I am currently working with an unknown and from the IR spectrum I was able to figure out that the compound I am working with is salicylic acid.
I am looking at the H NMR, but I am a bit confused on the deshielding, integration, and upsheilding of the hydrogens that are near the $\ce{OH}$ and $\ce{COOH}$ groups on the compound. Can someone explain this to me:
Attached is a picture
Since you have identified your unknown as salicylic acid (2-hydroxybenzoic acid), I'd help you understand how easily understand the substituents of aromatic nucleus induce shielding-deshielding effects. The following diagram illustrate the $\mathrm{^1H}$-$\mathrm{NMR}$ assignments of salicylic aromatic protons:
A lot of experts do not like explaining shielding-deshielding effects using electron densities around the sought atom. I agree with them, but a student of your capacity could easily related to that explanation. A good rule of thumb is o,p-directing electron donating groups (coursing positive mesomeric or resonance effect) usually shield the relevant protons at o,p-positions (inducing $\delta^-$-charge at the positions). Similarly, m-directing electron withdrawing groups (coursing negative mesomeric or resonance effect) usually deshield the relevant protons at o,p-positions (inducing $\delta^+$-charge at the positions). Accordingly, $\ce{OH}$-group shields $\ce{H_a}$ and $\ce{H_c}$ while $\ce{COOH}$-group deshields $\ce{H_b}$ and $\ce{H_d}$.
Thus, it is safe to say that the most up-field aromatic signals (sheilded) at $\delta \ 6.97$ (apparent $triplet$) and $\delta \ 7.03$ (apparent $doublet$) are due to $\ce{H_a}$ and $\ce{H_c}$. The assignments of $\delta \ 6.97$ is belong to $\ce{H_c}$ (two vicinal protons) and $\delta \ 7.03$ is belong to $\ce{H_a}$ (one vicinal proton) are solely due to the number of neighboring (vicinal) protons to each. Keep in mind that chemical shifts of these two signals can be moved to up- or down-field positions depending on the using locking solvent. For example, the locking solvent used to get present $\mathrm{^1H}$-$\mathrm{NMR}$ spectrum here is apparently $\ce{CDCl3}$ (based on solvent residue at $\delta \ 7.28$). However, two signals have been moved to more up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 6.92$ ($t,\: ^3\!\!J=\pu{7.4 Hz}$) and $\delta \ 6.96$ ($d,\: ^3\!\!J=\pu{7.5 Hz}$) are assigned for $\ce{H_c}$ and $\ce{H_a}$, respectively.
Similarly, the most down-field aromatic signals (desheilded) at $\delta \ 7.55$ (apparent $triplet \ of \ doublet$) and $\delta \ 7.96$ (apparent $doublet \ of \ doublet$) are due to $\ce{H_b}$ and $\ce{H_d}$ (Note: The splitting of each signal with a small coupling constant is due to meta-or $\omega$-coupling generated by the presence of proton at meta to each; This phenomenon is generally called $^4\!\!J$-coupling). The assignments of $\delta \ 7.55$ is belong to $\ce{H_b}$ (two vicinal protons) and $\delta \ 7.96$ is belong to $\ce{H_d}$ (one vicinal proton) are also due to the number of neighboring (vicinal) protons to each. Similar to $\ce{H_a}$ and $\ce{H_c}$, these two signals have also been moved to more some what up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 7.52$ ($t,\: ^3\!\!J=\pu{7.7 Hz}$) and $\delta \ 7.82$ ($d,\: ^3\!\!J=\pu{7.7 Hz}$) are assigned for $\ce{H_b}$ and $\ce{H_d}$, respectively.
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