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3

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 ...


2

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!


-1

They are the same If you draw all bonds of the left diagram and rotate the first sigma bond containing the aldehyde group, the right diagram will pop out The conventional Fischer protection is the right one. The left one seems to be a bit condensed and needs to be expanded


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