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Ch.4 Introduction to Organic Compounds

Chapter 1, Problem 4.43c

Lewis structures, condensed structural formulas, and skeletal structures are used to represent the structure of an organic compound. Each of the following compounds is shown in one of these representations. Convert each compound into the other two structural representations not shown.


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Welcome back, everyone. The structures of organic compounds can be represented using Louis structures, condensed structural formulas and skal structures. The compound given below is illustrated using one of the mentioned structural representations through the compound. Using the other two representations. Note for the condensed structural formula, we present the parent chain using the condensed formula and show the bonds for the branches. If applicable. First of all, we have to understand that the given structure is a scalar structure. That's, that's simply because we only see bond line representation which has no carbons, no hydrogens. And we basically see only solid lines in a zigzag form. So if we have a skeletal structure, this means that we want to draw a Louis structure and a condensed one, which is technically not really condensed because it states that we want to show the bonds or the branches. For the condensed formula, we essentially want to show no bonds and write it as a text. But in this case, we will modify it slightly. So for the Lewis structure, we first will want to identify all of the carbon atoms, a draw them. So we have five carbon atoms, 12345 in the parent, we're going to draw a double band and then we are going to add the final carbon, which is the muscle group. And now let's add the missing hygen. Every carbon must have four bonds. So starting with carbon on the right, it needs three additional hys. Then another carbon needs one hydrogen because it has three bonds or the muscle group. Once again, we need wi hygen followed by one hygiene. We can see three bonds once again, one hygen, we can see three bonds and three hygen, we can see only one bond. So now for the loose structure, we are essentially going to illustrate a structure where we are depicting all of the bonds given to us because they are will be many bonds. We actually want to draw it in a straight line or the carbon chain, right? One c 34 and five carbon atoms. Now, every carbon must have four bonds. So we can immediately add all of the solid lines. And now let's see what we have. So the first carbon has three hydrogens in in a loose structure. We are showing all of the bonding and non bonding electron pairs. Now, carbon number two has only one hydrogen and it has a double bond. So we can essentially eliminate that remaining bond followed by carbon number three, which also has a double bond and only one hygen. Carbon number four is bonded to hydrogen. It's bonded to another carbon C three. So we can immediately add three hydrogen atoms around that carbon. And now finally, the next carbon has CH three, which means, well, we simply want to add we hydrogens. So now let's check if everything is consistent, CH three CH CH ch which has two C three groups on it said well done. Now for the condensed structure, we can simply write a text. So we are going to begin with Siege three. And we have to recall that in a condensed formula, we're not showing any bonds unless there is branching. So CH three followed by CH and another C, we don't need to show that double bond. And then we have C and this is really important. We can treat that specific Siege three group as a branch and state and branches that they siege three followed by CH three or basically just draw a branch as the problem suggests and finish with sh three. But this is technically not accurate in a condensed formula. We can treat those two metal groups as two subscriptions bonded to the tertiary carbon atom. So we're not technically considering any branching here. We can just state that it is C in Brents C three, two times which essentially condenses it, right? It makes it more readable and easier to read in a single line of text. We can interpreted as C three F by CH CH. And the final C simply has two metal groups attached to it and those would be the final answers for this problem. Thank you for watching.