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2-methyloctahydro-1H-indene

Since there is a plane of symmetry passing through the middle of the compound, the C (chiral centres) at 2 and 3 will rotate the light through angles opposite in magnitude, say, +A and -A respectively (since they have 4 same valencies). Hence 1 is always acting as a chiral centre in this compound, then why do we classify it as pseudo chiral and not purely chiral?

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    $\begingroup$ There is not always a plane of symmetry which you mentioned. Also, precisely when there is a plane, the carbon 1 can't be considered chiral, since the entire molecule isn't. $\endgroup$ Mar 30, 2023 at 7:48

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The methyl hydrindane you have drawn lacks stereochemistry and the question you ask cannot be answered as such. There are four possible diastereomers (red arrows) of the hydrindane: 1, 2, 3a(4a) and 3b(4b). Diastereomers 1 and 2 both have a horizontal plane of symmetry, a cis ring juncture and they are meso compounds. The ring fusion carbons are stereogenic and chirotopic (R/S) while the carbon bearing the methyl group is stereogenic and achirotopic (r/s).
Hydrindanes 3a(4a) and 3b(4b) have a trans ring juncture and they are enantiomers of each other. The carbon bearing the methyl group is both non-stereogenic and achirotopic. Hence, the lack of a stereochemical descriptor.
For related discussions on this topic on ChemSE, go here, here, here or here.


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We do not have a plane of symmetry in a typical conformation. If we assume a typical chair conformation for the cyclohexane ring, then the hydrogen atoms attached to the brodgehead carbons are oppositely positioned above or below the plane of the paper, which then causes both bridgehead carbons to have R orientation or both S. Either the R, R or S, S configurations or the locations of the hydrogen atoms above or below the plane of the paper is then not invariant upon reflection through the proposed (but demonstrated false) mirror plane.

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