Posted by: Mark Foreman | January 31, 2010

Alkyl halides II

More on alkyl halides 

To help us explore this idea of the steric effect we will consider the reaction of dimethyl sulphide (smells like decaying cabbage) with a series of alkyl iodides. The first reaction will be of methyl iodide with this sulphur compound. I have used molecular mechanics to get the shape of the transition state. Molecular mechanics is a computer simulation which is a POSH version of a lot of springs and metal balls; it gets us a shape based on steric and other effects due to sigma bonds only. 

TS of methyl iodide and dimethyl sulfide

 

For the transition state we get the following shape, the S-C distance is 2.044 Å and the C-I distance is 2.441 Å. The hydrogens of the methyl are all 2.063 Å away from each other. In terms of steric effects this is a nice low energy transition state. 

If any of you want to look at the TS for yourselves then here are the X, Y and Z coordinates of the atoms. The most easy way I have found to use a set of XYZ coordinates is to make a XYZ file and read it into ORTEP. For academic use ORTEP is free of charge and can be obtained from Dr Louis Farrugia’s web site at Glasgow.(http://www.chem.gla.ac.uk/~louis/software/). You will need a free license to run ORTEP but I hope it will give you years of viewing pleasure. I will sometimes be posting XYZ files for you to use. 

14
XYZ file : TS of Me2S with CH3I
C   -1.382701    1.252459   -0.121315
S   -0.551204   -0.191540    0.611947
C   -0.764212   -1.640372   -0.468829
C    1.446250    0.239636    0.579932
I    3.831916    0.755882    0.541505
H   -2.477383    1.060969   -0.176849
H   -0.987713    1.431410   -1.146253
H   -1.196854    2.150868    0.508484
H   -1.846066   -1.888612   -0.548902
H   -0.362028   -1.415250   -1.481816
H   -0.217533   -2.509285   -0.039029
H    1.739675   -0.822180    1.035750
H    1.455292    0.400913   -0.600856
H    1.312561    1.175102    1.306232

When we change the hydrogens of the methyl iodide to make a tert butyl iodide and repeat the calculations we get a very different answer. Now the C-S distance is 2.077 Å and the C-I distance is now 2.511 Å. This might seem a large change but only a small change in bond length will have a great effect on the strength of a bond. The more important thing is that the C(CH3)3 group is now being distorted, the C-C distance in the group is now 1.598 Å which compares with 1.531 Å for the C-C single bond in ethane as calculated by the same MM software. It is clear to me that the C-C bonds in the C(CH3)3 unit are being elongated to keep the methyls (CH3 groups) from bumping into each other. 

TS of the impossible reaction of tert-butyl iodide and dimethyl sulphide

 

To me the elongation of the bonds suggests that this second transition state will have a very high energy. The fact that it has a high energy will mean when molecules hit each other in solution it will be unlikely that they will have the energy required to get past the energy barrier and enter the transition state. 

For those who want a look for themselves 

23
XYZ file : Me2S CMe3 I TS
C   -1.854561   -1.451604   -0.424704  
S   -1.084095    0.194682   -0.447982
C   -1.836715    1.302038    0.781240   
C    0.960817    0.075309   -0.105918
C    1.157037    1.604670   -0.568131  
C    0.728951   -0.487079    1.386211
C    1.203718   -0.953960   -1.303625 
I    3.397720    0.062724    0.501394  
H   -2.932311   -1.337783   -0.677799  
H   -1.785291   -1.935530    0.572231
H   -1.387844   -2.114089   -1.185302
H   -2.917970    1.416253    0.545011 
H   -1.742270    0.893214    1.809963 
H   -1.354281    2.303036    0.740225 
H    0.371070    2.043956   -1.218267 
H    2.040184    1.762548   -1.225576 
H    1.233939    2.271848    0.321716
H   -0.269531   -0.871765    1.663699
H    0.926289    0.309391    2.141291  
H    1.326963   -1.398211    1.611703 
H    1.114791   -2.001984   -0.936390
H    0.517158   -0.829588   -2.169211
H    2.186232   -0.858076   -1.811779 

Which means for the tert-butyl iodide no SN2 reaction will occur. We will come back to tert-butyl iodide later and see what happens if we try to do a SN2 reaction with it. But in the mean time lets go back to the reaction of methyl iodide and dimethyl sulphide. The organic product of the reaction is trimethyl sulphonium iodide. This is a polyatomic cation which is made of three methyls bonded to a sulphur atom. The charge on the sulphur is predicted by extended Huckel calculations to be plus 1.18. Below is shown a picture of the cation. The more red an atom the more positive it is and also the more blue an atom the more negative it is. 

  

Me3S cation

 

If we compare this with the dimethyl sulphide, in this molecule the sulphur is predicted to have a charge of 0.09. So it is clear that the model predicts that the positive charge at the sulphur atom increases as a result of the SN2 reaction. Quite reasonable ! 

Dimethyl sulphide

 

When the same calculations are done for methyl iodide we get a charge of 0.037 on the iodine, we know that in the final product that the charge on the iodine becomes -1. This change of charge is perfectly reasonable. 

Let us now look at the charges on the atoms in the transition state. We now find that the sulphur has a charge of 0.92 on it while the iodine has a charge of -0.74 on it. It clearly appears that the atomic charges are partway between those on the atoms at the start and those on the atoms at the end. Great ! The big take home message is the the SN2 transition state is the highest energy point in a one step reaction which converts reactants into products. 

The transition state showing the charges

 

While we are here we should look at a series of methyl halides. 

Methyl fluoride

Atom Charge
   
C 0.168375
F -0.226034
H 0.0192209
H 0.0192197
H 0.0192181

 Methyl chloride

Atom Charge
   
C -0.0293756
Cl -0.0657289
H 0.0316952
H 0.0317014
H 0.0317079

 Methyl bromide

Atom Charge
   
C -0.0734434
Br -0.0318401
H 0.0350852
H 0.0350963
H 0.035102

 Methyl iodide

Atom Charge
   
C -0.15373
I 0.0374188
H 0.0387644
H 0.0387702
H 0.0387765

Looking at the numbers in the tables you should notice that as the halogen becomes more heavy, it is less negative. This is due to the fact that the later halogens are less electronegative than the early ones. As the atomic number of the halogen increases the electronegativity goes down. As the halogens become less electronegative the carbon becomes less positive. I mean less positive in terms of charge not in terms of outlook on life !

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