Skip to main content
Ch.9 - Chemical Bonding I: The Lewis Model

Chapter 9, Problem 0.89

Carbon ring structures are common in organic chemistry. Draw a Lewis structure for each carbon ring structure, including any necessary resonance structures. d. C6H6

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
1248
views
Was this helpful?

Video transcript

Hello everyone. So in this video we're trying to provide resonant structures for R. C. Six H 50. H molecule. So that is is so the structure for C six H 50. H. Is going to be where we have a ring, six member ring with alternating pi bonds and then one of the carbons is going to have this O. H. Group attached. Of course we need to put the pears as well. Alright, so this is the structure and this structure could also be one of the resonant structures. So let's go ahead and start that then. Alright, of course, putting in by resonance structure arrows. So the additional resident forms, we can draw this by creating a double bond with oxygen using its long pairs and just alternating the carbon carbon double bond. So going ahead to draw my arrow, pushing are my lecture on pushing arrows. What I'm gonna do is go ahead and use one of the long pairs to create a pi bond. And then one of these pi bonds will go ahead and form a bomb pair on this carbon. So drawing that structure out then again, we have the ring, the left pipe on does not the bottom right bond does not move. We have created a double bond oxygen with just one long period. Now giving this a plus one for more charge. And this carbon right over here that we have placed the lone pair on. Well now have a negative one formal charge. So the next right in the structure again, using my arrow pushing, we're basically going to use this long pair to create a pipeline. And this pipeline breaks to create a long pair. Again, we have our six members ring stone. This pipe on does not move. We still have our double bond oxygen With one long pair giving this a plus one formal charge. Then we have brought this one pair down to create a pi bond. Let's go ahead and put that pi bond. And now this carbon will have a long pair giving a negative one formal charge and basically doing the same exact thing. We're gonna go ahead and move this little pair to create a pi bond. This pipe on bricks to create a long pair on that carbon, Join my six member ring here. So we have a double volunteer double volunteer oxygen does not change Plus one formal charge. Long pair goes on this carbon giving that carbon a negative one formal charge. And that's the end of our resonant structures. So, the problem asking us the reason structures for this molecule and here it is, all of this is going to be my final answer for this problem. Thank you all so much for watching