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If I react the following with Raney nickel and $\ce{H2}$,
$\hspace{65 mm}$
then I am getting a meso compound, as the $\ce{H-}$ anion will have anti addition, right? One $\ce{H}$ will attack on right $\ce{C}$ from bottom and the other on the left $\ce{C}$ from the top. That makes it a meso. But in the book it says, there is racemization. How?
The hydrogenation of alkenes using heterogenous catalysts such as Raney nickel is syn-stereospecific; i.e. both hydrogen atoms are added to the same face of the double bond.
Therefore, the hydrogenation of (2E)-(2,3-2H2)but-2-ene yields
(2R,3R)-(2,3-2H2)butane and
(2S,3S)-(2,3-2H2)butane.
Good old paper drawing has blindsided you. Think of the $\ce{H2}$ as adding on top of the molecule as you have drawn it on paper. Not all the molecules know that they are supposed to land "right-side" up.
Some will land up side down.
You get stereoisomers when there are four different groups bonded to a single carbon atom. Such a mixture will effect polarized light. One configuration will twist the polarized light positive and the other negative. An equal mixture of such stereoisomers is a racemic mixture and is optically inactive.
Right-side up you get (H)(D)(CH3) going clockwise looking down the C-C axis that was the C=C double bond. Upside-down you get (H)(CH3)(D) going clockwise.
