ok guys for today's problem we have a compound here, and we're gonna do two
things. number one: we want to be able to draw the Newman projection of going down
this bond in this direction here as is marked, and number two: you want to be able
to name this molecule including stereochemistry, so if any of this is
confusing you're gonna wanna check out my tutorials on Newman projections as
well as nomenclature and Cahn-Ingold-Prelog convention, how to assign R and S.
the first thing we wanted to be able to do was the draw the Newman projection
for down this bond, so I've drawn that here, let's take a look what's
going on. remember that when we draw Newman projections we want to
imagine ourselves in the plane of the board or the page so I'm here in the
board and i'm looking at this molecule this way, that means that
everything that is in the plane of the board is above or below me, or just right
here in this plane, and then things that are on wedge and dash bonds, those
are to the left or the right
accordingly so if I'm right here on this front carbon, remember that this front
carbon is represented by the circle and then the bond to the back carbon and the
back carbon itself can't be seen, those are behind that circle and so if I'm right
here i'm looking up and I'm seeing an isopropyl group above me, directly above
so on the front carbon here is that isopropyl, and then this hydroxyl group
that's on the wedge bond, that's out here, sticking out like that. so if I'm in the board
I see that down to my left, so that's down to my left. and then the implied
hydrogen which would be on the dash bond, that's this way, that's past the board
and that's down and to my right. now on the back carbon going down we have a propyl
group, that's directly down from here, so there's the propyl group, on the
dash bond, which is going this way there's that ethyl group, that is up
and to my right there, and then the implied hydrogen
which would be on the wedge bond, that is going to be up and to my left, so there is
the correct Newman projection for what we have drawn. now we also want to name
this so whenever we're gonna name a molecule the first thing we need to do is number
the parent chain and so for the hydroxyl group, the hydroxyl group is the
highest priority group on this molecule so we need to number in such a way so as to
give the hydroxyl occurring soonest on this molecule. that is going to be right
to left, so that is going to be your numbering scheme there because if we
went left to right the hydroxyl would be occurring later than if we went this way
and now we have to recognize what substituents are on the molecule, so clearly
we have a methyl group on carbon 2 and we have an ethyl group on carbon 4
and those are gonna be listed alphabetically so E ethyl before M methyl
let's go ahead and list the majority of this molecule, we know that this is going
to be 4-ethyl-2-methyl, and then the hydroxyl group is occurring on carbon 3 and this is
a 7 carbon parent chain so that is a heptanol, so that's that's the
majority of it but we're not quite done yet because we said we also want the
stereochemistry so let's see which which carbons are chiral centers. now
going from the right to the left
oftentimes we might be tempted to to say that that is a chiral center, a lot of
times students will see, just, there's anything on a wedge or dash bond so that's
a chiral center, but we have to really examine and say are there
fourr different groups projecting from this carbon and here there are not
because these two methyl groups are identical so this is not a stereocenter
which we could see if we just rotate around this bond, this carbon could
could be up here, that carbon would be down there, it would look different so
that is actually not a chiral center. over here though
number 3 we do have a chiral center and so we want to assess the priority of
these groups and so from here we've got carbon, carbon, implied hydrogen
and oxygen, so let's draw in that implied hydrogen, so the implied hydrogen is clearly
number 4, the oxygen is clearly number 1. now let's look at these two carbons
now these two carbons are identical so we have to see what they are attached to
this carbon is attached to carbon, carbon, hydrogen
this one is attached to carbon, carbon, hydrogen, so that actually is still
a tie so let's see what those are attached to. this carbon
is attached to hydrogen, hydrogen, hydrogen, this one is hydrogen, hydrogen, hydrogen
and then over here this one is attached to one carbon, as well as this, so because
these two carbons are both attached to one carbon whereas these carbons are
attached to all hydrogens, this one ends up taking priority and so that is going
to be number 3. and so we do have higher priority away from us so we
can look at this and go 1, 2, 3
that is R, so number three, carbon 3 is an R stereocenter. now let's do the
same for this one here so this one is actually attached to
we've got this implied hydrogen which is going to be priority 4 but then other
than that we have three carbon atoms that are all identical. carbon, carbon
carbon. well let's see what they're attached to. once again this carbon is
attached to an oxygen atom which is heavier than any other atoms in the
molecule so that is going to end up being priority 1 this direction is
priority 1 for that carbon. now we look at these two carbons, this is
attached to hydrogen, hydrogen, carbon, this is attached to hydrogen, hydrogen
carbon. so again these two positions tie. now we're over here, now this carbon is
attached to hydrogen, hydrogen, hydrogen, this one is attached to hydrogen
hydrogen, carbon, so this one actually ends up winning, that's going to be number
2 and then this one is going to be number 3. now here
we are in a situation where the lowest priority group is actually towards us and
so what we need to be able to do is one of two things: either view the molecule from
the other side which many students have difficulty with
so instead let's do the trick that we mentioned, let's pretend that the
hydrogen is on the dash bond and the ethyl is on the wedge bond, so if the ethyl
was on the wedge bond, we would simply go 1, 2, 3 like that and we would say R but
because we inverted the stereochemistry to be able to view it that way we
must then invert our answer to reflect the actual molecule so we will call that S
so now the way we list the stereochemistry there is we simply say
we have 3R,4S and then the rest of the molecule. so the complete
name will be (3R,4S)-4-ethyl-2-methyl-3-heptanol
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