Secrets of a university tutor: dissection of a reaction mechanism.

Its a bit like a jigsaw puzzle in reverse, finding out to disassemble a chemical reaction into the pieces it is made from, and learning the rules that such reaction jigsaws follow. The following takes about 45-50 minutes to follow through with a group of students.

The problem is initially posed as the above (ignore the wavy bonds for now). The challenge is to identify the basic components that the reaction is built from and the rules these follow. It can be usefully salami-sliced as follows

  1. You are told the puzzle may consist of one or more (consecutive) pericyclic reactions. This should load up in your mind (from lecture notes) the various basic types of such reactions (the basic shapes of the jigsaw puzzle if you like).
  2. Rules from other areas of chemistry may be needed. Thus from your knowledge of the chemistry of benzene and its aromaticity, you need to remind yourself that there are two resonance forms (the Kekule forms) which are entirely equivalent. Problems such as the above may however be posed using either one or both of these forms. We will find out if this matters or not shortly.
  3. We need to clearly identify exactly what changes when the reaction occurs. To do this, it is useful to number what you think might be the key atoms.
  4. Notice that some atoms are not numbered. It keeps things simple, but in fact numbering them all will not do any damage. The atoms not numbered are the methyl groups (it does seem as if they emerge from the reaction unchanged) and the benzo group on the left. Only time will tell if this scheme needs changing.
  5. And now we are in a position to create a checklist of changes that occur during the reaction.
    1. A σ-bond between 1-6 clearly forms
    2. A π-bond between 5-6 decreases to a σ
    3. The π-bonds in the (un-numbered) benzo group rotate. We recognise this as a benzene resonance rather than a (pericyclic) reaction.
    4. And now for the elephant in the room, the atoms that we (as chemists) know are there, but which are not explicitly shown. These are the hydrogens. We know a rule for this, which is that any structure shown without hydrogens is assumed to have as many attached as are required to achieve a four valent carbon. This is in fact a fuzzy rule, because some carbons can be divalent (carbenes) and some trivalent (carbocations). Normally the former have a : glyph appended to them, and the latter a + charge, and we can see neither here so our rule stands. Time to count the elephants, and to draw the significant hydrogens explicitly (drawing them all would only clutter). We only select those hydrogens that appear to have moved during the reaction. Thus:
    5. A σ-bond between 5-7 clearly forms
    6. A σ-bond between 1-7 clearly breaks
  6. We have four significant bonds that change, 1-6, 5-6, 5-7 and 1-7. The task now is to partition them into groups that might correspond to one of the basic types of pericyclic reaction, and these tend to be defined by how many σ-bonds make or break during the reaction
    1. Thus an electrocyclic reaction either forms or breaks just one σ-bond
    2. A cycloaddition forms two (or more) σ-bonds and its reverse, a cyclo-elimination breaks two (or more) σ-bonds
    3. A sigmatropic reaction forms one σ-bond and breaks another.
    4. Ene reactions break at least one σ-bond and form at least one other, but in unequal numbers that distinguish them from a sigmatropic reaction.
  7. Juggling with these pieces soon reveals that items 5.5 and 5.6 above can comprise a sigmatropic reaction, and that item 5.1 above constitutes an electrocyclic reaction. Item 5.2 above, involving only a π-bond is not counted.
  8. The next task is to decide which comes first! To do this, we need to again recollect carbon tetravalency, and the sacrosanct need not to exceed it. Clearly forming the 1-6 bond as our first action would violate this rule by creating a pentavalent carbon atom. So this leaves 5-7/1-7 as our first action, which is going to be a sigma tropic reaction.
  9. We might recognise at this point that 5-7/1-7 share a common atom (7). We can probably pencil in that this sigma tropic reaction is going to be of the type [1,?] from this observation. From the numbering above (which in fact was deliberately chosen to achieve this effect) we infer that hydrogen 7 moves along a chain of 5 carbon atoms, and so our nomenclature is complete; it is going to be a [1,5] hydrogen migration or sigmatropic shift. Had the numbering been different, we would have had to spot that the non-common bonds differed by five atoms.
  10. The arrow pushing to achieve this transformation is shown below. Notice that the arrows rotate anti-clockwise. It is a feature of pericyclic reactions that it does not matter which clock-direction they rotate in (mostly). Hence pushing them the other way would achieve exactly the same result.
  11. This brings a surprise; we needed five arrows, or ten electrons. Is that a unique solution? Well no. Had we remembered point 5.3 above, then another initial resonance form for the benzo-ring is possible, and this form requires us to push only three arrows, or six electrons.
  12. Is there a common factor between 6 and 10 electrons? Yes, it is the famous Hückel aromaticity 4n+2 rule, for which n =1 or 2. So we get the result we really wanted, which is does not matter which of the two resonance forms for the benzo group we start with, we end up with arrow pushing that either way merely conforms to the 4n+2 rule. In other words, the transition state for this first reaction is aromatic. The stereochemistry implied by this result is going to be deferred to a second tutorial on this topic (and this is where the wavy lines will also come in).
  13. There is another observation we can make. The product of the [1,5] sigmatropic hydrogen shift no longer carries an aromatic ring on the left. We might infer that it will only be a transient intermediate, and will be very inclined to restore the aromaticity at the first opportunity.
  14. We are now in a position to create the 1-6 bond without violating the valency of either atom.
  15. The arrows shown above are two (black) to which can be followed either one more (green) or three more (red), making two possibilities carrying either 6 or 10 electrons. Again, both conform to the 4n+2 rule and so it does not matter which set is followed; the electrocyclic reaction will have an aromatic transition state (again we ignore stereochemistry for the time being).
  16. And hey, we have also recovered the aromaticity of our benzo group on the left.
Well, it is now time to finish up this first tutorial on the topic. In the follow up, I will show these aromatic transition state I have referred to here, and also include discussion of the stereochemistry.

 

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3 Responses to “Secrets of a university tutor: dissection of a reaction mechanism.”

  1. Nathanael says:

    Is it not possible that the reaction proceed by the following mechanism in a kind of SN2 hydride shift?

    http://imageshack.us/photo/my-images/21/mechanism.png/

    Or would this not be allowed because of symmetry, evidence from stereochemistry or the high activation energy necessary for the double bond to go across such a large distance?

    Regards,
    Nathanael Hsueh

  2. Henry Rzepa says:

    Its an interesting suggestion. Nominally, it would be classified as a 4-electron Ene reaction, and 4n reactions would require an antarafacial component. The alternative suprafacial 4n+2 [1,5] shift is probably going to be lower in energy. But worth checking by calculation!

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