Draw all internal symmetrical alkynes with the molecular formu...

Question

Draw all internal symmetrical alkynes with the molecular formula C₁₀H₁₈.

Accepted Answer

Hey everyone and welcome back in today's video. We are going to draw all the possible alkin isomers with the molecular formula C five H eight and assign common and systematic names to each. First of all. Let's see if we only have one triple bond in the structure. And to do that, we are going to calculate the hygiene deficiency index, which is given as two multiplied by the number of carbon atoms plus two plus the number of ngen atoms minus the number of hydrogen atoms and minus the number of halogens. Everything would be divided by two. In this case, we have two times five plus two plus zero minus eight minus zero, divided by two. If we do the math, we have 10 plus two, right? That's 12 minus eight, that's +44, divided by two is two because the problem suggests that this must be an alkin. It makes sense if you have a hydrogen deficiency index of two, it might correspond to one triple bond or two double bonds or two rings or a ring and a double bond, right? So you have many possibilities and a triple bond is one of them. So the easiest way to solve this problem is just to start by drawing your triple bond. And then since you need a total of five carbon atoms, you will add three more. So basically, you have 12345. So that's the first structure you can get, you can now start thinking about changing the position of your triple bond. What if your triple bond starts at position two? So you can draw your triple bond at a muscle group on one side, you have three carbon atoms, you need to add two more. So that means you're going to add an ethyl substation on the other end. Now, the question is, can we add our triple bond at position three? If we essentially move it to position three, we will get the same structure. Notice that your triple bond will essentially start at position three and extend to position four, but you will change your numbering and it will say that it's actually the same structure where you have the double, the triple bond starting at carbon two. So we cannot shift our triple bond anymore. We can also not branch our chains here. We cannot branch a metal group, we cannot branch a an group. So we can actually start thinking about branching the chain of the first structure, right? We can think about branching our propal substituent. That's how we can visualize it. And we can basically draw our triple bond if we branch it let's suppose that we add this carbon at the first one in our substitution or basically, let's make prop an isopropyl group, right? So it will look as follows. So now we got our branch chain and that would be it. There are no more possibilities for us to analyze. Let's simply name them to name all of these structures. We are, first of all going to separate them a bit. So we have sufficient space. Let's start with, are you pack names? So are you pack and to give them I names? Let's remember that because those are Alkins, they will all have suffixes of iron. We are going to start with selling that there are five carbon atoms in the pan. So we are going to say right, the position of the triple bond starts at carbon number one. So we're going to say T one and we will finish it with an appropriate suffix I. So this is the name for the first one. The IPA name for the second one will also be pant because we have five carbon atoms in the longest chain. But now it would be pant two im because the triple bond starts at carbon two and for the file structure, the parent has 1234 carbon atoms. So we're going to say that this is time. But in addition to that 123 at position three, we have our metal subs positions who are going to say three metal. And only now we're going to say that this is but one I right, we have a four member kind. The position of the triple bond starts at carbon number one. Now, we simply wish to finish with the common names. So let's add common and to give them common names. Let's recall that each structure or basically, you're going to think about each triple bond as part of acetylene. Remember that acetylene or basically your triple bonded carbon atoms with two hyen atoms, that's acetylene, right? And every common name will be based on acetylene, but you're going to replace your hygienes with alki groups and you're going to name those alki groups followed by acetylene. So for example, if we look at the first name, I would be our 123, right? That's our prop group. So it makes sense to give it a name prophyl acetylene. In other words, we're saying that we are replacing one hygiene from acetylene with a acetylene. Similarly, for the second structure, we are now replacing two hydrogen atoms. So we have two substitutions. One of them is an ethyl group. Another one is a metal group, we want to arrange them alphabetically and e gets the priority alphabetically. So we're going to say that's our first substi the second one is me and that's it acetylene. So notice how if it's just one substi you're just adding one of them and you're still using acetylene, it automatically implies that on the other end of your triple bond, you have hydrogen prop acetylene, you have a prop group bonded to your acetylene fragment. So it implies that there is this one terminal and where you have hydrogen ethyl methyl acetylene means that you have an internal alk where the two alkie groups are bonded along your triple bond. And for the pound structure, we have our isopropyl group, that's how we call it, right? And we don't have any other group. On the other end, we have hydrogen. So we can simply call it isopropyl acetylene. And that would be it. We can label these as our final answers and conclude that the correct answer to the multiple choice question is a. Thank you for watching.

Draw all internal symmetrical alkynes with the molecular formula C₁₀H₁₈.

Key Concepts

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Symmetry in Organic Molecules

Molecular Formula Interpretation

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Draw all internal symmetrical alkynes with the molecular formula C₁₀H₁₈.