# How to reduce propyne selectively for the trans product

In my textbook it is given that when alkynes react with sodium in liquid ammonia, hydrogenation takes place and a trans isomer is formed.

It is also given (in another part of the chapter) that when terminal alkynes react with sodium in liquid ammonia, they form a salt with sodium. I assume the first mentioned reaction doesn't take place in the case of terminal alkynes.

Then, how do we, for example, add deuterium to propyne so as to form trans 1,2-dideuteropropene?

• How about $\ce{Na}$ and $\ce{ND3}$? May 8 '17 at 2:56
• I am confused if that would result in alkene or sodium alkynide May 8 '17 at 4:00

As you point out, a dissolving metal reduction using $\ce{Na}$ and $\ce{NH3}$ in the presence of a proton source such as $\ce{tBuOH}$ is able to reduce a disubstituted alkyne to the corresponding trans alkene.
Both protons on the product come from the $\ce{tBuOH}$, rather than the $\ce{NH3}$ as suggested in the comments, so in theory, using $\ce{tBuOD}$ would give you the desired product.
One issue with this however, is the ease of exchange between $\ce{H}$ and $\ce{D}$– any residual water present would be able to convert $\ce{tBuOD}$ to $\ce{tBuOH}$, which would then give you the protonated rather than deuterated product. Experimentally, you'd probably end up getting a mixture of products with varying degrees of deuteration (mono and di).