(i) Which of the following pairs of compounds would you expect...

Question

(i) Which of the following pairs of compounds would you expect to have different physical properties?

(ii) What is the relationship between each of the pairs?

(iii) Assign the absolute configuration of each stereocenter to confirm your answer.

(c)

Accepted Answer

Welcome back, everyone. Given structures, one and two determine the following. Number one, if they are different or identical in physical properties. Number two isomeric relationship between the given pair of structures and number three absolute configurations of each stereo center. Let's begin with the first part. That's all what we notice about the two structures is that they look quite similar. They definitely have the same molecular formula if we count the number of carbon atoms and hydrogen atoms. But we can visually determine that the question is, are they identical in physical properties? That would be correct if these compounds are aware the same structure. Now, are they? Well, essentially we noticed that each structure has two chiral carbon atoms, those are carbon atoms with four different substituents. And the two structures are arranged in a way such that they are mirror images of each other. Right now, the question is if they are mirror images of each other, are they super imposable? If they are, it's a meso compound. If they are not, they are in anti ier essentially, we can immediately say that they are super imposable because the structures have an internal plane of symmetry right. And this tells us that the structure is a meso structure. We can also essentially flip one of them and superimpose with another. However, it's sufficient to identify the internal plane of symmetry. So the two structures are identical, meaning they will have identical physical properties. Number two, as we have just said, those structures are identical and they represent a meso structure. A meso compound has chiral carbon atoms, but essentially, it doesn't have an in Antium repair because it contains an internal plane of symmetry. Number three, this is where we have to apply the Khan Ol pre-law rules. So let's go ahead and do that. We want to assign the absolute configurations of each stereo center. So let's consider the first structure. Let's start with the first chiral carbon. We know that there's an implicit hydrogen atom pointing up on a wedge which gets priority number four because it has the lowest atomic number. Now, let's consider the remaining substituents on the left, we have carbon that is bonded to another carbon and to implicit hydrogens, that's on the left, right, on the right, we have carbon that is bonded to another carbon and to implicit hydrogens. On top, we have a metal G, there are three hydrogens, right. So the metal group will get the second lowest priority because each hydrogen is lower in priority than carbons between the two remaining substitutions. We have to consider what's next. And essentially later on, we have another carbon that is bonded to carbon and to implicit hydrogens versus carbon that is bonded to two carbons and one implicit hydrogen. So that extra carbon gives us a high priority, right? Priority number one, priority number two, what we're going to do from here is just assign the rotation, 123, we are essentially going clockwise, right. But we have to flip it to a counterclockwise rotation because priority number four is pointing up on a wedge. So instead of having a clockwise rotation, we're going to have a counterclockwise rotation. So that would be OK. And now what we want to consider next, that would be the absolute configuration of the remaining chiral carbon atoms. So let's go ahead and erase all of the information that we have so that we don't get confused, we can immediately label the first stereo center ss right. And now considering the remaining one, we already know what the priorities are. So we can essentially just draw that implicit hygiene atom give it priority number four, the muscle group gets priority number three, the carbon under white that has branching gets priority number one on the left, we get carbon number two. And now we trace our path 123. This gives us counterclockwise rotation, but it's actually r because priority number four is pointing up. So we have to reverse the assignment. 123, counterclockwise, make it clockwise, right. Because priority number four is pointing up. So we have S and R configurations. We can essentially assign the configurations of the remaining carbon atoms for the second structure. What can we tell? Well, essentially it is an identical structure, right? If it is an identical structure, the corresponding chiral carbon atoms will have the same configuration. So what can we sell? Well, essentially, we can even prove that once again, we can show an implicit hydrogen a muscle group. Priority number 41, I'm sorry. Priority number four, we one nt 123, we're getting a counterclockwise rotation, but it must be clockwise because hydrant is pointing up, right. Priority number four is pointing up. So we get R and now on the bottom, we will have s right because it's the flipped form of the previous compound. Not only it is a mirror image, right. Essentially, if we flip it relative to the mirror plane, we will have two wedges instead of two dashes, meaning the corresponding chiral carbon atoms must have opposite configurations. So the identical carbon atoms, if we want to label them would be number one for S number two, for R, right? And this is what we get when we superimpose them. And that's it. We have the desired configurations. We can essentially state that for number three, we have assigned those and we can conclude the video that will be it for today. And thank you for watching.

(i) Which of the following pairs of compounds would you expect to have different physical properties?
(ii) What is the relationship between each of the pairs?
(iii) Assign the absolute configuration of each stereocenter to confirm your answer.
(c)

Key Concepts

Stereochemistry

Chirality

Absolute Configuration

Watch next