Draw the structures that correspond to the following names.

Question

(c) (Z)-2-chloro-7-methyloct-2-en-4-yne

Accepted Answer

Welcome back, everyone. Draw the corresponding structure for the Iupac name E two, Chloro six methyl hept two en four I. Our first step is to start with drawing the parent chain in order to determine the parent chain. Notice the ending of the Iupac name where we have Hept two E four I. So the hept portion is telling us that we have a seven carbon parent chain. We're at carbon number two, we have the en indication telling us that we have a double bond. And at carbon number four, we should have a triple bond indicated by the ion ending. So let's represent the parent chain where we have seven carbons. We're going to use a bond line diagram. So we have 1234 and five. And then we would again at carbon number two, create a double bond. So we'll say carbon number two is at this position. And then at carbon number four, we should have a triple bond. So this is carbon number three, then this is carbon number four. So we will create a triple bond, meaning that we've got a total count of 1234567 carbons making up this parent chain. Now that we have our parent chain, let's add any substituents. So looking at the beginning portion of the name from the prompts, we have two chloral and then six methyl. So just as before the beginning with the two chloro, at least, we would be able to understand that at carbon number two, we should have a Chloro group or a chlorine group. And so drawing that out, we would just extend a bond so that we have 1/4 bond going to chlorine at carbon number two. And next, we need to indicate the six methyl substituent. And so as we stated, we have carbon number 12345. And then at carbon number six, we're going to include that methyl group. Now, the last thing to focus on is the e indication in our Iupac name. So this speaks to the stereo chemistry of our molecule. We're going to need to recall the rules for EZ labeling. So we need to analyze specifically the double bond and the groups that are surrounding this double bond or in other words, the groups attached to the carbons making up each side of this double bond. Recall that the higher the atomic number of the atoms in the groups that will correspond to a higher priority because we need to assign priorities to the group surrounding the carbons of this double bond. And if we have a tie between the atomic numbers we would just compare atomic mass. And specifically, in the case of double bonds versus triple bonds, we would just count the atomic mass twice for surrounding groups of the double bond. And we would count the atomic mass thrice for the groups on the triple bond. And then another rule we should recall is that if we have matching atoms or identical atoms, we would just compare adjacent Adams. So looking at our double bond, we have carbon number two and we observe that it is attached to this Chloro group. And another carbon atom recall that atomic mass increases as we go towards the bottom left of the periodic table. And so therefore, chlorine being across period three, which is further below than carbon across period two, chlorine has the higher atomic mass. So we will assign it priority number one and then we will assign the carbon bonded to carbon number two with priority number two. Now, looking at the other side of our double bond, we observe that this carbon has an implied bond to hydrogen. We can draw that in just for clarity and it has a bond to the carbon making up that triple bond A K A carbon number four. Now hydrogen is in period one. And so therefore, it has the lower atomic mass. So we would assign it with priority number two and the carbon on carbon number four, priority number one. Again, that carbon is a part of the triple bond. So now we're going to focus on whether our carbons making up our double bond have the highest priority groups on either the same or opposite side. And because our highest priority group being the atoms which we labeled with, the number one priority are actually on opposite sides of this double bond. So note down that the number ones on opposite sides, then therefore, we can confirm that we have an E stereo isomer, which does align with the given Iupac name from the prompt. And again, that is because the priority number one atoms were on opposite sides of the double bond. And so overall, to determine this, we were able to use the Kangol pre log rule successfully to determine that the E stereo isomer is the correct indication for the drawn structure that we have corresponding to the Iupac name given in the prompt. And so for our final answer just to clean things up, I'll redraw it. We should have the following bond line structure where we've got the highest priority groups. Again, on the opposite side of that double bond being the chlorine at carbon number two, and the carbon at carbon number four, again, on opposite sides of this double bond. And this entire structure is going to be our final answer to complete this prompt as the corresponding structure for E two Chloro six methyl Hep two en four, I I hope that this made sense. And let us know if you have any questions?

Draw the structures that correspond to the following names.
(c) (Z)-2-chloro-7-methyloct-2-en-4-yne

Key Concepts

Stereochemistry

Nomenclature

Functional Groups

Watch next