# Application of the Woodward-Hoffman rules to a [14+2] cycloaddition

I am just doing some pericyclic practice problems and I am just trying to understand the Woodward Hoffman rules.

The attached picture is a problem I attempted that was checked against the answers but I am confused how the symmetry is allowed.

When I add the (4n + 2) and 4n together I get 4. Doesn't n have to be an integer and the sum an odd number for the symmetry to be allowed?

EDIT: I copied the answer below with incorrect stereochemistry. Please refer to the second image for the correct stereochemistry EDIT: Correct stereochemistry below

Before getting into the actual Woodward Hoffman 'analysis', one can quickly check any proposed cycloaddition reaction against the following table of general results.

The table is essentially an overview of the Woodward Hoffman rules that doesn't require any consideration of orbitals/orbital symmetry. It can often be useful to check against a table like this to ensure you've done the real Woodward Hoffman analysis correctly.

\begin{array}{ccc} \hline {\text{Number of }\\\text{Electron Pairs}} & \text{Photochemical} & \text{Thermal}\\[2ex]\hline \text{even}~(4n) & \text{suprafacial-suprafacial} & \text{suprafacial-antarafacial} \\[1ex] \text{odd}~(4n + 2) & \text{suprafacial-antarafacial} & \text{suprafacial-suprafacial} \\ \hline \end{array}

From your example, we can count $14 + 2$ electrons in the π system, giving a total of $16$ which is a $4n$ integer (where $n = 3$). We also know from the reaction scheme given that the cycloaddition is taking place under photochemical conditions. We can, therefore, conclude that the reaction is allowed photochemically if both components react suprafacially.

### The actual Woodward Hoffman analysis

Note: I'm doing this for the thermal $[14 + 2]$, firstly because its a little more interesting in terms of orbital overlap, but secondly to give you the opportunity to do the analysis properly for your problem. The approach taken here is the one taken by Ian Fleming in his Molecular Orbitals and Organic Chemical Reactions book.

First: Draw a diagram of the reaction showing the orbitals involved. In this case we are interested in the π system of the heptafulvarene and the π system of the tetracyanoethene.

When drawing a diagram, no orbital phases are needed- this is already taken into account by the generalised Woodward Hoffman rule. (Also note that its not important how you define the components, the 14π system could be considered as a series of seven 2π systems).

Second: Consider how each component is reacting. In a cycloaddition, each component may either react suprafacially (the reaction takes place on the same side of the π system) or antarafacially (the reaction takes place from opposite sides of the π system) (also note that although this is theoretically possible, it is geometrically impossible in some cases due to the inability to get into the correct orientation for orbital overlap).

Third: Apply the Woodward Hoffman rule, as given below.

A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

In the example above, the 14π component is taken to be reacting antarafacially (denoted π14a), and the 2π component suprafacially (denoted π2s). When the Woodward-Hoffman rule is applied, this gives an odd number of components, making the reaction thermally allowed.

We could, however, also have drawn the reaction as having the 14π component reacting suprafacially (denoted π14s), with side on overlap of the 2π component reacting antarafacially (denoted π2a). Both are equally valid.

From that,you should now be able to consider your photochemical reaction. One slight difference exists in the formulation of the Woodward-Hoffman rules for photochemical reactions, taking account of the excited state that a component exists in. Namely, the total is now even, not odd, for an allowed reaction.

A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is even.