Diversion from paragraph 2B: examples for path or grid searches

  1. The Menshutkin reaction: NH3 + CH3Cl -> CH3NH3+ + Cl-

    A possible input file is:

     AM1 T=1000 STEP1=0.06 STEP2=0.06 NOINTER NOXYZ
     menshutkin reaction
    
     N              0.000000  0    0.000000  0    0.000000  0    0    0    0
     C              1.400250 -1    0.000000  0    0.000000  0    1    0    0
     H              1.117146  1  110.724835  1    0.000000  0    1    2    0
     H              1.117140  1  110.728184  1  120.007164  1    1    2    3
     H              1.117038  1   90.735924  1   59.998414  1    2    1    4
     H              1.117174  1   90.707078  1 -179.995793  1    2    1    4
     H              1.117173  1   90.708093  1  -60.006254  1    2    1    4
     H              1.000000  1  107.350748  1  179.993734  1    1    2    5
     Cl             1.900000 -1   90.949851  1  181.532073  1    2    7    1
     0              0.000000  0    0.000000  0    0.000000  0    0    0    0
    
    
    This reaction, leading to charged species, is of course very sensitive to solvent effects.
    The grid search should be repeated with the keyword EPS=80.0, which will result in a totally different TS geometry.
    For clues, check Acc.Chem.Res.1996, 29, 303.

  2. Electrophilic aromatic substitution

    Presented here is an input file for the nitration of toluene at the para position:

     AM1 T=600  CHARGE=1
     nitration, ch3 group at c4
     
      C    0.0000000  0      0.000000  0      0.000000  0    0    0    0 
      C    1.4866445  1      0.000000  0      0.000000  0    1    0    0 
      C    1.3609792  1    121.785393  1      0.000000  0    2    1    0 
      C    1.4247329  1    121.366721  1      1.708758  1    3    2    1 
      C    1.4253078  1    118.933406  1      3.116765  1    4    3    2 
      C    1.3619835  1    121.561128  1     -1.691121  1    5    4    3 
      H    1.1523008  1    106.306450  1    109.399799  1    1    2    3 
      H    1.1081283  1    120.197983  1   -177.993070  1    3    2    1 
      H    1.1078886  1    118.454679  1    177.279576  1    5    4    3 
      H    1.1089069  1    121.515790  1    177.593819  1    6    5    4 
      N    1.5597307 -1    112.024479  1   -136.310653  1    1    2    3 
      H    1.1089687  1    117.148607  1   -178.690208  1    2    1    3 
      O    1.1882627  1    116.241220  1   -114.983546  1   11    1    2 
      O    1.1889221  1    117.350866  1     65.066008  1   11    1    2 
      C    1.4624542  1    120.713889  1   -179.007872  1    4    3    2 
      H    1.1208300  1    112.274336  1     22.155616  1   15    4    3 
      H    1.1300888  1    108.933195  1    -96.972421  1   15    4    3
      H    1.1227933  1    111.537208  1    145.154482  1   15    4    3  
    
     1.60  1.70  1.80  1.90  2.00  2.10  2.30  2.50  3.00  3.50  4.00  5.00
    
    Other substituents than methyl can be introduced by editing the last four lines, which describe the methyl group.
    For meta substitution, the atom path ( 4 3 2) for C15 should be changed to 5 4 3, and for the connected H's to 15 5 4.
    Don't forget to change the path for the hydrogen at C5 (H9) in the opposite way (so to 4 3 2). The values for the variables remain the same.

    The result will be a curve with a maximum, the TS, and a minimum at the other side of this hill. The HoF difference between these two points is a measure for the activation energy of the reaction.
    Comparing these values with the result for unsubstituted benzene reveals the substituent effects, which could be compared to a table with sigma+ values.

  3. Addition of HX to double bond, Markovnikov's rule.

    The first step in this reaction is the protonation of the double bond. In isobutene two products can be formed: the tert.butyl cation (CH3)3C+, and the primary cation (CH3)2CH-CH2+.
    In the following input file a proton is transferred from a water molecule, fixed at 6 A from the central C, towards this central C:

     AM1 T=600 CHARGE=1 NOINTER
     water fixed at 6A, H13 moving towards C2
     
      C    0.0000000  0      0.000000  0      0.000000  0    0    0    0 
      C    1.4835371  1      0.000000  0      0.000000  0    1    0    0  
      C    1.3361502  1    122.405331  1      0.000000  0    2    1    0    
      C    1.4832593  1    115.169374  1   -179.847320  1    2    1    3    
      H    1.1186971  1    110.066005  1    119.962428  1    1    2    3   
      H    1.1170753  1    111.613479  1     -0.481234  1    1    2    3    
      H    1.1187939  1    110.030834  1   -120.941510  1    1    2    3   
      H    1.0972207  1    122.420340  1      0.350476  1    3    2    1   
      H    1.0972191  1    122.432461  1   -179.844665  1    3    2    1   
      H    1.1188524  1    110.062024  1     58.717790  1    4    2    1   
      H    1.1187756  1    110.064097  1    -60.325243  1    4    2    1  
      H    1.1171908  1    111.497898  1    179.186134  1    4    2    1  
      H    4.0       -1     80.00      1     90.000     1    2    3    8
      O    6.0        0     90.000     0     90.00      0    2    3    8
      H    1.10       1    120.0       1     0.000      1   14    2    1
      H    1.10       1    120.000     1     0.000      1   14    2    4
    
    3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.1 2.0 1.9 1.8 1.7 1.60 1.5 1.4 1.3 1.2 1.12 
    
    The z-matrix is obtained from a mopac calculation on isobutene within Sybyl, the H3O+ is added manually.
    The result of this calculation is interesting: it is observed that the proton approaches the terminal carbon first, to form the more stable tert.butyl cation; in the end it is forced towards the central carbon with an increase in energy.
    This illustrates Markovnikov's rule. For an explanation as to why the tertiary cation is more stable, the MO's of the cations could be compared.

    Now let's move to allene: C=C=C
    For an electrophile there are two options: attack of the central carbon, or at the terminal carbon. The first would lead to an allyl cation, the second to a vinyl cation. The first is clearly more stable (why?).

    1. Construct the input file, starting from the file above
    2. Perform the calculation or
    3. Check the result from here.
    4. Why is the allyl cation not formed immediately?
    5. Explain the experimental observation (REFLIB search) that e.g. hydration in dilute sulfuric acid appears to occur via the vinyl cation, leading to an enol, and subsequently a keto group at the central carbon, while other additions yield the allyl-substituted product.
    Literature: March, Advanced Organic Chemistry, 3rd Ed., p. 675.


Back to paragraph 2B.