Draw and name all the chlorinated benzenes having from one to ...

Question

Draw and name all the chlorinated benzenes having from one to six chlorine atoms.

Accepted Answer

Everyone and welcome back in today's video, we are going to number one draw the structures of all brominated benzene is having from 1 to 6 bromine atoms and number to name all the structures from part number one. Now because our goal is to control all the possible brominated ben scenes, it's essentially easier for us if we draw a benzene ring because that would be our parent. Right? And let's do that. That'd be our benzene ring. And thinking about the strategy in this problem because every benzine ring will be substituted with bromine atoms. We will just start with one bromine atom as the problem suggests and we will keep adding more and more bromine atoms up to six until we are out of any opportunities to draw more. So if we draw just a mono substituted Benzene, we can only add brahman see this one position over here because even if you added at any different position, your numbering system according to the pack rules rules will change and roaming will always occupy position number one. Right? Because there are no other substitue ints that'll be exactly the same compound where you may simply rotated and you will notice that you're still using the same compound for your rotation. But you visually see that a different position even though it's still your carbon number one, meaning there's only one possibility to get a mono substitute at como Benzene and therefore the name of this compound would be bromo Benzene right? Because there are no other substitutes. If it's bonded to position number one, we don't need to write that. So the name for the structure is bromo Benzene if this is the only possibility we are ready to move on. And now let's think about t bromo substitue ints the easiest strategy is to add them adjacent to each other. So first of all let's add two bromo substitue ints adjacent to each other. So one and then the other one on the right hand side. Now it doesn't really matter if you add it on the right or on the left, right because he always control that vertical line of symmetry crossing your benzine ring. And you will notice that either you had this room in here here doesn't matter because this will still give you one comet to dye bromo Benzene. Right, so we can give it the name, one comment to die bro mo Benzene. Right. So now what if we draw, what if we keep the top bromine atom fixed and shift the other one downwards. They would not be adjacent to each other, meaning that works for us and that would correspond to the constitutional is more of the previous structure. So we may simply add the first bromine atom over here and then the next one Over here. We may notice that this is 1:03. Right? So that'd be 1:03 Die because there are two promo substitutions bromo Benzene. Okay now what if we keep the top roaming fixed but shift the other one downward. Once again, basically we're going clockwise. Right, So it would appear at position number four. That would be a different name. Right? Position number one and position number four they would be occupied by bromine atoms meaning those two structures would be different. So now we have position number one and position number four, meaning this compound would be one comma 1234. Right, so that be one comma four dye bromo Benzene. Now can we get any more structures from for a di substituted benzene? Actually we cannot because even if you shift this room into that position, if you still keep going clockwise, you may think that you may possibly get one comma five promo Benzene but this is not really possible according to the pack rules, we're trying to minimize our coefficients of the substitutions meaning instead of going clockwise we would name it counterclockwise and that would be one comma three robot Benzene or basically this structure. And you may notice that even if you flip it with respect to the vertical line, this ramos obsession will appear here. So that's actually your position number three. Not position number five, meaning we are out of opportunities to draw any more di substituted Benzene. So that essential implies we are ready to move on to three bromine atoms. And just as we did it before we may simply draw three adjacent bromo substitutions. So let's suppose we add one over here, another one And then the 3rd 1. First of all we give it a name. So that would be our position number one. The second one. The third one. So we call it one comma two comma three. We had a prefixed. Try because there are three substations. Try bromo Benzene. Now from here our strategy is to keep one of the actually two of the bromo substitutions fixed and let's keep changing the position of the remaining substitution. So let's suppose that we threw our benzene ring, we keep two of them fixed. So we are going to draw one of them the second one and we are just going to make sure that this roaming is not adjacent to this roaming anymore. So we are going to get it over here and then we need to think about the correct name of this structure. So if we start with this roman atom that'll be 124. Right? If we go clockwise, if we go counterclockwise, let's suppose that we start with this one. That'll be 134. That doesn't work first because the coefficients are bigger, right? 12124 is better than 134. Right? Because instead of having three we have a lower coefficient of two. So that means the name of this structure would be one comma two comma four. We use a prefix try to represent three bromo. So that would be try bromo. Benzene. Okay, well done. Now. What if we add this promo substitution to this position. Let's still keep these two bromo substitutions fixed and that would actually correspond to the same compound. Right? You would have one comma two comma 34. Right? So even if you had this prominent position over here that we 1 to 4. So that means we're out of opportunities. We even if we added here we would get one comma two comma three. So we are done wet this part, right? We cannot say we are done with we bromo substitutions because we only made sure that we keep two bromo substitutions fixed. Right? But what if we think about the structure in which none of the bromo substitue ints are adjacent to each other. Let's suppose that we essentially shift this Romo substitution downwards, that would be different structure. So we may notice that none of the blooming atoms are adjacent to each other, they're all separated. So this one And we are shifting this 11 unit downwards. So we can add it over here and you may notice that it's pretty symmetrical, it has a lot of lines of symmetry. So we can name it from here and go clockwise. So that be one comma three comma five. Right, so let's name it. One comma three comma five. There are three bromo substitutions so that would be try Romo Benzene. And finally let's think if there are any other chances to distribute three blooming atoms. Even if we shift this one over here we would get two adjacent ones and one separated, right that we still one comma two comma four. There are no other ways to get a new compound. So we are done with Three bromo substitutions, meaning we are ready to look at four bromo substitutions. If we think about four bromo substitue ints we are going to start our problem by drawing four adjacent ones because once again that's easier for us to visualize and to make sure that we're not missing anything. So we may just add or bmo substitutions that B one, the next one, the third one And the 4th 1. Okay, so we may immediately name it if we go clockwise because apparently that would correspond to the lowest possible coefficients in our name. That would be 1234, one, comma two comma three comma four, we are going to use the brief etcetera Because there are four substitutes that trouble bromo Benzene which is our parent. So that's the name of the structure and now we can think about it this way. What if we keep these three bromo submissions fixed and only change the position of the fourth roaming? Let's add it over here. This would indeed give us a different structure, right? Because the four bro mean Adams would not be adjacent to each other. So let's add three of them, one. See you The 3rd Romain. And let's shift this room into that position and we can now give it a name, that would be one comma two comma three comma five. And even if we go in a different direction, let's suppose we go this direction. That'd be one comma two comma three comma five. Right, So it's still the same name, no matter which way you go because it has a line of symmetry. So let's give it a name that be one comma two comma three comma five. We're going to add Tetra because there are four substitutions. Tetra bromo Benzene. Okay, well done. Now, as we previously said, there is no other way to keep three bromo suppositions fixed. Right? Even if we add this one over here, that would correspond to the first structure. If we added here, that would correspond to the first structure. So that's the only way to make sure that the three bromo suppositions are adjacent to each other, meaning we have another possibility. What if we keep only two bromo substitutions fixed at the original positions one and two and we shift the other one. We can also make sure that this one stays at its original position within the structure. So we can shift only that roaming downwards. This would allow us to achieve two bromo substations adjacent to each other and another pair of two bromo substitutions adjacent to each other, that would definitely be a different structure. So let's see how that looks like now physicians one and 2 are occupied then we are essentially shifting this bombing downwards So it would appear over here and then there'll be another one over here. So if we look at its name, let's try to name it. That would be one comma two comma whereas three is empty so we have substitutions at positions four and five so 1245. We can name it one comma two comma four comma five which is indeed a different name. So it doesn't correspond to any other names that we got and therefore this is indeed and the original structure we're going to give it a prefix etcetera because there are four substitutions. Tetra bromo Benzene. Okay now are there any other possibilities? Actually there aren't whenever you try to shift any of your blooming atoms you either get three bromo substations adjacent to chatter, right this one or that one. In each case we have three bromo substitutes adjacent to each other or four bromo substitutes adjacent to each other. So we are out of opportunities for for bromine atoms meaning we are ready to move on and troll summers 45 romain atoms or Penta substituted Benzene. We may again begin by drawing five bromo substitutions consecutively. So one two bromine atoms three four and five. So now let's give it a name. What would the name for this structure be? One comma two comma three comma four comma 5. 10 to bromo Benzene. Now let's think about it. It's actually the only structure that we can get because let's suppose that you eliminate one of the roman atoms. Let's randomly choose this roman atom and added over here at the remaining position it would still give us five adjacent roman atoms. So no matter which one bro mean you eliminate and changes positions to the remaining M. T. One, you still end up with five adjacent roman atoms, meaning it's the only possible structure. And therefore because it's the only possible structure, we don't need to add coefficients and we are just going to call it panta because we have five positions substituted by roman Adams. So that'll be pensa como Benzene and we are done And finally we need six. So if we draw a Benzene ring, let's add sex roman atoms. That means we're occupying every position of the Benzene ring. And that would again correspond to the only possibility because there are no vacant spaces for us to use to change positions of our booming atoms, meaning that this is the only possible distribution of six. Booming atoms to occupy every position within the benzene ring. And therefore again we don't need to use coefficients to represent their positions because this is the only possible arrangement. The name would be Hexham meaning we have six substitutions. Hex up bromo Benzene. Now we are done. Right because we started with one bromo substitution and we went up to six. We noticed that we used all of our possibilities. We may also noticed that the total number of eyes. Summers actually, my apologies that they wouldn't be all eyes on us, right. They're actually different structures in many cases because we have a different number of grooming atoms in some of them. So we can say that there will be total of 12 structures to draw Brominated Benzene that contains from one up to six bombing substitue ints. And therefore because we see total of 12, let's label all of them starting from the beginning, starting from here And then going downwards. We noticed that we have a total of 12 and this would correspond to the answer C in the small cultural question. However, if you have any questions, don't hesitate to leave your comment down below in the comments section and thank you for watching

Draw and name all the chlorinated benzenes having from one to six chlorine atoms.

Key Concepts

Chlorinated Benzenes

Substitution Reactions

Isomerism in Chlorinated Compounds

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