Draw and label the E and Z isomers for each of the following: ...

Question

Draw and label the E and Z isomers for each of the following:

d. Structural representation of E and Z isomers with chemical formulas and bonds labeled.

Accepted Answer

Hey everyone. And welcome back. The structure below has E N Z is MERS drawn label as en Ci Zimmer's E N Z is Mars are applicable whenever we are dealing with Elkins. It's a type of ism tourism where we have our highest priority groups arranged in a specific way with respect to the double bond, the double bond is highlighted in red. And before we analyze the MGS Mars, let's recall priorities based on con Ingle pre log rules. So we have to analyze our first carbon atom within the double bond. And if we look at this carbon atom, we have to identify two functional groups monitored. We have a key tone and we have an ether. So essentially, if we just show these two groups separately, we can understand that we have our metal group bonded to that carbon steel group, right? This is our first substitution and our next substitution is that ether group. So, c three bonded to oxygen siege to and this carbon is bonded to the carbon that we are analyzing. In order to understand priorities. We first, we have to compare the first atoms bonded to our carbon. So essentially we have two carbon atoms, right, we can label them in green, this is carbon and this is carbon as well. According to the con Ingle pre log rules, you want to find an atom that is greater in atomic number. So it gets the higher priority. But because these two atoms are exactly the same, we have to think about the atoms bonded to them. So if we expand them, first of all, this is carbon which is bonded to oxygen be a double bond. But for simplicity, we are going to show to single bonded oxygen atoms. And then we are going to say that this carbon is bonded to another carbon. Now, for this one carbon is bonded to two hydrogen atoms and one oxygen based on commingle pre log rules, we can eliminate equivalent parts, we can eliminate these t oxygen's right. And then we have a comparison between CH and C C for example, C H and C E O, right. So in each case, this one wins, right? Because oxygen is higher and atomic number than hydrogen and carbon is higher and atomic number than hydrogen. So our key tone guest priority number one And our Easter Group gets priority # two. Now, similarly, we have to think about our second double bonded carbon atom. It has two groups, essentially we have our Elkins. So we're going to say that we have this one and then we have our al kind. So we have this one similarly to what we did before, we have to look at the first atoms bonded to our double bonded carbon atom. So in each case, that's carbon right there, exactly same. So we have to expand all of the other atoms bonded to them. If we look at our elkin that we see, right? It has hydrogen and then it's double bonded to another carbon. So we can say that we have two carbon carbon bonds for simplicity. Based on that rule. Now, for this one, we have carbon which is triple bonded to another carbon. So we can say that there might be three carbon carbon bonds. So based on a single pre law rules, we're going to eliminate equivalent parts, carbon, carbon, carbon, carbon. And then we have a competition between carbon, hydrogen and carbon carbon because carbon is greater in priority than hygiene. And that's because carbon is hard in atomic number than hygiene is then we assign a higher priority to our alcohol. And we are going to say that this is priority number one, this is priority number two. And now here where we start thinking about E N Z I Zammar is um because the highest priority groups, number one on the left and number one on the right, they're on opposite sides of the double bond. This is called an E I Zimmer. Remember that E means on opposite sides, right? Essentially the highest priority groups are on opposite sides of the double bond. And now simply to get see which implies that the highest priority groups are on the same side of the double bond. We simply need to make sure that this Al Qaeda is pointing up. So priorities number one will be on the same side. This will give us our Ci Zimmer. So we got our E and we can label this structure S E for our answer. We simply need to draw our Ci Zimmer. So we're going to throw our double bond, we're going to leave the left side as it is. So we're going to simply add our ketone, our ether. And here we want to make sure that priority one is pointing up. So we're going to add our alkaline and here we're going to add our L king. So now you may notice clearly that this is your priority number one. This is your priority number one. They are on the same side of the double bond. And because they're on the same side of the double bond, this would be your C I Zimmer. Let's label this is our final answer and that would be it based on our answers. We may conclude that the correct answer to the multiple choice question is D Thank you for watching.

Draw and label the E and Z isomers for each of the following:
d.

Key Concepts

E/Z Isomerism

Cahn-Ingold-Prelog Priority Rules

Stereochemistry

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