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Organic chemistry
Course: Organic chemistry > Unit 3
Lesson 4: Conformations of cycloalkanesPolysubstituted cyclohexane
How to draw the chair conformations for menthol.
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- Why equatorial position is more stable for bulky groups? From the video I can see that at the 2nd variant oh-group and isopropyl group are closer together then in 1st variant...(6 votes)
- Equatorial is more stable because when the substituents are in the equatorial group their electron clouds are farther away from the other large electron clouds in the ring. We know from General Chemistry that electrons want to get away from other electrons (negative doesn't like negative) so since equatorial is promoting a farther distance from other electron clouds it is more stable. If you have a substituent like a methyl group in the axial position, and you draw out your hydrogens on the other carbons, you will notice that the methyl is eclipsed with the hydrogens. This causes what chemists call a 1,3-diaxial interaction. Think of it like this. In the equatorial position the substituents are sticking out farther away from the ring. This means they are farther from what is in the ring, which mean the electron clouds are farther away, which is good.(18 votes)
- Are there any quizzes for Organic Chemistry? These quizzes are essential for learning and memory retention. If not, are they any plans to add quiz questions? This would complement a well-designed set of videos.(10 votes)
- How do you determine whether a starting group say on carbon one is axial or equatorial, or is it arbitrary when you're first drawing one configuration?(5 votes)
- wouldn't the oxygen of the OH group interact with the hydrogens on the adjacent carbon and form intermolecular hydrogen bond?(4 votes)
- Yes you can have "intramolecular" hydrogen bonding (hydrogen bonds within the same molecule) as well as "intermolecular hydrogen bonding (hydrogen bonds between two or more disubstitutedcyclohexane molecules)....Check it out.....
https://en.wikipedia.org/wiki/A_value(5 votes)
- Have we talked about Bol-Hee-Siv substituents before?(0 votes)
- Wrong subtitles. There should be "bulky".(9 votes)
- Wouldn't carbon 1 be the carbon closest to the OH group because menthol is an alcohol?(2 votes)
- Yes, the "official" name of menthol is 5-methyl-2-(1-methylethyl)cyclohexanol
In an alternative "methane" nomenclature, the numbering starts with C-1 being the carbon bearing the methyl group.(4 votes)
- Why is it okay to number the carbons in our own way?(2 votes)
- The numbering here is only for your benefit in putting the groups on the right carbons, so do it however you want.(2 votes)
- Why didn't he follow IUPAC nomenclature rules to draw the chair molecule?(1 vote)
- He wasn't literally assigning them the numbers as per IUPAC naming, he's using them to help draw the chair form of that molecule from the line structure that's given. The numbers he is using is to help ensure each group ends up on the right carbon.(2 votes)
- Does anybody know what is Bol-Hee-Siv means at? 8:30(1 vote)
- For the second chair he draws (the flipped one), did he number the carbons wrong? If we flipped the first chair, he should have named the carbon above #1 2, right? Is his chair then still correct?(1 vote)
Video transcript
- [Voiceover] Here, we have
the dot structure for menthol, which is a famous compound found in various mint oils like peppermint. Our goal is to draw
both chair conformations for menthol, and then we're gonna choose the more stable one. The first thing we have
to do is assign numbers to our substituents on the ring. and how you assign numbers does not have to follow IUPAC nomenclature. For example, I'm gonna
call this carbon one, this, carbon two, this, carbon three, and this, carbon four. And if you already know
how to name this compound, you'll know that's not how to number it according to IUPAC nomenclature. But this numbering system is just to help us draw our chair conformations. So, let's start with
one chair conformation. We've already seen how to
draw chair conformations, you start with two parallel lines, one that's offset from the other. So, here is our first parallel line. And then here is the other one. Next, we draw two horizontal lines. So, this horizontal lines intersects with the top point on the top line. So, something like that so
it comes close to this point. And then we draw another one down here that intersects with the bottom point on the bottom line like that. Next, we draw a line
from the top dotted line down to the bottom, and then
we draw another line over here in parallel with the
one that we just drew. So, something like that. Next, you put in your last
set of parallel lines. So, from this point to this point. And then from this point to this point. So now, we have our carbon skeleton. Now, we have to put in our bonds. And we call this carbon one. And we know we start
axial up at carbon one. So, there's axial up,
carbon two is axial down, we just keep alternating. Carbon three is axial up. Carbon four would be down. Carbon five is axial up. And then carbon six is axial down. Next, we put in equatorial. So, at carbon one, this
would be equatorial down. Carbon two would be up. Carbon three would be down. Carbon four would be up. Carbon five, down. And finally, carbon six would be up. Let's look at our groups on the ring. So, we have a methyl group at carbon one, and since this is a wedge, that means it's going up in space. So, we're gonna put the
methyl group going up relative to the plane of the ring. So that must meant the ethyl, the, sorry, the methyl group is axial because axial's the only one that's
going up at carbon one. Next, we look at the OH. The OH is going up at carbon three, so we need to, let's go ahead
and number our ring here. So, this will be carbon one, carbon two, and then carbon three. So, we need to put the OH
going up at carbon three. And the only way we can do that is by putting the OH on axial. So, we're gonna put the
OH here on carbon three, so going up in space. And then finally, we have our
isopropyl group at carbon four and this is a dash. So, this is going away from us in space, or down relative to the plane of the ring. So over here, I'm gonna go
ahead and number carbon four. Alright, we have two choices. So, where do we put our isopropyl group? It must be going down relative
to the plane of the ring which must mean it's axial. Again, that's the only
one that's going down. So, let's draw in our
isopropyl group at carbon four. So, there's one chair
conformation for menthol. Alright, we know this chair conformation is an equilibrium with our
other chair conformation, so let's go ahead and draw that. So, two parallel lines that
are offset from each other. So, something like that. And then we draw in our
dotted lines right here just as guidelines to
help us as we're drawing our chair conformation. We go from this point
down to our bottom line, we go from our bottom
line up to our top line. And then we need to connect the dots to finish our chair conformation. We know this is carbon one now, so we start axial down at carbon one. So, it's the opposite of the
other chair conformation. Our carbon two would be this
point, so that's axial up. Carbon three would be axial down. Carbon four would be axial up. Carbon five would be down. And six would be up. Next, equatorial, so at carbon one, this would be up, and then at carbon two, it would be down. So, we just keep alternating. At carbon three, it would be up. At carbon four, down. Carbon five, up. And carbon six, down. So, let me redo that carbon six one, that wasn't very good. So, let me draw that one in again. Now, carbon one, this time, we know when this undergoes a ring flip, this is carbon one, and
we'll see this in the video that I'll show you in a few minutes. So, that's carbon one, this is carbon two. This must be carbon three,
and then this is carbon four. So, let's put in our groups. We know at carbon one,
we have a methyl group. And this methyl group is up relative to the plane of the ring. We know when this undergoes a ring flip, the methyl group has to stay up. So, the methyl group goes up axial to up equatorial when this
undergoes a ring flip. Next, let's look at the OH. So, the OH is also up relative
to the plane of the ring. So, it's gonna go, on carbon
three, it's gonna go up relative to the plane of the ring. So, it's up axial for the
chair conformation on the left and it's up equatorial
for the chair conformation on the right. Finally, we look at carbon four. We have our isopropyl group which is down relative to the plane of the ring. So, it's gonna stay
down, but it's gonna go from axial to equatorial because that makes it down right here. So, there's our isopropyl group going down relative to the plane of the ring here. Sometimes, it's really hard
to tell if an equatorial bond is up or down relative to the plane. And the way to tell is
to look at the axial one. So, this axial one at carbon
four here is very obviously up, which must mean that
this one is going down. So, that's a little trick to help you if you're stuck with those bonds. Alright, lemme go ahead
and put in hydrogen's. So, I went ahead and drew in every bond, so I might as well put in hydrogens on all these chair conformations. So, I'll leave off the hydrogens
on the isopropyl group, so we put those in. You don't have to put in these hydrogens when you're drawing your
chair conformations. Actually, it's probably easier
to see if you leave them out. But if it helps you, if it helps you to get it in the drawing
chair frame of mind, you can go ahead and do that. So now, we have both chair conformations and let's analyze them
in terms of stability. But first, let's check out the video to make sure that we drew the
proper chair conformations. So, here we have one chair
conformation for menthol, and at carbon one, you can
see we have a methyl group going up axial. If that's carbon one, the
this is carbon two right here, and then at carbon three, we
have an OH going up axial. Then at carbon four, we
have our isopropyl group. So, down axial. If we undergo a ring flip,
so if I rotate this carbon up in space, and then if I
rotate this other carbon down, and if we turn it a little bit, we'll be able to see our
chair conformation better. You can see, all those
groups we just talked about went from axial to equatorial, which is the more stable
place for these Bol-Hee-Siv, relatively Bol-Hee-Siv stituents. Let's compare the chair
conformations that we drew with what we saw in the video. So, in the video, this is carbon one, and you can see this
methyl group is up axial. And that's what we have here,
our methyl group is up axial. This would be carbon two,
this is carbon three, we have our OH up axial
just like we have it here. So, up axial. Then at carbon four, we
should have an isopropyl group going down in space,
and this is carbon four. You can't really see the bond, but you can see that this
isopropyl group is going down. When this chair conformation
undergoes a ring flip, all of those axial bonds go equatorial. So, on the right, we can see that. This would be carbon
one now and the mehtyl is still up relative to
the plane of the ring, but it is equatorial. And the hydrogen that was,
that was equatorial right here in our drawing, you can
see, now it's gone axial just like we have here. This is carbon two, this is carbon three, so at carbon three, we
have our OH up equatorial, just like we drew it up here. And then at carbon four,
we have our isopropyl group down equatorial. So here it is, down. We know that the equatorial position is the more stable
position for a relatively Bol-Hee-Siv stituent. So, if all three substituents
are out to the sides, this is the more stable conformation. So, this is an easy one to figure out, the more stable chair conformation. Sometimes it's a little bit harder and you might have to consult a table, and look at some of the energy differences between the two positions.