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Topic: Nmr Proton Equivalence  (Read 3489 times)

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Offline offlinedoctor

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Nmr Proton Equivalence
« on: May 11, 2013, 12:48:05 AM »
I'm having trouble understanding how the following side chain has three non equivalent protons;

H    H
|     |
C = C - H

I thought if they were on the same carbon, then they would be chemically equivalent.

Offline sjb

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Re: Nmr Proton Equivalence
« Reply #1 on: May 11, 2013, 01:59:56 AM »
I'm having trouble understanding how the following side chain has three non equivalent protons;

H    H
|     |
C = C - H

I thought if they were on the same carbon, then they would be chemically equivalent.

One way to look at this is that if you replaced the proton with something else, would the name of the resulting compound be the same?

So for styrene, and replacing with iodine
  • c1ccc(cc1)/C=C/I
  • c1ccc(cc1)/C=C\I
  • C=C(c1ccccc1)I
; are these compounds the same?

Offline Sophia7X

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Re: Nmr Proton Equivalence
« Reply #2 on: May 11, 2013, 03:08:56 PM »
I'm having trouble understanding how the following side chain has three non equivalent protons;

H    H
|     |
C = C - H

I thought if they were on the same carbon, then they would be chemically equivalent.
That is generally the case with H's on sp3 carbons, but this has a double bond. So you want to think geometric isomerism if you were to use the replacement test like the example above
Entropy happens.

Offline Corribus

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Re: Nmr Proton Equivalence
« Reply #3 on: May 11, 2013, 05:04:57 PM »
To elaborate on the previous post, in this case you do not have "free" rotation around a double bond as the experiment temperature, so the two protons are in different chemical environments. 
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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