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Pfints) for the molecule below; draw the Newman projection of the highest encrgy and lowst energy contormation_ BrWould you expect this molecule to have higher stra...

Question

Pfints) for the molecule below; draw the Newman projection of the highest encrgy and lowst energy contormation_ BrWould you expect this molecule to have higher strain (cither torsional or steric) in a conformation thanis presentin thebuune molecule? Ulese bricfly Epluin in the box below:

pfints) for the molecule below; draw the Newman projection of the highest encrgy and lowst energy contormation_ Br Would you expect this molecule to have higher strain (cither torsional or steric) in a conformation thanis presentin thebuune molecule? Ulese bricfly Epluin in the box below:



Answers

Draw Newman projections for the staggered and eclipsed conformations of pentane for rotation about the $\mathrm{C} 2-\mathrm{C} 3$ bond. Which conformation is lowest in energy?

Let's try Newman projection for the lowest energy confirmation of two metal hack scene along the C two c three bond. So two methods vaccine is this structure here, and we have carbon to and Carbon three. So let's draw Newman production projection in the staggered and eclipsed confirmation along the C two c three Double bonds So we will have CH three age three c and each each eight and C three h seven. And this is the eclipsed confirmation. Let's draw the stagger confirmation exceed three h three c h h h c three h seven and we can see that, uh, there is less strain in the staggered confirmation as the functional groups air farther apart so we can see that this would be the lowest energy confirmation. And this would be our Newman projection in the lowest energy confirmation and the staggered confirmation for to method heck sane along the C two c three bond

Let's draw the Newman projections for the standard and Eclipse confirmations for two metal panting for rotation about the carbon to and carbon three bond. So let's take a look at to muffle Pantene and identify carbons two and three So we'll have that h three c c h ch three ch two ch two ch three. And if we identify, this is the carbon to carbon three. So we're gonna identify the rotation about the carbon, too. Carbon three bond. So let's start by drawing the eclipsed confirmations. We have ch three each three each. One, 23 and the eclipse configuration. Yeah, a little bit better here with one to three coming across this way. Here each H c h two c h three. This will be ch three h three c h. They're equipped with be each he ch two ch three and the third eclipsed confirmation will be each age ch two c h three are staggered confirmations way will have th 38 h three c and then they're staggered. We will have sure h h ch two ch three h three h d c c This confirmation here would be h each ch two ch three and ch 38 age three C that each each c two c h three. So there would be our eclipsed and staggered confirmations for two methyl painting about the C two C three bond rotation. Which confirmation is lowest and energy. So the staggered confirmation in which one methyl group is opposite the folk group. Yeah. Mm hmm. Um, em mother group is nothing. Nothing. Um, in both, um ghost and And he there's two of them are the most stable and the lowest energy confirmations. And those two staggered confirmations that would be the most stable would be these two here.

Whereas to draw the most stable confirmation of the given molecule or molecules. Here is a try substituted cyclo hack scene, and we can see here that we have, uh, to roaming Adams on one side. Ah, of the ring and one on the opposite side of the ring. So we draw our sick lack seen here. One confirmation is we're gonna have a pro mean pro mean and bro mean have a ring flip here. This would give us pro mean pro mean in bro. Mean So looking at these two confirmations in the chair confirmation, Um, I'm sorry we went back up here. The confirmation in which the to groaning atoms are on the equatorial bonds is more stable as it suffers from less 13 di axial interactions. So the most stable confirmation would be the structure here on the right. And we have to of the Bruning atoms in the equatorial position. And this would be the most stable confirmation of this Try substituted, uh, cyclo hexane

For these. Try substituted cyclo hexane mean we see that were given this confirmation here were asked to draw the most stable confirmation I have from the way that the structure is drawn, we can see that all bro Mean Adams are on the same side of the ring, so we control. Aw, our first confirmation. We have a pro mean pro means. And Ramin when I have a ring flip and we'll have, bro. Mean I mean and bro mean So looking at these two confirmations, um, the structure containing the roaming atoms in he acts your position, which is? The structure on the left suffers from 13 di oxy of repulsion zones, whereas the structure containing the bro mean Adams in the equatorial position, which is a structure on the rate is free from such repulsion ins. Therefore, the structure with he bro Ming Adams, right in the equatorial position is the most stable. And this is the one on the right. So going back, we can highlight this structure here on the right, and this would be the most stable confirmation off the given molecule.


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