(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?

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(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?

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Answer 1 · 2018-07-01T18:57:53

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

Explanation:

And of course a $meso compound$ $"meso compound"$ is an $R S$ $RS$ compound, such that it contains an internal plane of symmetry. And the goto example is $dimethylene glycol$ $"dimethylene glycol"$ , i.e. $H_{3} C - . C H (O H) . C H (O H) C H_{3}$ $H_3C-dotCH(OH)dotCH(OH)CH_3$ ...the chiral centres are indicated.

Now such a structure gives rise to TWO enantiomers, i.e. $RR$ $"RR"$ , and $SS$ $"SS"$ ...and the $RS meso$ $"RS meso"$ structure...i.e. here $R S \equiv S R$ $RS-=SR$ ...

For $1,1 - B r_{2} C_{5} H_{8}$ $"1,1"-Br_2C_5H_8$ ...THERE IS NO GEOMETRIC/STEREOCHEMICAL dimension to the structure. And certainly its mirror-images are superposable...

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(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?

1 Answer

Explanation:

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