16
$\begingroup$

2,3,7,8-tetramethylspiro[4.4]nonane

The above compound – 2,3,7,8-tetramethylspiro[4.4]nonane – clearly exhibits both optical and geometrical isomerism. The question is, how many stereoisomers does it have in all?

As the number of possibilities are many, I couldn't really visualise the molecule's all possible stereoisomers properly. Could someone please help me count the total number of stereoisomers?

I figured that the carbon common to both rings is sp3 hybridised (tetrahedral geometry), and the two rings don't lie in the same plane.

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
16
$\begingroup$

The compound 2,3,7,8-tetramethylspiro[4.4]nonane has five potential stereogenic or pseudoasymmetric centres: the four C atoms with the methyl substituents and the spiro atom. In total there are $2^5=32$ possible configurations. However, many of these combinations are actually identical compounds.

A simple and obviously exhaustive (but also exhausting) approach is to draw all 32 configurations and name them. (This approach may also be used to check the results if someone finds a more elegant solution.) The resulting names for the 32 combinations are:

  1. (2​R,3​R,7​R,8​​R)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)
    (2R,3R,7R,8R)-2,3,7,8-tetramethylspiro[4.4]nonane
  2. (2​R,3​R,7​R,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)
    (2R,3R,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane
  3. (2​R,3​R,7​S,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)
    (2R,3R,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane
  4. (2​R,3​S,5​r,7​R,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)
    (2R,3S,5r,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane
  5. (2​R,3​S,5​s,7​R,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)
    (2R,3S,5s,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane
  6. (2​R,3​S,7​S,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)
    (2R,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane
  7. (2​S,3​S,7​S,8​S)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)
    (2S,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane

Therefore, there are seven different combinations.

Improve this answer
$\endgroup$
3
  • 6
    $\begingroup$ It might be more meaningful to draw them using Newman projections, much as you would for an allene, just a thought. $\endgroup$
    – ron
    Commented Apr 27, 2018 at 19:04
  • $\begingroup$ For the middle spiro carbon, why did you draw the wedge and dash in cases 4 and 5, but not in any of the other cases? Wouldn't in all cases the carbon atom be sp3 hybrid, and hence, need a wedge and dash bond also? $\endgroup$
    – Gaurang Tandon
    Commented Apr 28, 2018 at 2:12
  • 4
    $\begingroup$ @GaurangTandon In the cases 1, 2, 3, 6, and 7, the spiro atom is nonstereogenic; i.e. flipping of the spiro atom portion of the structure changes nothing otherwise. Single bonds attached to nonstereogenic atoms should normally be drawn as plain bonds (neither as solid wedged bonds nor as hashed wedged bonds). The use of stereobonds at nonstereogenic atoms should generally be avoided. $\endgroup$
    – user7951
    Commented Apr 28, 2018 at 6:30
13
$\begingroup$

Here is another approach to @loong's solution to the seven stereoisomers of this tetramethyl spiro[4.4.0]nonane. [This question is akin to one recently asked about tetramethylspiropentanes].

enter image description here

Using the arbitrary numbering scheme shown above and assigning C1 as having the R-configuration, the eight permutations for C1-4 are listed on the left of the diagram. Only RSRS and RSSR have stereochemical designations at the spiro carbon (C5). Inside the black box are the eight structures. The R/S designations are in the order 1234(5). The three blue structures, RRRS, RRSR and RSRR, are identical and its enantiomer is RSSS, aka SSSR. The black RRRR isomer’s enantiomer is the SSSS isomer shown on the left. The green pair are enantiomers. The red isomer is a meso structure.
Summing up, there are three pairs of enantiomers and one meso compound. There are seven stereoisomers.
ADDENDUM: Structure A (the red meso isomer RRSS) is identical to its mirror image B by reflection and rotation. There is no plane of symmetry.
enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ The red structure is only a meso compound if the rings were flat, or if you'd allow the molecule to vibrate/relax into the same structure again. $\endgroup$
    – Martin - マーチン ♦
    Commented May 2, 2018 at 11:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.