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Ch.23 - Organic and Biological Chemistry

Chapter 23, Problem 23.57a

Give line drawings for each of the following molecular formulas. You may have to use rings and/or multiple bonds in some instances.

(a) C2H7N

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Welcome back everyone. What is the line angle structure with the molecular formula? C three H nine N hint the compound may contain ranks and or double bonds. First of all, let's identify the number of possible ranks or pi bonds using the hydrogen deficiency index. It is calculated as one half multiplied by two, multiplied by the number of carbons. That's three plus two minus the number of hydrogens, which is nine plus the number of nitrogen, which is one. Let's evaluate the result. Six plus two gives us eight plus one, gives us nine minus 90, divided by two, gives us zero. Meaning the structure doesn't contain any py bonds or ranks. OK. So now if it doesn't contain any py bonds or ranks, what can we have alkyl groups and nitrogen? So we can essentially consider amines that only contain single bonds for this problem. So we want to think of possible amines. Now, if we have three carbons, we can just throw a three member chain and we can think of the addition of nitrogen to different carbons, we can add nitrogen at the end of the chain bonded to carbon. Number one, we can also add nitrogen to the metal carbon carbon number two, right? And what else can we get? Well, essentially we can incorporate nitrogen within our carbon chain. So instead of having a three member continuous chain, we can just have carbon bonded to another carbon bonded to nitrogen that is bonded to carbon. So those are the only possibilities for us. And now we simply want to add the missing hydrogens. Every carbon must have four bonds. So for the first structure, the first carbon would have three hydrogens, the second one would have two, the third one would have two and nitrogen would be three bonded. Meaning we just need to add two hydrogens. And if we count hydrogens, we get 246 plus three, that gives us nine, which is consistent with the molecular formula. For the second structure, we can just look at the carbons, they would have three hydrogens each, those are our metal groups and the middle carbon is we bonded. So it needs one hydrogen and nitrogen of course needs two hydrogens because it is we bonded once again, we have six seven. So, so this structure is sensible as well. And now for the third one, we need three hydrogens for the first carbon, two hydrogens, for the second one, only one hydrogen for nitrogen because it's already double bonded, it would be more accurate to say it has two single bonds. So it needs one more, right? That's why we're adding one hydrogen. And at the end of the chain, we need three hydrogens for carbon, we plus two gives us +56 and then nine hygen. So this is also a sensible structure. And now let's turn these into line angle formulas where each corner and vertex represents an implicit carbon. So carbon one carbon, two carbon three, 12 NH two NH hour first structure. Now for the second one, we want to draw a three member chain to carbon 123 and then NH two group bonded to the metal carbon. And finally, for the third structure, we have 12 carbon atoms. So 12 bin it to NH that is bonded to CH three. So we just want to draw a single line because at the end of that line, there's carbon with three implicit hydrogens and this is how we have managed to get three line angle structures for the given compound. Thank you for watching.