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Science Forum Index » Chemistry Forum » stability of carbokation(?) as in SN1 Reactions
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| Johannes Babach |
 Posted: Sun Jan 04, 2004 8:31 am |
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Hello,
first, I am sorry for my bad english, but however i hope that you understand my problem.
Can anyone please sort the following molecules in their stability in the transition state (don't know if that is the right word),
with the first number representing the most stabil carbokation. (Br- is thought to be splitted from the molecul and there should
stay a positive charge)
1 h-CH2Br
2h-CHBr-CN
3:Tert-Butylbromide
I suppose that it is 1,2,3 but it could also be 1,3,2 or 2,1,3 after what i know.
Concering molecule 2, I just know that a triple bond stabilizes a neighboured carbonsextet less good than a twice bond, but this
actually has to mean that it is still better for the carbonsextet than just a proton as in molecule 1. But also the nitrilgroup
(hope thats the name of -CN group in english) withdrawes electrons, which is destabilizing the carbonsextet. So i don't know what is
more important for the stability, the two occupied pi-orbitals or the -I effect of the nitril group. If the latter, than it could
also be possible that the molecule 3 would react faster than 2, if the pi-orbitals would be more important than molecule 2 could
perhaps react faster than molecule 1.
Any help is very welcome, most welcome would be some general rule of the importance of the different effects, not just for the
nitril group.
Johannes
note: i also posted this to sci.chem.organic.synthesis but i hope to find more readers here  |
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| Mark Tarka |
| Posted: Mon Jan 05, 2004 11:31 am |
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"Johannes Babach" <jo-news@chaotisch.com> wrote in message news:<bt94k8$u3g$04$1@news.t-online.com>...
Quote: Hello,
[impossibly long sentances snipped...]
1 h-CH2Br
2 h-CHBr-CN
3:Tert-Butylbromide
[ditto...]
Is this question coming from an undergraduate
or graduate student, or a practicing chemist?
Given a lack of context, I feel quite confident
in saying that the relative order of stability
would be 2 < 3 < 1 for gas-phase ions formed
in a vacuum. You might check the mass-spec
literature -- there have been studies made of
a wide range of ions. As a practical matter,
in vitro, you'd have to consider temperature,
solvent, and other reactants, as well as any
catalysts -- I wouldn't even try to guess at
relative stabilities in those situations.
Mark (So Saddam gets busted with a list of
baddies in his hand, Iran and Libya throw
in their nuclear cards, Russia ups the
bet by announcing a new generation of
nuclear-tipped missiles with a range of
6000 miles, and George goes to St. Louis
to explain to 4th graders why they're going
to be left behind. Have I got that right?) |
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| Johannes Babach |
| Posted: Mon Jan 05, 2004 12:55 pm |
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Mark Tarka wrote:
Quote: "Johannes Babach" <jo-news@chaotisch.com> wrote in message
news:<bt94k8$u3g$04$1@news.t-online.com>...
[most stabil carocation?]
1 h-CH2Br
2 h-CHBr-CN
3:Tert-Butylbromide
Is this question coming from an undergraduate
or graduate student, or a practicing chemist?
Undergraduate student I think, dont know how to convert it from the german system to your system.
Quote: Given a lack of context, I feel quite confident
in saying that the relative order of stability
would be 2 < 3 < 1 for gas-phase ions formed
in a vacuum. You might check the mass-spec
literature -- there have been studies made of
a wide range of ions.
Well the context is just an excercise. There are no precise instructions in which solvent or at what temperature and so on. The only
question ist, which of the molecules would react fastest in a Sn1 reaction. (which means which of the carbocations is the most
stable because of the Bell-Evance-Principle or what it's called :/ )
http://web.chem.ucla.edu/~harding/tutorials/cc.pdf
On page 4, paragraph 2:
"In actuality, resonance usually (but not always) outweighs other factors."
So this probleme here is one of the "not always"?
Because if not, then the order would have to be 2,1,3 with 2 having the most resonance.
Thanks for your previous answere, Johannes.
Hope I didnt make my sentences too long. |
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| Mark Tarka |
| Posted: Mon Jan 05, 2004 6:34 pm |
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"Johannes Babach" <jo-news@chaotisch.com> wrote in message news:<btc8fd$icm$05$1@news.t-online.com>...
Quote: Mark Tarka wrote:
"Johannes Babach" <jo-news@chaotisch.com> wrote in message
news:<bt94k8$u3g$04$1@news.t-online.com>...
[most stabil carocation?]
1 h-CH2Br
2 h-CHBr-CN
3:Tert-Butylbromide
Is this question coming from an undergraduate
or graduate student, or a practicing chemist?
Undergraduate student I think, dont know how to convert it from the german system to your system.
Given a lack of context, I feel quite confident
in saying that the relative order of stability
would be 2 < 3 < 1 for gas-phase ions formed
in a vacuum. You might check the mass-spec
literature -- there have been studies made of
a wide range of ions.
Well the context is just an excercise. There are no precise instructions in which solvent or at what temperature and so on. The only
question ist, which of the molecules would react fastest in a Sn1 reaction. (which means which of the carbocations is the most
stable because of the Bell-Evance-Principle or what it's called :/ )
http://web.chem.ucla.edu/~harding/tutorials/cc.pdf
On page 4, paragraph 2:
"In actuality, resonance usually (but not always) outweighs other factors."
So this probleme here is one of the "not always"?
Because if not, then the order would have to be 2,1,3 with 2 having the most resonance.
If it's just a paper exercise, then, assuming
that in each case the leaving group is the Br(-),
you are probably correct (I would question putting
a positive charge on the nitrogen, but I'm out of
practice).
In electron impact mass-spectrometry, if
I'm reading correctly, a neutral HCN molecule
(maybe a radical) could be split-out of 2,
or maybe HBr. HCN, HF and HCl are listed in Table
A.5. of F.W. Mclafferty's _Interpretation of Mass
Spectra_, 3rd Edition, University Science Books,
Mill Valley, California, 1980. Our mass-spec
guru might be around here with an answer. Laser
induced ionization should also be similar.
There have been mass-spec studies examining the
gas-phase acidities and basicities of molecules
in the gas phase. I know little about it, but
would expect that stability of an ion once formed
would be a factor. This is probably far beyond
what you are looking for to answer your question,
but is interesting stuff if you have the curiosity.
I would guess that a good understanding at the
atomic and molecular levels of ionization might
prove useful in material sciences at the least.
Mark |
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| GeneralChemTutor |
| Posted: Thu Jan 08, 2004 10:33 pm |
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Joined: 08 Jan 2004
Posts: 41
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I am thinking...from the least stable to the most stable...2,1,3
Assuming that we are using relatively non-nucleophilic polar solvent at around relatively low temperatures (at higher temperatures E1, E2 reactions are predominant):
2: I am thinking that by having a strong electronegative agent such as CN- next to the developing carbocation, that such a electron withdrawing group would not serve to stabilize but rather intensify the developing positive charge. Also, think about the product...an OH group next to a CN group.
1:a primary carbocation having only one neighboring carbon for a electron releasing group. In constrast to 3...
3: Which has three carbons to serve as local electron contributors to somewhat delocalize the charge and thus lower the activation energy of the transition state. |
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Help with college chemistry (organic and general chemistry)
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| Mark Tarka |
| Posted: Mon Jan 12, 2004 4:15 pm |
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GeneralChemistryTutor@hotmail-dot-com.no-spam.invalid (GeneralChemTutor) wrote in message news:<3ffe2b2c$1_1@Usenet.com>...
Quote: I am thinking...from the least stable to the most stable...2,1,3
Assuming that we are using relatively non-nucleophilic polar solvent
at around relatively low temperatures (at higher temperatures E1, E2
reactions are predominant):
2: I am thinking that by having a strong electronegative agent such as
CN- next to the developing carbocation, that such a electron
withdrawing group would not serve to stabilize but rather intensify
the developing positive charge. Also, think about the product...an
OH group next to a CN group.
1:a primary carbocation having only one neighboring carbon for a
electron releasing group. In constrast to 3...
3: Which has three carbons to serve as local electron contributors to
somewhat delocalize the charge and thus lower the activation energy
of the transition state.
Allow me to bring back a portion of the original post:
Quote: "Johannes Babach" <jo-news@chaotisch.com> wrote in message
news:<bt94k8$u3g$04$1@news.t-online.com>... [most stabil carocation?]
1 h-CH2Br
2 h-CHBr-CN
3:Tert-Butylbromide
This is not an easy problem...arguments can be made for
and against the stability of each. And not the sort
of question asked of U.S. undergrads (well, there
might be exceptions, but generally speaking...).
It's true, 1 does form a primary carbocation,
but it is also true that it is a benzylic ion,
stabilized to some extent by the aromatic
ring. In mass spectrometry, it is believed
that a rearrangement of the system to a
seven-membered ring occurs, giving the
tropylium ion (GOOGLE: ~710 hits).
Allylic stabilization (C=C-C+) of this sort
is well known in organic chemistry.
I still think that we'd best be getting
an answer to this from a practicing organic
chemist...and even then, he might be vague...
it depends...it depends.
Mark (How did -OH get into the picture?) |
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