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Using the R- & S- nomenclature assign a configurational symbol to decalin 4-methyl (above the plane), 4-hydroxy (below the plane), 7-isopropyl (above the plane), 1,9 diene. (Numbering the carbons in decalin ring-system: ring-junction carbons are 1 (top), 6 (bottom))

2-methyl-8-(propan-2-yl)-1,2,3,4,7,8-hexahydronaphthalen-2-ol

Carbon 7—holding the isopropyl group: I think this is S (isopropyl $ > \ce{C-C=C} > \ce{CH2} > \ce{CH}$).
Carbon 4—holding $\ce{CH3}$ & $\ce{OH}$: I think this is S ($\ce{OH}> \ce{CH3} > \ce{CH2=CH2}$).

Is my reasoning correct?

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  • $\begingroup$ I have added a guessed structure diagram based on your description. However, your numbers of locants are not in accordance with this structure. In particular, the bridgehead atoms of decalin are actually not numbered 1 and 6. $\endgroup$ – Loong Oct 2 '16 at 17:06
  • $\begingroup$ Guess things have changed in 30-years. Numbering decalin, as I did, the 1-double bond is on the ring junction, otherwise the diagram is correct. Don't have a drawing package; and, given the silly prices of Chemdraw (a monthly subscription) am not likely to acquire one. All I require is the comfort of knowing my assignments are correct--if they are. Cheers. $\endgroup$ – tony Oct 2 '16 at 17:25
  • $\begingroup$ @tony I changed the picture. Please check and confirm or rollback my edit. $\endgroup$ – Loong Oct 2 '16 at 17:44
  • $\begingroup$ Loong:yes, that's it: guess I'd better do something about presentation! Thanks. Cheers. $\endgroup$ – tony Oct 2 '16 at 19:01
  • $\begingroup$ There are a number of free chemical drawing programs; just check the corresponding Wikipedia sites (if need really be I can find a link). But even if they are too much of a hassle, drawing on paper and scanning/taking a photo works, too. $\endgroup$ – Jan Oct 2 '16 at 20:59
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I don’t really know how you arrived at your reasoning, but it seems to fail as soon as double bonds are involved. If you encounter a double bond, it must be expanded with ghost atoms to two single bonds in the following manner:

$$\ce{(R2)C=C(R'2) -> (R2)C(C)-C(C)(R'2)}$$

The carbon atoms in bracket are ghost atoms not connected to anything else. These ghost atoms are treated just like normal atoms when determining priorities. You can read more on Wikipedia, which also includes a better graphic example. (But since your case is rather simple, we can do without.)

Let’s first determine the tertiary alcohol’s carbon’s absolute configuration:

  1. $\ce{O}$ from the hydroxy group.

  2. $\ce{C}$ (along the ring to the left)

    1. $\ce{C}$ (the quarternary carbon)

      • $\ce{C}$
      • $\ce{C}$ ← This is the first point of difference.
      • $\ce{(C)}$ — this is the double bond’s ghost atom.
    2. $\ce{H}$

    3. $\ce{H}$
  3. $\ce{C}$ (along the ring upwards)

    1. $\ce{C}$
      • $\ce{C}$
      • $\ce{H}$ ← This is the first point of difference.
      • $\ce{H}$
    2. $\ce{H}$
    3. $\ce{H}$
  4. $\ce{C}$ from the methyl group.

    1. $\ce{H}$
    2. $\ce{H}$
    3. $\ce{H}$

Thus, the priorities are $\ce{OH} > \ce{CH2C_{(q)}} > \ce{CH2CH2R} > \ce{CH3}$, because the quarternary carbon that one of the two $\ce{CH2}$ groups is bonded to beats the secondary carbon at the top of the right-hand ring. That makes that carbon’s absolute configuration (S).

Similarly, for the carbon attached to iso-propyl:

  1. $\ce{C}$ (the quarternary carbon)

    1. $\ce{C}$
    2. $\ce{C}$
    3. $\ce{C}$
  2. $\ce{C}$ (the iso-propyl group)

    1. $\ce{C}$
    2. $\ce{C}$
    3. $\ce{H}$
  3. $\ce{C}$ (the secondary carbon)

    1. $\ce{C}$
    2. $\ce{H}$
    3. $\ce{H}$
  4. $\ce{H}$

As you can see, we only need go down to the second level to find a difference: $\ce{C(C3)} > \ce{C(C2H)} > \ce{C(CH2)} > \ce{H}$ Thus, that carbon’s absolute configuration is (R).

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I think the carbon with the isopropyl group on it should be (R). The carbon at the ring fusion has highest priority because it has two real carbons and a phantom carbon from the double bond. This is higher priority than the two carbons and a hydrogen of the isopropyl group.

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