How many atoms share the unpaired electron in semiquinone?

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Hey everyone, let's do this problem. It says determine how many atoms share the free radical electron in the given species. And we have a fin oxy radical which is a benzene ring attached to an oxygen atom and that oxygen has the radical in this given structure. So how do we determine how many atoms share the free radical electron? Well, we can draw the resonance structures and then simply count all of the atoms that held the radical. Okay. So a few things to remember about drawing resonance forms is that first, we can simply draw the Lewis structure and that will be our first residents form and they already gave one to us. Then we want to only move the lone pairs or pi electrons, never the sigma electrons and never Adams, right? Then those electrons can only move to certain places. We only want to move them to sp or sp two atoms, never sp three atoms. And lastly the total number of electrons and therefore the net charge is always going to remain. All right. So another important thing, this is sort of a special situation because we have a radical, right? So in a usual or more common resonance form question. When we're moving a pair of electrons, we use a double headed arrow, right? But since we're using or moving a radical, we use 31 headed arrows. So the one headed arrow represents one electron, right. So let's apply this to our problem. So I'm going to first redraw the fin oxy radical that was given to us. So we have our benzene ring, our oxygen with the radical and we want to move our electron are radical to our S P 123 sp two carbon towards that pi electron or towards the pi bond rather. So we're going to break this pi bond, one electron will go towards that carbon oxygen bond and one electron will go in the other direction towards the next carbon in the benzene ring to create a new radical. So now we have a carbonell group there and our new radical is now in the ring. So continuing this, we take the neighboring pi bond break that with two arrows going in either direction. One two, that's not how you draw an arrow. And then the radical will come and meet that pi electron to form a new double bond. Okay. Drawing what stayed the same first and then drawing our new double bond in our new radical. So same thing our pi electrons are moving counterclockwise towards the radical. Well, one of them is and the radical is moving clockwise To meet that one pi electron. And we basically do this until we can't do it anymore. So we can do it one more time because we're back up at the oxygen and the pi electrons from our new carbon group, we'll meet with that radical sending the other pi electron back onto that oxygen. So it looks similar to the first structure, except our pi electrons have rotated around the ring to give us a different structure. Okay. So the question asked how many atoms share the free radical electron? So let's count the oxygen is one atom. Let's name these carbons going around the ring. Well, carbon one, I can actually write these. So carbon 123456, starting with carbon one would be the carbon that has the oxygen atom. So the radical goes from the oxygen, two, carbon 22, carbon 42 carbon six and then back to the oxygen. So since the oxygen is doing it has the radical two times, but that's the same atom. So we'd have 1234 original atoms holding the radical. So the correct answer is B four atoms. That's it for this problem. I hope this video is helpful and thanks for watching.

How many atoms share the unpaired electron in semiquinone?

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