4 added 55 characters in body edited Apr 27 '12 at 13:00 F'x 18.7k66 gold badges8080 silver badges139139 bronze badges Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3CH3 groups are three bonds away from the hydrogens of the CH2CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3CH3 group are magnetically equivalent, due to the fast rotation along the C-CC–C bond, and the two CH3CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2CH2 hydrogens. The result of that is a splitting into a septet (7). The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra. Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra. Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra. 3 added 571 characters in body edited Apr 27 '12 at 5:58 Mad Scientist 4,85311 gold badge2929 silver badges5555 bronze badges Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra. Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra. 2 added 382 characters in body edited Apr 26 '12 at 12:43 Mad Scientist 4,85311 gold badge2929 silver badges5555 bronze badges Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $$\mathrm{^3J_{HH}}$$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $$\mathrm{^3J}$$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $$\mathrm{^4J}$$-couplings or more. So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way. The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other. In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C-C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7). 1 answered Apr 26 '12 at 12:34 Mad Scientist 4,85311 gold badge2929 silver badges5555 bronze badges