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

Question

Draw the condensed and skeletal structures for each of the following:
d. sec-butylisobutylacetylene

Accepted Answer

Hey everyone and welcome back through the skeletal and condense structures that correspond to the name given below. And beetle Isopropyl a settling. Now, first of all, we are going to think about this name. Let's begin with the very end. It says a settling which is our parent. Let's recall that the Ceta lean is the simplest al kind that we have, which corresponds to two carbon atoms bonded by a triple bond. And then we have two implicit hydrogen atoms. So we can essentially state that these two carbon atoms in blue are bonded V A triple bond and then we have two hydrogen atoms bonded to each carbon. Now, the difference is that this structure is and beetle isopropyl beautiful and isopropyl are two different submissions that we are going to bond to our settling by replacing these hydrogen atoms. The order in which we replace them doesn't really matter. We will still get the same structure. So now let's remember what an beetle is and beautiful or simply speaking, normal, beautiful just corresponds to a straight four carbon chain. Remember that beautiful group essentially means four carbon atoms. So that'd be 1234. Now we want to add the number of hydrogen atoms to each hour of our carbon. So we must have four bonds for each carbon. Meaning we will have C three ch two, we have two bonds. Siege two, we have two bonds. And here we are going to create our linkage to the actual parent. So that means this carbon atom must have four minus one, that's three minus one additional amount. That's two hydrogen atoms. That's our normal beetle group. For carbon atoms. That's our l kill group straight chain because it's normal. Now, for isil pro pill, we have to remember that normalize a proposal is just a three member chains. We can just exclude the siege to and we will have our eyes, our propel group normal proposal. But when it says, so we are going to create a branched Three member chain. In other words, you're going to add three carbon atoms. But when you think about the connectivity to your parents, you're going to connect it at position number two, because that would create a branched substitution instead of connecting it at any of the two ends. If you connect it at any of the two ends, you will get your normal pro bill. But this is isopropyl meaning it's branched. So we will have a connectivity here. We may simply add the missing hydrogen atoms. Keeping in mind that every carbon must have four bonds. We have our normal beetle. We have our isopropyl groups and we are ready to draw the structure. First of all, let's draw a partial expanded structures. We understand how it looks like we can draw our triple bond at two carbon atoms. Now, on the left, we will add our isopropyl group. So we will add that carbon which has two carbons bonnets, it, but these two carbon atoms are actually methyl groups. Now, we have to be careful because for the isopropyl group, we only have one hydrogen bond. It's the central carbon, right, we must have four bonds. So let's fix that. There would only be one Haijun. Okay. Now, thinking about the normal beetle group, we are just going to add our normal beetle group to the other carbon within the triple bond. So that besieged to Bonded to Ch two bonded to another CS two and bonded to Ch three. Okay. Now it's clearly illustrate connectivity. Sometimes it's more important to show hide regions below carbon atoms, right? So that we clearly see that those are carbon atoms that are connected to each other. So let's just draw our carbon chains that B CH two CH two ch two and then we have CH three. So that's our partial expanded structure which will allow us to get our condensed form. So first, if we start from left to right, we can throw our condensed form as follows. We are going to start with the fact that there are two metal groups bonded to CH So we're going to say C H three, two times, right? Because we have two of them. And then we're going to show which carbon, their bonnets were using parentheses to clearly illustrate that the next carbon will be the carbon that has both of them. So that's our ch okay. Now, we want to show our next carbon, which is this carbon that is triple bonded. And because al Qaeda is our parents, we have to expand that triple bond. And this is the only bond that we are going to show because it corresponds to the main functional group. Then we have carbon, which is followed by three CS two groups. So we're going to add CH two, ch two again, CH two. And finally, we have C3, that's our condensed form. And now we are ready to draw the skeletal form. We call that joy skeletal form. It's really easy if you understand how many carbon atoms you need to add in your longest chain. So if we go from bottom to top, we can we get 12345678, we are going to draw An eight member chain. So first of all, let's draw carbons number one and T. So that's one T now we approach carbon number three, right? Every verdicts corresponds to a, an implicit carbon atom. And at position number three, this is where our triple bond starts. So we want to make sure that we draw it in a linear form because those are sp hybridized carbons, that's 1234 carbon atoms, disposition four. And then we move on to position number five. Once again, this must be linear, right? And then this is our carbon five 678. Now, the world last thing we want to do is just to check the structure. So if we add the numbers that we added previously, that we want to 345678, notice how carbon number two still has that metal group that we haven't added. So let's add that metal group by drawing a solid line which corresponds to one carbon atom with three hydrogen atoms, right? Because we only see one bond and that's it. As you can see, it might be simpler to draw a skeletal structure. So let's call it skeletal than it is to do a condensed one because we simply want to find our longest chain through it and then add the remaining substitutions. The hydrogen atoms that are implicit are automatically taken care of using these structures. For example, if you take carbon six, you will essentially notice that it must have two hydrants. And that's true because you see two bonds. So there must be implicit to hydrants to correspond to four bonds for carbon. Meaning you don't need to worry about the high genes that are implicit. He simply wants to draw your parents at the substitutions and you are done. So let's label, this is our final answer. We have our condemned structure. We have our skeletal structure and the correct answer to the multiple trades question is c thank you for watching.

Draw the condensed and skeletal structures for each of the following: d. sec-butylisobutylacetylene

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