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Neuroprotectin D1 (NPD1) is synthesized in the body from highly unsaturated essential fatty acids.NPD1 is a potent natural anti-inflammatory agent.a. Label each dou...

Question

Neuroprotectin D1 (NPD1) is synthesized in the body from highly unsaturated essential fatty acids.NPD1 is a potent natural anti-inflammatory agent.a. Label each double bond as conjugated or isolated.b. Label each double bond as $E$ or $Z$.c. For each conjugated system, label the givenconformation as s-cis ors-trans.

Neuroprotectin D1 (NPD1) is synthesized in the body from highly unsaturated essential fatty acids. NPD1 is a potent natural anti-inflammatory agent. a. Label each double bond as conjugated or isolated. b. Label each double bond as $E$ or $Z$. c. For each conjugated system, label the given conformation as s-cis ors-trans.



Answers

Neuroprotectin D1 (NPD1) is synthesized in the body from highly unsaturated essential fatty acids.
NPD1 is a potent natural anti-inflammatory agent.
a. Label each double bond as conjugated or isolated.
b. Label each double bond as $E$ or $Z$.
c. For each conjugated system, label the given
conformation as s-cis ors-trans.

It began with part B and drawing the chair confirmation of Compound B. Which should be this if you assume that the Green Adams Ark. Laurean's. So let's go back to part a label. Each substitute is axillary territorial. Starting with this chloride, the hydrogen is going straight up. Is AC sealed? So the chlorine is equatorial? It's not going straight up or straight down like this. Chlorine is Therefore it is Axl, and so is theme. Ethel Group is also axial. This is how you would draw the three ring system and to define the cyst versus Trans of the three ring system. Here we can see that both of the hydrogen are going up relative to the other two groups attached. So here the carbon group's going down here. The carbon group's going down relative to the hydrogen Sze. Therefore, this is assists fusion both of the hydrogen czar on the same side. For this ring fusion, we have one method group going up in a hydrogen going down, which is a transfusion so trans, because the two groups are on opposite sides

This is the answer to Chapter 10. Problem number 47 Fromthe Smith Organic chemistry textbook. Ah, and this problem presents us with a late ic acid, which has an e double bond. Um, and we're asked to predict how the melting point of late IC acid compares with the melting points of ste Eric and oleic acids on. They could be found in table 10.2. I've drawn them here. Um, and so we should remember that melting point is going to be inversely correlated with the number of double bonds. So the more double bonds that there are, the lower the melting point is, um and that's because, uh, double bonds impact the ability of these long chains to pack together closely. Um, when? When there's double bond, the chains cannot pack as closely together. Um, And so, of course, Z double bond really kind of kinks the chain. And so that's gonna have ah greater effect than an e double bond would, um, and so to put these in order here, stare a gas, it is gonna have our highest, uh, melting point because it has no double bonds. So the next between the late IQ and oleic. We're comparing an E, uh, double bond and a Z double bond. Ah, and so the e double bond is gonna have less of an effect. So this is gonna be our intermediate fuel. A tick s. So it's, um, in between the two on and a lake will be our lowest melting point. So, uh, that's really all that this problem asked us, um, is to put these in order. And again, we need to remember that double bonds are going to impact the ability of these chains to pack together closely. Um, and of course, Z double bond will have a greater effect than an e double bond. Ah, and so stare. It is gonna be our highest boiling point. A lake is gonna be our lowest boiling point. Um, we're sorry. Melting point melting point. Um, it's hysterical. Be our highest melting point. A lake will be our lowest melting point. Ah, and allay, tick. I will be intermediate to those two. And that's the answer to Chapter 10. Problem number four

Report, they will see that this compound is sorrow that has a following. Um, structure. It has the construction, many of them. And there are also keep going. Keep going through which are found in stores for part B, We have a false a lipid. This is because we see a diagram right present. We also see too long things of carbon. Thanks.

This question access to describe what about the structure of a body asset of the track Mr. Reid that Lends to its physical characteristics. Here I have three the 3 um fatty acid types. Um you can see that they all have a slightly different structure in terms of the kinks in the chain. And that literally has to do with the double bonds. So here we have the saturated battery acid. Notice it's a straight chain. That's because there's all single bonds between the carbons. Next we have in the middle of the unsaturated fat. Well, actually both of these are unsaturated fats. However, this one is a trance mhm fat and then this one insists and it has to do with where the double bond lies. So the hydrogen is on the double bond our place. So if we look at the cysts unsaturated fat, the double the hydrogen czar on both sides. So the kink, it's a perfect L looking king. And then if we look at the trans fat, it's just a little less of a bend because the hydrogen Zarin opposite side. So the physical characteristics has everything to do with where the double bonds are. So the answer for that to me would be see.


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