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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?

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  • $\begingroup$ How about $\ce{Na}$ and $\ce{ND3}$? $\endgroup$ – Pritt Balagopal May 8 '17 at 2:56
  • $\begingroup$ I am confused if that would result in alkene or sodium alkynide $\endgroup$ – Govind Balaji S May 8 '17 at 4:00
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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.

The mechanism is understood to go via the following stepwise mechanism:

Warren/Clayden: mechanism of trans alkyne reduction

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).

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