# Stereochemistry of hydrogenation with Raney nickel

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?

• check edit @bon – Shubham Nov 6 '15 at 14:35

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.

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.

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