[narrator] Let’s start our discussion of alkyne reductions with the reaction we learned
from our alkenes chapter where we just take many atmospheres of hydrogen gas and use a
metal catalyst.
We’re showing palladium doped with carbon.
This is a non-selective reduction and we’re going to go all the way to the alkyne.
We go from 2-butyne to butane.
If we want to stop at the alkene and we want a cis-syn addition for the alkene we use Lindlar’s
catalyst, which is derived from a lead doped compound, and, don’t worry about the structure
we don’t need to know that we just need to know Lindlar’s catalyst--there are other
catalysts that can do the same thing, Lindlar’s being one of them, it adds in a syn fashion,
forms a cisalkene and stops, right there.
We end up with cis, or z, 2-butene.
If we want the transalkene, and it is a true trans, we could name it with trans if we wanted
to because we have hydrogens on both sides that are the same group, even if we had other
groups that weren’t the same instead of these methyls here.
We could have any groups we want but as long as we have two groups on opposite sides of
the double bond that are the same we could name it as a true trans.
I like to use E or Z instead of cis and trans for these.
So, (E)-2-butene with what we call dissolving metal reduction.
Lithium also works for this.
We’re--radical donor here.
Dissolving metal reduction.
Let’s look at the mechanism for dissolving metal reduction.
The reaction we see, right here.
Well, like we said, we have a lithium or sodium metal that is going to donate its radical.
We’re going to use a half arrow, with only a half arrow at the end, because only one
electron moves, here.
Then it knocks--or it forces the pi bond, one of the pi bonds of the triple bond, to
break leaving a lone pair on the carbon.
This is, in essence, an anti-addition of an electron, here, but it’s really--we’re
not adding anything except this electron.
The most stable intermediate is going to be the trans version that we see here.
We’ve got a radical next door to an anion.
Then we can use our solvent, liquid ammonia.
This is a--at very low temperature, below 0 °C, around -32 for ammonia.
Acid-base reaction and we’re getting closer.
We satisfy that anion.
We’re still left with the radical in this orbital.
Then another one of our metal atoms, plenty of them available, will donate an electron
to that orbital, will spin pair, and now we will form another anion.
Another carbanion.
Our plenty solvent molecules will then undergo another acid-base reaction to form the transalkene.