Stereoisomers of 1,3,4,6-tetramethylcyclohex-1,4-diene

How many stereoisomers are possible for following compound (1,3,4,6-tetramethylcyclohex-1,4-diene)?

My Approach
the 1,4 methyl groups lie in the plane, whereas 3,6 are possible chiral centers. However it doesn't seem like enantiomers are possible due to centre of symmetry, one structure can be meso. Apart from this I don't see any possibility of Geometric isomers as well. However the answer given is:

3

What am I missing in this?

• The isomers are RR, SS, and the third is RS which is also SR – Maurice May 27 at 19:28
• @Maurice so RR,SS are unique because there is no plane of symmetry and RS,SR are the same because of centre of symmetry. Is that the case? – Ashish May 27 at 19:56
• I think I might have taken RR, SS to be the same structure in my working which is what went wrong – Ashish May 27 at 19:57
• @Ashish yes, RR and SS have no centre of symmetry and RS and SR are same due to centre of symmetry. – S R Maiti May 27 at 20:41
• For $n$ stereogenic centres, there are up to $2^n$ stereoisomers. As shown e.g., with meso-compounds, not all of these realizations mathematically possible are chemically unique. – Buttonwood May 28 at 8:40

Your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, does not have line symmetry (plane of symmetry), and hence $$\ce{C}$$3 and $$\ce{C}$$6 are chiral centers. Thus, maximum possible stereoisomers are $$2^2 = 4$$. However, there is a symmetry element in the structure, which is order 2 ($$180^\circ$$) rotational symmetry (also called point symmetry). Therefore, one stereoisomer, $$(3S,6R)$$ would be superimposable with it's mirror image $$(3R,6S)$$, thus it is a meso-isomer:
To understand this, I put "$$\bullet$$" symbol to indicate relevant hydrogen is above the plane of paper. And, "$$\bf{.}$$" symbol to indicate relevant hydrogen is below the plane of paper. Thus, first two structures have both 3,6-dimethyl groups below the plane of the paper while the third structure has 3-methyl group below and 6-methyl group above the plane of the paper. Accordingly, all chiral centers have CIP nomenclature as indicated under each structure:
• The first structure is $$(3S,6S)$$ and the second structure is $$(3R,6R)$$, thus they are non-superimposable mirror images as indicated by the mirror. Therefore, they are enantiomers. To make all the groups to coincide, you may need to rotate the second structure $$180^\circ$$ horizontally along the axis going through $$\ce{C}$$3 and $$\ce{C}$$6. However, after the rotation, the two methyl groups at $$\ce{C}$$3 and $$\ce{C}$$6 will be now above the plane. Therefore, these two structures cannot be superimposable.
• On the other hand, the third structure is $$(3S,6R)$$ ($$\ce{C}$$3 methyl is below the plane and $$\ce{C}$$6 methyl is above the plane). To the mirror image of this molecule, if you follow the same operations as previous steep and do another $$180^\circ$$ rotation on the plane of the paper (or vise versa), you'd get the same compound. That means, the third compound is not optically active and called meso-isomer. This can be found easily by looking at the two chiral centers: They are mirror images of each other, $$(3S)$$ and $$(6R)$$ or $$(3R)$$ and $$(6S)$$.