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Below is the structural formula of a compound known as the queen substance, which is secreted in the mandibular gland of queen honey bees (Apis mellifera). The quee...

Question

Below is the structural formula of a compound known as the queen substance, which is secreted in the mandibular gland of queen honey bees (Apis mellifera). The queen substance inhibits the development of ovaries in worker bees, prevents queen cell formation, and attracts male bees (drones) to virgin queens for the purpose of mating. Write the IUPAC name of this compound. Note that in the IUPAC system, when the carboxylic acid also contains an aldehyde or ketone group in the molecule, the presenc

Below is the structural formula of a compound known as the queen substance, which is secreted in the mandibular gland of queen honey bees (Apis mellifera). The queen substance inhibits the development of ovaries in worker bees, prevents queen cell formation, and attracts male bees (drones) to virgin queens for the purpose of mating. Write the IUPAC name of this compound. Note that in the IUPAC system, when the carboxylic acid also contains an aldehyde or ketone group in the molecule, the presence of the aldehyde or ketone group is indicated by the prefix - $0 x$ o.



Answers

Identify each organic compound as an alkane, alkene, alkyne, aromatic hydrocarbon, alcohol, ether, aldehyde, ketone, carboxylic acid, ester, or amine, and provide a name for the compound.

So now we'LL work on Problem ninety from Chapter twenty one here were asked to identify all the functional groups in the molecules or the the functional group that's present in each structure and then provide the name of each compound. So let's go out and write the structure were given for part A. That's C H three h scene double wand C C. H. Three in Let's Head in three method groups. So first we've identified the functional group that we have here, and that's an AL Keen because of the double bond. And then we want to name our carbons number are carbons and we'LL start on the side That's closer to the double bond. So one, two, three, four and five So now we can list our substitute mints and their position. So two, three, four, Try me Ethel, and we have five carbons and a double bond at position, too. So it's two Penn team, so that's moved Part B where were given the following structure in drawing our substitute mints. This is in Al Kane to the fact that we have a saturated hydrocarbon, so let's go ahead and name number are carbons. And since we have? No, Those bonds are functional groups. We will start from the side that has more substance. One too. Three, four, five and six. So we number our substitute mints and get two, two, four Try methanol and then we have a six carbon chains. So just heck sane so we could move on Art See, here we are given the following structure and we know, you know, subsist mint here and we know that this double bond Oh h is indicative of a car back cilic acid. And you know, five carbon chain ah which will give rise to the P and T prefix Will number are carbon starting from the karmic selic acid three four And we can read her name numbering the substations first. So three metal pen to know no position there Penta Noah Cassidy The car looks other gas It has to be on the end. So let's go to her. Dino we were given the following structure so we see that this nitrogen here with the three subsidiary it's connected is an amine and so we just need to list the substitutes in alphabetic order and then next to a mean annette naming of this compound will be complete. So we see that on the left we have an isopropyl group and on the right, we have a you told group. So we combine these to get the name of our compound, which is beautiful s o approval. I mean, so let's go ahead and move onto her e where we have C h three ch ch too ch ch three. I'm drying it exactly as it's written in the book C H two O. H. And then we have two carbons over here. No. So we see that the always group here's our functional group, So we haven't alcohol now. The longest carbon chain here is tricky. It would be tempting to say that it's this five carbons going straight across. However, we can see that the longest carbon chain is actually Miska here. So once on, then we conserve from the alcohol to name them a number of them. So we see when we number like us, we actually have two metal groups and a six carbon chain Versace. What we would have had if we had just gone straight from here to here. So this is a question where they're trying to trick you with the way that they're drawings of a structure. So we have a method group of position to and position for in metal. And we have a alcohol at position one and a six carbon chain. So to Ford, I'm Ethel. One external now for part F structure that were given is ch three ch too. See? It's two again No carbon Neil ch ch three seeing a JJ three So we see here that this is our longest current chain We see that the functional group is a key tone because of the carbon you'LL hear on the interior for molecule. If we draw our start numbering from the side of thie key tone, we get one, two, three for five and six. So our answer our named for the structure is two methods followed by the position of the key tone at three hex for the length of the carbon chain and own for taquito

This is the answer to Chapter 21. Problem number three from the Smith Organic Chemistry textbook. On this problem, we're given three Aldo hides and were asked to come up with the eye impact name for each Alba. Hi. Ah. And so the first thing that I would do here, um, is a is given. Ah, as, ah lettering structure. I would rewrite that. A skeletal structure. It makes it much easier to see on. And then we need to keep in mind our rules for naming. So A and B are, um open chain Aldo hides. Ah, and see is a ring. So the rules are gonna be a little different s o for A and B for the open chains, we need to find the longest chain that contains the Alba hide. Um, and named The molecule is a derivative of that. So, for example, for a, um, we can number this as, uh also will make the carbon with the alga hide carbon number one. So then we have to three, four and five. And so when we number it that way, it's a painting. Derivatives. So it's gonna be pent a pent. And now, um because remember, we changed the ending to Al Tau to signify that it's an Alba hide. Ah, and then it looks like we have, um, two sets of two method groups each on carbons three and four. And so, um, altogether this becomes, uh, three three for for tantrum, Ethel Pent. And now Okay. Um, all right, so then moving on to be, um sorry. Okay, So moving on to be we'll do the exact same thing. So we need to determine the longest chain here on name the molecule as a derivative of that longest chain. Ah, and just change the ending to al to signify that it's an Alba hide. And so in this case, um, it's going to be a knock Octonal. So on octane derivative. So again, we make the carbon with the Alba hide Carbon number one, two, 34 five, 67 eight. So there we go. It's gonna be a nocturnal derivative. As I said, um, and then we just need to name, uh, the substitute. So it looks like we have method groups at carbons. 25 and six. Ah, and so this will be too. 56 try. I'm Ethel. Doctor now. Well, that s O c. Is a ring. So as I said, that's a little bit different. Um, basically, we just need Thio. Name the ring and add the Suffolk scar. Valda, hide. Um, and we need to number the ring such that the Alba hide is attached at carbon number one. Um and so that's gonna look like this, Then 123 four or weaken number at the other way. It's not gonna matter. In this case, the chlorine is there gonna be on carbon three either way. Ah, and so this becomes three three dyke Loro Psychlo Beauty in Karbala. Hide dyke. Loro Psychlo, you teen carb Alba. Hide. Okay. On DSO. That's it. Um, thes naming rules could be found immediately above this problem in textbook. Um, and it's their consistent with value pack roles that we've seen in earlier chapters. We're giving the AL Jihad priority so that it gets numbered. Carbon number one, uh, finding the longest chain numbering accordingly and making that the base chain on, then just naming the substitute NSA's we've seen before. And that's the answer to Chapter 21. Problem number three

This is the answer to Chapter 21. Problem number five from the Smith Organic Chemistry textbook. In this problem, we're asked Thio come up with the AIPAC name for these three key tones that we've been given. Um, okay. And so the rules for naming key tones. Um, so for a and C, the open chain key tones, um, we need to find the longest chain that contains the key tone on change. The ending to own, um and then you need to number, uh, that change so that the key tone has the lowest number. Okay, um and so for a that's going to look like this. So 123 So the key tones of carbon 34 56 78 So this is gonna be an Octa known. Ah, and it looks like we have an ethel group at carbon five and a metal group at carbon four. Um, and so when we put all of that information together, this becomes five Ethel for meth A ll. Because, remember, it's in alphabetical order for substitute mints. Ah, so five ethel form Ethel three Octa known and again that O N e ending signifies key tone okay. S o B is a cyclic key tone so that the rules a little different. We just need to name the ring. Um, and so this is a five member ring, so it's gonna be a cyclo plantain, but again will change the ending to own, So it's gonna be Psychlo pent and known. Um, and then we need to number this ring such that the key tone is carbon number one. Um, and then we want the substitue INTs. The other substitutes toe have the lowest possible numbers. So that is gonna look like this Carbon one carbon to carbon three. Ah, And then remember, for alphabetical purposes, we ignore, uh, TERT. So, uh, the the beautiful takes precedence. So this is gonna be three hurt beautiful to me, Ethel Psychlo pent and own. So three turns beautiful to me, Ethel Psychlo Penton own. Okay. Ah, and then for see, we're back to a, uh, an open chain key tone. Um, see was given to us in ah, lettering notation. So I would always just rewrite that in skeletal notation because it's so much easier to see in to work with, um And so since this molecule was cement. You're gonna symmetric through the key tone. We can start numbering at either side. It doesn't matter it Either way that we do it. The key tones gonna be a carbon three. It's gonna be a pain tain derivative on. We're gonna have to method groups at carbons two and four, uh, or rather to metal groups each at carbons tuned for. And so this is gonna be As I said, a paint ain't derivative. So it'll be Penton known. Little free time to known because that's where the key toe news, Um and then we just need to account for those four methods. Substitute INTs. So it's gonna be too, too. Four four. Hedstrom. Ethel. Three. Pent unknown. Okay. All right, so that's it. Um, and these rules, if you need to review them, can be found immediately above this problem in the textbook. Ah, and that's the answer to Chapter 21. Problem number five

In this caution. We just have to draw the structure off. Algy Hyde or Taquito. So option A, we have property. Now that is an alley hide. Here is the structure in case too. We have a painting on which is Taquito. So here to represents deposition off double bonded oxygen on a second carbon atoms. So we have first carbon second carbon atom with oxygen toad 4th 5th Moving towards option C. We have team a time to to beauty Known again We have aqui TEM. So to beauty known means that the position off key tone oxygen will be on second carbon atom while the position off me tight group will be on third carbon atom. So marking the carbon we have 1234 carbon atom On the second we have key tone oxygen while on the throat Carbon Adam, we have a metal group. And lastly for option D, we have to Mikhail butin a ll. It's an alley hide having a meeting group on second Garvin toe the title


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