hey guys for today's practice problem we're going to want to look at this
reaction here this is a hydrogenation over platinum metal and what we're gonna
want to do is be able to draw both products that will result from this
state their relationship, that's going to be their stereochemical relationship, and
then we're going to want to be able to state which one is more
thermodynamically favorable of the two. if any of this looks confusing to you
you're gonna wanna go back and check out my lectures on addition reactions
specifically hydrogenation as well as aspects of stereochemistry and as a hint
to do #3, you're going to need to be able to draw cyclohexane chair
conformations so watch that tutorial as well, so give it a shot
okay guys let's take a look at the answer, now the first thing we want to
understand is the stereospecificity of hydrogenation over platinum metal so
remember that this is heterogeneous catalysis, so the way this works is that
we have a solid surface of platinum and hydrogen atoms are sticking out like
this. when the molecule lowers down it is necessarily the case that the hydrogen
atoms are added in syn fashion from the same side of the molecule ok, so that
means that we have two options because this portion of the molecule is planar
they can either add like that or from behind the board like that so let's say
that they add from behind the board, we've got the two hydrogens coming in, they're
gonna be on the other side that means this methyl is gonna get pushed out. now
the two hydrogen atoms that we added those are implied so we haven't drawn
them in, but if we were to, those would be on the dash bonds right here
hydrogen, hydrogen, and that's why the methyl got pushed out. now the other
option is if they were added from this side of the molecule, like that. that
means the implied hydrogens would both be wedges right there. and that means that
the existing methyl had to get pushed back. so here are the two
resulting compounds, and so we have a cis-1,3-dimethylcyclohexane and we have a
trans-1,3-dimethylcyclohexane. now these are chiral, actually this one is not chiral
and this one is chiral, but we we have to understand that those are
diastereomers, diastereomers because this stereocenter has the same configuration
but this stereocenter has opposite configuration and so anytime molecules
differ in one or more but not all of the stereo centers they're going to be
diastereomers. so that is the stereochemical relationship between them. now
if we want to talk about the thermodynamic favorability we're going
to have to draw the chair conformations. so I've drawn the two chair confirmations
for each molecule, this is the cis, now remember for cis what we're seeing here
is if these are both up, they're both on the same side of the ring, both of the
methyl groups are up, that means that whichever chairs you've chosen to draw
the methyl groups have to be on the same side of the ring, so we've arbitrarily
chosen these two carbons here so let's say that that I'm calling this carbon
one and I'm calling that carbon one and then I'm going around like this
2, 3, then that would be 2, 3, right here, right? now when this chair flips this is
going to be the resulting chair because remember the leftmost carbon remains the
leftmost carbon and you continue to number in the same way that you did. so
there's our 1, 3, there's our 1, 3, and now the thing is here this one is diaxial
which makes this one diequatorial. now, and that is the case because any
substituent that is axial in one chair will become equatorial in the other chair and
vice versa so we have our diaxial conformation we have our diequatorial
conformation, that is for the cis molecule. now for the trans molecule
we've got one axial, one equatorial. that means when we flip it we're gonna get
that one equatorial and that one axial. now we know that the axial position is going
to be less thermodynamically favorable because of the diaxial interactions
that are happening, so the equatorial position is the more favorable one of
the two. that means that this is the most favorable conformation because both of
the methyl
groups are in the equatorial position, and that means that the cis compound
is the more thermodynamically favorable of the two in terms of conformation
thanks for watching, guys. subscribe to my channel for more tutorials, and as always feel free to email me