(1R,3S)1,3-dibromocyclopentane (or 1s,3r?) and 1,1-dibromocyclopentane both have internal planes of symmetry but why is only the first one a meso compound?

Question

Ava Andrews · Accepted Answer

Because by definition, a #"meso compound"# contains TWO potential chiral centres..

And of course a #"meso compound"# is an #RS# compound, such that it contains an internal plane of symmetry. And the goto example is #"dimethylene glycol"#, i.e. #H_3C-dotCH(OH)dotCH(OH)CH_3#...the chiral centres are indicated. Now such a structure gives rise to TWO enantiomers, i.e. #"RR"#, and #"SS"#...and the #"RS meso"# structure...i.e. here #RS-=SR#... For #"1,1"-Br_2C_5H_8#...THERE IS NO GEOMETRIC/STEREOCHEMICAL dimension to the structure. And certainly its mirror-images are superposable...

Easton Jones · Answer

undefined (1R,3S)1,3-dibromocyclopentane is a meso compound because it has an internal plane of symmetry that divides the molecule into two identical halves, making it superimposable on its mirror image. 1,1-dibromocyclopentane lacks this internal symmetry, preventing it from being a meso compound.