Organic Chemistry 2 · Aromaticity and cyclic conjugation
Cyclooctatetraene + 2 K: the COT dianion
Conditions: Cyclooctatetraene, 2 equiv K metal, THF
Draw this problem yourself
Drag from the electrons that move to where they go, or tap one and then the other. Every arrow is checked; Hint helps when you're stuck. No account needed.
The question
What forms when tub-shaped cyclooctatetraene takes up two electrons from potassium?
Reveal the answer and explanation
Answer and explanation
Correct answer A planar, aromatic dianion with 10 π electrons and equal C–C bonds
Correct: 10 is a 4n+2 number.
Not this one A tub-shaped, nonaromatic dianion with alternating bonds
With 10 π electrons, planarity is rewarded, so the ring flattens.
Not this one A planar antiaromatic dianion with 8 π electrons
Two added electrons give 10, not 8.
Not this one Cyclooctane, from complete reduction
Potassium transfers electrons; it does not add H2.
Worked solution
Each potassium atom gives one electron, so the ring gains two electrons and holds 10 π electrons (4n+2, n = 2). The dianion flattens to a planar regular octagon with equal C–C bonds, because planarity now gains aromatic stabilization instead of antiaromatic destabilization. This explains why the reduction occurs easily.
What you'll practice
Predict the structure of the product formed when cyclooctatetraene is reduced by potassium.
You just pushed every arrow of this one.
Every one of the 5349 Orgo 1 and Orgo 2 problems works like this: feedback on each arrow, hints when you're stuck, a no-hint quiz mode. Free with an account, no card.
Stuck on your own homework?
Snap it — first scan free, no sign-up
Scanning every week? Scan Standard: for about problems