H NMR of 2-bromobutane

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Why does 2-bromobutane have five different types of hydrogens?

By replacing the two hydrogens on C3, one at a time, with an imaginary group A you can compare the resulting imaginary compounds to see if they are identical.

The hydrogens on C3 produce diastereomers. The diastereotopic hydrogens will have different, but very similar chemical shifts. When we look at bromoethane using the same technique as above:

The resulting imaginary compounds are enantiomers and the hydrogens have identical shifts. Does anyone know why the enantiotopic hydrogens have identical shifts while the diastereotopic hydrogens have different but very similar chemical shifts?

Since all the hydrogen atoms in 2-bromobutane are bonded to carbon I should have been able to count the unique carbons and then answer which of those has hydrogen bonded to it. It should have 4 unique carbons of which 4 have hydrogen. Only by employing the above technique am I able to see that there are 5 different types of hydrogens in 2-bromobutane. What exactly am I missing?

 

[Czarcasm]

Why does 2-bromobutane have five different types of hydrogens?

Gosh, I totally forgot about this crap, lol. I was told a few years ago for my first take that this wasn't a scenario commonly presented. Generally, there's only really two scenarios where diastereotopic hydrogens are considered: alkenes and the example above, where 2 chiral centers can co-exist, by replacing a hydrogen with a different atom.

If they're gonna ask about diastereotopic hydrogens, they'd most likely present it with an alkene, since it's more obvious I guess. But in this scenario, the correct answer is 5 signals and not 4, for the reasoning you mentioned above -- at least that's the way I learned it. There's a bunch of other concepts regarding J coupling and all that jazz and to be honest, I brain dumped that a long time ago so I'm not sure if that's even relevant tbh. I'm curious what others have to say.