Determine the degree of unsaturation and then draw possible st...

Question

Determine the degree of unsaturation and then draw possible structures for noncyclic compounds with the following molecular formulas:

b. C₃H₄

Accepted Answer

Hey everyone. And welcome back through the possible structures of cyclic compounds with the molecular formula C four H six based on the degree oven saturation, the degree oven saturation or the hygiene deficiency index can be calculated using the following formula. It says that we have to multiply two by the number of carbon atoms, then we have to add to, then we add the number of nitrogen atoms, we subtract the number of hydrogen atoms and we subtract the number of intelligence acts in our structure. And after that, we essentially have to divide our result by two. This gives us our degree of saturation or hygiene deficiency index which tells us a lot of important information. So essentially let's calculate the degree of in saturation for the structure. Now, essentially we have to multiply two by the number of carbon atoms. We have four in our structure. We are adding to their are no nitrogen atoms, right. So we are not adding anything. The number of nitrogen atoms is zero, the number of halogen atoms zero. We essentially want to subtract the number of hydrogen atoms that will be six. So we are subtracting six and essentially from here, we are going to divide our result by two. Okay. So two times four is 88 plus two is 10, 10 minus six is four. So we have 4/2 and that gives us two. Now, the degree of saturation sells us a lot of information. If it's equal to t we can have the following possibilities to double bonds in our structure than one triple bond in our structure than two rings in our structure. So let's suppose that in the general case, we draw rings as three members rings, but it can be any type of ring, right? Or basically, we can just say that there can be two rings Or we can also have one ring And one double bond. Those are our possibilities because the problem tells us that the structures are cyclic compounds. We are going to think about the one that has either two rings or one ring and one double bond, right? So it must be cyclic. So we can consider these two. Now, can we actually have two rings if we have four carbon atoms, the smallest ring we can have, must have three carbon atoms. So we cannot make a second ring, right? We can essentially have only one additional carbon and that's not enough for us to make another ring. Therefore, we are going to exclude that option as well. And the problem becomes really simple. We exclude all of these. We can say we exclude see double bonds, we exclude a triple bond, we exclude two rings. And we're only considering one ring, one double bond. So let's begin our problem. The smallest ring we can get is a three member ring. So we are going to draw a three member grain. And now we need to think about the number of carbon atoms we need to add, we have 123. So we're going to add one more carbon atom. Okay. So we have four carbon atoms. We don't need to worry about calculating the number of hydrogen atoms. We already understand that our structure will only have one ring and one double bond. The number of hydrogen atoms in that structure will always be six. We took care of that in our calculation, we just need to think about what are the possibilities to add our double bonds. Let's suppose your double bond is within the ring. So you have cheap possibilities to add your double bond. You can add that between your ch two group and this ch right or you can add it between your two ch two groups. So your second structure would essentially be a three member ring where you're connecting to CH two groups. Now, if you add your double bond here, you will get the same structure structure. Number one, this is your number one, this is your number two because essentially if you flip it with respect to the vertical line, you will end up with a double bond at this position. So it's not unique, meaning we are done. However, you may also add your double bond over here outside of your ring. So you can draw your 33 member ring and instead of having your double bond within the ring, you can draw it outside of the ring and you can make a double bond here. So that this is your sp two carbon atom and this is an sp two carbon atom, you have ch two bonded to your ring, right? So now we have our third structure and we are done. There are no other possibilities to analyze our three member ring. Now, we are going to expand it into a four member ring and that would involve the presence of all four carbon atoms. So now what are the possibilities to add a double bond here? Do we have four possibilities? Actually, we only have one, right? If you add a double one here that you are one and only structure, you can rotate that square looking structure. And you will notice that if you are rotated 90°, you will end up with a double bond at this position, another 90°, it will be over here and another 90° will be over here. So essentially, you will get exactly the same structure with a different position of the double bond just visually, but it's still exactly the same position. Number one, you're just changing the way it looks. So if you draw something like this, it's basically the same structure which is flipped, right, essentially with respect to the vertical and so on. So basically, we only have one possibility here to add our double bond. This is our structure for. And we understand that those are the four structures that we control. We cannot have a bigger ring. We are already limited to four carbon atoms. So that's the biggest ring we can get. There is only one chance to add a double bond and that will be it. Based on our answer, we may conclude that the correct options. The multiple choice question is d Thank you for watching.

Determine the degree of unsaturation and then draw possible structures for noncyclic compounds with the following molecular formulas:
b. C₃H₄

Key Concepts

Degree of Unsaturation

Structural Isomers

Noncyclic Compounds

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