{"id":482,"date":"2015-01-24T21:33:32","date_gmt":"2015-01-24T21:33:32","guid":{"rendered":"http:\/\/www.studyorgo.com\/blog\/?p=482"},"modified":"2015-02-03T07:22:25","modified_gmt":"2015-02-03T07:22:25","slug":"deciphering-1h-nmr-spectra","status":"publish","type":"post","link":"https:\/\/www.studyorgo.com\/blog\/deciphering-1h-nmr-spectra\/","title":{"rendered":"Deciphering 1H NMR Spectra"},"content":{"rendered":"

Deciphering 1<\/sup>H-NMR Spectra<\/strong><\/p>\n

One of the most important concepts taught in organic chemistry is the method for determining the chemical structure of newly synthesized or unknown compounds. In this article, we will summarize the concept of proton NMR, the most common NMR information acquired by organic chemists.<\/p>\n

While proton NMR is used every day in the real world by organic chemists, it is also tested every day in the real world by organic chemistry professors. That is why we are bringing you this exclusive StudyOrgo.com<\/a> simplified review on NMR.<\/p>\n

Before you read on, if you like what you see, but are looking for explanations on different organic chemistry topics and reactions, check out how our signature online organic chemistry program works <\/a>to help you earn higher grades.<\/p>\n

Now if you are ready to read on about NMR- here ya go:<\/p>\n

Placing an unknown sample in a strong magnetic field allow 1<\/sup>H nuclei (99.98% abundance) \u00a0to \u201cresonate\u201d, which is when their nuclear spins flip at a unique electromagnetic (EM) frequency (Hertz, Hz).\u00a0 The instrument detects this and plots it on a graph in units of ppm.\u00a0 This value is relative to an internal standard, tetramethylsilane (TMS) which is arbitrarily set at 0 ppm. The signals originating from your unknown sample are recorded as a net difference from the TMS reference.\u00a0 The ranges for common functional groups are shown in the graph below.<\/p>\n

\"Deciphering
\nNotice that protons (in red) neighboring or attached to electronegative elements (N,O,S,Cl,Br,I) have a large chemical shift (d).\u00a0 This is colloquially referred to as \u201cdownfield\u201d on the spectrograph. This is due to the deshielding effect, which is simply that the electron cloud around protons near electronegative or electron-withdrawing groups are smaller, so less EM radiation is required to resonate the nucleus.\u00a0 Conversely, when the proton is connected to carbons, the chemical shift is <2 ppm.\u00a0 This is because these proton nuclei are \u201cshielded\u201d from the magnetic field and higher energy is required for them to resonate and referred to as \u201cupfield\u201d on the spectrograph.\u00a0 Once a NMR spectrograph is recorded, 4 pieces of information can be determined from the data as long as the chemical formula of the compound is known.<\/p>\n

To illustrate the points, we will consider the following 1<\/sup>H-NMR <\/strong>spectrum of the C5<\/sub>H10<\/sub>O.<\/p>\n

\"\"<\/p>\n

    \n
  1. Signal Count \u2013 Number of unique hydrogens<\/em><\/strong><\/li>\n<\/ol>\n

    This is the easiest to interpret.\u00a0 The number of peaks correspond to the number of unique, or chemically indistinguishable, hydrogen nuclei.\u00a0 There are two peaks on the graph, therefore of the 10 hydrogens in the molecule, there are two types.<\/p>\n

      \n
    1. Chemical Shift \u2013 Identity of neighbors<\/em><\/strong><\/li>\n<\/ol>\n

      The two peaks on the spectrum are located at (d)2.42 and (d)1.07.\u00a0 Remembering that the chemical formula includes 5 carbons and one oxygen, it is clear there is a good deal of symmetry to the molecule.\u00a0 Furthermore, the only proton near an oxygen that exhibits a chemical shift ~ 2.5 with is neighboring a carbonyl group HCRC=O.\u00a0 So far, we know our compound is symmetric and has a carbonyl group, therefore only 4 carbons can have carbons.<\/p>\n

        \n
      1. Integration \u2013 Number of Equivalent Hydrogens<\/em><\/strong><\/li>\n<\/ol>\n

        Integration is the calculation of the area beneath the peaks.\u00a0 This information is determined by the computer software and would not be required for you to determine from visual inspection.\u00a0 But if it is provided, it will instantly determine the number of \u201cequivalent\u201d (or chemically indistinguishable) hydrogens in your graph.\u00a0 Looking at our spectra, we have integration of 33 + 48 = 81 cart units in total.\u00a0 The integration fraction can be multiplied by the total number of hydrogens from the chemical formula to determine number of equivalent hydrogens.\u00a0 For instance, at (d)2.42 [33\/81 = 0.40 * 10 total hydrogens = 4 hydrogens at (d)2.42].\u00a0 Therefore, we know that the protons near the carbonyl total 4.\u00a0 Given the symmetry from the low peak count, we know these hydrogens must be two CH2 groups.\u00a0 The same for the peak at (d)1.07 [48\/81 = 0.60 * 10 total hydrogens = hydrogens at (d)1.07].\u00a0 Since only 2 carbons are left, we must conclude these are terminal CH3 groups.<\/p>\n

          \n
        1. Signal Splitting \u2013 Number of Non-equivalent Hydrogen Neighbors<\/em><\/strong><\/li>\n<\/ol>\n

          Signal splitting occurs from a phenomenon of coupling, which occurs when NON-equivalent neighboring protons interfere with the resonance of a proton nuclei.\u00a0 The degree of splitting occurs in an N+1 rule.\u00a0 For instance, at d1.07 we have a triplet.\u00a0 This indicates that there are 2 protons neighboring the CH3 group that are different.\u00a0 This would have to be the protons from the CH2 group at (d)2.42.\u00a0 Similarly, the peak at (d)2.42 as a quartet, indicating that 3 protons are neighboring the CH2 group that are different.\u00a0 This would have to be the protons from the CH3 group.<\/p>\n

          And voila! \u00a0The chemical structure from the spectrum must be 3-propanone.\u00a0 This is an example of the clear-cut explanations you will receive with your membership with StudyOrgo<\/a>. With practice and help from StudyOrgo, you\u2019ll be solving your organic reaction problems in no time!<\/p>\n

          \"3\"<\/p>\n

           <\/p>\n

          Want to try it about before you purchase? No problem- check out our sample reaction flashcard page<\/a>.<\/p>\n

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           <\/p>\n","protected":false},"excerpt":{"rendered":"

          Deciphering 1H-NMR Spectra One of the most important concepts taught in organic chemistry is the method for determining the chemical structure of newly synthesized or unknown compounds. In this article, we will summarize the concept of proton NMR, the most common NMR information acquired by organic chemists. While proton NMR is used every day in […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-482","post","type-post","status-publish","format-standard","hentry","category-organic-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/posts\/482","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/comments?post=482"}],"version-history":[{"count":10,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/posts\/482\/revisions"}],"predecessor-version":[{"id":512,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/posts\/482\/revisions\/512"}],"wp:attachment":[{"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/media?parent=482"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/categories?post=482"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.studyorgo.com\/blog\/wp-json\/wp\/v2\/tags?post=482"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}