Match four of the following UV absorption maxima (λmax) with the ...

Question

Match four of the following UV absorption maxima (λmax) with the corresponding compounds: (1) 232 nm; (2) 256 nm; (3) 273 nm; (4) 292 nm; (5) 313 nm; (6) 353 nm. (a)

Four chemical structures labeled (i) to (iv) for matching with UV absorption maxima.

(b)

(c)

(d)

Accepted Answer

Hey everyone. Let's do this problem and says determine the UV absorption maxima or lambda max for compounds 123 and four. The UV absorption maximum possible values are listed below. So the UV absorption maxima for each structure can be predicted using Woodward Fight Sir. Rules. And according to these rules, the linda max equals the sum of the base value. So the absorption for the base of the structure plus the sum of substitue in contributions. So the absorption for the substitue ints and the sum of other contributions. So the absorption from other things in the structure. So let's review some of the absorption values for different bases and different substitue ints and other things that will contribute to that maximum absorption. So first starting with different types of bases, we have hetero hetero annular and homo annular dying's. So hetero annular diogenes look like this. Two rings and they each have a double bond and notice that they are trans and alternating. So they're conjugated. Right? This value will also represent a trans oid. So if it's not in a ring it's the same sort of structure. And this has a base value of nm. All right. A homo annular dying, we'll have both of the double bonds in the same ring looking something like this and this has a base value Of 253 nm. So I'm going to color code, my base values in blue. Now looking at some substitue in contributions and other contributions. We could have additional double bonds Which will be 30 nm added to the total. I'll highlight those in green. We have alcohol or ring residues. I'll abbreviate that as a slash or or And that will be plus five nanometers. And we'll underline these in black. And lastly we can have X. O. Cyclic double bonds. And those will be also plus five nanometers and we will underline these in red or use red to show them on the structure. So I forgot to mention alcohol. Ring residues are basically our groups coming off the main path of conjugated double bonds. And exocet click double bonds. Or when you have a double bond that is outside of the ring in question. But it's still connected, right? It's on the ring but it's just directed outwards. So let's go through our structures and it'll probably make more sense. So, starting with structure number one, what is our base? Do we have a hetero annular dying. A homo annular dying. I see a homo annular dying right here. So that's going to be 253 nm. Alright, do I have additional double bonds outside of this ring? Yes, we have two additional double bonds. Two. I keep wanting to draw peas instead of these today double bonds. So that will be 30 times two, which is 60 nm. Alright, do we have alcohol or ring residues? Let's see. So looking at our chain of conjugated double bonds that goes this way. Right, so do we have our groups coming off of this chain. Well, let me follow along and see if there's another pathway, yep. There's one here, keep going, keep going. Here's another fork. It's almost like a fork in a road like you're driving down the road. And now there's a fork, right, you can go two ways, so that's another one. Keep going, keep going, Here's a third one, keep going, There's a 4th 1 And there's a 5th 1 coming off the other end of that road. Okay, so that is five Alcohol or ring residues which would be five times 5 is nanometers. And anything else do we have exhaust cyclic double bonds. So if we're looking at our rings 123 here that have double bonds. So let's say we're looking at ring number one, right here at this carbon, there's a double bond touching the ring but it's pointing outward. Right? So that is an exocet click double bond. Now looking at ring number two, this is a double bond touching the ring but pointing outwards so that is exocet click and looking at ring number three. This double bond is connected to the ring but pointing outward. So those are three exo cyclic. I'm going to write X. O double bonds which is nanometers, right? Five times three. So all of these added together Gives a value of 353 nm for the UV absorption maxima for this compound, which is choice f in our list. Right, Alright, moving on to the second structure. What is our base? We have another homo annular dying. So base is also 253 nm. Do we have additional double bonds? Yes, there is one double bond. one double bond, which would be 30 nm. Okay. Do we have alcohol ring residues? Hmm. Yes, we do. If we go along our path. So it starts here and ends here. Right. So one going along too. Going along three and 4. Alright, so four alcohol ring residues for AR. ARR will be 20 nm. And do we have any exocet click double bonds. Yes, we do. So, if we're looking at ring one, this double bond is touching but pointing outwards. Right. And let's see if we have ring number two and three. Ring number three. This double bond is touching the ring but pointing outwards. So that is two X. O double bonds which would be five times two is 10. So, adding all of these values up 2:53, 30, 20 and 10. We get 3 13 nanometers which is E on our list. All right, compound number three. What is our base? Well, I don't explicitly see either one of these type of bases but I see this trans oid. Right. We call this a trans oid which will count as the same as the hetero annular dying. So this is our base here. So that will be a base of 217 nm. And let's see, do we have any alcohol or ring residues? We're looking at our chain which starts here and ends here. We have one going along going along too. And that's it. So too, Alcohol ring residues which is two times 5 is 10 nm. And do I have any exocet click double bonds? Yes, in this ring we'll call it ring number one. This double bond is touching the ring but it's facing outwards. So that is one exocet click Double bond which is five nm and that's it for this structure. So adding up these values, We get 232 nanometers which is choice A In our list given to us In our last structure structure number four we have a base here, homo annular dying. 253 nm. And let's see we have another double bond right here outside of the ring. So that is one double bond. 30 nanometers. And what else do we have? Any alcohol ring residues. Looking at our chain as we can call it. Starting from here to here, our chain of conjugated double bonds. We have one, two In three alcohol ring residues three A R R is five times 3 15 nanometers and any exocet click double bonds. Well, let's look at our rings, one in two. If we look at ring number one, there are no double bonds touching the ring. But pointing outward, if we look at ring number two, this double bond here is touching the border of the ring, but pointing outward so we have that one exhaust cyclic double bond. Oh, I lost my color coordination along the way So that one exhaust cyclic double bond is going to be five nm. So adding up these values, We get 303 nm infrastructure for. So with all that being said, that makes a the correct answer for this problem. That's it for this video. I hope this was helpful and thanks for watching.

Match four of the following UV absorption maxima (λmax) with the corresponding compounds: (1) 232 nm; (2) 256 nm; (3) 273 nm; (4) 292 nm; (5) 313 nm; (6) 353 nm. (a)
(b)
(c)
(d)

Key Concepts

UV-Vis Spectroscopy

Conjugation and λmax

Structure-Activity Relationship (SAR)

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