Draw and name all five isomers of formula C₃H_5...

Question

Draw and name all five isomers of formula C₃H₅F.

Accepted Answer

Welcome back everyone to another video show and name all five isomers with the molecular formula C three H five CL use the Iupac naming system. First of all, since we the molecular formula, we once calculate the hygiene deficiency index, which is also known as the degree of saturation, so we will have one half multiplied by two, multiplied by the number of carbon atoms plus two plus the number of nitrogen atoms minus the number of Hagens minus the number of halogens X. And in this case, we have one half multiplied by two, multiplied by three carbon atoms plus two, we don't have any nitrogen atoms. So we are just subtracting five hydrogens and we're subtracting one chlorine. So if we evaluate the result, six plus two is 88 minus X is two and two divided by two is one. So the hydrogen deficiency index of one essential, it tells us that we either have one double bond on our structure or a ring. And this allows us to narrow down our possibilities. Essentially, we can start by drawing a ring structure and we have to remember that chlorine will form one bond only, meaning we can only have a three member ring of carbon atoms. And then we're essentially going to bond chlorine to any of our carbon atom speakers. Our ring is symmetrical. Now, we want to name it, but we can actually do it at the end, right. So let's focus on the strategy behind drawing these isomers. So we got our one and only ring structure because the smallest ring that we can form is a three membered ring. And chlorine cannot be part of that ring, meaning we simply move on to our straight chain alkyl chains. So now we can also add we member chains. So that would be 123 carbon atoms. And then we can bond chlorine at the end of the chain because once again, chlorine can form only single bonds in this case. So we can add our pie bond. Now, because we are done with our rings, we can add that pie bond at the end of the chain or we can also add it between carbons two and one, right, if we count from right to left. So if we add that double bond between carbons, one and two, we can get our next structure. And what we understand is that this structure will actually have a dias right? Because we have an L king, we can have our highest priority groups on the same side of the double bond or the opposite sides. So that means the next one will essentially correspond to chlorine and methyl being on the same side of the double bond. Now, we have four of them. This means that there's an additional one. Now, what if we draw a three membered carbon chain, add a double bond at the end of the chain and bond chlorine to carbon? Number two, this gives us our final possibility, right? And now that we have our five structures, we can essentially name them separately. Starting with the first one. We can say that this is cyclo because it's a cyclic structure, right. So we can say it's cyclopropane and cyclopropane is our parent. We just want to specify our substituent which is chloral. So the name of the structure would be chloral cyclopropane. We're not specifying the position of the chloral substi because it automatically defines carbon number one and all of our carbon atoms are equivalent. So we have choc cyclopropane. Now let's think about the second structure. It doesn't have any E or C isomers because the terminal alkin has two equivalent hygen at the end of the chain. And this means that we are going to give it a name of we f laurel probe one in the L kin gets priority. So we enumerate it from left to right. The double bond starts at carbon number one and carbon number three has that chloral group. So we're saying three chloral probe one in. Now, let's consider our third structure. And simultaneously we're going to consider the fourth one. First of all, we understand that if we correctly enumerate our parent, we want to minimize the lo of the double bond. And this means that the double bond will start at carbon number one, if we go from right to left. So that would be prop one in again. And the chloral group is fed to the same carbon number one. Meaning we're going to say one chloral probe one in. Similarly, we will have a very similar name were the first structure. And the main difference is their configuration. We want to remember that if the highest priority groups are on the same side of the double bond, then it would be C and the reason why the fourth structure Z is simply because we have two implicit hydrogen atoms with the lowest priorities possible on the same side of the double bond for the first structure that will be E one chloral probe one in because the lowest priority hydrogen atoms are on opposite sides of the double bond. And finally, we also want to name our fifth structure. Once again, we have to remember that the al king gets priority. So we enumerate it from the top to the bottom 123. So that means the chloral group is bonded to carbon number two, we're going to say two chloral probe one in. So we have shown and we have named all the five eyes emerged with the molecular formula of C three H five cl, we're going to label them as our final answer and conclude the video. Thank you for watching.

Draw and name all five isomers of formula C₃H₅F.

Key Concepts

Isomerism

Structural Isomers

Nomenclature in Organic Chemistry

Watch next

Draw and name all five isomers of formula C3H5F.

Key Concepts

Isomerism

Structural Isomers

Nomenclature in Organic Chemistry

Watch next

Draw and name all five isomers of formula C₃H₅F.