Q15-7P - Organic Chemistry (Mcmurry)

Question

 Cyclooctatetraene readily reacts with potassium metal to form the stable cyclooctatetraene dianion, ${\mathbf{\text{C}}}_{\mathbf{8}}{{\mathbf{\text{H}}}_{\mathbf{8}}}^{\mathbf{2}\mathbf{-}}$. Why do you suppose this reaction occurs so easily? What geometry do you expect for the cyclooctatetraene dianion?

Step-by-Step Solution

Answer

Cyclooctatetraene gains two electrons to become an anion because its anion is aromatic in nature and hence, more stable than its neutral form.

Step 1:4n+2 rule

It is the rule to check the aromaticity of the compound by knowing the number of delocalized electrons like in cyclodecapenatnene has 6 double bonds which means the then electron and n=2 so that the system is aromatic because n is a whole number.

$\begin{array}{c}\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{6}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{6}\mathbf{-}\mathbf{2}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{4}\\ \mathbf{n}\mathbf{=}\mathbf{1}\end{array}$

Step 2: Non-aromatic nature of cyclooctatetraene

Cyclooctatetraene is have four double bonds that it has 8 pi electrons which show delocalization so as per the 4n+2 rule n value is in a fraction which is it is anti-aromatic in nature is less stable than the aromatic nature.

$\begin{array}{c}4n+2=8\\ 4n=8-2\\ 4n=6\\ n=\frac{2}{3}\end{array}$

Step 3: Aromatic nature of cyclooctatetraene anion

After gaining the two electrons the cyclooctatetraene has 10 electrons and now as per the 4n+2 rule the n=2 that’s a whole number so that the anion is aromatic in nature and is much more stable than the anion form as shown: