{"id":1742,"date":"2017-10-10T15:46:26","date_gmt":"2017-10-10T15:46:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1742"},"modified":"2018-10-05T19:44:02","modified_gmt":"2018-10-05T19:44:02","slug":"complex-spin-spin-splitting-patterns","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/complex-spin-spin-splitting-patterns\/","title":{"raw":"Complex Spin-Spin Splitting Patterns","rendered":"Complex Spin-Spin Splitting Patterns"},"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>explain how multiple coupling can give rise to complex-looking <sup>1<\/sup>H NMR spectra.<\/li>\r\n \t<li>predict the splitting pattern expected in the <sup>1<\/sup>H NMR spectrum of an organic compound in which multiple coupling is possible.<\/li>\r\n \t<li>interpret <sup>1<\/sup>H NMR spectra in which multiple coupling is evident.<\/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\nMake certain that you can define, and use in context, the key term below.\r\n<ul>\r\n \t<li>tree diagram<\/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<div id=\"note\">\r\n<h3 class=\"boxtitle\">Study Notes<\/h3>\r\nWe saw the effects of spin-spin coupling on the appearance of a <sup>1<\/sup>H NMR signal. These effects can be further complicated when that signal is coupled to several different protons. For example, BrCH<sub>2<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>Cl would produce three signals. The hydrogens at C<sub>1<\/sub> and C<sub>3<\/sub> would each be triplets because of coupling to the two hydrogens on C<sub>2<\/sub>. However, the hydrogen on C<sub>2<\/sub> \u201csees\u201d two different sets of neighbouring hydrogens, and would therefore produce a triplet of triplets.\r\n\r\nAnother effect that can complicate a spectrum is the \u201ccloseness\u201d of signals. If signals accidently overlap they can be difficult to identify. In the example above, we expected a triplet of triplets. However, if the coupling is identical (or almost identical) between the hydrogens on C<sub>2<\/sub> and the hydrogens on both C<sub>1<\/sub> and C<sub>3<\/sub>, one would observe a quintet in the <sup>1<\/sup>H NMR spectrum. [You can try this yourself by drawing a tree diagram of a triplet of triplets assuming, first, different coupling constants, and then, identical coupling constants.] Keep this point in mind when interpreting real <sup>1<\/sup>H NMR spectra.\r\n\r\nAlso, when multiplets are well separated, they form patterns. However, when multiplets approach each other in the spectrum they sometimes become distorted. Usually, the inner peaks become larger than the outer peaks. Note the following examples:<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87104\/13-12a.png?origin=mt-web\" alt=\"two multiplet NMR signals brought closer together become distorted\" \/>\r\n\r\n&nbsp;\r\n\r\nAromatic ring protons quite commonly have overlapping signals and multiplet distortions. Sometimes you cannot distinguish between individual signals, and one or more messy multiplets often appear in the aromatic region.\r\n\r\nIt is much easier to rationalize the observed <sup>1<\/sup>H NMR spectrum of a known compound than it is to determine the structure of an unknown compound from its <sup>1<\/sup>H NMR spectrum. However, rationalizations can be a useful learning technique as you try to improve your proficiency in spectral interpretation. Remember that when a chemist tries to interpret the <sup>1<\/sup>H NMR spectrum of an unknown compound, he or she usually has additional information available to make the task easier. For example, the chemist will almost certainly have an infrared spectrum of the compound and possibly a mass spectrum too. Details of how the compound was synthesized may be available, together with some indication of its chemical properties, its physical properties, or both.\r\n\r\nIn examinations, you will be given a range of information (IR, MS, UV data and empirical formulae) to aid you with your structural determination using <sup>1<\/sup>H NMR spectroscopy. For example, you may be asked to determine the structure of C<sub>6<\/sub>H<sub>12<\/sub>O given the following spectra:\r\n\r\n<strong>Infrared spectrum:<\/strong> 3000 cm<sup>\u22121<\/sup> and 1720 cm<sup>\u22121<\/sup> absorptions are both strong\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th><sup>1<\/sup>H NMR<\/th>\r\n<th><em>\u03b4<\/em> (ppm)<\/th>\r\n<th>Protons<\/th>\r\n<th>Multiplicity<\/th>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>0.87<\/td>\r\n<td>6<\/td>\r\n<td>doublet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>1.72<\/td>\r\n<td>1<\/td>\r\n<td>broad multiplet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>2.00<\/td>\r\n<td>3<\/td>\r\n<td>singlet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>2.18<\/td>\r\n<td>2<\/td>\r\n<td>doublet<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo answer this question, you note that the infrared spectrum of C<sub>6<\/sub>H<sub>12<\/sub>O shows $\\ce{\\sf{C-H}}$ stretching (3000 cm<sup>\u22121<\/sup>) and $\\ce{\\sf{C-O}}$ stretching (1720 cm<sup>\u22121<\/sup>). Now you have to piece together the information from the <sup>1<\/sup>H NMR spectrum. Notice the singlet with three protons at 2.00 ppm. This signal indicates a methyl group that is not coupled to other protons. It could possibly mean the presence of a methyl ketone functional group.\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87105\/13-12b.png?origin=mt-web\" alt=\"methyl ketone group\" \/>\r\n\r\nThe signal at 1.72 ppm is a broad multiplet, suggesting that a carbon with a single proton is beside carbons with several different protons.\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87106\/13-12c.png?origin=mt-web\" alt=\"lone proton on carbon\" \/>\r\n\r\nThe doublet signal at 2.18 ppm implies that a $\\ce{\\sf{-CH2-}}$ group is attached to a carbon having only one proton.\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87107\/13-12d.png?origin=mt-web\" alt=\"carbon with two hydrogens attached to carbon with one hydrogen\" \/>\r\n\r\nThe six protons showing a doublet at 0.87 ppm indicate two equivalent methyl groups attached to a carbon with one proton.\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87108\/13-12e.png?origin=mt-web\" alt=\"carbon with one proton and two methyl groups\" \/>\r\n\r\nWhenever you see a signal in the 0.7-1.3 ppm range that is a multiplet of three protons (3, 6, 9) it is most likely caused by equivalent methyl groups.\r\n\r\nUsing trial and error, and with the above observations, you should come up with the correct structure.\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87109\/13-12f.png?origin=mt-web\" alt=\"4-methyl-2-pentanone\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_1\">\r\n<h3 class=\"editable\">Complex coupling<\/h3>\r\nIn all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to <em>just<\/em> <em>one<\/em> neighboring set of hydrogens. When a set of hydrogens is coupled to <em>two or more<\/em> sets of nonequivalent neighbors, the result is a phenomenon called <strong>complex coupling<\/strong>. A good illustration is provided by the <sup>1<\/sup>H-NMR spectrum of methyl acrylate:\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\/05154511\/image084.png\" alt=\"image084.png\" width=\"713px\" height=\"420px\" \/>First, let's first consider the H<sub>c<\/sub> signal, which is centered at 6.21 ppm.\u00a0 Here is a closer look:\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\/05154514\/image086.png\" alt=\"image086.png\" width=\"442px\" height=\"265px\" \/>\r\n\r\nWith this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? H<sub>c<\/sub> is coupled to both H<sub>a<\/sub> and H<sub>b<\/sub> , but with <em>two different coupling constants<\/em>.\u00a0 Once again, a splitting diagram (or tree diagram) can help us to understand what we are seeing.\u00a0 H<sub>a<\/sub> is <em>trans<\/em> to H<sub>c<\/sub> across the double bond, and splits the H<sub>c<\/sub> signal into a doublet with a coupling constant of <sup>3<\/sup>J<sub>ac<\/sub> = 17.4 Hz. In addition, each of these H<sub>c<\/sub> doublet sub-peaks is split again by H<sub>b<\/sub> (<em>geminal<\/em> coupling) into two more doublets, each with a much smaller coupling constant of <sup>2<\/sup>J<sub>bc<\/sub> = 1.5 Hz.\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\/05154515\/image088.png\" alt=\"image088.png\" width=\"252px\" height=\"274px\" \/>\r\n\r\nThe result of this `double splitting` is a pattern referred to as a <strong>doublet of doublets<\/strong>, abbreviated `<strong>dd<\/strong>`.\r\n\r\nThe signal for H<sub>a<\/sub> at 5.95 ppm is also a doublet of doublets, with coupling constants <sup>3<\/sup>J<sub>ac<\/sub>= 17.4 Hz and <sup>3<\/sup>J<sub>ab<\/sub> = 10.5 Hz.\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\/05154518\/image090.png\" alt=\"image090.png\" width=\"348px\" height=\"269px\" \/>\r\n\r\nThe signal for H<sub>b<\/sub> at 5.64 ppm is split into a doublet by H<sub>a<\/sub>, a <em>cis<\/em> coupling with <sup>3<\/sup>J<sub>ab<\/sub> = 10.4 Hz. Each of the resulting sub-peaks is split again by H<sub>c<\/sub>, with the same <em>geminal<\/em> coupling constant <sup>2<\/sup>J<sub>bc<\/sub> = 1.5 Hz that we saw previously when we looked at the H<sub>c<\/sub> signal.\u00a0 The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the <em>cis<\/em> interaction.\u00a0 Here is a blow-up of the actual H<sub>b<\/sub>signal:\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\/05154520\/image092.png\" alt=\"image092.png\" width=\"348\" height=\"221\" \/>\r\n<div>\r\n<div id=\"example\">\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nConstruct a splitting diagram for the H<sub>b<\/sub> signal in the <sup>1<\/sup>H-NMR\u00a0 spectrum\u00a0 of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).\r\n<h3>Solution<\/h3>\r\n[reveal-answer q=\"863938\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"863938\"]<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6480\/image273.png?revision=1\" alt=\"image272.png\" width=\"431\" height=\"288\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div style=\"margin: auto\">\r\n\r\nWhen constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).\r\n\r\n<\/div>\r\nWhen a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `<em>n<\/em> + 1 rule` of non-complex splitting.\u00a0 In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for H<sub>b<\/sub> to be split into a triplet by H<sub>a<\/sub>, and again into doublets by H<sub>c<\/sub>, resulting in a 'triplet of doublets'.\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\/05154522\/image094.png\" alt=\"image094.png\" width=\"252px\" height=\"79px\" \/>\r\n\r\nH<sub>a<\/sub> and H<sub>c<\/sub> are not equivalent (their chemical shifts are different), but it turns out that <sup>3<\/sup>J<sub>ab<\/sub> is very close to <sup>3<\/sup>J<sub>bc<\/sub>.\u00a0 If we perform a splitting diagram analysis for H<sub>b<\/sub>, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the <em>n<\/em> + 1 rule.\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\/05154524\/image096.png\" alt=\"image096.png\" width=\"398px\" height=\"401px\" \/>\r\n\r\nFor similar reasons, the H<sub>c<\/sub> peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined H<sub>b<\/sub> and H<sub>d<\/sub> protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.\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\/05154527\/image098.png\" alt=\"image098.png\" width=\"672px\" height=\"375px\" \/>\r\n<div style=\"margin: auto\">\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nWhat splitting pattern would you expect for the signal coresponding to H<sub>b <\/sub>in the molecule below? Assume that J<sub>ab<\/sub> ~ J<sub>bc<\/sub>.\u00a0 Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.\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\/05154530\/image100.png\" alt=\"image100.png\" width=\"183px\" height=\"91px\" \/>\r\n<h3>Solution<\/h3>\r\n[reveal-answer q=\"374592\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"374592\"]\r\n\r\nWe can think of this signal as being a triplet of triplets, but because the two coupling constants are very close, what we would actually see is a 1,2,3,2,1 pentet.\r\n<p align=\"center\"><img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6482\/image275.png?revision=1\" alt=\"image274.png\" width=\"341\" height=\"203\" \/><\/p>\r\n<p align=\"center\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div style=\"margin: auto\">\r\n\r\nIn many cases, it is difficult to fully analyze a complex splitting pattern.\u00a0 In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for H<sub>b<\/sub> to be a doublet, H<sub>d<\/sub> a triplet, and H<sub>c<\/sub> a triplet.\r\n\r\n<\/div>\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154532\/image102.png\" alt=\"image102.png\" width=\"625px\" height=\"395px\" \/>\r\n\r\nIn practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult.\u00a0 In this case, we would refer to the aromatic part of the spectrum as a <strong>multiplet<\/strong>.\r\n\r\nWhen we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments.\u00a0 Quite simply, the stronger our magnet is, the more resolution we get in our spectrum.\u00a0 In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide.\u00a0\u00a0 In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider.\u00a0 In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)\r\n\r\n<\/div>\r\n<div id=\"section_2\">\r\n<div class=\"textbox exercises\">\r\n<div id=\"section_2\">\r\n<h3 class=\"editable\">Exercises<\/h3>\r\n<ol>\r\n \t<li>Given the information below, draw the structures of compounds <em>A<\/em> through <em>D<\/em>.\r\n<ol>\r\n \t<li>An unknown compound <em>A<\/em> was prepared as follows:<strong><img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87110\/13-12-1a-exercise.png?origin=mt-web\" alt=\"3-chloro-2-(chloromethyl)propene\" \/>Mass spectrum:<\/strong>base peak <em>m<\/em>\/<em>e<\/em> = 39\r\nparent peak <em>m<\/em>\/<em>e<\/em> = 54<strong><sup>1<\/sup>H NMR spectrum:<\/strong>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th><em>\u03b4<\/em> (ppm)<\/th>\r\n<th>Relative Area<\/th>\r\n<th>Multiplicity<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>1.0<\/td>\r\n<td>2<\/td>\r\n<td>triplet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5.4<\/td>\r\n<td>1<\/td>\r\n<td>quintet<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Unknown compound <em>B<\/em> has the molecular formula C<sub>7<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>.<strong>Infrared spectrum:<\/strong>3200 cm<sup>\u22121<\/sup> (broad) and 1747 cm<sup>\u22121<\/sup> (strong) absorptions<strong><sup>1<\/sup>H NMR spectrum:<\/strong>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th><em>\u03b4<\/em> (ppm)<\/th>\r\n<th>Protons<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>6.9<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>7.4<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9.8<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>10.9<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Hint:<\/strong> Aromatic ring currents deshield all proton signals just outside the ring.<\/li>\r\n \t<li>Unknown compound <em>C<\/em> shows no evidence of unsaturation and contains only carbon and hydrogen.<strong>Mass spectrum:<\/strong>parent peak <em>m<\/em>\/<em>e<\/em> = 68<strong><sup>1<\/sup>H NMR spectrum:<\/strong>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th><em>\u03b4<\/em> (ppm)<\/th>\r\n<th>Relative Area<\/th>\r\n<th>Multiplicity<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>1.84<\/td>\r\n<td>3<\/td>\r\n<td>triplet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2.45<\/td>\r\n<td>1<\/td>\r\n<td>septet<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Hint:<\/strong> Think three dimensionally!<\/li>\r\n \t<li>Unknown compound <em>D<\/em> (C<sub>15<\/sub>H<sub>14<\/sub>O) has the following spectral properties.<strong>Infrared spectrum:<\/strong>3010 cm<sup>\u22121<\/sup> (medium)\r\n1715 cm<sup>\u22121<\/sup> (strong)\r\n1610 cm<sup>\u22121<\/sup> (strong)\r\n1500 cm<sup>\u22121<\/sup> (strong)<strong><sup>1<\/sup>H NMR spectrum:<\/strong>\r\n<table style=\"width: 254px\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 64px\"><em>\u03b4<\/em> (ppm)<\/th>\r\n<th style=\"width: 103px\">Relative Area<\/th>\r\n<th style=\"width: 87px\">Multiplicity<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 64px\">3.00<\/td>\r\n<td style=\"width: 103px\">2<\/td>\r\n<td style=\"width: 87px\">triplet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 64px\">3.07<\/td>\r\n<td style=\"width: 103px\">2<\/td>\r\n<td style=\"width: 87px\">triplet<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 64px\">7.1-7.9<\/td>\r\n<td style=\"width: 103px\">10<\/td>\r\n<td style=\"width: 87px\">Multiplets<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h3>Solutions<\/h3>\r\n[reveal-answer q=\"781180\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"781180\"]\r\n\r\n1. Methylenecyclopropane 2. 4-hydrobenzaldehyde 3. bicyclo(1.1.1)pentane 4. 1,4-diphenyl-1-butanone[\/hidden-answer]\r\n<div id=\"s61718\">\r\n<div id=\"section_36\">\r\n<h3 id=\"Questions-61718\">Question<\/h3>\r\nIn the following molecule, the C2 is coupled with both the vinyl, C1, and the alkyl C3. Draw the splitting tree diagram.\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\/05154536\/13.12.png\" alt=\"\" width=\"311\" height=\"147\" \/>\r\n<h3>Solution<\/h3>\r\n[reveal-answer q=\"484749\"]Show Answer[\/reveal-answer]\r\n\r\n<\/div>\r\n<div id=\"section_37\">\r\n\r\n[hidden-answer a=\"484749\"]\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\/05154538\/13-12-1sol.png\" alt=\"\" width=\"185\" height=\"233\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_3\">\u00a0[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_3\">\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>explain how multiple coupling can give rise to complex-looking <sup>1<\/sup>H NMR spectra.<\/li>\n<li>predict the splitting pattern expected in the <sup>1<\/sup>H NMR spectrum of an organic compound in which multiple coupling is possible.<\/li>\n<li>interpret <sup>1<\/sup>H NMR spectra in which multiple coupling is evident.<\/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>Make certain that you can define, and use in context, the key term below.<\/p>\n<ul>\n<li>tree diagram<\/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<div id=\"note\">\n<h3 class=\"boxtitle\">Study Notes<\/h3>\n<p>We saw the effects of spin-spin coupling on the appearance of a <sup>1<\/sup>H NMR signal. These effects can be further complicated when that signal is coupled to several different protons. For example, BrCH<sub>2<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>Cl would produce three signals. The hydrogens at C<sub>1<\/sub> and C<sub>3<\/sub> would each be triplets because of coupling to the two hydrogens on C<sub>2<\/sub>. However, the hydrogen on C<sub>2<\/sub> \u201csees\u201d two different sets of neighbouring hydrogens, and would therefore produce a triplet of triplets.<\/p>\n<p>Another effect that can complicate a spectrum is the \u201ccloseness\u201d of signals. If signals accidently overlap they can be difficult to identify. In the example above, we expected a triplet of triplets. However, if the coupling is identical (or almost identical) between the hydrogens on C<sub>2<\/sub> and the hydrogens on both C<sub>1<\/sub> and C<sub>3<\/sub>, one would observe a quintet in the <sup>1<\/sup>H NMR spectrum. [You can try this yourself by drawing a tree diagram of a triplet of triplets assuming, first, different coupling constants, and then, identical coupling constants.] Keep this point in mind when interpreting real <sup>1<\/sup>H NMR spectra.<\/p>\n<p>Also, when multiplets are well separated, they form patterns. However, when multiplets approach each other in the spectrum they sometimes become distorted. Usually, the inner peaks become larger than the outer peaks. Note the following examples:<img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87104\/13-12a.png?origin=mt-web\" alt=\"two multiplet NMR signals brought closer together become distorted\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Aromatic ring protons quite commonly have overlapping signals and multiplet distortions. Sometimes you cannot distinguish between individual signals, and one or more messy multiplets often appear in the aromatic region.<\/p>\n<p>It is much easier to rationalize the observed <sup>1<\/sup>H NMR spectrum of a known compound than it is to determine the structure of an unknown compound from its <sup>1<\/sup>H NMR spectrum. However, rationalizations can be a useful learning technique as you try to improve your proficiency in spectral interpretation. Remember that when a chemist tries to interpret the <sup>1<\/sup>H NMR spectrum of an unknown compound, he or she usually has additional information available to make the task easier. For example, the chemist will almost certainly have an infrared spectrum of the compound and possibly a mass spectrum too. Details of how the compound was synthesized may be available, together with some indication of its chemical properties, its physical properties, or both.<\/p>\n<p>In examinations, you will be given a range of information (IR, MS, UV data and empirical formulae) to aid you with your structural determination using <sup>1<\/sup>H NMR spectroscopy. For example, you may be asked to determine the structure of C<sub>6<\/sub>H<sub>12<\/sub>O given the following spectra:<\/p>\n<p><strong>Infrared spectrum:<\/strong> 3000 cm<sup>\u22121<\/sup> and 1720 cm<sup>\u22121<\/sup> absorptions are both strong<\/p>\n<table>\n<tbody>\n<tr>\n<th><sup>1<\/sup>H NMR<\/th>\n<th><em>\u03b4<\/em> (ppm)<\/th>\n<th>Protons<\/th>\n<th>Multiplicity<\/th>\n<\/tr>\n<tr>\n<td><\/td>\n<td>0.87<\/td>\n<td>6<\/td>\n<td>doublet<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>1.72<\/td>\n<td>1<\/td>\n<td>broad multiplet<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>2.00<\/td>\n<td>3<\/td>\n<td>singlet<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>2.18<\/td>\n<td>2<\/td>\n<td>doublet<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To answer this question, you note that the infrared spectrum of C<sub>6<\/sub>H<sub>12<\/sub>O shows $\\ce{\\sf{C-H}}$ stretching (3000 cm<sup>\u22121<\/sup>) and $\\ce{\\sf{C-O}}$ stretching (1720 cm<sup>\u22121<\/sup>). Now you have to piece together the information from the <sup>1<\/sup>H NMR spectrum. Notice the singlet with three protons at 2.00 ppm. This signal indicates a methyl group that is not coupled to other protons. It could possibly mean the presence of a methyl ketone functional group.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87105\/13-12b.png?origin=mt-web\" alt=\"methyl ketone group\" \/><\/p>\n<p>The signal at 1.72 ppm is a broad multiplet, suggesting that a carbon with a single proton is beside carbons with several different protons.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87106\/13-12c.png?origin=mt-web\" alt=\"lone proton on carbon\" \/><\/p>\n<p>The doublet signal at 2.18 ppm implies that a $\\ce{\\sf{-CH2-}}$ group is attached to a carbon having only one proton.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87107\/13-12d.png?origin=mt-web\" alt=\"carbon with two hydrogens attached to carbon with one hydrogen\" \/><\/p>\n<p>The six protons showing a doublet at 0.87 ppm indicate two equivalent methyl groups attached to a carbon with one proton.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87108\/13-12e.png?origin=mt-web\" alt=\"carbon with one proton and two methyl groups\" \/><\/p>\n<p>Whenever you see a signal in the 0.7-1.3 ppm range that is a multiplet of three protons (3, 6, 9) it is most likely caused by equivalent methyl groups.<\/p>\n<p>Using trial and error, and with the above observations, you should come up with the correct structure.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87109\/13-12f.png?origin=mt-web\" alt=\"4-methyl-2-pentanone\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_1\">\n<h3 class=\"editable\">Complex coupling<\/h3>\n<p>In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to <em>just<\/em> <em>one<\/em> neighboring set of hydrogens. When a set of hydrogens is coupled to <em>two or more<\/em> sets of nonequivalent neighbors, the result is a phenomenon called <strong>complex coupling<\/strong>. A good illustration is provided by the <sup>1<\/sup>H-NMR spectrum of methyl acrylate:<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154511\/image084.png\" alt=\"image084.png\" width=\"713px\" height=\"420px\" \/>First, let&#8217;s first consider the H<sub>c<\/sub> signal, which is centered at 6.21 ppm.\u00a0 Here is a closer look:<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154514\/image086.png\" alt=\"image086.png\" width=\"442px\" height=\"265px\" \/><\/p>\n<p>With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? H<sub>c<\/sub> is coupled to both H<sub>a<\/sub> and H<sub>b<\/sub> , but with <em>two different coupling constants<\/em>.\u00a0 Once again, a splitting diagram (or tree diagram) can help us to understand what we are seeing.\u00a0 H<sub>a<\/sub> is <em>trans<\/em> to H<sub>c<\/sub> across the double bond, and splits the H<sub>c<\/sub> signal into a doublet with a coupling constant of <sup>3<\/sup>J<sub>ac<\/sub> = 17.4 Hz. In addition, each of these H<sub>c<\/sub> doublet sub-peaks is split again by H<sub>b<\/sub> (<em>geminal<\/em> coupling) into two more doublets, each with a much smaller coupling constant of <sup>2<\/sup>J<sub>bc<\/sub> = 1.5 Hz.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154515\/image088.png\" alt=\"image088.png\" width=\"252px\" height=\"274px\" \/><\/p>\n<p>The result of this `double splitting` is a pattern referred to as a <strong>doublet of doublets<\/strong>, abbreviated `<strong>dd<\/strong>`.<\/p>\n<p>The signal for H<sub>a<\/sub> at 5.95 ppm is also a doublet of doublets, with coupling constants <sup>3<\/sup>J<sub>ac<\/sub>= 17.4 Hz and <sup>3<\/sup>J<sub>ab<\/sub> = 10.5 Hz.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154518\/image090.png\" alt=\"image090.png\" width=\"348px\" height=\"269px\" \/><\/p>\n<p>The signal for H<sub>b<\/sub> at 5.64 ppm is split into a doublet by H<sub>a<\/sub>, a <em>cis<\/em> coupling with <sup>3<\/sup>J<sub>ab<\/sub> = 10.4 Hz. Each of the resulting sub-peaks is split again by H<sub>c<\/sub>, with the same <em>geminal<\/em> coupling constant <sup>2<\/sup>J<sub>bc<\/sub> = 1.5 Hz that we saw previously when we looked at the H<sub>c<\/sub> signal.\u00a0 The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the <em>cis<\/em> interaction.\u00a0 Here is a blow-up of the actual H<sub>b<\/sub>signal:<\/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\/05154520\/image092.png\" alt=\"image092.png\" width=\"348\" height=\"221\" \/><\/p>\n<div>\n<div id=\"example\">\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Construct a splitting diagram for the H<sub>b<\/sub> signal in the <sup>1<\/sup>H-NMR\u00a0 spectrum\u00a0 of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).<\/p>\n<h3>Solution<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q863938\">Show Answer<\/span><\/p>\n<div id=\"q863938\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6480\/image273.png?revision=1\" alt=\"image272.png\" width=\"431\" height=\"288\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div style=\"margin: auto\">\n<p>When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).<\/p>\n<\/div>\n<p>When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `<em>n<\/em> + 1 rule` of non-complex splitting.\u00a0 In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for H<sub>b<\/sub> to be split into a triplet by H<sub>a<\/sub>, and again into doublets by H<sub>c<\/sub>, resulting in a &#8216;triplet of doublets&#8217;.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154522\/image094.png\" alt=\"image094.png\" width=\"252px\" height=\"79px\" \/><\/p>\n<p>H<sub>a<\/sub> and H<sub>c<\/sub> are not equivalent (their chemical shifts are different), but it turns out that <sup>3<\/sup>J<sub>ab<\/sub> is very close to <sup>3<\/sup>J<sub>bc<\/sub>.\u00a0 If we perform a splitting diagram analysis for H<sub>b<\/sub>, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the <em>n<\/em> + 1 rule.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154524\/image096.png\" alt=\"image096.png\" width=\"398px\" height=\"401px\" \/><\/p>\n<p>For similar reasons, the H<sub>c<\/sub> peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined H<sub>b<\/sub> and H<sub>d<\/sub> protons. Technically, this &#8216;sextet&#8217; could be considered to be a &#8216;triplet of quartets&#8217; with overlapping sub-peaks.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154527\/image098.png\" alt=\"image098.png\" width=\"672px\" height=\"375px\" \/><\/p>\n<div style=\"margin: auto\">\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>What splitting pattern would you expect for the signal coresponding to H<sub>b <\/sub>in the molecule below? Assume that J<sub>ab<\/sub> ~ J<sub>bc<\/sub>.\u00a0 Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154530\/image100.png\" alt=\"image100.png\" width=\"183px\" height=\"91px\" \/><\/p>\n<h3>Solution<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374592\">Show Answer<\/span><\/p>\n<div id=\"q374592\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can think of this signal as being a triplet of triplets, but because the two coupling constants are very close, what we would actually see is a 1,2,3,2,1 pentet.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6482\/image275.png?revision=1\" alt=\"image274.png\" width=\"341\" height=\"203\" \/><\/p>\n<p style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div style=\"margin: auto\">\n<p>In many cases, it is difficult to fully analyze a complex splitting pattern.\u00a0 In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for H<sub>b<\/sub> to be a doublet, H<sub>d<\/sub> a triplet, and H<sub>c<\/sub> a triplet.<\/p>\n<\/div>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05154532\/image102.png\" alt=\"image102.png\" width=\"625px\" height=\"395px\" \/><\/p>\n<p>In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult.\u00a0 In this case, we would refer to the aromatic part of the spectrum as a <strong>multiplet<\/strong>.<\/p>\n<p>When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments.\u00a0 Quite simply, the stronger our magnet is, the more resolution we get in our spectrum.\u00a0 In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency &#8216;window&#8217; in which we can observe proton signals is 1200 Hz wide.\u00a0\u00a0 In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz &#8211; five times wider.\u00a0 In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)<\/p>\n<\/div>\n<div id=\"section_2\">\n<div class=\"textbox exercises\">\n<div id=\"section_2\">\n<h3 class=\"editable\">Exercises<\/h3>\n<ol>\n<li>Given the information below, draw the structures of compounds <em>A<\/em> through <em>D<\/em>.\n<ol>\n<li>An unknown compound <em>A<\/em> was prepared as follows:<strong><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/87110\/13-12-1a-exercise.png?origin=mt-web\" alt=\"3-chloro-2-(chloromethyl)propene\" \/>Mass spectrum:<\/strong>base peak <em>m<\/em>\/<em>e<\/em> = 39<br \/>\nparent peak <em>m<\/em>\/<em>e<\/em> = 54<strong><sup>1<\/sup>H NMR spectrum:<\/strong><\/p>\n<table>\n<tbody>\n<tr>\n<th><em>\u03b4<\/em> (ppm)<\/th>\n<th>Relative Area<\/th>\n<th>Multiplicity<\/th>\n<\/tr>\n<tr>\n<td>1.0<\/td>\n<td>2<\/td>\n<td>triplet<\/td>\n<\/tr>\n<tr>\n<td>5.4<\/td>\n<td>1<\/td>\n<td>quintet<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Unknown compound <em>B<\/em> has the molecular formula C<sub>7<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>.<strong>Infrared spectrum:<\/strong>3200 cm<sup>\u22121<\/sup> (broad) and 1747 cm<sup>\u22121<\/sup> (strong) absorptions<strong><sup>1<\/sup>H NMR spectrum:<\/strong><br \/>\n<table>\n<tbody>\n<tr>\n<th><em>\u03b4<\/em> (ppm)<\/th>\n<th>Protons<\/th>\n<\/tr>\n<tr>\n<td>6.9<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>7.4<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>9.8<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td>10.9<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Hint:<\/strong> Aromatic ring currents deshield all proton signals just outside the ring.<\/li>\n<li>Unknown compound <em>C<\/em> shows no evidence of unsaturation and contains only carbon and hydrogen.<strong>Mass spectrum:<\/strong>parent peak <em>m<\/em>\/<em>e<\/em> = 68<strong><sup>1<\/sup>H NMR spectrum:<\/strong><br \/>\n<table>\n<tbody>\n<tr>\n<th><em>\u03b4<\/em> (ppm)<\/th>\n<th>Relative Area<\/th>\n<th>Multiplicity<\/th>\n<\/tr>\n<tr>\n<td>1.84<\/td>\n<td>3<\/td>\n<td>triplet<\/td>\n<\/tr>\n<tr>\n<td>2.45<\/td>\n<td>1<\/td>\n<td>septet<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Hint:<\/strong> Think three dimensionally!<\/li>\n<li>Unknown compound <em>D<\/em> (C<sub>15<\/sub>H<sub>14<\/sub>O) has the following spectral properties.<strong>Infrared spectrum:<\/strong>3010 cm<sup>\u22121<\/sup> (medium)<br \/>\n1715 cm<sup>\u22121<\/sup> (strong)<br \/>\n1610 cm<sup>\u22121<\/sup> (strong)<br \/>\n1500 cm<sup>\u22121<\/sup> (strong)<strong><sup>1<\/sup>H NMR spectrum:<\/strong><\/p>\n<table style=\"width: 254px\">\n<tbody>\n<tr>\n<th style=\"width: 64px\"><em>\u03b4<\/em> (ppm)<\/th>\n<th style=\"width: 103px\">Relative Area<\/th>\n<th style=\"width: 87px\">Multiplicity<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 64px\">3.00<\/td>\n<td style=\"width: 103px\">2<\/td>\n<td style=\"width: 87px\">triplet<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 64px\">3.07<\/td>\n<td style=\"width: 103px\">2<\/td>\n<td style=\"width: 87px\">triplet<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 64px\">7.1-7.9<\/td>\n<td style=\"width: 103px\">10<\/td>\n<td style=\"width: 87px\">Multiplets<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h3>Solutions<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q781180\">Show Answer<\/span><\/p>\n<div id=\"q781180\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Methylenecyclopropane 2. 4-hydrobenzaldehyde 3. bicyclo(1.1.1)pentane 4. 1,4-diphenyl-1-butanone<\/p><\/div>\n<\/div>\n<div id=\"s61718\">\n<div id=\"section_36\">\n<h3 id=\"Questions-61718\">Question<\/h3>\n<p>In the following molecule, the C2 is coupled with both the vinyl, C1, and the alkyl C3. Draw the splitting tree diagram.<\/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\/05154536\/13.12.png\" alt=\"\" width=\"311\" height=\"147\" \/><\/p>\n<h3>Solution<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484749\">Show Answer<\/span><\/p>\n<\/div>\n<div id=\"section_37\">\n<div id=\"q484749\" class=\"hidden-answer\" style=\"display: none\">\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\/05154538\/13-12-1sol.png\" alt=\"\" width=\"185\" height=\"233\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_3\">\u00a0<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_3\">\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":12,"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-1742","chapter","type-chapter","status-publish","hentry"],"part":29,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1742","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":13,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1742\/revisions"}],"predecessor-version":[{"id":2357,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1742\/revisions\/2357"}],"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\/1742\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1742"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1742"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1742"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1742"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}