How many equivalent hydrogens are there in octane?

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The discussion centers around a discrepancy in the number of non-equivalent hydrogen atoms in n-octane as presented in a Kaplan test for the DAT. The original poster believes there are four types of non-equivalent hydrogens, while others clarify that for NMR shifts, only the nearest-neighbor groups matter, leading to the conclusion that n-octane actually has three non-equivalent hydrogens. Despite some agreement on the theoretical basis for four types of hydrogens through a substitution test, practical considerations regarding NMR resolution suggest that distinguishing between the shifts of hydrogens on the third and fourth carbons may not be feasible. Ultimately, the consensus is that Kaplan's answer is incorrect, affirming the presence of three non-equivalent hydrogens in n-octane.
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The answer to my Kaplan test (studying for the DAT, just FYI) says 3, but I think it's 4?

Am I wrong?
 
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To which structural isomer of octane are you referring? And do you mean non-equivalent hydrogens?

Assuming you mean n-octane, I think you are correct in principle that there are 4 kinds .. in other words, if you use the substitution test, you can create 4 distinct molecules by substituting a single H-atom with an F-atom. However, for the purpose of determining NMR shifts in alkanes (which I assume is what you are asking about), I am fairly sure that only the nearest-neighbor groups matter. Assuming that is true, can you now see why n-octane has only 3 non-equivalent types of hydrogens?
 
SpectraCat said:
To which structural isomer of octane are you referring? And do you mean non-equivalent hydrogens?

Assuming you mean n-octane, I think you are correct in principle that there are 4 kinds .. in other words, if you use the substitution test, you can create 4 distinct molecules by substituting a single H-atom with an F-atom. However, for the purpose of determining NMR shifts in alkanes (which I assume is what you are asking about), I am fairly sure that only the nearest-neighbor groups matter. Assuming that is true, can you now see why n-octane has only 3 non-equivalent types of hydrogens?

I'd be inclined to disagree there, even for NMR, though for e.g. distinguishing between 3- and 4-halooctanes there may not be that much of a difference in shift, in principle this still exists.
 
sjb-2812 said:
I'd be inclined to disagree there, even for NMR, though for e.g. distinguishing between 3- and 4-halooctanes there may not be that much of a difference in shift, in principle this still exists.

I agree .. in practice it's all a question of resolution. The fact is the the OP is right, and strictly speaking there are 4 non-equivalent types of hydrogens in n-octane. However I am not sure if there is a strong enough magnet to distinguish the difference in chemical shifts of H-atoms on the 3rd and 4th carbons.

However given the phrasing of your post, I think it's important to point out that we are not talking about halo-octanes at all, but just normal n-octane (at least I think so, the OP has still not confirmed that is the correct structural isomer). So, there is no heteroatom in the molecule we are considering .. the substitution test I mentioned is just a thought-experiment to determine chemical non-equivalence. I do agree that if there were a heteroatom in the molecule, the dependence of the chemical shifts of H-atoms on the distance from the substitution site would be more pronounced.
 
Yes, I was talking about n-octane and yes, I meant non-equiv hydrogens.

Thanks guys, I guess Kaplan is wrong here.
 
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