{"id":25610,"date":"2022-02-08T14:42:45","date_gmt":"2022-02-08T20:42:45","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=25610"},"modified":"2025-03-21T14:46:44","modified_gmt":"2025-03-21T19:46:44","slug":"c13-nmr-how-many-signals","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2022\/02\/08\/c13-nmr-how-many-signals\/","title":{"rendered":"13-C NMR – How Many Signals"},"content":{"rendered":"

Chemical Shift Equivalence – How Many Unique Signals? <\/span>13<\/sup><\/span>C NMR (proton-decoupled)<\/span><\/strong><\/p>\n

Just like with <\/span>1<\/sup><\/span>H NMR, chemical shift equivalence applies to <\/span>13<\/sup><\/span>C NMR. A spectrum produced by a <\/span>13<\/sup><\/span>C NMR experiment may not always display a 1:1 ratio of signals to individual carbons atoms. When two or more carbon atoms in a molecule have chemically equivalent nuclei, instead of producing two or more signals on a <\/span>13<\/sup><\/span>C NMR spectrum, they will produce one signal at a specific chemical shift that will represent those equivalent carbons. In order for atoms to be chemically equivalent, their nuclei must be interchangeable through the performance of symmetry operations (planes of symmetry) or rapid intramolecular processes (bond rotation or tautomerization). Carbon atoms with chemical equivalence may be homotopic (identical) or enantiotopic (equivalent in achiral solvents).<\/span><\/p>\n

For the purposes of this topic, we will be dealing with proton-decoupled <\/span>13<\/sup><\/span>C NMR in order work with spectra consisting of single line signals rather than multiplets corresponding to specific carbon atoms.<\/span><\/p>\n

Table of Contents<\/strong><\/p>\n

    \n
  1. \n
      \n
    1. Methane: One Carbon, One Signal<\/a><\/li>\n
    2. Ethane Has Two Carbons, But Only One Signal Due To Symmetry<\/a><\/li>\n
    3. Pentane (5 Carbons) Has Only Three 13C NMR Signals Due To Symmetry<\/a><\/li>\n
    4. Symmetry In Aromatic Rings<\/a><\/li>\n
    5. Some Quick Examples<\/a><\/li>\n
    6. Slight Disruptions To Symmetry Can Have A Large Impact On The Number Of Signals In A 13<\/sup>C NMR Spectrum<\/a><\/li>\n
    7. Ortho, Meta, and Para Di-Substituted Aromatic Rings: Look Out For Planes Of Symmetry!<\/a><\/li>\n
    8. The Effect Of Chirality Centers On 13C NMR Spectra<\/a><\/li>\n
    9. Carbons That Are Equivalent Through Fast Bond Rotation Will Have Their Signals Averaged Together<\/a><\/li>\n
    10. Some More Practice Examples<\/a><\/li>\n
    11. Summary<\/a><\/li>\n
    12. Notes<\/a><\/li>\n
    13. Quiz Yourself!<\/a><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
      \n

      <\/a>1. Methane: One Carbon, One Signal<\/h2>\n

      To start visualize this concept, let\u2019s start with the basic example of methane. Methane contains one carbon atom. A <\/span>13<\/sup><\/span>C NMR experiment will produce one signal on a spectrum at a specific chemical shift, corresponding to the one carbon in methane.<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      <\/a>2. Ethane Has Two Carbons, But Only One Signal Due To Symmetry<\/h2>\n

      How about ethane? Ethane contains two carbons. Does this mean that two signals will be produced from a <\/span>13<\/sup><\/span>C NMR experiment?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      While ethane does contain two carbon atoms, two signals will NOT be produced. We can see by looking at the molecule that ethane contains a mirror plane of symmetry, bisecting it in half. If we perform a symmetry operation by flipping ethane 180\u00ba at a time (along its C<\/span>2<\/sub><\/span> mirror plane axis), we can see that the molecule\u2019s carbon atoms are identical in their environments, making them interchangeable. This results in the chemical equivalence of the two carbon atoms of ethane, producing a single signal in a <\/span>13<\/sup><\/span>C NMR spectrum.<\/span><\/p>\n

      <\/a>3. Pentane (5 Carbons) Has Only Three 13<\/sup>C NMR Signals Due To Symmetry<\/h2>\n

      To expand on this, let\u2019s take a look at pentane. How many signals would you expect to see on its <\/span>13<\/sup><\/span>C NMR spectrum?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      Three signals! Just like in the ethane example, pentane has a mirror plane straight down the middle. If we flip pentane 180\u00b0 at a time, we can see three types of carbon atoms present in the molecule. The pairs of carbons 1 and 5, as well as 2 and 4, are homotopic, while carbon 3 is distinct. The two homotopic pairs will produce one signal each, while carbon 3 will produce one for itself. This results in a total of 3 <\/span>13<\/sup><\/span>C NMR signals present despite pentane containing five carbon atoms.<\/span><\/p>\n

      <\/a>4. Symmetry In Aromatic Rings<\/h2>\n

      These rules of symmetry in alkanes apply to alkenes and alkynes the same. Does toluene have any carbon atoms that are equivalent? How many signals would it produce in a <\/span>13<\/sup><\/span>C NMR experiment?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      Toluene contains a mirror plane. If we identify that mirrorplane and perform rotational operations about it, we can see that two pairs of equivalent carbon atoms are present in addition to three non-equivalent ones. This adds up to a total of 5 different carbon atoms that will produce signals at 5 separate chemical shifts on a spectrum. Despite the molecule containing seven carbon atoms total, only 5 signals will be produced.<\/span><\/p>\n

      <\/a>5. Some Quick Examples<\/h2>\n

      Let\u2019s try a few examples based on what we\u2019ve gone over so far. How many signals would each molecule below produce in a <\/span>13<\/sup><\/span>C NMR experiment?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      Each of these compounds produce 3 signals! Like the previous examples, we can confirm carbon atom equivalence through observing a molecule\u2019s planes of symmetry.<\/span><\/p>\n

      <\/a>6. Slight Disruptions To Symmetry Can Have A Large Impact On The Number Of Signals In A 13<\/sup>C NMR Spectrum<\/h2>\n

      What happens when the symmetry of a molecule is thrown off? If we add a methyl group to pentane, how many signals would you expect to see on a <\/span>13<\/sup><\/span>C NMR spectrum of 2-methylpentane?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      The addition of this methyl group to pentane will shift the symmetry the molecule. Instead of observing a bisecting mirror plane, we see a mirror plane that runs straight through the compound\u2019s aliphatic chain, splitting between its branching carbons. Symmetry operations can now be performed by flipping the molecule 180\u00b0 at a time along this new axis. In doing so, we can see the two carbons of the isopropyl group are interchangeable and chemically equivalent. Thus, that pair of homotopic branching carbons will produce one single signal. With the addition of the 4 signals produced by the remaining non-equivalent carbons in the chain, five signals total will be present on a <\/span>13<\/sup><\/span>C NMR spectrum of 2-methylpentane.<\/span><\/p>\n

      As we can see above, molecules with slight structural differences can result in totally different symmetries that drastically change the number of signals present on a <\/span>13<\/sup><\/span>C NMR spectrum.<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      These two ethers that contain the same number carbon atoms will yield a different number of signals on a <\/span>13<\/sup><\/span>C NMR spectrum. Dibutyl ether contains eight carbons but will produce a total of 4 signals due to its bisecting mirror plane that results in each carbon atom being part of a homotopic pair. Pentyl propyl ether also contains eight carbons, but the mirror plane present (running through the chain of atoms) does not bisect the molecule to make any of the carbon atoms equivalent. This results in pentyl propyl ether producing a total of 8 signals in a <\/span>13<\/sup><\/span>C NMR spectrum that represent each of its 8 individual carbon atoms.<\/span><\/p>\n

      <\/a>7. Ortho, Meta, and Para Di-Substituted Aromatic Rings: Look Out For Planes Of Symmetry!<\/h2>\n

      How about some regioisomer examples to practice?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      While regioisomers contain the same number of carbon atoms, they differ in that their substituents are positioned in different chemical environments. These structural differences result in differing symmetries among the molecules, which correspond to each molecule having different pairs of homotopic carbons. A difference in chemical environments will cause a difference in chemical shifts, while a difference in quantities of equivalent and non-equivalent carbons cause a change in the number of signals.<\/span><\/p>\n

      <\/a>8. The Effect Of Chirality Centers On 13C NMR Spectra<\/h2>\n

      Let\u2019s take a look at a chiral molecule. How many signals would you expect to be produced from the compound below?<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      Looking at the molecule, we can see there is a chiral center present at the carbon atom with a chlorine substituent. The presence of this chiral center eliminates the possibility of a plane of symmetry that would make any of its carbon atoms equivalent. Therefore, each of the eight carbons in the compound are distinct, producing 1 signal each on a <\/span>13<\/sup><\/span>C NMR spectrum, totaling to 8 signals.<\/span><\/p>\n

      <\/a>9. Carbons That Are Equivalent Through Fast Bond Rotation Will Have Their Signals Averaged Together<\/h2>\n

      As we can see in the example above, the stereochemistry of a compound can affect the number of signals produced on a spectrum. What about conformers? Would different conformations of a molecule always affect the number of signals? Lets take a look at 2,2,3,3-tetramethylbutane.<\/span><\/p>\n

      \"\"<\/a><\/p>\n

      Just as we observe in <\/span>1<\/sup><\/span>H MNR, the same chemical equivalence through rapid bond rotation occurs in <\/span>13<\/sup><\/span>C NMR. While the rotations of t<\/i>-butyl groups occur about the central C-C bond of the compound, they are fast enough relative to the NMR \u201cshutter speed\u201d that the carbon atoms become equivalent, blurring together like the spinning of a bicycle wheel or blades of a fan. The resulting signal produced on the spectrum that corresponds to these atoms will represent the average of the carbon atoms in the different conformers<\/strong>. So in this case, conformers can be ignored and a total of two signals will be visible on a spectrum. It is important to remember that unlike <\/span>1<\/sup><\/span>H NMR, <\/span>13<\/sup><\/span>C NMR peak ratios are not to be used quantitatively. This spectrum only supplies the qualitative information that there are two types of carbon atoms present in the molecule. It does not specify any quantities of its primary and tertiary carbons.<\/span><\/p>\n

      <\/a>10. Some More Practice Examples<\/h2>\n

      Let\u2019s apply the information we\u2019ve learned so far to some practice problems.<\/span><\/p>\n

      \"how<\/a><\/p>\n

      <\/a>11. Summary<\/h2>\n

      Just as in <\/span>1<\/sup><\/span>H NMR, a molecule\u2019s symmetry will greatly affect the number of signals it produces on a <\/span>13<\/sup><\/span>C NMR spectrum. Finding any mirror planes present and performing the necessary symmetry operations greatly simplifies predicting the outcomes of this spectroscopic method.<\/span><\/p>\n

      \n

      This article was contributed by Nick Tiedemann. Thanks Nick!<\/p>\n

      <\/a>Notes<\/h2>\n

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