For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds. To confirm, I have optimized the molecular geometry on different levels of quantum ...


Meso compounds have chiral centre , but they also have a plane of symmetry making them optically inactive....or u can say meso compounds are optically inactive due to internal compensation. Whereas enantiomers taken together ie. formation of racemic mixture (50:50) is optically inactive as a whole due to external compensation.(They can be optically active ...


It looks to me like the left one is just wrong. It looks like who made the image took D-glucose and just flipped the 5-OH. The result? The molecule at left is not glucose, but L-Idose!

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