Draw the condensed and skeletal structures for each of the follow...

Question

Draw the condensed and skeletal structures for each of the following:
a. 1-chloro-3-hexyne

Accepted Answer

Hey, everyone, let's do this problem and says, draw the skeletal and condensed structures that correspond to the systematic name given below. We have one bromo three hex sign. So let's review some steps in order to solve this problem. So first, we can actually draw the Lewis structure of the compound. And then from there, we can get the condensed structure or the condensed formula, we'll call it structure. So we do this by taking the longest continuous carbon chain from our Lewis structure. And then in between the carbons, we add the atoms that are bonded to each carbon and we want to make sure that those atoms are touching the carbon that they are bonded to. And it's usually written after the carbon that is bonded to. Alright, then we can draw the skeletal skeletal structure sort of going back to the Lewis structure. So first, we'll draw that structure in a zigzag shape and then we'll remove all the carbons and hydrogen ins in order to get its skeletal structure. So let's do these steps. So first we start with the Lewis structure and our systematic name of our compound is one bromo three hex sign So breaking this down, we have the root is hex, which we know is six carbons. The ending line will let me draw one step at a time. Okay. Six carbons 12345, hoops, 56 Ein is a triple bond ending, right? For an Al Keen. And we have the three out here which tells us that the triple bond is on carbon three. So carbon 123, triple bond between carbons three and four. And we have BROMO which is a bromine atom on carbon number one. Okay. So that is our almost our structure. Now, we just need to add hydrogen is to complete the octet. So on carbon number one, we have two hydrogen, two hydrogen and carbon two carbons three and four have a full octet because of the all kind. Carbon five has 2 hydrogen ions and carbon six has three hydrogen. Alright. So now from this, let's draw the condensed structure. So we'll take our longest continuous carbon chain which is 123456 carbons, 12, oops, no bonds, 123456. Okay. Now we're going to add our triple bond between carbons three and four. And now we want to add the groups that are born into each carbon and like I said, we usually add them after the carbon that it's bonded to. So carbon one has to hydrogen and brahmins. So h to be our carbon two has two hydrogen, carbon three has the triple bond to carbon four. Now that carbon is bonded to carbon five, which has two hydrogen and carbon six, which has three hydrogen. Okay. So that is our condensed structure. So now we're going to revisit the Lewis structure and draw our skeletal structure. So we're going to draw this in a zigzag. So grooming 1234, oh wait, we have a triple bond that's going to affect our zigzag, right? Because triple bonds are always linear. So 123, then we have our triple bond. Then the next bond also has to still be linear for five. Wait, 123456. Somehow I miscounted those. Okay. And then we'll have our hydrogen all on our carbons, but we're going to remove the hydrogen. I'll just draw them really quick. Like so condensed, now we remove all the carbons and the hydrogen is to have our shape of our compound. So we still write our bro mean 123, Triple Bond, 456. So that is our skeletal structure which makes a the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching.

Draw the condensed and skeletal structures for each of the following: a. 1-chloro-3-hexyne

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