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If I react the following with Raney nickel and $\ce{H2}$,

$\hspace{65 mm}$alkene

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?

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  • $\begingroup$ check edit @bon $\endgroup$ – Shubham Nov 6 '15 at 14:35
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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 (2​E)-(2,3-2H2)but-2-ene yields (2​R,3​R)-(2,3-2H2)butane and
(2​S,3​S)-(2,3-2H2)butane.

hydrogenation

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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.

right sideup drawing

Some will land up side down.

right sideup drawing

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.

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  • $\begingroup$ sorry, didnt understand what you mean by (H)(CH3)(D) going clockwise $\endgroup$ – Shubham Nov 6 '15 at 16:29
  • $\begingroup$ so, is my book correct ? $\endgroup$ – Shubham Nov 6 '15 at 16:32
  • $\begingroup$ Yes. Edited my answer to make that point more clearly. $\endgroup$ – MaxW Nov 6 '15 at 16:57

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