From owner-chemistry -8 at 8- ccl.net Wed Sep 11 06:22:28 1996 Received: from bedrock.ccl.net for owner-chemistry;at;ccl.net by www.ccl.net (8.7.5/950822.1) id FAA07232; Wed, 11 Sep 1996 05:54:42 -0400 (EDT) Received: from goliat.ugr.es for aentrena%!at!%goliat.ugr.es by bedrock.ccl.net (8.7.5/950822.1) id FAA06260; Wed, 11 Sep 1996 05:54:39 -0400 (EDT) Received: from quimfarm2 (quimfarm2.ugr.es [150.214.65.52]) by goliat.ugr.es (8.6.10/8.6.12) with SMTP id LAA02976 for ; Wed, 11 Sep 1996 11:58:45 +0100 Message-ID: <323706A8.2C0 # - at - # goliat.ugr.es> Date: Wed, 11 Sep 1996 11:36:24 -0700 From: "Dr. Antonio Entrena" Organization: Dpto. Quimica Organica. Fac. Farmacia. Universidad de Granada. 18071 GRANADA (sPAIN) X-Mailer: Mozilla 2.0 (Win16; I) MIME-Version: 1.0 To: Computational Chemistry List Subject: TS on a SN2 reaction Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Dears CCL users: Last week I post a question regarding to a TS on a SN2 reaction. I have received some responses and here is the summary: My original question was: ********************************** I need to calculate the TS for a variety of SN2 reactions using semiempirical Mopac calculation and I have some problems. I will be very greatfull for every help that you can gave me. The strategy that I have used is the following: I have define a dummy atom in order to fix the geometry of the nuclephile and the leaving group as in the following scheme: Du..........C | | Nu..........C---L I have defined the distance between the dummy atom (Du) and the upper carbon atom about 10 A and between the nucleophile (Nu) and the lower carbon atom about 5 A. The Du-C-C-Nu and Du-C-C-L dihedral angles have been defined as 0 and 180 degrees, respectively. I used the symmetry keyword (option 17) in order to varies the C-C-Nu bond angle as 180 degrees - C-C-L bond angle during the calculation. I believe that with those definitions, the nucleophile, the carbon atom and the leaving group (L) must be in straight line when the distances Nu-C is decreased at 0.1 or 0.05 A intervale. I expect that when the nucleophile is located at a apropiate distance the distance C-L begin to increase so that the geometry of higher energy could be used to optimized the TS of the reaction, but this does not occur. When the nuclephile is about 2.4-2.2 A, the leaving group is inmediately ejected to a distance of 3.5 A or more. Can anybody with experience in this type of calculations gave me some orientations to resolve this problem?. Please, e-mail directly to me and I will summarized to the net. Sincerely A. Entrena ************************************** And the answers are the following: ********************************************* From: ansu;at;trout.csb.ki.se (Ansuman Lahiri) Dear Dr. Entrena Did you try the SADDLE keyword in mopac? What I usually do is to choose two configurations (in one of which the leaving group is bonded and in the other the attacking group is bonded to the central group) and then run SADDLE. The approximate transition state obtained from SADDLE can then be further refined using TS. Hope this helps Good luck! Ansuman ***************************************** From: "Anatoli Korkin" Hi! Fixing C-Nu eagual C-L in optimization, you will approach TS closely. Then compute frequencies and use the resulting force field along with opt=ts option. But I have feeling, that you guy hardly undertand what are you doing and why. regards, Anatoli Korkin ******************************************************** From: Alan.Shusterman.,at,.directory.Reed.EDU (Alan Shusterman) --- You wrote: I expect that when the nucleophile is located at a apropiate distance the distance C-L begin to increase so that the geometry of higher energy could be used to optimized the TS of the reaction, but this does not occur. When the nuclephile is about 2.4-2.2 A, the leaving group is inmediately ejected to a distance of 3.5 A or more. --- end of quoted material --- I am not sure what Nu/L combinations you are using, but this may be reasonable behavior. First, if either of the reactants or products are charged, then electrostatic interactions (ion-dipole) will be very strong and a true transition state may not even exist. Consider chloride attacking methyl chloride (degenerate rxn): as chloride approaches the energy falls (favorable ion-dipole interaction), then sharply rises to give a symmetric TS, then falls again to give the product ion-dipole complex. The AM1 transition state is only a kcal/mol or two higher than the separated reactants. Now consider hydroxide attacking methyl chloride: as hydroxide approaches the energy falls steadily without intervention of a transition state. This reaction is very exothermic, and bond energy changes plus electrostatic interactions erase the transition state. To summarize: 1) you may have trouble locating transition states, 2) minima may correspond to ion-dipole complexes in which reactants approach to within LESS than the sum of the nonbonded radii, 3) TS, when they do exist, should be viewed as resonance hybrids of the reactants and products, and therefore reasonably short CN and CL bond distances are expected. Finally, least motion paths are interesting, but may not tell the full story. For example, hydroxide attack on methyl chloride initially gives an ion-dipole complex in which the chloride is close to the methyl hydrogens of methanol. This is probably not the global minimum, and may not even be a local minimum. The best structure is probably one in which a hydrogen bond, Cl..HOCH3, holds the ion and dipole together. Alan Shusterman Department of Chemistry Reed College 3203 SE Woodstock Blvd Portland, OR 97202 ************************************ From: Hens Borkent Dear Dr. Entrena, There is no reason why the leaving group should stay in the neighbourhood of the carbon atom; once you're over the 'top' (TS), the structure will minimize the C-L distance. And moreover, there is no reason why the N-C-L angle should be 180 degrees, unless your carbon skeleton is symmetrical (methyl, t-butyl). The savest thing would be to do a grid search, in which you vary the C-N distance in the range 2.0 2.8, and the C-L distance something similar, depending on the nature of the leaving group. In this way you fix both values and a saddle point should appear in the grid. This point should be optimized using the TS keyword. Have a look at: http://www.caos.kun.nl/tutorials/camm/mopac/gsm.html Sincerely, -- ***** J.H. (Hens) Borkent, CAOS/CAMM Center, *CAOS * P.O. Box 9010, 6500 GL Nijmegen, The Netherlands * / * Tel 0031 24 36 52137 Fax 0031 24 36 52977 * CAMM* e-mail: borkent(+ at +)caos.kun.nl ***** http://www.caos.kun.nl/staff/borkent.html **************************************************** From: Jose Ignacio Garcia He leido tu pregunta en la CCL. Puede haber varios motivos para el comporta- miento que describes. En primer lugar, la forma de la superficie de energia potencial depente por supuesto de la reaccion, de forma que podria suceder que en la reaccion particular que estas estudiando, haya una zona en la que el sistema evolucione muy rapidamente. La solucion en este caso es estudiar mas en detalle dicha region, modificando la distancia de ataque Nu-C con intervalos mas cortos. Algo parecido sucede cuando se comparan las reacciones Cl(-) + CH3Cl --> ClCH3 + Cl(-) y F(-) + CH3Cl --> FCH3 + Cl(-) En el primer caso, se obtiene un diagrama de reaccion "de libro", pero en el segundo, dependiendo de la geometria inicial y el intervalo utilizado puede obtenerse un diagrama en el que aparentemente la energia disminuye siempre, sin que aparezca barrera de activacion. Resulta necesario estudiar muy en detalle la region critica para localizar dicha barrera. Por otra parte, creo que utilizas demasiadas restricciones en el acercamiento de Nu, lo cual puede dar problemas. Ten en cuenta que cuantas mas restricciones utilices, menos representativa sera la superficie de energia calculada con respecto a la "verdadera". Nosotros solemos emplear el siguiente esquema: Nu ------C-------L Donde el diedro se fija en 180 grados y los angulos | Nu-C-Du y Du-C-L se fijan en 90 grados. Es estricta | logica, no deberia fijarse ninguna restriccion, pero Du si Nu o L llevan carga neta, la tendencia a dar reacciones "exoticas", como arrancar un proton cercano es bastante grande. Con las restricciones indicadas, se fuerza a que el ataque de Nu y la salida de L se produzcan a lo largo de una linea recta, como usualmente se asume en las SN2. Espero que estas observaciones te sean de alguna ayuda. Un cordial saludo. Jose Ignacio -- -------------------------------------------------------------------------------- Dr. Jose Ignacio Garcia-Laureiro Phone : 34-(9)76-762077 Departamento de Quimica Organica 761210 Instituto de Ciencia de Materiales de Aragon Fax : 34-(9)76-761159 C.S.I.C.-Universidad de Zaragoza e-mail: jig %-% at %-% qorg.unizar.es E-50009 ZARAGOZA (SPAIN) jig#* at *#msf.unizar.es jig ":at:" posta.unizar.es -------------------------------------------------------------------------------- "And all this science I don't understand it's just my job five days a week..." ELTON JOHN - Rocket man ------------------------------------------------------------------------- ********************************************** Thanks to all af them for their help. Sincerely A. Entrena