Subject: G94: G94/DFT/Tr.State Problem

From: ipcakc@Postoffice (Dr.A.K.Chandra)
Date: Mon, 22 Jul 1996 19:54:20 -0500

Dear Netters,
I found some funny problem using B3LYP/6-31G** keyword for the following
transition state. It is a H-transfer process between two Cl atoms forming
a symmetric transition state

		Cl---------H---------Cl

with equal bond lengths. The Geometry optimisation leads to a perfectly
symmetric structure (B3lyp/6-31G**/UHF). The system is a doublet and is
a radical-molecule reaction.  Now, the fun is that the eigenvalue corresponding
to the imaginary frequency at the saddle point is ONLY ONE. But, if the
frequencies are seen, there are 3 IMAGINARY FREQUENCIES. The Keyword I have
used for optimisation of transition state is OPT=(TS,EF,CALCALL), hence
frequency calculation is done along with the optimisation.  The output is 
attached below.  

Can anyone help me in interpreting this result? 

Sincerely,
V. Sreedhara Rao.

		--------------------------------------------
			       V. Sreedhara Rao
			       Department of IPC,
			  Indian Institute of Science,
			   BANGALORE - 560 012. INDIA.

		   E-mail : [email-masked]
		--------------------------------------------

------------------------The output follows ----------------------
 **********************************************
 Gaussian 94:  IBM-RS6000-G94RevC.2  5-Aug-1995
                  12-Jul-1996
 **********************************************
 -----------------------------------------------------------
 #uhf 6-31g** opt=(ef,ts,calcall) b3lyp pop=full den=current
 -----------------------------------------------------------
 EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-
 ------------------------------------------------------------------------
                         Z-MATRIX (ANGSTROMS AND DEGREES)
 CD Cent Atom  N1     Length/X     N2    Alpha/Y     N3     Beta/Z      J
 ------------------------------------------------------------------------
   1   1  Cl
   2   2  H     1   1.497666(  1)
   3      X     2   1.000000(  2)   1   90.000(  4)
   4   3  Cl    2   1.497673(  3)   3   90.000(  5)   1  180.000(  6)   0
 ------------------------------------------------------------------------
 SCF Done:  E(UB+HF-LYP) =  -920.925868655     A.U. after   13 cycles
             Convg  =     .7439D-08             -V/T =  2.0030
             S**2   =    .7571
 Annihilation of the first spin contaminant:
 S**2 before annihilation      .7571,   after      .7500
 EIGENVECTOR FOLLOWING TRANSITION STATE SEARCH
 ANALYTICAL SECOND DERIVATIVES USED
 ITERATION   3
 EIGENVECTORS AND EIGENVALUES
                           1         2
     EIGENVALUES --     -.02252    .16489
       1  R1            -.70710    .70711
       2  R2             .70711    .70710
 TS SEARCH. TAKING P-RFO STEP
 SEARCHING FOR LAMDA THAT MAXIMIZES ALONG THE LOWEST MODE
 VALUE TAKEN    LAMDA=    .00000000
 SEARCHING FOR LAMDA THAT MINIMIZES ALONG ALL OTHER MODES
 VALUE TAKEN    LAMDA=    .00000000
 STEP TAKEN. STEPSIZE IS   .000014
 CURRENT PARAMETER VALUES (INTERNAL COORDINATES)
    I        X         GRADIENT    DISPLACEMENT       NEWX
    1    2.830179      -.000001       .000014      2.830193
    2    2.830192      -.000001       .000001      2.830193
         ITEM               VALUE     THRESHOLD  CONVERGED?
 MAXIMUM FORCE             .000001      .000450     YES
 RMS     FORCE             .000001      .000300     YES
 MAXIMUM DISPLACEMENT      .000014      .001800     YES
 RMS     DISPLACEMENT      .000010      .001200     YES

 *************************************************
 **  CONVERGENCE CRITERIA APPARENTLY SATISFIED  **
 *************************************************

                       ----------------------------
                       !   Optimized Parameters   !
                       ! (Angstroms and Degrees)  !
 ----------------------                            ----------------------
 !      Name          Value   Derivative information (Atomic Units)     !
 ------------------------------------------------------------------------
 !       R1          1.4977   -DE/DX =    0.                            !
 !       R2          1.4977   -DE/DX =    0.                            !
 ------------------------------------------------------------------------
 PREDICTED CHANGE IN ENERGY    .000000

 EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-
 ------------------------------------------------------------------------
                         Z-MATRIX (ANGSTROMS AND DEGREES)
 CD Cent Atom  N1     Length/X     N2    Alpha/Y     N3     Beta/Z      J
 ------------------------------------------------------------------------
   1   1  Cl
   2   2  H     1   1.497666(  1)
   3      X     2   1.000000(  2)   1   90.000(  4)
   4   3  Cl    2   1.497673(  3)   3   90.000(  5)   1  180.000(  6)   0
 ------------------------------------------------------------------------
  Exact polarizability:  14.928    .000  14.928    .000    .000  53.183
 Approx polarizability:  15.126    .000  15.126    .000    .000  79.255
 Rotating electric field derivatives to standard orientation.
 Rotating nuclear coordinate derivatives to standard orientation.
 Full mass-weighted force constant matrix:
 Low frequencies ----1517.7789-1517.7789-1094.5308    -.0078    -.0064    -.0064
 Low frequencies ---    7.4380    7.4380  352.9932
 ******    3 imaginary frequencies (negative signs) ****** 
 Harmonic frequencies (cm**-1), IR intensities (KM/Mole),
 Raman scattering activities (A**4/AMU), Raman depolarization ratios,
 reduced masses (AMU), force constants (mDyne/A) and normal coordinates:
                     1                      2                      3
                    PI                     PI                     SG
 Frequencies -- -1517.7789             -1517.7789             -1094.5308
 Red. masses --     1.0219                 1.0219                 1.0219
 Frc consts  --     1.3870                 1.3870                  .7213
 IR Inten    --   460.3973               460.3973               446.2129
 Raman Activ --      .0000                  .0000                  .0000
 Depolar     --      .0000                  .0000                  .0000
 Atom AN      X      Y      Z        X      Y      Z        X      Y      Z
   1  17     -.01    .00    .00      .00   -.01    .00      .00    .00   -.01
   2   1     1.00    .00    .00      .00   1.00    .00      .00    .00   1.00
   3  17     -.01    .00    .00      .00   -.01    .00      .00    .00   -.01
                     4
                    SG
 Frequencies --   352.9932
 Red. masses --    34.9689
 Frc consts  --     2.5672
 IR Inten    --      .0000
 Raman Activ --      .0000
 Depolar     --      .0000
 Atom AN      X      Y      Z
   1  17      .00    .00    .71
   2   1      .00    .00    .00
   3  17      .00    .00   -.71

 -------------------
 - Thermochemistry -
 -------------------
 Temperature   298.150 Kelvin.  Pressure   1.00000 Atm.
 Atom  1 has atomic number 17 and mass  34.96885
 Atom  2 has atomic number  1 and mass   1.00783
 Atom  3 has atomic number 17 and mass  34.96885
 Molecular mass:    70.94553 amu.
 Principle axes and moments of inertia in atomic units:
                           1         2         3
     EIGENVALUES --      .00000 560.19741 560.19741
           X             .00000    .00000   1.00000
           Y             .00000   1.00000    .00000
           Z            1.00000    .00000    .00000
 THIS MOLECULE IS A PROLATE SYMMETRIC TOP.
 ROTATIONAL SYMMETRY NUMBER  1.
 WARNING-- ASSUMPTION OF CLASSICAL BEHAVIOR FOR ROTATION
           MAY CAUSE SIGNIFICANT ERROR
 ROTATIONAL TEMPERATURE (KELVIN)       .15461
 ROTATIONAL CONSTANT (GHZ)            3.221616
    3 IMAGINARY FREQUENCIES IGNORED.
 Zero-point vibrational energy       2111.4 (Joules/Mol)
                                     .50463 (Kcal/Mol)
 WARNING-- EXPLICIT CONSIDERATION OF   1 DEGREES OF FREEDOM AS
           VIBRATIONS MAY CAUSE SIGNIFICANT ERROR
 VIBRATIONAL TEMPERATURES:    507.88
          (KELVIN)
 
 Zero-point correction=                            .000804 (Hartree/Particle)
 Thermal correction to Energy=                     .003523
 Thermal correction to Enthalpy=                   .004467
 Thermal correction to Gibbs Free Energy=         -.023207
 Sum of electronic and zero-point Energies=           -920.925064
 Sum of electronic and thermal Energies=              -920.922346
 Sum of electronic and thermal Enthalpies=            -920.921402
 Sum of electronic and thermal Free Energies=         -920.949076
 
                         E                  CV                 S
                      KCAL/MOL        CAL/MOL-KELVIN    CAL/MOL-KELVIN
 TOTAL                    2.210              6.537             58.245
 ELECTRONIC                .000               .000              1.377
 TRANSLATIONAL             .889              2.981             38.695
 ROTATIONAL                .592              1.987             17.019
 VIBRATIONAL               .729              1.569              1.153
 VIBRATION  1              .729              1.569              1.153
                       Q            LOG10(Q)             LN(Q)
 TOTAL BOT        .472556D+11         10.674453         24.377872
 TOTAL V=0        .110751D+12         11.044347         25.229583
 VIB (BOT)        .521657D+00          -.282615          -.650746
 VIB (BOT)  1     .521657D+00          -.282615          -.650746
 VIB (V=0)        .122258D+01           .087278           .200965
 VIB (V=0)  1     .122258D+01           .087278           .200965
 ELECTRONIC       .200000D+01           .301030           .693147
 TRANSLATIONAL    .234881D+08          7.370848         16.972005
 ROTATIONAL       .192837D+04          3.285191          7.564431

 Internal  Forces:  Max      .000001377 RMS      .000000714
 1\1\GINC-IBM580_7\Freq\UB3LYP\6-31G(d,p)\Cl2H1(2)\IPCVSR\12-Jul-1996\1
 \\#UHF 6-31G** OPT=(EF,TS,CALCALL) B3LYP POP=FULL DEN=CURRENT\\clhcl-1
 .6 ang.\\0,2\Cl\H,1,R1\X,2,1.,1,90.\Cl,2,R2,3,90.,1,180.,0\\R1=1.49766
 611\R2=1.49767315\\Version=IBM-RS6000-G94RevC.2\State=2-SG\HF=-920.925
 8687\S2=0.757\S2-1=0.\S2A=0.75\RMSD=7.439e-09\RMSF=5.912e-07\Dipole=0.
 ,0.,0.0000045\DipoleDeriv=-0.3424881,0.,0.,0.,-0.3424881,0.,0.,0.,0.33
 71667,0.684973,0.,0.,0.,0.684973,0.,0.,0.,-0.6743387,-0.3424849,0.,0.,
 0.,-0.3424849,0.,0.,0.,0.3371721\Polar=14.9278148,0.,14.9278148,0.,0.,
 53.1825654\PG=C*V [C*(Cl1H1Cl1)]\NImag=3\\-0.02161655,0.,-0.02161655,0
 .,0.,0.07118866,0.04330625,0.,0.,-0.08661238,0.,0.04330625,0.,0.,-0.08

From [email-masked] Jul 27 11:24:34 1996
Date: Fri, 12 Jul 1996 17:12:21 +0100 (NFT)
From: Van Keer Annik <[email-masked]>
To: "Park, Tae-Yun" <[email-masked]>
Subject: transition state
Status: RO
Content-Length: 12566

On Wed, 10 Jul 1996, Park, Tae-Yun wrote:

> Dear all,
> 
> I've received a lot of helpful messages from many CCLers, 
> concerning my earlier questions on how to get the structure 
> of transition state of the reaction,
> 
> 
> 
>    H              H     H                       H H H
>    |               \   /                        | | |
>  H-C(+)       +     C=C     ---->(TS?)---->   H-C-C-C(+)
>    |               /   \                        | | |
>    H              H     H                       H H H
> 
>  gas-phase                                     gas-phase
>    methyl         ethylene                   primary propyl
> carbenium ion                                 carbenium ion
> 
> 
> ************************************************************** 
> 
> SADDLE DENSITY GRAPH XYZ CHARGE=1 PRECISE
> Me--->p methylation activated complex
> 
> C    0.000000  0    0.000000  0    0.000000  0    0   0   0
> C    8.471619  1    0.000000  0    0.000000  0    1   0   0
> C    1.325912  1  162.328125  1    0.000000  0    2   1   0
> H    1.108597  1   65.592545  1 -150.347214  1    1   2   3
> H    1.113968  1  120.001495  1 -151.401566  1    1   4   2
> H    1.108612  1  119.996246  1 -179.367538  1    1   4   5
> H    1.098282  1  122.722214  1   57.159393  1    2   3   1
> H    1.098328  1  122.715210  1  179.999634  1    2   3   7
> H    1.098267  1  122.715210  1   -0.316483  1    3   2   7
> H    1.098206  1  122.715210  1 -179.684021  1    3   2   7
> 0    0.000000  0    0.000000  0    0.000000  0    0   0   0
> C     .000000  0     .000000  0     .000000  0    0   0   0
> C    1.506600  1     .000000  0     .000000  0    1   0   0
> C    1.418947  1  116.688263  1     .000000  0    2   1   0
> H    1.118077  1  111.116249  1  -59.952415  1    1   2   3
> H    1.119762  1  109.783371  1 -179.449097  1    1   2   3
> H    1.118210  1  111.069687  1   61.123405  1    1   2   3
> H    1.150598  1  110.672989  1  174.990601  1    2   1   4
> H    1.150955  1  110.650314  1   64.874481  1    2   1   4
> H    1.109621  1  121.247902  1 -179.729874  1    3   2   1
> H    1.110257  1  121.556961  1     .300391  1    3   2   1
> 0     .000000  0     .000000  0     .000000  0    0   0   0
> 
,
Hello Park,

I donnot know if I can help you with your Mopac version because I am
a Gaussian user.  Firstly I do not understand why the bond between C1 and 
C2 is so large: 8.47 A? If you take 3 A or 3.5 A as distance between these 
bonds than you also have two reactants without any interaction. I guess 
that any change in geometry than  will earlier lead to a transition state 
for which I guess that the initially formed CC bond is about 1.8 A. 
Secondly, I do not know which optimization procedure you are using, but 
as I see two inputs one for the reactants and another one for the products I 
guess 
your program is looking for a TS via a LST path. In this case you have to 
identify two identically z-matrices, with of course different parameters, 
which you didn't.
 Thirdly, it is possible that the addition of CH3(+) to C2H4 is a barrier 
free proces. I know that the addition of triplet CH2 to C2H4 is barrierfree.
I hope that you can use my suggestions.

Good luck,

Annik

 ******************************************************************

 Annik Van Keer 
 Katholieke Universiteit Leuven
 Laboratorium voor Quantumchemie
 Celestijnenlaan 200F
 3001 Heverlee, Belgium

 Tel:00-32-16-327384
 Fax:00-32-16-327992
 E-mail:[email-masked]

-------------------------------------------------------
PART V
      Subject: Help!: IRC by mopac
-------------------------------------------------------
####################(questions)########################

Dear all,

I'm searching the structure of the transition state with 
mopac for the reaction type,

    olefin + carbenium ion ---> (TS?) ---> carbenium ion  (1)

My previous try for the reaction,

   H              H     H                       H H H
   |               \   /                        | | |
 H-C(+)       +     C=C     ---->(TS?)---->   H-C-C-C(+) 
   |               /   \                        | | |
   H              H     H                       H H H

 gas-phase                                     gas-phase
   methyl         ethylene                   primary propyl
carbenium ion                                 carbenium ion

has been found that there is no transition state, thanks to 
great help from many CCLers, since the gas-phase methyl 
carbenium ion is too unstable.

Actually I'm dealing with more than 100 reactions which have
the reaction type written in (1).  Therefore, I tried with 
other reaction, which contains more stable species, i.e.,

                    H  
                    |                      H
    H         H   H-C-H                   HCH H H H
  H | H        \   /                     H |  | | |  
 HC-C-CH   +    C=C    ---->(TS?)---->  HC-C--C-C-C-H
  H + H        /   \                     H |  | + |
              H     H                      H  H   H
secondary
  propyl      propene                   secondary 2-M
carbenium                              pentyle carbenium

Searching the TS in this reaction, I followed the suggestion
by many CCLers, i.e., try to get the TS structure by the 
following steps instead of using SADDLE calculation:

1. Try the reacion path calculation in mopac and save the 
   structure that contains the highest energy level.

2. Refine the structure at highest energy level with key 
   word, NLLSQ, or SIGMA, or TS.

3. FORCE calculation with the refined structure to check 
   if there is one and only one negative frequecy value.

4. Do IRC(=1 and -1) calculation to check if the refined 
   TS gives the reactants and product.

My result is that until step 3, everything seems OK.  Refinning
the TS by NLLSQ/SIGMA/TS gives no significant change of the TS
that obtained from step 1, probably because I used the key word
PRECISE in step 1.  The FORCE calculation showed that the refined
structure contains only one frequency values.

My problem is in the last step.  When I do the IRC calculation
and try to look at the result structure with "molden" software,
it gives exactly same as the structure of TS I input.

I wonder if (i) my TS is not correct, or (ii) the IRC calculation
doesn't give the molecular structure result from the calculation
so that molden only edit the structure I input.  In the latter
case, however, I don't know how to edit the molecluar structure
generated by IRC calculation.

Could anyone PLEASE help me how to confirm my TS structure by IRC?

I enclosed the input/result file at the end of this message.

Thank you very much in advance.

				Sincerely,

				     Park, TAE-YUN    

Data Files for the 4 steps.  Please note that the result file of 
step 1 is the input file for step 2... etc.

----------------------------------------------------------------
Step 1 input file:

CHARGE=1 PRECISE GRAPH DENSITY
r3_s-o3.dat

C    0.000000  0    0.000000  0    0.000000  0    0   0   0
C    1.437805  1    0.000000  0    0.000000  0    1   0   0
C    4.501968 -1   96.018692  1    0.000000  0    1   2   0
C    1.331116  1   87.080215  1 -155.227356  1    3   1   2
C    1.476334  1  124.262665  1 -135.173477  1    4   3   1
C    1.437424  1  123.306213  1  133.821869  1    1   2   3
H    1.135651  1  110.106583  1  123.092896  1    2   1   6
H    1.120895  1  114.528610  1    0.000000  1    2   1   6
H    1.135742  1  110.066360  1 -123.042191  1    2   1   6
H    1.115219  1  118.018661  1  -56.365555  1    1   2   7
H    1.097427  1  122.281586  1 -179.999634  1    3   4   5
H    1.097809  1  122.792145  1    0.000000  1    3   4   5
H    1.103470  1  120.865265  1    0.000000  1    4   3  11
H    1.118958  1  110.075104  1 -120.477097  1    5   4   3
H    1.119080  1  110.076859  1  120.510315  1    5   4   3
H    1.117584  1  111.867340  1    0.000000  1    5   4   3
H    1.121948  1  113.719040  1 -179.999634  1    6   1   2
H    1.134903  1  110.648621  1   57.360474  1    6   1   2
H    1.134933  1  110.636383  1  -57.367462  1    6   1   2
0    0.000000  0    0.000000  0    0.000000  0    0   0   0
4.0 3.5 3.0 2.8 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.55

----------------------------------------------------------------
Step 2 input file(for NLLSQ):

NLLSQ CHARGE=1 PRECISE DENSITY GRAPH
Optimizing ts

 C     .000000 0     .000000 0     .000000 0       0    0    0
 C    1.493971 1     .000000 0     .000000 0       1    0    0
 C    2.400000 1   99.829562 1     .000000 0       1    2    0
 C    1.366739 1  107.774987 1 -156.603212 1       3    1    2
 C    1.495494 1  127.295132 1  -95.474920 1       4    3    1
 C    1.493103 1  122.565420 1  111.465448 1       1    2    3
 H    1.108922 1  112.541937 1  169.169618 1       2    1    6
 H    1.109391 1  113.333661 1   45.130877 1       2    1    6
 H    1.118316 1  107.399773 1  -73.220364 1       2    1    6
 H    1.098555 1  116.705711 1  -32.527608 1       1    2    7
 H    1.092716 1  119.490590 1  165.857707 1       3    4    5
 H    1.091757 1  122.354434 1    9.374023 1       3    4    5
 H    1.098702 1  118.584661 1  -13.736600 1       4    3   11
 H    1.111796 1  109.120250 1 -115.208112 1       5    4    3
 H    1.110446 1  110.429073 1  126.570168 1       5    4    3
 H    1.107787 1  113.319142 1    4.958871 1       5    4    3
 H    1.108134 1  112.416348 1 -172.721816 1       6    1    2
 H    1.118042 1  107.755456 1   69.492728 1       6    1    2
 H    1.109678 1  113.288218 1  -48.800221 1       6    1    2

----------------------------------------------------------------
Step 3 input file (for NLLSQ):

FORCE CHARGE=1 DENSITY GRAPH ISOTOPE
TS checking by FORCE calculation

 C     .000000 0     .000000 0     .000000 0       0    0    0
 C    1.494700 1     .000000 0     .000000 0       1    0    0
 C    1.493679 1  122.365784 1     .000000 0       2    1    0
 C    2.380560 1  102.504395 1  110.467232 1       2    3    1
 C    1.368080 1  107.971291 1   74.388344 1       4    2    3
 C    1.495277 1  127.264130 1  -95.821228 1       5    4    2
 H    1.108898 1  112.496826 1  169.901352 1       1    2    3
 H    1.109529 1  113.296120 1   46.004711 1       1    2    3
 H    1.117890 1  107.496994 1  -72.487534 1       1    2    3
 H    1.098734 1   86.146576 1  -42.602459 1       2    4    5
 H    1.092931 1  119.365471 1  165.091675 1       4    5    6
 H    1.092000 1  122.211273 1    9.421019 1       4    5    6
 H    1.098602 1  118.611084 1   84.593826 1       5    4    2
 H    1.111888 1  109.074448 1 -114.666496 1       6    5    4
 H    1.110686 1  110.417656 1  127.198174 1       6    5    4
 H    1.107639 1  113.346970 1    5.549870 1       6    5    4
 H    1.108230 1  112.441048 1 -172.787689 1       3    2    1
 H    1.117836 1  107.759140 1   69.464531 1       3    2    1
 H    1.109580 1  113.282127 1  -48.948452 1       3    2    1

-----------------------------------------------------------------
Step 4 input file (for NLLSQ and for backward reaction, i.e, IRC=-1)
       This step is restarted immediately after the step 3.

IRC=-1 RESTART CHARGE=1 DENSITY GRAPH ISOTOPE
TS checking by IRC calculation

 C     .000000 0     .000000 0     .000000 0       0    0    0
 C    1.494700 1     .000000 0     .000000 0       1    0    0
 C    1.493679 1  122.365784 1     .000000 0       2    1    0
 C    2.380560 1  102.504395 1  110.467232 1       2    3    1
 C    1.368080 1  107.971291 1   74.388344 1       4    2    3
 C    1.495277 1  127.264130 1  -95.821228 1       5    4    2
 H    1.108898 1  112.496826 1  169.901352 1       1    2    3
 H    1.109529 1  113.296120 1   46.004711 1       1    2    3
 H    1.117890 1  107.496994 1  -72.487534 1       1    2    3
 H    1.098734 1   86.146576 1  -42.602459 1       2    4    5
 H    1.092931 1  119.365471 1  165.091675 1       4    5    6
 H    1.092000 1  122.211273 1    9.421019 1       4    5    6
 H    1.098602 1  118.611084 1   84.593826 1       5    4    2
 H    1.111888 1  109.074448 1 -114.666496 1       6    5    4
 H    1.110686 1  110.417656 1  127.198174 1       6    5    4
 H    1.107639 1  113.346970 1    5.549870 1       6    5    4
 H    1.108230 1  112.441048 1 -172.787689 1       3    2    1
 H    1.117836 1  107.759140 1   69.464531 1       3    2    1
 H    1.109580 1  113.282127 1  -48.948452 1       3    2    1

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