classify-each-pair-of-compounds-as-isomers-or-resonance-structures

Name: classify each pair of compounds as isomers or resonance structures
Uploaded: 2019-08-05
Duration: 2 min 37 s
Channel: Numerade
Description: Classify each pair of compounds as isomers or resonance structures.

Question

Classify each pair of compounds as isomers or resonance structures.

Ian Kaigh · Accepted Answer

This is the answer to Chapter one problem number nine from the Smith Organic Chemistry textbook. In this problem asks us to, uh characterized two pairs of molecules a CZ Either I simmers or residents structures. And so remember, In order to do this, we need to remember the two different definitions Eso ice summers are going to have the same formula but different connectivity. Um, and residents structures are going to have the same formula. The same connective ity of the bonds, the lone pairs and the charges are going to be arranged in different ways. Um and so for a what we have is a pair of eye simmers. Ah, and it&#x27;s easy to tell that because in the 1st 1 you have nitrogen connected to carbon are connected to oxygen. And in the 2nd 1 you have nitrogen in the middle of carbon connected to nitrogen, connected the oxygen. So the formula here is the same. But the connectivity is different on therefore, by the definition that I just reviewed their eye summers, um, so and be what we have is a pair of residents structures, uh, and you can see that they&#x27;re residents structures because the connectivity is exactly the same. So we have a central carbon in each surrounded by three oxygen&#x27;s, one of which has a hydrogen. Um, the only thing that changes is the location of the double bond and the location of the negative charge. So, uh, in the structure on the left, we have a single bond to the oxygen on the right in native charge on that same oxygen, Um, and then in the structure to the right, we have the single bond instead. Toothy top oxygen. Ah, and that oxygen also has the negative charge on it there. I mean, so the exact same connectivity, but different, um, the bonds and the charges were different. And so that&#x27;s going to be a pair of resident structures. And that&#x27;s the answer to Chapter one problem number nine.

Chloe Schroeder · Answer

constitutional eyes. Others have the same look of away, but a different connectivity. Stereo chemical I summers actually have the exact same connective ity, but different spatial orientation. So, um, stereo chemical I summers will have the exact same name except for the position of sis versus trans or R vs Ass stereo chemistry for this first compound, the longest chain is one, two, three, four, five compound or five carbon atoms. Same here. One, two, three, four, five. The difference isn&#x27;t where the methyl group is placed. So here it&#x27;s on carbons two and four here some carbons, two in three. So these have physically different names for their compounds. Therefore, there I summers of one another, their constitutional icers For part B. This is probably easiest to look at. What functional group resides in both these compounds? This one is an ether where this one is an alcohol. Therefore, it must be a constitutional iceman. For this compound, we have the one one cyclo are one one dimethyl cyclo butane. Whereas here we have a one to diamond full cycle butane. So the names air physically different, therefore, is a constitutional ice mma. And finally, for the last one, we have a one to dimethyl cycle butane. In both cases, the only difference is, in this case, the two meth of groups, or sis, where, as here they are trans to one another. Therefore, these air stairs

Chloe Schroeder · Answer

classify each pair of compounds. His constitutional I summers are identical molecules. So I brought up this page for reference. Just so we draw things the same, I&#x27;m going to start with this isopropyl change. So we have to ch three groups attached to CH and that is bonded to the central carbon here, which is also bonded to an ethnic group. So there&#x27;s the EPA group and a hydrogen. So these are the three groups attached to the central carbon. And then we&#x27;re looking down this bond in this direction attached to the back carbon are to hide visions and a method. So is this the same compound as this next one? Let&#x27;s do the exact same thing. So to method groups and hydrogen attached the first carbon. And then we&#x27;re looking down this bond and on the back carbon we had and Ethel group a hydrogen hand and foot. So the question is, Are these two compounds the same? Well, let&#x27;s go ahead and to leave the hydrogen is then I make it a little easier to look at. These are indeed the same compound. We have an isopropyl group here with two F&#x27;s group stashed to the other carbon. So these air the same compound, just looking down a different bond for this next one. Let&#x27;s do the same thing. Let&#x27;s go ahead and start with the front carbon. So we have two methyl groups and a hydrogen attached to the first carbon. I&#x27;m gonna go ahead and get rid of the hydrogen. I don&#x27;t think at this point in time it&#x27;s helpful to visualize that with the product. I just want you to recognize what I&#x27;m looking at. Then we&#x27;re looking down this bond so the other carbon has a metal and isopropyl group attached. So Ethel Hydrogen and Isopropyl Group. So the question is, are these two compounds the same? Well, they are not. May take a little time to see this, but this compound has two isopropyl groups attached to it. This one only has one. So one branching point with two metals coming off of this carbon. So it&#x27;s like moving this metal onto this carbon would give you the same compound thes heir, Not the same

Ian Kaigh · Answer

after four. Problem number 61 Fromthe Smith Organic chemistry textbook. Ah, And this problem gives us, um, or gives us eight pairs of molecules and asks us whether they are constitutional. I simmers stereo. I simmers. Ah, identical molecules or not Ice Immers. Um, okay. And so when we look at a, um, these two molecules have the same molecular formula, but a different connectivity. So these are constitutional. I simmers so constitutional. Okay, same formula, different connectivity. When we look at B by, these have the same formula and connectivity, but a different arrangement in space. And so these are stereo. I simmers when we look at sea. Um, so we have one metal group on each side of the ring here in the first molecule. And then when we look at the second molecule, we see ah, the same arrangement. So for C, these are actually the same molecule, so I&#x27;ll put same um, so then, looking at D, obviously, uh, each of these is a cyclo hack saying with two Ethel substitue INTs. But the Ethel substitue ins are on different carbons. So in the first molecule, there&#x27;s a 13 relationship in the second molecule there&#x27;s a one to relationship. I said that these are definitely not the same molecule or stereo I summers, but they are constitutional. I summers because their formulas are the same, but they differ in connective ity. Um, okay, so then moving on to e um e they don&#x27;t have the same formula on. And so right there we can stop and say that these are not ice Immers because remember, Ah, the same formula is a requirement for any type of ice over. So then looking at F, um, we have in the first structure, we have a trans relationship and then thinking about, um, the second structure. So if we were to look at this chair structure from above were to look at it like this from that angle, both of these Ethel groups are going to be facing away from us on dso Uh, that&#x27;s gonna be assists relationship since they&#x27;re both on the same back side of the chair here. I&#x27;m and so these are gonna be stereo I summers. So the connectivity is the same, but the one on the left has a trans relationship, and the one on the right has a cyst relationship So their stereo I Summers s so then moving on to G um, again if we look at these shares from above. So in the 1st 1 if we look us at this angle, um, we see that both of these meth a group or both of these ethel groups are coming at us. Um, and then if we go ahead and look at the second chair also from above again, both of these ethnic groups are coming at us. So the connectivity here is the same. Um, the cyst relationship in both of these is the same. Um, and, uh, indeed, these are the same molecule. Um, they&#x27;re identical molecules. Ah, and then lastly for H. Um, obviously, the connectivity is different here. Um, So, uh, we need to consider them, like formula. Both of these structures are of the molecular formula. See eight, h 18. And so these are going to be another pair of constitutional ice mers to show? No. Okay, I&#x27;m in. So that&#x27;s the way to approach this problem. We just need to remember the different definitions of constitutional. Ah, I simmer stereo. I simmers. Um, what would make something the same molecule and what would make a pair of molecules not I summers at all on, and then just apply those roles to each of these pears molecules. And that&#x27;s the answer to Chapter four problem number 61.

Chloe Schroeder · Answer

to classify whether a pair of compounds are ice memories. Resident structures were neither. The first step is to determine their molecular formulas, so I&#x27;ve done that for you already. A is C to H for O two BC to H for oh to C C to H five o two and D C to H for Oh, too. I&#x27;m just adding up the number of atoms within each of those structures. It wants us to compare A and B to start with. Well, first, we need to know what a nice summer is, what a resident structure is, and then we can determine neither from there. So when I stammer is two compounds that have the same molecular formula. But they&#x27;re connected in a different fashion, so the atoms are actually connected differently. Nothing about the actual member bonds or number of electrons, but different connectivity of physical Adams. Resident structures have the same molecular formula and the same Adam can activity. You&#x27;re just pushing electrons, pushing pie bonds, pushing electrons, and then neither, of course, probably is just not gonna have the same molecular formula. So let&#x27;s compare Ambi Well, they have the same molecular formula, so it has to be either a nice summer or resident structure. So the next thing you want to look at is how the atoms are actually connected. We have CH three C In this case, it&#x27;s bonded to two Oxygen&#x27;s One of them&#x27;s an alcohol ch three psi oxygen Ohh So they had the same connective ity. You&#x27;re just pushing electrons instead. So this they must be resident structures. And at the end, we&#x27;LL withdraw Samaras to prove that Compare A and C well, it hasn&#x27;t like their formula of sea to H for to see has him like your form of eight C to H five o two already that tells us that can&#x27;t be I sinners or resident structures. So the answer is neither compare Andy have the same liquid formula, so it could be a nice summer or a resident structure. Now we have to look at connectivity. Well, this carbon is attached to two oxygen&#x27;s in a carbon this carbons attached to a hydrogen and oxygen and apartment. So already we have different connective ity. You&#x27;ve physically moved the location of the atoms s so here we have a ch too attached to an oxygen and carbon here we have a CH three so physical, different connective ity, but with the same molecular formula is a constitutional iceman. Last one they want us to consider is B and D. So we said A and B, we&#x27;re residents forms. They have the same molecular formula. B and E also had the same molecular formula, but they&#x27;re connective ity again is different, just like with a So this carbon has attached to two oxygen&#x27;s another carbon. This carbon is attached to a carbon and oxygen and hydrogen, so therefore, it also must be a nice summer. The last point I want to make is drawing the physical, resonant structures. So let&#x27;s just do that quickly to go from this residence structure to this one. What happened was this. Oxygen has changed in order to form a new double bond between the carbon, and it&#x27;s using a lone pair of electrons to do that one Dr Resident structures. You want to actually push electron density that&#x27;s forming this new bond. This carbon cannot have five bonds. You have to break one. Easiest bond to break is a pie bond, so we can push that piland up as a long pair of electrons onto this oxygen, and I gives us our resident structure if you want to go the opposite direction, push alone Perrot of former New Pie Bond. This carbon can&#x27;t have five bonds. Easiest way too easy. Spawned A break is the pipe bomb, so that could be pushed up on the oxygen.

Chloe Schroeder · Answer

So we need to compare these compounds of a constitutional isom Our story. Why summers or not ice murders at all? So for this first one, we have the same number of atoms. So four carbons here, four carbons here, one oxygen and the same number of hydrogen zzzz. Well, so they have the same like regular formula. Therefore, they have to at least be I summers, but they actually have a physical Conectiv ity differences. So this wasn&#x27;t either. This one&#x27;s a key tone. Therefore, there constitutionalism er&#x27;s different Adam connective ity is a constitutional I smirk, Part B. These are both key tones, but they&#x27;re also constitutional Ice summers, I believe one, two, three, four No, these are actually not ice in words of one another. And the quick way, I can tell is based on the number of unsaturated ins. So rings and pie bonds count for unsaturated ins, and every molecular formula has the same number of ice of unsaturated ins. That&#x27;s how I quickly saw this. But you could count up all the atoms and you would see that the molecular formulas are actually different. So here we have one, two, three, four, five carbons three, six seventy ten hydrogen and one oxygen. Here we have two, four, six, eight Hydrogen Sze. So they&#x27;re physically different molecular formulas. Therefore, they&#x27;re not ice mers. And I smart has the same molecular formula for part. See, we&#x27;re comparing this flat structure with the chair come from er in the chair. Both of the methyl groups are going up relative to the hydrogen. Is that air attache in that carbon? So these metal groups are sis. If we look at the flat version while one is going up and the other method group is going down, this is a translated relationship. Therefore, these air stereo I summers finally for the last compound, we have physical, different connectivity of the methyl groups. So here you have a one three relationship. Here you have a one one relationship as to where the methyl groups are. Therefore, these air constitutionalism er&#x27;s

Problem

Considering structures $\textbf{A-D}$, classify e…

Problem 9 Easy Difficulty

Classify each pair of compounds as isomers or resonance structures.

Answer

a. The compounds are isomers.
b. The compounds are resonance structures.

Related Courses

Organic Chemistry

Related Topics

Discussion

Top Organic Chemistry Educators

Bobby B.

Lizabeth T.

Julie G.

Nicholas M.

Organic Chemistry Courses

Introduction to Organic Chemistry

Structure and Bonding - Overview

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Janice Gorzynski Smith

Organic Chemistry

Bobby B.

Lizabeth T.

Julie G.

Nicholas M.

Introduction to Organic Chemistry

Structure and Bonding - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a. The compounds are isomers.b. The compounds are resonance structures.

Organic Chemistry

Bobby B.

Lizabeth T.

Julie G.

Nicholas M.

Introduction to Organic Chemistry

Structure and Bonding - Overview

Janice Gorzynski SmithOrganic Chemistry

Bobby B.

Lizabeth T.

Julie G.

Nicholas M.

a. The compounds are isomers.
b. The compounds are resonance structures.

Janice Gorzynski Smith

Organic Chemistry