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Chemistry 2010 Activity Week 2. Build the following models:Where pi bonds are involved build one model with bowed bonds, buildanother using paddles.IUPAC NameChemic...

Question

Chemistry 2010 Activity Week 2. Build the following models:Where pi bonds are involved build one model with bowed bonds, buildanother using paddles.IUPAC NameChemical formula 2pts eaStructural formula 2pts eaLine Structure 2pts eaCondensed Structure 1pt eaMM show calculations 1pt ea.butanebut-2-enebut-1-ynebenzeneXXXXXXXXXXXcyclohexaneXXXXXXXXXXX

Chemistry 2010 Activity Week 2. Build the following models: Where pi bonds are involved build one model with bowed bonds, build another using paddles. IUPAC Name Chemical formula 2pts ea Structural formula 2pts ea Line Structure 2pts ea Condensed Structure 1pt ea MM show calculations 1pt ea. butane but-2-ene but-1-yne benzene XXXXXXXXXXX cyclohexane XXXXXXXXXXX



Answers

Draw all possible isomers having molecular formula C$_4$H$_8$ that contain one $\pi$ bond.

So here, a draw the extremist content, structural formula for the Alkins or the skills affording for a cycle can So for a propane. Here we have siege three connected to siege to connected. To see each three part B, we have X sane, which is ch three connective siege to make a dizzy age too connected to ch two connected ch two connected to ch three for part C, we have happened Siege three connected to siege One second. You know, we can actually condense this further ch two And of these, there will be five in the middle and then connected to a last CH three. Never for Partney cycle approach, propane or painting, it would just be a Pentagon.

So I've rear end out the equations for the ice immersion problem. Seven You were asked to draw the skeletal structures, starting with part a bridge over the aeons that I summers of these compounds. If I like to start by drawing the carbon to carbon double bond, we look at the carbon on the left. So I'll be this one of the double bond you have with two carbons involved in a double bond. Carbon on the left is attached to a hydrogen which I will draw, and then a two carbon chain. So 12 carbon on the right. Also, it has to a hydrogen and a one carbon chain. So since the two high priority groups here and here are on opposite sides of the double bond, this is E configuration. He also right in these air to hide regions and those air lower priority. I will not be driving hydrogen for the said configuration again, we're going to start with the carbon carbon double bond, keep one side the same and then changed the direction of another group. So this ch three group on the right put it facing down. So no, both of the high priority groups are on the same side of the double bond. So this is that configuration and our two hydrogen will be up there like that for her p. Again, I like to start with the carbon carbon double bonds. That's thes two carbons here, carbon on the left attached to a two carbon chain and also a chlorine carbon on the right attached to a hydrogen and also another two carbon sheen. So for the carbon on the left, the group with the highest party with the highest priority is the chlorine. Since it has a higher atomic number and the carbon on the right, the group of the highest priority will be that to carbon group and not the hydrogen was attached to. Since the carbon has a higher atomic number, it wasn't the high party groups from all sides of the double want this has he configuration to draw the side configuration. You will keep one side the same and once I difference will keep the left side the same. Then the groups on the right will switch their positions so the hydrogen will be the top. There are two carbonated will be at the bottom now are too high. Priority groups are on the same side of the double bond, and this house, that configuration for Part C again start with the carbon to carbon double bond. So these two carbons here carbon. The left has a four carbon chain, so I was like this one, too 34 And it's also attached to re to carbon chain carbon on the right. It's attached to a carbon, which is attached to a chlorine and also a carbon here that it attached to two other CH three groups So I would look like this. We look at the carbon on the left of the double bond. This will be the higher priority group, since the longer carbon chain and the group with Higher party on the right will be this group with the chlorine, since they're identical up until this carbon over this curb on the top is attached to to Harmon's in a hydrogen carbon. On the bottom is it have to two headed ins and chlorine in Florida's higher told me number, so it has a higher priority. So since the two high priority groups were on opposite sides of the double bond, this is your configuration and to draw Does that configuration. We will keep one side the same so to the left side and put the chlorine at the top in this group of the bottom. And now the high priority groups are on the same side of the double bond. So this is that configuration massacre party. This is a carbon carbon double bond. Draw that here. Carbon on the left is attached to a to carbon chain 12 and also in O H group and those attached to a carbon which is double bonded to an oxygen coming on the right is it has to a carbon that is triple bonded to another carbon and a carbon thought is it has should three ch three groups looking at the carbon on the left of the double bond, the group of the highest priority. You will look and see that both of them were touched. A carbon Suherman here in carbon here over the karma at the top here is born into to hide regions and another carbon group. The bottom is attached to one hydrogen and double wanted to an oxygen. So since the oxygen has a higher Thomas number than the carbon it will. This will be the high priority group on the left high priority group. On the right will be the carbon carbon triple bond, since triple bonds have higher party than the carbon in carbon single bonds. So for this, the high party groups run opposite sides of the double bond. Has he configuration to draw those that configuration? We will just change one side. People outside the same switch, the location of the groups on the right. Now we're high priority groups from the same side at the bottom, and this will be set configuration.

So now we're getting into more complicated molecules. So we look at the skeletal structure thing to give us an idea of how the molecules laid out his first one. Age bonded and I wanted to and spawned its age. But the skeletal structure doesn't tell us necessarily how many bonds we have knows at least one. So Nitrogen wants to form three bonds. It's on ly formed to so far someone to give it a second one Here. Most of the collect can't give the second one the hydrogen. It can only form one Paul. Now we shouldn't forget that alone. Paris Here waken count the number of things around the nitrogen. It's going to be one two three. That makes it F P two hybrids, both of them. You can see that there's symmetric. He could imagine playing a mirror in between them. And because there s p two hybrids, that means that they have a pure metal. It's going to be bonding. And that's where second bond comes from. The Reds are Pete first next one and two h four. We can see the skeletal structure. Looks like I'LL draw over here. Two hydrogen sze bonded to a nitrogen. That's what the H two n means and then the and and then the age to its two more hydrogen, both nitrogen also have a loan pair is because they have five valence electrons. They've only formed three bonds. They also already have all the three bonds that make them happy, which means they're gonna be F p three. We've got four things around their three bonds and bow. Wanna let Trump Air Chris next one. We've got a little bit of, ah, offside molecules who's got each three seats. I like to draw that, starting with the carbon and then adding on the three hydrogen sze and the next one in line is nitrogen, and it's funded to to hide regions. And if we do a quick trick, everything has. The number of bonds that likes hydrogen will have one. Carbon has four nation has three and we just need to add His girl appeared there. This is we've got no double bonds and you were here and no expanded up tests. Everything is F p three

Let's forget the problem from the chapter Introduction to Organic chemistry. This problem is based on the structural formula or the skeletal formula of the organic compounds. Here we are given some compounds which are real gains sort like real games. We have to draw the condensed structure formula for the l canes in the skeletal family, for the cycle begins. So in the condensed structural formula relied carbon and it's attached to hydrogen as group. So first of all, we have the structure of the protein. Butane contains four carbon attempts in the street chain and it's condensed structure from like in Britain. As I see it three single bond see it too, single bond, CH two, Single bond, see it three. Then we have cyclo eggs. In cyclone vaccine contains six carbon atoms arranged in the form of a cycling ring. So it's skeletal formula and Milton as shown here. Then we have the compound known in the donation compound contains nine carbon atoms in the street chain. It's condensed structure from like and billiton as CH three, Then CH two repeated seven times, Then at the last CH three. Then we have the next compound, 18. The compound 18 contains two carbon atoms, and it's condensed structure from Lincoln bilton, as CH three single bond, CH three. So these are the structural formula and the skeletal formula of the given respective compounds.


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