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Solution: Draw each of the following compounds: (c) 2,2,4,4-Tetraethy...

Question

Draw each of the following compounds:

(c) 2,2,4,4-Tetraethylbicyclo[1.1.0]butane

Video Transcript

Alright guys. So, for this compound it give us this name that may have looked a little bit strange to you at first, right? It has tetra ethyl bicyclo butane and then it gives you the numbers at which our ethyl groups are located, at two there's two of them, and at four there's another two, and then it gives you these numbers inside are inside these brackets, right? So let's break this down, we know it's going to be a butane, right? Because it tells us and we know it's going to be some kind of bicyclo ring where it's just fused because notice that this number, our last number is zero, so what tells us is we have a fused bicyclic, right? If it had a number there that would mean it's bridged, so we're just going to have two rings that are fused together and now this actually tells us how many carbons are in our ring. So, carbons in ring. Now, it's going to be four carbons in total but an each ring is only going to be one carbon, right? So, let me just draw this out to, it just going to explain a lot better than I can through words. Notice that here we have two different rings. Now, how many carbons is there in total?Well, 1, 2, 3 and 4, so that's where we get our butane. Now, how many carbons are in each ring that aren't part of this right here? Well in this ring there's one and then this ring there's one, okay? So we've got two there so far. Now, what are we going to do next all we need to do is add in our substituents. Notice that we're going to start numbering this right here at these, right here where it's fused and then go to the larger ring and the smaller ring but here they're the same size, so now that we've numbered it like this accordingly, right? We can just add in those substituents and they're all the same substituents, so let me just go ahead and redraw this over here and what we're going to do is add in all these ethyl groups which are just two carbon chains and if we do so we would actually arrive at our correct structure. So, first this part over here, everything over here just talked about this structure down here and then once, we added in the substituents we were able to get to our final answer for this which was this structure right here, so if you're confused by anything we did, definitely review your videos on bridges and bicyclic bridge fuse and bicyclic molecules, right? there's fuse and others bridge, so it should view that whole section come back and then hopefully this will make more sense guys. Alright but if you have any questions comment below and now let's move on to your next question.