My teacher said that from $3$ to $6$ member cycloalkanes having any number and type of substituents we can always draw the planar structure and decide the specifics of stereochemistry like:
- Are they optically active or not?
- Are they meso compunds or not?
- Do they have a plane of symmetry/centre of symmetry/axis of symmetry/alternating axis of symmetry or not?
But I feel that the molecules are hardly in planar form. It would be incorrect to assume they are planar and find the specifics of stereochemistry. Only for cyclopropane the points should be correct as it is always planar. Maybe my teacher said that mistakenly in a hurry.
So regarding points $(1),(2)$ and $(3)$ who is correct? Why? Or are we both partially correct? I need a slightly detailed answer.
Answer
For a compound to be optically active it must not be superimposable with its mirror image. It's as simple as that. Also note that stereogenic centres are not always required for chirality. There are other types of chirality such as axial and planar chirality where a compound possesses either an axis or plane of chirality. Sometimes it can be the case that just the sterics of a molecule does not allow for free rotation of bonds and the molecule is locked one configuration. I think in the simplest cases of open chain molecules, you could indeed use the structural formula, but with rings which adopt specific 3D configurations, you must refer to the 3D structure.
Again, as said in 1., to determine if a molecule is chiral, it's better to refer to its 3D structure, especially when with meso, when you are dealing with more than one stereogenic centre. For a compound to be meso, it must not be chiral, but posses stereogenic centeres (look up the definition).
No, you cannot use the simple structure to determine symmetry elements present if that structure is not really planar. It's not necessary to build a model, but you must use wedged/dashed bonds to show stereochemistry.
If someone thinks differently, please comment and correct me.
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