For those compounds that can exist as cis and trans isomers, d...

Question

For those compounds that can exist as cis and trans isomers, draw and label the isomers.

3. Structural formula of butene showing cis and trans isomers with labeled carbon and hydrogen atoms.

4. Structural formula of butene showing cis and trans isomers with labeled carbon atoms.

Accepted Answer

Hey everyone and welcome back, joint label, CIS trans eyes MERS for the compound that may exist as CIS and trans eyes mars. So we are given two structures. Let's start with the first one. Our goal is to identify our double bond. Notice that it's over here, we're dealing with this transit is on tourism which is applicable to al kings and cycle al canes. So here we have our double bond, meaning we are going to think about the cis trans isomerization for Elkins. First of all, notice that the two bromine atoms are in parentheses. So we're just bonding two carbon atoms V A double bond and we are going to think about the substitutions present. Now see parentheses. BR essentially implies that this bromine atoms bonded to carbon, which is pointing outside of our line, right? It's basically pointing up. And similarly here we have Sea princess BR which means that even though this carbon is double bonded, it still has that blooming substitution bonded to it. Now, if we look on the left side, we have CH right, we're going to draw CH fragment or basically H C but it would even be better to draw this hydrogen below carbon, right? Because we want to make sure that our bonding is consistent ch and essentially it's a C 32 times and Francis, it basically implies that there are two methods, groups bonded to your carbon underwrite. And essentially that makes sense, right? Because carbon must have four bonds and you're seeing only two. So you want to draw too method groups bonded to it. Now, on the right, we have our double bonded carbon with roaming bonded said and then it must be bonded to another carbon, right? Because you have three bonds. So there must be another bond and that carbon has hydrogen and again to metal groups. So we're going to add two metal groups bonded to that carbon. And let's see if everything is consistent. So if we go from left to right, there are two metal groups right to method groups bonded to something and that something is carbon with hydrogen, which is bonded to another carbon. That makes sense. And that carbon Hasbro means. So we added booming, then we have our double bond between the two carbon atoms and we just have exactly the same fragment as we analyzed on the left carbon bonded to booming bonded to carbon, that carbon has hydrogen and Francis here indicate that there are two methods, substitutions bonded to that carbon. Now, we want to think about the system trans exam or is um so first of all, if we think about the structure. Now that we clearly see it, we can simply through our double bond through the two carbon atoms. Cis trans means that if we analyze this, this means on the same side of the double bond. So if you want to Joyce, this is Emma, we want to make sure that our highest priority groups, these two bromine atoms or basically same atoms are on the same side of the double bond. So let's choose two bromine atoms to be pointing up with respect to our double bond. So we will have our bloomin pointing up here, another broom in pointing up and now they're on the same side, right? So we're going to call this one sis and then we will have our remaining alcohol groups, right? We have our isopropyl groups which will essentially look like this ch with two methyl groups. Bonnett said, and we have exactly the same thing over here. We're going to have ch With two method groups bonded to that carbon. Now, the reason why it says once again is because we have two equivalent bromine atoms pointing it pointing up right there on the same side of the double bond. So that's our society. And for the trans, when you just want to make sure that your two bromine atoms are on opposite sides, so we can just swap these two submissions and this is how we will get our trans eyes more. So we're going to draw our double bond, we're going to draw two carbon atoms immediately, we can draw one broom in pointing up the other broom in pointing down with respect to the double bond. And here we are going to draw our isopropyl substitutions with that BCH with two method groups. And same thing goes here, right? We have ch With two method groups. So that's our trans because the two bromine atoms are on opposite sides with respect to the double bond, well done. So let's move to the next one. And for the next one, the structure might seem a bit easier to draw. Number two, we have two bromine atoms br two bonded to carbon. So we have carbon atom that is bonded to another carbon and carbon on the left has to broom of substitutions. That makes sense because this carbon must have four bonds. Right. Now, this other carbon atoms double bond, it has a metal substitution bondage yet, right? It's inferences and then it has an ethyl substitution. And the reason why inferences is because these carbons, they do not create a straight chain instead they're separate substitutions who are going to add C three, which is inferences and then just continue with our ethel substitution. So we're going to draw CS two C three. Now, this is our structure. This is our L king. And the question is, does it exhibits this translates Tamar is um the answer is no. If you look at our structure, it doesn't matter which one you choose. In each case, one brimming with be pointing up, the other one will be pointing down their equivalent atoms, right. So you cannot have cis trans isomerization if you have to exactly same groups bonded to one of your double bonded carbon atoms, right. So there is no sis trans for this one because essentially you have two exactly same bromine atoms bonded to the same carbon atom. So essentially swapping these two substitutes will not change your structure at all. So you don't have any cis trans. So we can conclude that cis trans isomerization is only applicable structure. Number one, we have drawn our structure where the two bromo substitute prints are on the same side of the double bond. And we have drawn our trans structure where the two bromo submissions are on opposite sides of the double bond. And for structure to, we don't have any trans exam arisen because we have to exactly same Romo substitutions bonded to that carbon atom based on our answers. We may conclude that the correct answer to the multiple choice question is A B. Thank you for watching.

For those compounds that can exist as cis and trans isomers, draw and label the isomers.
3.
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Key Concepts

Cis-Trans Isomerism

Alkenes and Double Bonds

Drawing Structural Formulas

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