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Examine the geomety ofthe molecule in the Window;ball & stlckIbalsThe molecule has carbony| groupRCHO RCOR" whete and R* are carbon groups. RCOOH ROHA genc...

Question

Examine the geomety ofthe molecule in the Window;ball & stlckIbalsThe molecule has carbony| groupRCHO RCOR" whete and R* are carbon groups. RCOOH ROHA gencral funetional group Fepresentntion Of the molecule iThe compound is a(n) aldohyde

Examine the geomety ofthe molecule in the Window; ball & stlck Ibals The molecule has carbony| group RCHO RCOR" whete and R* are carbon groups. RCOOH ROH A gencral funetional group Fepresentntion Of the molecule i The compound is a(n) aldohyde



Answers

Consider the following organic substances: ethylethanoate, ethylmethylether, hexanol, and propanone. (a) Which of these molecules contains three carbons? (b) Which of these molecules contain a $\mathrm{C}=$ O group?

All right, So this question asks us to identify the Carol Carbone in this two. Beautiful. So I'm gonna go ahead and draw to you to know, just so we can see what it looks like. We have the O age group on that second carbon right there, and then we could just finish the rest of the compound. So when we go ahead and look the Onley carbon that has four different substitute groups coming off of it would be this second compound. And we have our substance humans being the first being this O H group, the second being this hydrogen, the third being this metal group in the fourth being this Ethel group right here. So this carbon that second carbon is our Cairo carbon. Now, if we were to draw a mirror image of this molecule, we can pretend that our mirror is right here and we're just flipping it over so we can go ahead and start with our ch three whenever ch two. Then we have that see with the oxygen group attached to your the hydroxy group attached to it. And then finally, our last ch three. Now these two molecules, this one right here in this one right here are actually non superimposed a bull. Because if you were to stack these two molecules on top of each other, they would not produce the same image. So these two molecules are non super imposible on to one another.

Hillary won. Today we're doing Chapter 19 problem eight, and this room asked to identify the structure of a compound with a molecular formula of C four. It's 802 and with each anymore, spectra of a triple it. That's your 0.9 at double its or multiply, representing two. Hydrogen is at 1.6 of triple representing, two agents at 2.3 and a single. It's representing 111.8. So how we should go about solving this is first we need to turn the degrees on saturation. So how we do that is, well, the degrees on saturation or the number of double bonds equals the number of carbons minus number of hydrogen and highlights over too. So it's eight or to my X number of nitrogen is over to. That's your over two plus one. So you're four minus four equals zero plus one equals once we one degree oven saturation. Sometimes you know we have one double bond in this molecule, and if you look at the beach and more data, we see that we have one proton that's going to be a single it so it's not neighboring anything that can ah couple or split the signal at 11.8. So right off the back, we can say that the only way that we can have a proton that is not neighboring any other protons is that if it is on this oxygen, that is next to a cardinal group. So essentially, we have a carbon so acid. So you know that this proton is extremely acidic, mean that is going to fall off very fast. And it's also highly de shielded by the left arm. A drying groups, uh, adjacent to it, one being this option one being the carbonnel auction. So we can say that because it's gonna be highly dif dif. You listen to be significant on the left side of special spectrum and have a high ppm value such as the one that says 11.8. So we know that this proton is gonna have the highest PPL PBM value. So we know that that proton is taken care of and now we have this one carbon here and this should be a second carbon here. And then let's draw the other two cartons to satisfy the molecular formula with straw straight chains and see if this is true. So let's now work on the most shielded proton. So we should have three protons here. They're going to be the most shielded furthest away from these electronic drawing groups. And these three protons are neighboring two protons here. So we should see triplets. Or we should see these three protons being represented by two plus one peaks that equals triplet for three peaks. So three proton should be representing three peaks that should have the lowest ppm value. And that's actually what we see. We see that a 0.95 ppm. We have three protons that correspond to a triplet. So now what about these two protons here? Well, their neighboring three protons All right, but they're also neighboring these two protons on the left. So these two protons and blue, these two protons and blue are gonna be represented by a multiplex. So remember, you have your n plus one times you're n plus one. So you're gonna have four times three is. You might have 12 peaks in this one maximal, and this should have an intermediate value of P p. M. Somewhere between the Black Protons summer between the green proton. So these blue protons, if you look in between. So we have 0.991 point 65 and 2.3. So we should be interested in 1.65 And in fact, yes. This indeed is a multi planet representing two protons. So this supports this potential structural we drew here. Now, what about the green protons on the left? There's no neighboring. Protons are right. There's two. So these two protons to be represented by a triplet and they should have the highest value compared to other than 11.8. And that's going to be actually 2.3 triplets representing two protons. And that's exactly what we see in our questions them. So this is the proposed structure with given the data from this question.

We're problem. 36 they were given for molecules here. Part is your promo beauty. Um, part B is your elonis court. See, we have two hydroxy propane, Oleic acid gay. So, um, hopefully you see why that is? We have, um, three carbons, right? 123 And this makes you the pope. Phenolic acid. And we have the two hydroxy group it. And finally, we have our three metal hexane. Three methyl hexen have six carbons, and you have about methyl group on carbon three. All right, so this problem is asking for us to draw these molecules. So part everyone, John s n inch murky Part B. We want our part, c. We want our important TV want s So, um, the way we're going to do that as we were going to take these groups eso these groups hanging off of the mark. Okay, we're gonna take those and, um, draw them either going into the ward or out of the board in order to create an Antrim Earth that we're looking for. Okay, So part a here we want on s in an drummer. Okay. So, usually this not usually always so roaming in this case would be or, um, priority number one. This is your priority, number two. And this is your priority number three. OK, so if you do that, you're gonna have this clockwise direction. I'll just put it off to the side. You're gonna have this clockwise direction which will give you are in an Jamar, which is not what you want. So in order to see to flip that we want the hydrogen, Remember, there's a hydrogen that is with the bro me in here. Do you want this hydrogen to be going out of the board? Okay, so once it is going out of the board and it is priority for this are in an trimmer is going to turn into s So what I'm going to do here now is take that, bro mean and draw it, um, going back. So when it is going back, that means our hydrogen is going front. All right, so that were, um we're going to be using that very same reasoning for the rest of the three molecule escape. So we want on our an engineer or, um, part B. So this would be your priority one. Okay. And this would be your priority to when this would be a priority. Three. If you follow that, we're gonna have this counterclockwise direction, which is normally your s an and from ER so for the same reasons. Since we don't want to follow this, we want to flip it into an r. I'll take this. NH do, and I'll draw it. Go in back. Okay, So since I do it that way, that automatically means that h that hydrogen right is going, um, out of the port or in front. It's going in cake party. No, part C. We want on our in an Jamar. So again, let's set priorities 1st 1 to three. All right, So if we follow that, we would get a counter clockwise direction, which would be your s and Angela. Okay. And since we don't want s again, um, pretty much the same thing we take that which group and we draw it going into the board or going back. Finally, we have part B or last part here against, So we're setting priorities one more time. We have. Of course, this would be your priority one. It is a longer chain. Priority two and then three if we follow that direction, um, we'll have this counter clockwise direction. Okay, Which would be your s an instrument. And because we are this is already correct, right? We're looking for us. Um, we're going to draw this method group going out of the board or in front. Okay, so that would be your CHT. All right, so that's pretty much it for this problem.

And in this question we look at the amino asked pro lane. We can start by drawing out our structure of protein. This involves our carb oxalic acid and then we can go ahead and draw are coming which is a part of a ring in this case and that's our structure of protein. We can further label this amino acid based on the type of heterocyclic compound. It is so because this is an ami and we can label this as a hetero cyclic. Yeah I mean and we cannot distinguish this as an aromatic compound as our ring is not aromatic but we can label this as an al emphatic. I mean because it is a saturated heterocyclic compound meaning there are no double bonds. Then go ahead and add an additional description. Heterocyclic al emphatic. I mean thank you.


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