{"id":1706,"date":"2017-10-10T16:03:50","date_gmt":"2017-10-10T16:03:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&p=1706"},"modified":"2018-10-05T19:38:33","modified_gmt":"2018-10-05T19:38:33","slug":"integration-of-proton-nmr-absorptions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/integration-of-proton-nmr-absorptions\/","title":{"raw":"Integration of Proton NMR Absorptions","rendered":"Integration of Proton NMR Absorptions"},"content":{"raw":"
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Objectives<\/h3>\r\n
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\r\n\r\nAfter completing this section, you should be able to\r\n
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  1. explain what information can be obtained from an integrated 1<\/sup>H NMR spectrum, and use this information in the interpretation of such a spectrum.<\/li>\r\n \t
  2. use an integrated 1<\/sup>H NMR spectrum to determine the ratio of the different types of protons present in an organic compound.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    Study Notes<\/h3>\r\nThe concept of peak integration is that the area of a given peak in a 1H NMR spectrum is proportional to the number of (equivalent) protons giving rise to the peak. Thus, a peak which is caused by a single, unique proton has an area which measures one third of the area of a peak resulting from a methyl (CH3<\/sub>) group in the same spectrum.\r\n\r\nIn practice, we do not have to measure these areas ourselves: it is all done electronically by the spectrometer, and an integration curve is superimposed on the rest of the spectrum. The integration curve appears as a series of steps, with the height of each step being proportional to the area of the corresponding absorption peak, and consequently, to the number of protons responsible for the absorption.\r\n\r\nAs it can be difficult to decide precisely where to start and stop when measuring integrations, you should not expect your ratios to be exact whole numbers.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    Signal integration<\/h3>\r\nThe computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals.\u00a0 This is very useful, because in 1<\/sup>H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds<\/em>.\u00a0 The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons.\r\n\r\nTake a look next at the spectrum of para<\/em>-xylene\u00a0 (IUPAC name 1,4-dimethylbenzene):\r\n\r\n\"image034.png\"\r\n\r\nThis molecule has two sets of protons: the six methyl (Ha<\/sub>) protons and the four aromatic (Hb<\/sub>) protons.\u00a0 When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.\r\n\r\nThe integration function can also be used to determine the relative amounts of two or more compounds in a mixed<\/em> sample.\u00a0 If we have a sample that is a 50:50 (mole\/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons.\u00a0 If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.\r\n
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    Example<\/h3>\r\n
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    \r\n\r\nYou take a 1<\/sup>H-NMR spectrum of a mixed sample of acetone (CH3<\/sub>(CO)CH3<\/sub>) and dichloromethane (CH2<\/sub>Cl2<\/sub>).\u00a0 The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1.\u00a0 What is the molar ratio of the two compounds in the sample?\r\n

    Solution<\/h3>\r\n[reveal-answer q=\"319315\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"319315\"] the acetone: dichloromethane ratio is 0.76:1. If it was a 1:1 ratio, the signal integration ratio would be 3:1 because acetone has 6 protons while dichloromethane has 2. So you need to divide the 2.3 integral value by a factor of 3 to get 0.76.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    Example<\/h3>\r\nYou take the 1<\/sup>H-NMR spectrum of a mixed sample of 36%\u00a0 para<\/em>-xylene and 64% acetone in CDCl3<\/sub> solvent (structures are shown earlier in this chapter).\u00a0 How many peaks do you expect to see?\u00a0 What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)\r\n

    Solution<\/h3>\r\n[reveal-answer q=\"122489\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"122489\"]There are three peaks: two from para-xylene and one from acetone. The acetone peak and the para-xylene methyl peak both represent six protons, so the ratio of their integration values is simply 64 to 36 or 1 to 0.56. The ratio of the para-xylene methyl peak to the para-xylene aromatic peak is 6 to 4, or 0.56 to 0.37. So the final integral ratio of acetone:methyl:aromatic signals should be 1 to 0.56 to 0.37.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    Exercise<\/h3>\r\n
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    Question<\/h4>\r\nPredict how many signals the following molecule would have? Sketch the spectra and estimate the integration of the peaks.\r\n\r\n\"\"\r\n

    Solution<\/h3>\r\n[reveal-answer q=\"740722\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"740722\"]\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    Contributors<\/h3>\r\n