I just started learning about proton NMR. According to Molbase, the HNMR data for tert-amyl alcohol (2-methylbutan-1-ol) shows four kinds of protons at 0.9 ppm, 1.24 ppm, 1.44, and 3.65 ppm.

Structure of tert-amyl alcohol

However, it seems contradictory to what I learnt, since the protons on secondary carbon (1.44 ppm) are downfield to the protons on tertiary carbon (1.24 ppm), which confuses me, since I learnt that protons on tertiary carbons experience more de-shielding.

Is there something I have missed out in my thought process?

EDIT: Thanks to @orthocresol and @Buttonwood for pointing out my error. My revised question would then be: Between the two types of protons on secondary carbons (1.24 and 1.44 ppm), why is it that the one at 1.44 ppm is more deshielded?

From what I understand, it is because carbon is more electronegative than hydrogen, therefore causing the deshielding effect, but at the same time from organic chemistry I learnt that alkyl groups are electron donating, which should cause shielding. Is there some misconception in my thinking?

  • 2
    $\begingroup$ (1) Since you're just starting, it's a great time for you to unlearn the terms "downfield" and "upfield". These are ancient terms which come from the era of continuous-wave NMR and for some reason have not been erased from the literature yet. Stick to "deshielded" and "shielded"; they're much more easily understood. (2) In tert-amyl alcohol, I don't see any protons attached to tertiary carbons; only a secondary carbon, and two different types of methyl carbons. Could you clarify? $\endgroup$
    – orthocresol
    Aug 8 '20 at 15:04
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    $\begingroup$ In addition to @orthocresol's comment, your reference accounts for four different types of protons. Both in the visual representation as well as in the molecular structure-like one, there is the one shown for 3.65 ppm (which, depending on the experimental circumstances, either a) equally could be recorded at a different position, or b) could be invisible). $\endgroup$
    – Buttonwood
    Aug 8 '20 at 15:14
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    $\begingroup$ As a beginner to NMR, you should be aware of how important integration values in $\mathrm{^1H-NMR}$. If you knew and had them in your NMR, you could have easily identify these peaks. $\endgroup$ Aug 8 '20 at 16:09
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    $\begingroup$ I recommend you add to your question a list with all peaks, their integral, and the group it belongs to. The description in your text is not correct, the molecule has only one secondary carbon atom. $\endgroup$
    – Karl
    Aug 8 '20 at 16:54
  • 1
    $\begingroup$ There are only two kinds of protons: those are on primary carbon(s) and those on secondary carbon. $\endgroup$ Aug 8 '20 at 16:57

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