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Draw the arrows to show the two main propagation steps in the chlorination of propane to produce 2- chloropropane. Be sure to use the appropriate "fish-hook&qu...

Question

Draw the arrows to show the two main propagation steps in the chlorination of propane to produce 2- chloropropane. Be sure to use the appropriate "fish-hook" arrows in the appropriate patterns_

Draw the arrows to show the two main propagation steps in the chlorination of propane to produce 2- chloropropane. Be sure to use the appropriate "fish-hook" arrows in the appropriate patterns_



Answers

Draw curved arrows to show the flow of electrons responsible for the conversion of the reactants into the products:

So we were just considering a mechanism. So I've drawn out my starting materials and products and we will just be adding hours and then identifying or electric file on nuclear file. So here's my formal negative charge. So that means that my oxygen could be my nuclear file and it is going to attack a proton, which is my electric file. So I haven't our push in the direction where my electrons are going. We have the formal collapse here, and then we have a brew mean substitue in here that is forced to leave. Answer than this generates is Walter. We've got our AL Keen, and then we've got opera ammonium ion as a result.

So here we are, just looking at some residents structure. We have H C 03 miners, so I've drawn up some Lewis structures already on the screen. As you can see, I've included all of my single double bonds as well as the full display off atoms involved, as well as our lone pairs on formal charges. So this armor we have in center represents the transition between two residents states on what we're going to do here is draw arrows in order to link the two states that I've drawn up on the screen. So we take our loan per on one oxygen, and then we dump that electron density onto that single bond and then as a result of that that causes that are electron density within this double bond to knock up onto our oxygen. And so our loan per on formal negative charge is then sat on opposite oxygen. And now we have the double bond president where it used to be a single fund. So again, we've got this double headed arrow which represents the relationship between the two and then remember to use our Carly headed arrows where the arrowhead is pointing in the direction of where our electron density is moving towards

Hi, everyone. I hope you're doing well. My name is that so? We help you with problem to off Chapter 23 Carbon. Your conversation reactions from it. Organic chemistry textbook. So we're going to try to solve the push Aero pushing mechanism for four hydroxy for method to Penton on in base. We called a retro Aldo reaction and basically, what we're gonna do, more or less is the reverse of a carbon new condensation. So start with the whiteboard. You can see the molecule that would be pushed drawn here, but a baseball Just use a B with a negative charge. So this base, this negative church is going to attack the hydrogen on our hydroxy group here and then leaving the O with a negative charges. Next thing that will happen is that this oxygen is gonna form a double bond and that formation will push this bond back onto this side, pushing the double bonded electrons back onto this oxygen That's breaking this molecule into two different ones. As you can see here, this will form one of the two outdoors that's supposed to be in a product and this right here, the only form of this other now, in order to find finish, this problem well happens that we'll have the base bond to hydrogen and the negative charge. This oxygen is gonna push back down. And it's gonna pushes electrons here to grab this hydrogen from the base. And that will result in our second. Although and here at two elder equipments I want to see the mechanism again. Here it is. Thank you for listening and have a great day.

So in this podcast, we're taking a look at the flow of electrons to generate our products. So we'll draw out the first one. Uh huh. And so what we find is that the electrons move in the direction off the arrowhead. So we form a carbon. I'll on We have a leaving group as represented by those two arrows. In the next example. We're looking at a name ID. We have ch three C triple bond C H. We also have an AL kind. Then we have N h to the lone pair. And so what we attack is this hydrogen to this loan? Paris Attacking or hydrogen? We're shifting electron density onto that carbon. In the next example, we have aged three C h two c b r. This is attacked by O. C. H three minus. And our lone pair, which is a source of electron density attacks, is carbon on then, bro. Mean is a leaving group


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