I've been dealing with assigning stereochemistry and I seem to be doing okay, but I've come across a molecule that I've found a little tricky to deal with.

The molecule in question

I've identified three stereocentres, highlighted below. My first question would be to confirm that this is correct.

stereocentres highlighted

No. 1 is standard fare and I got that it was an R enantiomer.

But for 2 & 3, does the fact that both carbons are bonded to the oxygen have an effect on the rotation of the molecule? As in, when you go through the process of rotating the lowest priority group to the back and then tracing from first priority to third, does this affect the process? It may seem a silly question, but I don't have ball & stick models to easily visualise it so I'm struggling a bit. I treated them as normal and got that they were both R enantiomers, but I'm wondering if I've missed a trick.

  • 2
    $\begingroup$ Yes, 3 centers and they are all "R". You got it right. If you plan to take other chemistry courses get some models now. BTW, don't worry about rotating the lowest priority group to the back - if it's in the front just reverse the R/S assignment. $\endgroup$ – ron Apr 8 '15 at 0:55

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