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Topic: Instability of e.g. C(CH3)(OH)3  (Read 5840 times)

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Offline Mel22

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Instability of e.g. C(CH3)(OH)3
« on: February 07, 2006, 03:36:08 PM »
why does

C(CH3)4, C(OH)(CH3)3, C(NH)C(CH3)3

occur

whereas

C(CH3)2(OH)2, C(CH3)2(NH)2, C(CH3)1(OH)3, C(CH3)1(NH)3

is unstable? ???

Does anyone have an idea?
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Offline movies

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #1 on: February 07, 2006, 11:58:36 PM »
Think about pushing electrons to form double bonds.  What else has to happen for this to work?  
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Offline Mel22

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #2 on: February 08, 2006, 02:21:37 AM »
hmmm....that would make sense. Surely, if you have too much double bond character at one andthe same Carbon atom, the molecule will be totally unstable. But what else needs to happen in order for it to work?
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Offline Mitch

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #3 on: February 08, 2006, 04:54:59 AM »
Stability meaning prone to decomposition. You are comparing non-identical compounds and to say one is more stable than the other becomes problematic.
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Offline AWK

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #4 on: February 08, 2006, 07:28:56 AM »
C(CH3)2(NH)2 and C(CH3)1(NH)3 cannot exist, but C(CH3)2=NH and C(CH3)(=NH)(NH2) can, ant this is not a problem of stability !!!
« Last Edit: February 08, 2006, 07:29:50 AM by AWK »
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Offline Mel22

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #5 on: February 09, 2006, 10:51:57 AM »
So, basically, from what you've said, my understanding is the following: C is in the 4th group and can only build 4 bonds to other atoms. Hence, if there's a double bond (in e.g. C=NH) the molecule does not exist. But what about the OH's? Is it because they have a small proportion of double bond character? It might also be interesting to look at it in an orbital prespective...

Can anyone please tell me whether that is correct and if not explain what the answer is? I really want to know!  :P
Thanks so much!
« Last Edit: February 09, 2006, 11:06:09 AM by Mel22 »
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Offline Mitch

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Re:Instability of e.g. C(CH3)(OH)3
« Reply #6 on: February 09, 2006, 01:26:46 PM »
If you want to learn, then draw out the structures and post it.
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