A student was given the structural formulas of several compoun...

Question

A student was given the structural formulas of several compounds and was asked to give them systematic names. How many did she name correctly? Correct those that are misnamed.

a. 4-ethyl-2-pentyne

b. 1-bromo-4-heptyne

Accepted Answer

Hello everyone. So in this video, we have defined how many of the following systematic names are correct and to correct those that are incorrect. So for the first one here, this name is four propol pen two. I. So first, let's draw the parent chain of our molecule. We see that we have this prefix of pe N T, meaning that we have five carbon chain. So let's go ahead and start off the structure by drawing these five carbons. So 1234 and five, let's go ahead and number those as well. This is just a random numbering for now. All right. So we have completed the pen part of the name. Now continuing on, we have two Y N E. So the Y N E means that we have a triple bond and this two number here indicates that this is where our triple bond begins. So taking a look at this five carbon chain here at the second carbon is where the triple bond begins. So it's having a triple bond between our second carbon and third carbon. Now, considering our one substituent of four prop. So at the fourth position, we have a prop group. So that's three carbons total. All right. So now that we've completed all the necessary parts of the naming, we can now fill in the rest of the structure with the appropriate amount of hydrogens starting off with carbon. Number one, we need three more hydrogens for carbons two and three, it has fulfilled Octa already. So we don't need any more protons. So for carbon number four, we do need one more hydrogen. Carbon five is a terminal carbon. So that is three more carbons. Now, for our prop part of it, we have, let's see. So for the two in the middle, we need two more hydrogens. And then for our last one here, that's going to be a total of three hydrogens because it still is a terminal carbon. So this naming here of four purple P two I is actually going to be incorrect because the longest continuous chain of carbon atoms is incorrect. So let's see here. Let's see if our answer there is correct. If we have considered that the longest contained chain of carbon atoms is incorrect. Meaning that this full name that's given to us and the question text is incorrect. So let's let's actually use a system to name the structure appropriately. So first again, we have to find the longest continuous chain of carbon atoms that includes our triple bond. So let's see here. We're gonna go ahead and label this with a different color. All right. So let's see, we have and seven. We forgot to draw this bond here in between. All right, perfect. So we do actually have seven carbon chains to be the longest continuous chain that includes our triple bond. So like we said, we have seven carbons. So usually without the triple bonds, this will be a Hep chain. But because we do have this triple bond, this ending of A will change to Y and E. So we have a he I all right. Now, considering next is going to be the position of our triple bond. So let's see here. And our second carbon is where our trip of bond starts. So we have then our Hep two iron. All right. Now adding in, lastly our substituent and instead of the pro group being a substituent, we have a metal group located at carbon number four. So this over here. So if we add that in, then we get four methyl, then stepping on our parent chain, we have he P two I. So the final name for structure number one is going to be four methyl he P T two I, I'm moving on To Part two. Here. We have two chloral hept four iron. So now starting off with drawing the apparent chain, we have this H E P T prefix. This means that we have seven carbons. So let's draw that 1st and seven. Now, again, we have a triple bond starting off at the fourth position. Let's first number these carbons here in this parent chain. So 123456 and seven. Again, we said, sorry, not the fourth carbon. This is where our triple bond begins. So we add that in, we see a triple bond. Now between the 4th and 5th position. Next, we can go ahead and add in our substituent at the second carbon. That's a chlorine group. All right. So now let's see if this naming is correct. So again, we have to first identify the parent chain that includes our triple bond. So that is still the same. So we have 123456 and seven. So we do have seven carbons here because we have seven carbons just like what we did above. We have a HEP I identifying where our triple bond is, that's going to be between our starting off at carbon number three. So then that's going to be he three. I. Now lastly adding in our substituent group, that's going to be the chlorine group located at carbon number six. So then we will get six Chloro he P S3 I. And that's going to be my final answer for structure number two. Again, for structure number one, this is a four methyl HE PS two infrastructure number two, that is six Hept three. And those are going to be my final answer for this problem.

A student was given the structural formulas of several compounds and was asked to give them systematic names. How many did she name correctly? Correct those that are misnamed.
a. 4-ethyl-2-pentyne
b. 1-bromo-4-heptyne

Key Concepts

IUPAC Nomenclature

Alkynes

Substituent Positioning

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