Newman Projections to Bondline PDF

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Summary

Explanation on converting Newman projections to bondline structures with practice problems...

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Problem: Draw the following molecule in bond line.

Solution: The best way to learn how to do this is to take out a model kit and build the molecule above. Start by connecting the two central carbons (the one in the middle and the “invisible” carbon) and then add substituents. Once you have the whole thing assembled, turn it sideways and you will see the bond line structure. If that doesn’t work for you, or if you don’t want to use models because you won’t have them on the exam, I would recommend the following procedure. The goal is to find the parent chain on the Newman projection, and then add in the substituents. Here goes. First and foremost, you should ALWAYS begin these problems, whether you’re going from bondline to Newman or Newman to bondline, by counting the total number of carbons. Don’t forget about the hidden carbon. In this case, there are 8.

Next, find the longest continuous chain of carbons. You can use the other ethyl on the back carbon or the other methyl on the front carbon, but regardless, the parent chain is 5 carbons long. Now go ahead and draw pentane.

Numbering the carbons helps keep track of what goes where. Now start filling it in, first with the easy stuff. The Newman projection was looking down the 2 – 3 carbon bond, and carbon number 1 was a methyl sticking off of the central carbon, and carbons 4 and 5 were an ethyl coming off of carbon 3.

This leaves everything filled in except for the second and third carbon. Take a look at carbon 2. Notice that the methyl group is sticking straight up in the Newman projection. There is a hydrogen to the right, and a methyl to the left. If your eye was in the paper where I drew it in the picture above, the methyl group would be sticking up, as it was in the Newman projection. That means that the hydrogen should be sticking out of the paper (to the right of the eye’s perspective), and the methyl on a dash.

Now do the same thing with carbon number 3.

Now that everything is filled in, verify that there are 8 carbons in this structure. And you’re done! Now you can draw it normally.

Note: Look back at the Newman projection for a second. Notice that the parent chain that I chose involves the two substituents that are 180° apart, or in plane with each other. This makes the problem a lot easier, since when I went about filling in carbon 2 and 3, the methyl and the ethyl substituents were already drawn in their correct places and orientation. If you turn every problem like this so that the entire parent chain is in plain with itself, it will be easier.

Something I forgot to mention, a cyclic molecule can also be represented with a Newman projection. Check that up and see if you understand it....