http://chemwiki.ucdavis.edu/@api/deki/files/1131/=Mesopracticeprob.bmp
in the abv link.."D '' is a meso compound ..y ???
and "F" is not a meso compound ..y ???
please explain the reason..plzzz....proper explanation...with how u are drawing plane of symmetry in both...!! or v can even say its meso compound or not by just seeing the absolute configuration(R,S)???
in the abv link.."D '' is a meso compound ..y ???
and "F" is not a meso compound ..y ???
please explain the reason..plzzz....proper explanation...with how u are drawing plane of symmetry in both...!! or v can even say its meso compound or not by just seeing the absolute configuration(R,S)???
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D is a meso compound, because the configuration of the upper left C is (R) and of the lower right (S). Thus the two halves of the molecule are mirror images of one another, and you can infer a plane of symmetry.
F is not a meso compound. If you "grab" the bottom asymmetric carbon and give it a twist, The -CH3 will rotate to the rear to become |||CH3. The -CH2CH3 will rotate pointing to the lower right. The -H will come out of the paper to point to the lower left. Then you will see that top and bottom are not mirror images of one another, and the molecule lacks a plane of symmetry.
Pulling the same maneuver on C will show that it is a meso compound. The planes of symmetry are prettyy obvious in the two meso compounds A and E. B is not a meso compound; the two configurations are (S)(S).
F is not a meso compound. If you "grab" the bottom asymmetric carbon and give it a twist, The -CH3 will rotate to the rear to become |||CH3. The -CH2CH3 will rotate pointing to the lower right. The -H will come out of the paper to point to the lower left. Then you will see that top and bottom are not mirror images of one another, and the molecule lacks a plane of symmetry.
Pulling the same maneuver on C will show that it is a meso compound. The planes of symmetry are prettyy obvious in the two meso compounds A and E. B is not a meso compound; the two configurations are (S)(S).