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7.6Draw Kekule and skeletal structures from the names (E)-N-methylpent-3-en-2-amine 4methylcyclopent Z)-3-chloro-4 2-cnol -isopropylhept-3-ene 2,5-dlimethylhexa-2,...

Question

7.6Draw Kekule and skeletal structures from the names (E)-N-methylpent-3-en-2-amine 4methylcyclopent Z)-3-chloro-4 2-cnol -isopropylhept-3-ene 2,5-dlimethylhexa-2,4-diene (E)-+-nitrohex-2-ene trans- ~2-methylhex-3_ ene

7.6 Draw Kekule and skeletal structures from the names (E)-N-methylpent-3-en-2-amine 4methylcyclopent Z)-3-chloro-4 2-cnol -isopropylhept-3-ene 2,5-dlimethylhexa-2,4-diene (E)-+-nitrohex-2-ene trans- ~2-methylhex-3_ ene



Answers

Name the following alkenes, and convert each drawing into a skeletal structure:

Growth structures Responding to the foreign names a guy I support you mean well, I support true three. Try a little. I mean one a little No, A mill. A little in mithril in union. And you with one hard dejan substituted with missiles N f l M Mitchell cycle up into I mean cycle up until I m f l and Missiles and I support cycle Hexcel. I mean critics you and isopropyl in if if I roll Fertile and Ethel.

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.

The lower structures for the following compounds. Mhm. The H stands for the thing around the boat. Oh. Then goes problem. Age Reach for age group is detached. Can the rest of the 6 to 8 and san chain Yeah, me too. Yeah, Yeah. Shit. Thanks. If that's special Unconscious. Some people experience for running ch little hole in the side chain and or true Yeah, yes. Mhm to medical groups. That bitch. Third Carver then traveled. Long problem has ch, um, shears through. She's too. Don't Problem is the side. And then double bunk plus each and seek to h m p m. Yeah. You start writing from, uh, a comfortable in the knowledge base. It's fine garments. Uh, and this number that weren't true. 45 after common fire because it's five books. So there's that about the room and also in the position of fire. This is subversion. Is the fish okay? Mhm. No. Let's number carbons. Pure Channel one, Joe three. The deposition free. There is a night ago. That's their answer

Problem. 9.19 asks us to draw the compounds blow. We'll start with Compound a 33 dimethyl for off time. Notice that the parent chain is eight carbons with triple bonds on carbon four. So we'll start by drawing that. Now we'll add the two method groups on the third Carbon. This completes compound A. Now we'll look at Compound B, which has a parent chain of Decca trying with the triple bonds on carbons 16 and eight. Now we'll add the ethnic group on carbon three. So this carbon and then we'll draw the metal group on carbon five. This is Compound B now see as a parent chain of three X sign. So we'll begin by drawing that. And then it has metal groups on carbons to to let the group's on carbon. It's too, and to uncover in five, and this is 2 to 55 Test your mettle. Three. Heck, sign now for question. D. Question D has a cyclo deck kind, so we'll start by drawing that and member when county the account from the beginning of the triple bond towards the second end of the triple bond. So with two method groups on Carmen's three and four. We will count clockwise and then add in the method groups. This is 34 methylene cyclo. That kind Compound E has apparent chain. I've had to die in one E with double bonds on carbon Syrian five and a triple bond on the first carbon. We'll begin by drawing the parent gene and then drawing the double bonds and Compound F has a parent chain of known in 16 so we'll draw that. The double bond is on carbon one, and then the triple bond is on carbon six. The chloral group is on carbon three. There are two method groups on Carbon four. We already have the double bond with our parent, Jane and the drip of bond on carbon six. So this is three claro for four de metal, one noon and 16 now for a part G. Remember that a sec beautiful group is a beauty group connected to a chain on its second carbon, which would look like this. This would attach to the rest of the chain, so we'll begin by drawing the one help time at in the Triple bonds and then add in the second the SEC Bugle Group on the Third Carbon for the second Carbon, their carbon chain, and added to the second Carbon of the Bugle Group. And this is three sec beautiful one halftime now for compound each. The parent chain of this compound is three oc time. So we'll draw that first at the triple Bond and then add the tribute group. Remember that Turk Beautiful Looks like a cross as four carbons. It is attached in the middle. And then we have a metal on the second Corbin. This is five her beautiful to metal three off time.


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