By assuming that the heat of combustion of the cis isomer was larger than the trans, structural assignments were made many years ago for the stereoisomeric 2-, 3-, and 4-methylcyclohexanols. This assumption is valid for two of the stereoisomeric pairs but is incorrect for the other. For which pair of stereoisomers is the assumption incorrect? Why?
Hey everyone. So, for this question we're going to be talking about our methyl cyclohexanol. Now, what I drew is I here to those simple one, the one that this question is talking about, the 1-methyl cyclohexanol but now what we need to do is we need to take a look at this conformation, and you need to compare it to 2-methyl cyclohexanol 3-methyl cyclohexanol and 4-methyl cyclohexanol, so we're going to do is we're going to draw a bunch of these compounds and for one of them we're going to say that we're going to have a 2-methyl cyclohexanol, so we can draw a methyl group right here, let's just pick equatorial for the time being, another one that we can draw is we can have a 3-methyl cyclohexanol, let's include that methyl group right here and of course what we can do is, let's copy one of these and now paste it down here and now let's do the 4-methyl cyclohexanol, so what that means is our methyl group is going to move now to this position. Now, the assumption that this question gave us was that trans would be more stable, right? Because at the cis isomer, right? this is going to be for our cis isomer, if that has the higher heat of combustion that means going to have a higher energy, if it has a higher energy, we know is going to have a lower stability, so it turns out one of those assumptions is actually wrong. So, one of our trans isomer is actually going to be less stable here, so let's just go through these three, what's the relationship between these two groups? there's cis because they're facing the same direction, what about these two? Well, they're trans, they're facing opposite directions and what about these two? well for these two they're actually cis again, okay? Now, what we can do is let's draw the other isomer of them, so what we can draw it is for this one, let's copy it, paste it over here, let's copy this one, paste it over here and let's copy this last and now I'm going to jump out of the image for this one, okay? And now what we can do is that all of these we're going to change them. So, here I say the first one we have, this is going to become the trans isomer, this is going to become the cis isomer and this is going to become the trans isomer and the way we do that is let's erase these little circles here and the way we do that is by taking this OH group and making it face the opposite direction, taking this OH group make it face the opposite direction, taking this OH group and making it face the opposite direction. So, as you notice what we have is our cis isomer for our two methyl and the trans one as well, this is going to be in the middle, our 3-methyls, 3-methyl cyclohexanol, we have to bottom our 4-methyl cyclohexanol.
Now, remember the Assumption said that our cis isomer would be less stable, right? So notice that in our first one, which one's more stable? Well, the trans and the last one which one is more stable? the trans because they're both equatorial, so notice that only in our 3-methyl cyclohexanol does the trend vary and it varies because notice that here our cis isomer is actually more stable, right? this cis isomer is the one that has the lower heat of combustion, the lower energy and the greater stability and the main reason behind all that is because two of the groups are equatorial, both of the groups or equatorial, which is the more stable position, this that here, in trans, they are both equatorial and here in this trans they're both equatorial. So, is that this cis isomer is normally the less stable one but at 3-methyl cyclohexanol more specifically a cis 3-methyl cyclohexanol, this cis isomer breaks the trend and is actually more stable. Alright. So, let's take your knowledge of understanding chair conformations and then noting that the isomer of the 3-methyl cyclohexanol breaks that assumption that our question is talking about. Alright guys, so hopefully that made sense.