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I don't understand why this molecule is assigned S instead of R.

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The H is assigned a priority of 4; the methyl group on the chiral carbon gets 3; the left carbon gets 1 and the right carbon gets 2. The rotation should therefore be clockwise.

I guess what I have an issue with is assigning priorities in cases like these: what does moving away from the chirality center mean? Do I choose the H? The C? The other C? enter image description here

Do we just ignore the attached hydrogens? It seems like the hydrogen on the carbon left of the chirality center is ignored.

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And in the above case, why are we comparing an O with a C? The first atom to the left of the chirality center is a C, not an O.

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And in the last picture I give my configuration for the molecule in the first picture. Why doesn't the upper right carbon have a priority of 2? Double bonded carbon counts twice; the rest are hydrogens.

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The right carbon gets a "1", so the compound is "S". Attached hydrogens are not ignored. Here is a link to a nice set of rules. See rule #3 for handling double\triple bonds, that's where you made an error. It says, "Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar “phantom” atoms by single bonds. Note: “phantom” atoms are bonded to no other atoms". See the page for a clearer picture.

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In the case of the diol you show near the bottom of your post, the 2 carbons attached to the chiral center are equal, so we next look at, and compare, all of the substituents attached directly to them.

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  • $\begingroup$ Thank you. So the upper right carbon, by virtue of being part of a double bond, is considered by the R,S system to have two single-bonded carbons bonded to it? And the other carbon in the C=C double bond similarly has two single-bonded carbons attached? $\endgroup$
    – Dissenter
    Commented Jun 2, 2014 at 17:38
  • $\begingroup$ The carbon in the double bond closest to the chiral center is considered to have 3 carbons and a hydrogen attached to it, while the terminal double bond carbon (the one that is the "tie-breaker on this side of the moelcule") is considered to have 2 carbons and 2 hydrogens attached to it. $\endgroup$
    – ron
    Commented Jun 2, 2014 at 20:06
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As you move away from the chiral center along two chains you are comparing, for each atom in the chain consider the highest priority atom it is bonded to first.

See step 1, parts 2 and 3 in this reference: http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html

For double and triple bonds you have to add ficticous atoms before you evaluate.

See step 1, part 4 in the above reference.

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