n<td style=\"height: 71px\">3<\/td>\n<td style=\"height: 71px\">A : B : C<\/td>\n<td style=\"height: 71px\">2 : 2 : 6 (or 1 : 1 : 3)<\/td>\n<\/tr>\n<tr style=\"height: 77px\">\n<td style=\"height: 77px\"><img decoding=\"async\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87101\/13-8c.png?origin=mt-web\" alt=\"1,3-cyclohexadiene\" \/><\/td>\n<td style=\"height: 77px\">3<\/td>\n<td style=\"height: 77px\">A : B : C<\/td>\n<td style=\"height: 77px\">2 : 4 : 2 (or 1 : 2 : 1)<\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"height: 30px\"><img decoding=\"async\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87102\/13-8d.png?origin=mt-web\" alt=\"1-methoxy-2-butanone\" \/><\/td>\n<td style=\"height: 30px\">4<\/td>\n<td style=\"height: 30px\">A : B : C : D<\/td>\n<td style=\"height: 30px\">3 : 2 : 2 : 3<\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"height: 30px\"><img decoding=\"async\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87103\/13-8e.png?origin=mt-web\" alt=\"3-methylpyridine\" \/><\/td>\n<td style=\"height: 30px\">5<\/td>\n<td style=\"height: 30px\">A : B : C : D : E<\/td>\n<td style=\"height: 30px\">3 : 1 : 1 : 1 : 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<p>If all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, the information in the previous paragraph would be interesting from a theoretical standpoint, but would not be terribly useful to organic chemists. Fortunately for us, however, resonance frequencies are not uniform for all protons in a molecule. <em>In an external magnetic field of a given strength, protons in different locations in a molecule have different resonance frequencies, because they are in non-identical electronic environments.<\/em>\u00a0 In methyl acetate, for example, there are two \u2018sets\u2019 of protons. The three protons labeled H<sub>a<\/sub> have a different &#8211; and easily distinguishable \u2013 resonance frequency than the three H<sub>b <\/sub>protons, because the two sets of protons are in non-identical environments: they are, in other words, chemically nonequivalent.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154125\/image006.png\" alt=\"image006.png\" width=\"173\" height=\"79\" \/><\/p>\n<p>On the other hand, the three H<sub>a<\/sub> protons are all in the same electronic environment, and are chemically equivalent to one another.\u00a0 They have identical resonance frequencies. The same can be said for the three H<sub>b<\/sub> protons.<\/p>\n<p>The ability to recognize chemical equivalancy and nonequivalency among atoms in a molecule will be central to understanding NMR.\u00a0 In each of the molecules below, all protons are chemically equivalent, and therefore will have the same resonance frequency in an NMR experiment.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154127\/image008.png\" alt=\"image008.png\" width=\"612\" height=\"164\" \/><\/p>\n<p>You might expect that the equitorial and axial hydrogens in cyclohexane would be non-equivalent, and would have different resonance frequencies.\u00a0 In fact, an axial hydrogen <em>is<\/em> in a different electronic environment than an equitorial hydrogen.\u00a0 Remember, though, that the molecule rotates rapidly between its two chair conformations, meaning that any given hydrogen is rapidly moving back and forth between equitorial and axial positions.\u00a0 It turns out that, except at extremely low temperatures, this rotational motion occurs on a time scale that is much faster than the time scale of an NMR experiment.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154129\/image010.png\" alt=\"image010.png\" width=\"382\" height=\"110\" \/><\/p>\n<p>In this sense, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed &#8211; the result is a blurred image.\u00a0 In NMR terms, this means that all 12 protons in cyclohexane are equivalent.<\/p>\n<p>Each the molecules in the next figure contains <em>two<\/em> sets of protons, just like our previous example of methyl acetate, and again in each case the resonance frequency of the H<sub>a<\/sub> protons will be different from that of the H<sub>b<\/sub> protons.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154132\/image012.png\" alt=\"image012.png\" width=\"568\" height=\"155\" \/><\/p>\n<p>Notice how the symmetry of <em>para<\/em>-xylene results in there being only two different sets of protons.<\/p>\n<p>Most organic molecules have several sets of protons in different chemical environments, and each set, in theory, will have a different resonance frequency in <sup>1<\/sup>H-NMR spectroscopy.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154135\/image014.png\" alt=\"image014.png\" width=\"528\" height=\"192\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154139\/image016.png\" alt=\"image016.png\" width=\"676\" height=\"265\" \/><\/p>\n<p>When stereochemistry is taken into account, the issue of\u00a0 equivalence vs nonequivalence in NMR starts to get a little more complicated.\u00a0 It should be fairly intuitive that hydrogens on different sides of asymmetric ring structures and double bonds are in different electronic environments, and thus are non-equivalent and have different resonance frequencies. In the alkene and cyclohexene structures below, for example, H<sub>a<\/sub> is <em>trans<\/em> to the chlorine substituent, while H<sub>b<\/sub> is <em>cis<\/em> to chlorine.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154141\/image018.png\" alt=\"image018.png\" width=\"303\" height=\"86\" \/><\/p>\n<p>What is not so intuitive is that diastereotopic hydrogens (section 3.10) on chiral molecules are also non-equivalent:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154145\/image020.png\" alt=\"image020.png\" width=\"256\" height=\"154\" \/><\/p>\n<p>However, enantiotopic and homotopic hydrogens are chemically equivalent.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154147\/image022.png\" alt=\"image022.png\" width=\"487\" height=\"151\" \/><\/p>\n<div>\n<div id=\"example\">\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<div>\n<div id=\"example\">\n<p>How many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154150\/image024.png\" alt=\"image024.png\" width=\"503\" height=\"396\" \/><\/p>\n<h3>Solution<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q980947\">Show Answer<\/span><\/p>\n<div id=\"q980947\" class=\"hidden-answer\" style=\"display: none\">a) 8 b) 8 c) 5 d) 18 <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_1\">\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<div id=\"section_1\">\n<div id=\"s61718\">\n<div id=\"section_20\">\n<p>How many non-equivalent hydrogen are in the following molecules; how many different signals will you see in a H<sup>1<\/sup> NMR spectrum.<\/p>\n<p>A. CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>Br<\/p>\n<p>B. CH<sub>3<\/sub>OCH<sub>2<\/sub>C(CH<sub>3<\/sub>)<sub>3<\/sub><\/p>\n<p>C. Ethyl Benzene<\/p>\n<h3>Solutions<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14204\">Show Answer<\/span><\/p>\n<div id=\"q14204\" class=\"hidden-answer\" style=\"display: none\">\n<p>A. 3; B. 3; C. 5; D. 7<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_2\">\n<h3 class=\"editable\">Contributors<\/h3>\n<ul>\n<li><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>)<\/li>\n<li>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>)<\/li>\n<li><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)<\/li>\n<\/ul>\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-1700","chapter","type-chapter","status-publish","hentry"],"part":29,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1700","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":8,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1700\/revisions"}],"predecessor-version":[{"id":2349,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1700\/revisions\/2349"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/29"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1700\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1700"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1700"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1700"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}