The following compounds have the same molecular formula as ben...

Question

The following compounds have the same molecular formula as benzene. How many monobrominated products could each form?

1. HC≡CC≡CCH₂CH₃

2. CH₂=CHC≡CCH=CH₂

Accepted Answer

Hey, everyone. And welcome back, predict how many Moana brominated products each of the following compounds can produce. First, let's draw the given structure starting with A M, we have two metal groups bonded to a double bonded carbon atom. Let's begin simply by drawing that double bond. And let's add those two metal groups to that carbon. Now that carbon is double bonded to ch, right? And then we have CC triple bonds. So the only possibility for us to get it is just to make sure that we bond hydrogen to that carbon. So let's add hydrogen and then let's add that CC triple bond, right? That's what we're going to do. So if we draw a bond line structure, we can just say that there's that triple bonded carbon with an implicit hydrogen. So once again, let's see, we have two metal groups bonded to that carbon, then we have our double bond carbon with hydrogen. And then see, see that's exactly what we have one carbon atom, the other one. And there's an implicit hae Jin. The question is how many structures can we get for mono brominated products? So if you look at the structure we need to think about how many hydrants can we replace? First of all, we can replace one hydrogen within the metal group. So if we do that, we essentially add our bromo substitute print bonded to that carbon. Notice how we have replaced one of the hydrogen atoms we have seized to group here. So this is one possibility for us. Now, the other possibility will be to replace hydrogen from the other metal group. They're, they're not equivalent because the products produced with essential correspond to stereo is MERS notice how if we replace the other hydrogen from the adjacent metal group with broom in, we will get a stereo is more. So if we do that, we get the following structure, notice that they are not the same because the two metal groups, even though initially they were exactly the same. Now we get a pair of stereo is MARS hydrogen and metal group. They are lowest in priority and they're pointing in opposite directions with respect to the double bond. Meaning this is our E stereo is more because the higher priority groups, our cc right, triple bonded and our CH two BR they're pointing in opposite directions. So that's E and this one is actually see because the highest priority groups are on the same side of the double bond. So they are different structures. So we have already analyzed our two different hydrogen atoms that we can replace with booming and we can move on, we can also replace this hydrogen atom at the other carbon atom within the double bond. So if we do that, we get the following structure, let's replace that hydrogen with broom in there will be no stereo is um are produced, right? Because the two metal groups are exactly the same on the other end of the double bond. And therefore, if you have to exactly same groups, no stereo is mars are possible. And finally, that means you have your final possibility where you simply replace that hydrogen atom within your al kind. So that means you're going to replace that implicit hygiene within the triple bond with your bromo substitution. And this will be your force structure. There are no more hydrants to analyze and we see that we have a total of four structures and that will be our final answer for the first part for part A and now let's move on to part B. So for part B, We are essentially given the following structure. It says CH two ch double bonded. So we have our double bond. Now, ch has bonded to carbon or basically sees bonded to carbon that contains a triple bond. And then there's a metal group bonded to that sp hybridized carbon. Ch essentially means that we have one hygiene at that same position. In addition to that, we have to hide genes within the siege two groups. So overall, our structure looks like this siege to double bonded to CH, right? And then we have our CC triple bond with a metal group. How many possibilities do we have here? So let's replace the first hygiene with grooming. If we do that, we get our bromo substitution at position one right now, what if we replace the other hygiene? We indeed get a stereo Esma, right? So that's good for us because that would be a different structure. So we can also replace the bottom hygiene with grooming. The reason why these are stereo eyes versus because if you look at these two hydrants are on opposite sides of the double bond and in the second structure there on the same side, so that's our trans structure. This is our system. So we are good now we can replace our next hydrogen atom. Okay. So within the second double bonded carbon atom and let's see that. So if we do that, we leave our original hygiene at position number one, but now we are replacing the hydrogen atom at position number two. So that bromo substitution knows theory allies marks are possible, right? Because on the left, we have two exactly same hydrogen atoms and the world lost possibilities is to just replace one of the hydrogen atoms within the metal group. The three hydrogen atoms within the metal group are exactly the same. So we are going to replace one of them and we get the following. So let's draw it out. Now we are going to eliminate grooming and get our hygiene back. But now we are going to replace hydrogen from the metal group and this siege, we would now become CH two BR. So we're just going to indicate that c carbon bromine bonds, right? To indicate that we are placing one hydrogen with our bromo substitution. And now if you look at these structures, we got four of them as well. So let's label, this is our final answer. You may essentially neglect our model. This was the original structure. It does not belong to our mona ruminated products. And same thing goes to the first part, this was the original structure before we perform mon abomination. So for each part, we got four structures and based on our answer, the correct answer to the multiple choice questions, say, thank you for watching.

The following compounds have the same molecular formula as benzene. How many monobrominated products could each form?
1. HC≡CC≡CCH₂CH₃
2. CH₂=CHC≡CCH=CH₂

Key Concepts

Monobromination

Alkyne and Alkene Reactivity

Structural Isomers

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