Question
2) Draw the most favored chair configuration for the following molecules. Cl
2) Draw the most favored chair configuration for the following molecules. Cl
Answers
Consider the molecule shown. Draw both possible chair conformations for this molecule. Which of your drawings is expected to be most stable? Explain your reasoning.
Let's compare to molecules Got molecule A Got a message group there well nigh dine when a massive group there molecule B is similar, except for the placement of one of the metal groups, the muscle group in molecule B is positioned right here. There is less Derek Hindrance in molecule A because of the position of that metal group in molecule B. That mess Algroup leads to staircase hindrance because it's positioned in that axial position.
Australia the most stable khan former of the following molecule here we have. Yeah, mm hmm. Like. So just would give us Yeah. Mhm strips like so. And then we have a ring clip. Mhm. Yeah. This would give us, Yeah. Mhm. Yeah. Uh three Yeah. Three. Just home from me here is more stable. Mhm. Due to less 1, 3. Mhm. Diack cell interactions, Yeah. Okay.
Okay. So as to draw the chair confirmation of the following compounds and identify Carol centers and label them R R s. So the first compound is this? Yeah. Okay. We have a turf battle group. We're carbon bound to three methods here. 123 and then a carbon four. We have Hey, Mitchell. Yeah. Yes. So they draw a chair. Confirmation, use draw. Two lines go down and then up, then same here, too. So I'll put Axle in red. It's up. Down, up, down, up, down That equatorial and green. This is six. So we want to put the turbulent group in the equatorial position, and I'll explain why in a second? Yeah, so that's down. The methods should be down as well, so it'll be axial. So it doesn't matter whether it's axial or equatorial as long as there both pointing down or up. If they're assists. And if the trans they should be opposite, then the other chair confirmation, we can put it in the actual position. Yes, Mhm. So this is going to cause if we put the tribunal group of the axle position, it's going to interact with the hydrogen. Is that our three carbons away. So this one counting with red, we have one, 23 and then counting. With green, we have one, 23 So, yeah, this carbon, too. The hydrogen are going to interact with the whatever carbon group is in the actual position. And this is known as 13 dioxido stream. So it occurs when a carbon group is in the actual position and the hydrogen is on a carbon that's three away will interact with it to the second. So therefore, this is going to be one on the left is more stable. We also have to identify the carol centers, too. So, um, over here we have a very beautiful hydrogen, and these two sides are actually the same. So we don't have Carl center in this molecule, and it be the same with this, sir, is it? These two sides are the same. The second one is this compound to broom in here. Mhm. Okay, Yeah. Okay. So we can put Mm hmm. So it's number at first. This is one. This is two. This is three and four. If we did it the opposite way and read to 12345 maybe too many numbers. So let's put the broom in in the actual position. So it's up. Which means carbon four should be up as well. It'll be equatorial because there, sis. Mhm. And carbon two is down. Because there and it's anti from carbon one and four to carbon two is down an axle. And this is more stable. Yeah, I'll explain why in a second? Mhm. Yeah, it's Yeah. Mhm. Okay, so we can let's put the broom in down, which means this one has to be up. Mhm. So this is down because they're cysts. So you notice that these two Burmese are very close to each other. They interact, and this is going to cause strain. So therefore, this is less stable. It's due to the interaction. Okay, so let's look at the Carroll centers too, So we get rid of these, Okay? So all right. So looking at this one in red, we have a brimming. We also have a hydrogen are here. This side is that this side is different from this side. So this berman is one. The hydrogen is four, and now we're looking at ch br verse ch two ch two C H p R first ch two c h to the ch br was gonna win the Burmese a higher atomic number than the second hydrogen. So the starts to this is three. So going from one or two and then 2 to 3 is going to be our So you get there. It's pretty red. So this is our than this one again. The Bruins won. Hydrogen is four. So we've siege to siege too. And then it's a siege, too. This star is going to win. So if you look over here, we have ch two. And the second carbon has a a booming attached to it. But this side has We've siege too. And then a siege, too. Then the Berman wins against the second hydrogen. Yes. So the sides to over here this is three. So from 2 to 3 or from 321 And that one or two eventually skipping four. We have s they Finally, this this carbon and blue. This is one. The side over here is to weave a ch br instead of a siege too. This is too. The hydrogen is for and besides three so 1 to 2 to three or from 321 skipping four. It should be s but the hydrogen is on the on the wedge me redraw that it's on the wedge. So it's going to be instead of, um, yes, it's gonna be our So if we Okay, um it's everything. Yes, we are R. R. S. You want me to put the the Bethel here too?
Podcast. Siri's. We're taking a look at electron configuration. So the electron configuration, often element, is very useful at telling us chemical as well as physical properties that that element may possess and may present in reactions. So what we're doing here is writing out some electron configurations. So the fast on we have helium have a said value of too. We have won us to. Next. We have potassium, 19 electrons to fill one. Us too. To us to to be 63 s to three piece 64 s one Following this we have cobalt. That's 27. Want us to to us to two p six. Three s, 23 p 64 us to three deep. 10. Next we have sofa that it's 16. That is one US to to us to two p six, three years to three. Peaceful. Following. This we have. Chlorine is our last run where we have 17 electrons to fill. Want us to to us to be 63 s, two on three peat five. So halogen chlorine has a very strong tendency to gain an additional electron. Andi have the following electron configuration. Thus, one electron, we would have one s two to us to two p 63 s to three p six. And so what you can see is that we would have a full valence shell so we would have noble gas configuration.