Draw skeletal structures for the compounds in Problem 3, includin...

Question

Draw skeletal structures for the compounds in Problem 3, including any cis–trans isomers.

Accepted Answer

Everyone. And welcome back in today's video, we are going to draw the skeletal structures of the cis and trans is members of the given structure. Now, the given structure has four carbon atoms in its parent chain. So let's troll four carbon atoms and stuff that we are just going to add our lines because at each verdicts, we will have every single carbon atom shown in the structure. So specifically, if we have four carbon atoms, we're going to add one, 234, this corresponds to four carbon atoms. This is butane so far, right? Because we have our 1st, 2nd, 3rd and 4th verdicts at each verdicts, we have our carbon atoms. So we have four carbon atoms and we have to notice that the first carbon atom is a C three group. So there are implicit three high jeans because our carbon has four bonds total. The second carbon goes with Francis, it's a C princes. BR So that means that the second carbon, we have our Romo substitute print. Now we have our double bond and that double bond essentially is between carbon number two and carbon number three. So we're going to add our double bond. And now carbon # three goes with parenthesis, it's, it's CBR. So that means there's a bromo substitution at carbon number three. So we're going to draw a bond to be roaming. And finally, the last carbon is C three. And that's indeed see three already because we have only one bonds that carbon. So that means there are three implicit Hae Jin's. First of all, we have drawn one of our structures and this is actually trans I Zimmer because because if you recall, whenever you have an L keen with two equivalent groups on opposite sides, with respect to your double bond, that's trans. But if you essentially draw your two equivalent substitutions on the same side of that double bond, whether above or below, it doesn't matter that becomes sis in the given structure that we have. And if you look at the position of the double bond, one bromo group is pointing up, the other one is pointing down. So that must be trans. Now to get our structure, we are just going to redraw our parents, we are going to add the double bond and we are just going to make sure that we get to BROMO submissions on the same side of the double bond. So let's suppose that we swap this metal group in this promo substitution. So then if we draw two bromo substitutions pointing up, we will have two remaining methyl groups pointing down. And this will give us our sis I Zimmer. So we can now clearly see our system trans eyes MERS for the structure. And based on the answer that we have got here, we may conclude that the correct answer to the multiple choice question is the Thank you for watching.

Draw skeletal structures for the compounds in Problem 3, including any cis–trans isomers.

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