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1H-NMR of vanillin details in Nilered video
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Onorato Campopiano
1H-NMR of vanillin details in Nilered video
"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.
"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.
[...] but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a C-O bond, moving their chemical shift left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C and D are all next to a single proton, C is the odd one out.
First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:
Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$ while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.
When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta$\ce{CHO}$ group and ortho, para$\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.
Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.
[...] but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a C-O bond, moving their chemical shift left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C and D are all next to a single proton, C is the odd one out.
First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:
Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$ while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.
When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta$\ce{CHO}$ group and ortho, para$\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.
Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.
"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.
"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.
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First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:
Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$ while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.
When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta $\ce{CHO}$ group and ortho, para $\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.
Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.
First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:
Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$ while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.
When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta $\ce{CHO}$ group and ortho, para $\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.
Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.
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