CH 5-4 Practice-I: Converting a Newman Projection into a 3-D Line Structure Convert the following Newman Projection into a “zig-zag” line structure, and provide an accurate IUPAC name, labelling the configuration of any chiral centers. (1) Since we are looking down a C-C bond, lets start by drawing those two carbons. We are “looking” at the C with the “dot” in the Newman Projection. (2) Now find the bonds going straight up and down in the Newman Projection. These are to the Br atoms, so lets add those in next.

(3) Each C must have 4 bonds, so for the remaining two bonds on each C, one must be drawn “forward”, and the other “back”. (4) Next we add in the atoms and groups on the two C’s. On the C with the dot, we have the H on the right (going back) and the ethyl on the left (coming forward). (5) On the other C we see the ethyl on the right (going back) and the H on the left (coming forward).

(S) (R) (6) Now we can determine the configurations. The “front C is (S) and the “rear” C is (R). (7) The structure we drew is just fine, but lets draw the usual line structure. What is the longest C chain? (3R, 4S)-3,4-dibromo hexane (8) Finally, what is the IUPAC name?