Subject: ACTIVATION E AND STABILITY: RESPONSES

From: "E. Lewars" <[email-masked]>
Date: Fri, 2 Jul 1999 12:51:43 -0400 (EDT)

1999 July 2
REPLIES TO QUESTION ON ACTIVATION ENERGY AND STABILITY

Thanks very much to all (below) who responded to my question (below)

   E. Lewars
==============

QUESTION:

1999 June 24

Hello,
Does anyone know have some *experimental facts* for the relationship between
ab initio (or DFT, or semiempirical) calculated activation enewrgies and
stability? I mean simply the transition state energy minus the reactant energy,
for formally unimolecular reactions, e.g. isomerizations, or decompositions.
  How high does the calculated barrier have to be for the compound to be
isolable at room temperature, for example, from actual experience in the lab?

  Thanks
    E. Lewars
======================
ANSWERS:

#1
Date: Thu, 24 Jun 1999 17:47:36 -0400 (EDT)
To: "E. Lewars" <[email-masked]>
Subject: Re: CCL:CALC ACTIVATION E AND STABILITY

Hi,

  As in fact, I am modeling some system ivolving the isomerization and
reactions. Depending on the syetems you are interested, the barrier
varies from about 2 Kcal/mol to several decads Kcal/mol like 30 or 50.
Of course this also depending on the methods you use. Usually HF
overestimates the barrier(UMP2 for open shell too). While DFT (in general)
underestimates the barrier ( however depending on what kind of DFTs you
chose, hybrid DFT behave better than GGA and LSDA in many cases). There
are lots of works about this topic, you may want to check in Chem. Phys.
Lett., J. Phys. Chem. or J. Chem. Phys. or another European journals.

 Regards,

 Wei Quan Tian

 Univ. of Guelph
 
> ======================
Date: Thu, 24 Jun 1999 17:49:08 -0400 (EDT)
X-Sender: wtian@ccshst01

 Another point for stability.

  DFT is better than HF for stability.

  Wei Quan Tian
===============

#2
Date: Fri, 25 Jun 1999 01:31:39 +0200 (MET DST)
To: "E. Lewars" <[email-masked]>
Subject: Re: CCL:CALC ACTIVATION E AND STABILITY

Dear Mr Lewars,

        I am very much interested in the answers you will to this
question. Would summarize the responds you get or would it be possible to
forward those mails which are adressed directly to you and not to CCl.

Thank you very much for your efforts,

         Alexander

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                                              ("`-/")_.-'"``-._
Alexander Wittkopp                            (. . `) -._    )-;-,_()
Institut fuer Organische Chemie               (v_,)'  _  )`-.\  ``-
Georg-August Universitaet                     _;- _,-_/ / ((,'
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Germany            Phone: +49-(0)551-393290 Fax: +49-(0)551-399475

"eine ungerade Zahl, am besten prim hat immer guenstige Auswirkungen auf 
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#3
   Jun 25 Dr. Peter Burger   (61)   Re: CCL:CALC ACTIVATION E AND STABICommand: Read MessageMessage 5/5 from Dr. Peter Burger                        Jun 25 '99 at 9:37 am

Subject: Re: CCL:CALC ACTIVATION E AND STABILITY
To: [email-masked] (E. Lewars)

Hi,

Concerning activation energies from DFT etc: simply the transition state
energy minus the reactant energy - that's exactly, how it's done -
entropic factors, however are not taken care of this way, but you can do
that as well from a force calculation. We have one or two examples, where
there was a good match between the experimental data and theory.
 
Experimentally
at RT, 25 kcal/mol is usually what one considers OK and gives you a half
life of one day (unimolecular rxn). However, if required, experimental
chemist can go down to the 15+ kcal/mol range, to isolate compounds,
below this temperature things are getting more nasty, but it is possible,
too, by instant freezing techniques and assuming that the compound is
stable in the solid state. So just assume that you need in the range of
some minutes to an hour to handle things and then use the Eyring equation
to calculate the barrier at a given temperature!
Clearly, I am referring here to preparative techniques rather than matrix
isolation or gasphase studies.

Best wishes

Peter

-------------------------------------------------------
Peter Burger
University of Zuerich
============