Proton NMR chemical shift of water peak in different solventsWhat factors are important for quantitative analysis of a proton 1D-NMR spectrum?Which has the higher chemical shift E/Z alkene?1H NMR proton coupling1H NMR Broad peaksExplanation for these 1H NMR spectra?Benzene and proton in 1H NMRHow to explain two NMR spectra of the (allegedly) same compound with slightly different shifts?Possible Water peak in PMMA nmr spectrumProp-2-yn-1-ol NMR spectrum differencesWhat could cause a peak to split into 2 resonances as temperature increases in NMR spectroscopy

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Proton NMR chemical shift of water peak in different solvents


What factors are important for quantitative analysis of a proton 1D-NMR spectrum?Which has the higher chemical shift E/Z alkene?1H NMR proton coupling1H NMR Broad peaksExplanation for these 1H NMR spectra?Benzene and proton in 1H NMRHow to explain two NMR spectra of the (allegedly) same compound with slightly different shifts?Possible Water peak in PMMA nmr spectrumProp-2-yn-1-ol NMR spectrum differencesWhat could cause a peak to split into 2 resonances as temperature increases in NMR spectroscopy













5












$begingroup$


Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $ceCDCl3$ appears at 1.56 ppm.



What is the reason behind it? Why should the same species ($ceH2O$) give rise to two different chemical shift values?










share|improve this question









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    5












    $begingroup$


    Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $ceCDCl3$ appears at 1.56 ppm.



    What is the reason behind it? Why should the same species ($ceH2O$) give rise to two different chemical shift values?










    share|improve this question









    New contributor



    Latha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      5












      5








      5





      $begingroup$


      Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $ceCDCl3$ appears at 1.56 ppm.



      What is the reason behind it? Why should the same species ($ceH2O$) give rise to two different chemical shift values?










      share|improve this question









      New contributor



      Latha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $ceCDCl3$ appears at 1.56 ppm.



      What is the reason behind it? Why should the same species ($ceH2O$) give rise to two different chemical shift values?







      spectroscopy nmr-spectroscopy solvents






      share|improve this question









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      Latha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question









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      share|improve this question




      share|improve this question








      edited May 17 at 10:41









      orthocresol

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      asked May 17 at 10:08









      LathaLatha

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          2 Answers
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          $begingroup$

          Consider also water in addition to the solvents you mention:



          • water-d2 (or, $ceD2O$) is an extensively H-bonding polar solvent (dielectric constant $epsilon = 79$)


          • $ceDMSO-d6$) is polar ($epsilon= 47$) but aprotic (no-H-bond donation ability)

          • chloroform-d is an apolar aprotic solvent ($epsilon=4.81$)

          Now consider the state of a trace amount of $ceH2O$ in each of these solvents, giving rise to an NMR signal:



          • in $ceD2O$ the trace $ceH$ can exchange and forms $ceHDO$ which retains an extensive H-bond network. $ceH$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $ceH$ carrying a significant positive partial charge. The chemical shift is $approx pu4.7 ppm$

          • in DMSO-d6, $ceH$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $approx pu3.33 ppm$

          • in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $ceH$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $approx pu1.56 ppm$

          The following table shows shifts for residual water $ceH$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):



          $beginarrayc
          beginarrayc
          textsolvent &textshift(ppm)&epsilon\hline
          ceC6D6 &0.40 & 2.28\
          texttoluene-d8 &0.43 &2.38\
          ceC6D5Cl &1.03 &5.69\
          ceCD2Cl2 &1.52 &9.08\
          ceCDCl3 &1.56 &4.81\
          ceCD3CN &2.13 &36.64\
          textTHF-d8 &2.46 &7.52\
          ce(CD3)2CO &2.84 &21.01\
          ce(CD3)2SO &3.33 &47\
          textTFE-d3 &3.66&8.55 \
          ceCD3OD &4.87 &32.6\
          endarray
          endarray$



          Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
          Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
          Note also that the top three have an additional effect on water shifts due to the aromatic ring current.




          Reference
          [1] Fulmer et al. Organometallics 2010, 29, 2176–2179






          share|improve this answer











          $endgroup$




















            2












            $begingroup$

            Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.






            share|improve this answer










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              4












              $begingroup$

              Consider also water in addition to the solvents you mention:



              • water-d2 (or, $ceD2O$) is an extensively H-bonding polar solvent (dielectric constant $epsilon = 79$)


              • $ceDMSO-d6$) is polar ($epsilon= 47$) but aprotic (no-H-bond donation ability)

              • chloroform-d is an apolar aprotic solvent ($epsilon=4.81$)

              Now consider the state of a trace amount of $ceH2O$ in each of these solvents, giving rise to an NMR signal:



              • in $ceD2O$ the trace $ceH$ can exchange and forms $ceHDO$ which retains an extensive H-bond network. $ceH$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $ceH$ carrying a significant positive partial charge. The chemical shift is $approx pu4.7 ppm$

              • in DMSO-d6, $ceH$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $approx pu3.33 ppm$

              • in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $ceH$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $approx pu1.56 ppm$

              The following table shows shifts for residual water $ceH$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):



              $beginarrayc
              beginarrayc
              textsolvent &textshift(ppm)&epsilon\hline
              ceC6D6 &0.40 & 2.28\
              texttoluene-d8 &0.43 &2.38\
              ceC6D5Cl &1.03 &5.69\
              ceCD2Cl2 &1.52 &9.08\
              ceCDCl3 &1.56 &4.81\
              ceCD3CN &2.13 &36.64\
              textTHF-d8 &2.46 &7.52\
              ce(CD3)2CO &2.84 &21.01\
              ce(CD3)2SO &3.33 &47\
              textTFE-d3 &3.66&8.55 \
              ceCD3OD &4.87 &32.6\
              endarray
              endarray$



              Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
              Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
              Note also that the top three have an additional effect on water shifts due to the aromatic ring current.




              Reference
              [1] Fulmer et al. Organometallics 2010, 29, 2176–2179






              share|improve this answer











              $endgroup$

















                4












                $begingroup$

                Consider also water in addition to the solvents you mention:



                • water-d2 (or, $ceD2O$) is an extensively H-bonding polar solvent (dielectric constant $epsilon = 79$)


                • $ceDMSO-d6$) is polar ($epsilon= 47$) but aprotic (no-H-bond donation ability)

                • chloroform-d is an apolar aprotic solvent ($epsilon=4.81$)

                Now consider the state of a trace amount of $ceH2O$ in each of these solvents, giving rise to an NMR signal:



                • in $ceD2O$ the trace $ceH$ can exchange and forms $ceHDO$ which retains an extensive H-bond network. $ceH$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $ceH$ carrying a significant positive partial charge. The chemical shift is $approx pu4.7 ppm$

                • in DMSO-d6, $ceH$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $approx pu3.33 ppm$

                • in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $ceH$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $approx pu1.56 ppm$

                The following table shows shifts for residual water $ceH$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):



                $beginarrayc
                beginarrayc
                textsolvent &textshift(ppm)&epsilon\hline
                ceC6D6 &0.40 & 2.28\
                texttoluene-d8 &0.43 &2.38\
                ceC6D5Cl &1.03 &5.69\
                ceCD2Cl2 &1.52 &9.08\
                ceCDCl3 &1.56 &4.81\
                ceCD3CN &2.13 &36.64\
                textTHF-d8 &2.46 &7.52\
                ce(CD3)2CO &2.84 &21.01\
                ce(CD3)2SO &3.33 &47\
                textTFE-d3 &3.66&8.55 \
                ceCD3OD &4.87 &32.6\
                endarray
                endarray$



                Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
                Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
                Note also that the top three have an additional effect on water shifts due to the aromatic ring current.




                Reference
                [1] Fulmer et al. Organometallics 2010, 29, 2176–2179






                share|improve this answer











                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  Consider also water in addition to the solvents you mention:



                  • water-d2 (or, $ceD2O$) is an extensively H-bonding polar solvent (dielectric constant $epsilon = 79$)


                  • $ceDMSO-d6$) is polar ($epsilon= 47$) but aprotic (no-H-bond donation ability)

                  • chloroform-d is an apolar aprotic solvent ($epsilon=4.81$)

                  Now consider the state of a trace amount of $ceH2O$ in each of these solvents, giving rise to an NMR signal:



                  • in $ceD2O$ the trace $ceH$ can exchange and forms $ceHDO$ which retains an extensive H-bond network. $ceH$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $ceH$ carrying a significant positive partial charge. The chemical shift is $approx pu4.7 ppm$

                  • in DMSO-d6, $ceH$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $approx pu3.33 ppm$

                  • in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $ceH$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $approx pu1.56 ppm$

                  The following table shows shifts for residual water $ceH$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):



                  $beginarrayc
                  beginarrayc
                  textsolvent &textshift(ppm)&epsilon\hline
                  ceC6D6 &0.40 & 2.28\
                  texttoluene-d8 &0.43 &2.38\
                  ceC6D5Cl &1.03 &5.69\
                  ceCD2Cl2 &1.52 &9.08\
                  ceCDCl3 &1.56 &4.81\
                  ceCD3CN &2.13 &36.64\
                  textTHF-d8 &2.46 &7.52\
                  ce(CD3)2CO &2.84 &21.01\
                  ce(CD3)2SO &3.33 &47\
                  textTFE-d3 &3.66&8.55 \
                  ceCD3OD &4.87 &32.6\
                  endarray
                  endarray$



                  Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
                  Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
                  Note also that the top three have an additional effect on water shifts due to the aromatic ring current.




                  Reference
                  [1] Fulmer et al. Organometallics 2010, 29, 2176–2179






                  share|improve this answer











                  $endgroup$



                  Consider also water in addition to the solvents you mention:



                  • water-d2 (or, $ceD2O$) is an extensively H-bonding polar solvent (dielectric constant $epsilon = 79$)


                  • $ceDMSO-d6$) is polar ($epsilon= 47$) but aprotic (no-H-bond donation ability)

                  • chloroform-d is an apolar aprotic solvent ($epsilon=4.81$)

                  Now consider the state of a trace amount of $ceH2O$ in each of these solvents, giving rise to an NMR signal:



                  • in $ceD2O$ the trace $ceH$ can exchange and forms $ceHDO$ which retains an extensive H-bond network. $ceH$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $ceH$ carrying a significant positive partial charge. The chemical shift is $approx pu4.7 ppm$

                  • in DMSO-d6, $ceH$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $approx pu3.33 ppm$

                  • in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $ceH$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $approx pu1.56 ppm$

                  The following table shows shifts for residual water $ceH$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):



                  $beginarrayc
                  beginarrayc
                  textsolvent &textshift(ppm)&epsilon\hline
                  ceC6D6 &0.40 & 2.28\
                  texttoluene-d8 &0.43 &2.38\
                  ceC6D5Cl &1.03 &5.69\
                  ceCD2Cl2 &1.52 &9.08\
                  ceCDCl3 &1.56 &4.81\
                  ceCD3CN &2.13 &36.64\
                  textTHF-d8 &2.46 &7.52\
                  ce(CD3)2CO &2.84 &21.01\
                  ce(CD3)2SO &3.33 &47\
                  textTFE-d3 &3.66&8.55 \
                  ceCD3OD &4.87 &32.6\
                  endarray
                  endarray$



                  Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
                  Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
                  Note also that the top three have an additional effect on water shifts due to the aromatic ring current.




                  Reference
                  [1] Fulmer et al. Organometallics 2010, 29, 2176–2179







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited May 17 at 18:16

























                  answered May 17 at 15:17









                  Night WriterNight Writer

                  3,879527




                  3,879527





















                      2












                      $begingroup$

                      Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.






                      share|improve this answer










                      New contributor



                      sjb-2812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      $endgroup$

















                        2












                        $begingroup$

                        Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.






                        share|improve this answer










                        New contributor



                        sjb-2812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.





                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.






                          share|improve this answer










                          New contributor



                          sjb-2812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          $endgroup$



                          Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.







                          share|improve this answer










                          New contributor



                          sjb-2812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.








                          share|improve this answer



                          share|improve this answer








                          edited May 17 at 14:01





















                          New contributor



                          sjb-2812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          answered May 17 at 10:12









                          sjb-2812sjb-2812

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