Draw the condensed structural formulas for all the possible haloalkane isomers that have four carbon atoms and a bromine.

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Welcome back, everyone. Let's look at our next problem using condensed structural formulas show all possible. Halo Halo alkane isomers that satisfies the following descriptions. A total of five carbons contain chlorine and have the longest chain of four carbon atoms. So first of all, our problem says Halo alane. So we're going to assume there's just one halogen atom or it would say something about the fact there's more than one and our problem tells us that must be chlorine. So if we have a total of five carbons, but our longest chain is four carbon atoms, one of those carbons must then be in a methyl group. So I'm going to just draw four CS. I'm talking about condensed structural formula. So I'm going to have the CS drawn with dashes for bonds between them rather than in a skeletal structure where you don't specifically draw the CS, they are the ends of the line segments that indicate the bonds. I'm not drawing in the hydrogens yet. It will be easier to fill them in once I figure out how the bonds go. So I've got my four carbon base chain and then I need to add a methyl group, if I add a methyl group to either carbon one or four, it just makes a five carbon chain. When I add a carbon group to a methyl group to either two or three, it will make the same molecule since there's four carbons here. So there's only one arrangement possible. We'll put the, the methyl group on carbon two. So we'll go ahead and fill that in CH three. And now we know we have to have one chlorine. So we'll start by putting the chlorine on carbon one. So we'll add that there as the first atom in this structure make that a little tinier. So we have CL bond c. Now this, we know this carbon must have two hydrogens. So ch two, since it has a bond to chlorine and a bond to carbon, then we have our methyl group ch three branched off of carbon three, it has three bonds. So it has only one hydrogen. Then our next carbon has two bonds. So we'll have two hydrogens and our final carbon of course, will have three hydrogens. So it gives us a skeletal structure of clch two ch with ach three attached. Then ch two ch three get those condensed skeletal structures, use lines to show the connections between atoms, but show those ch groups as letter groups rather than showing the bond to every single hydrogen. So there's arrangement number one. So let's draw our four carbons again to look for a different arrangement. So four cars, we know our methyl group must be up top. Now, why wouldn't we add a chlorine to the methyl group? Well, look at the way this structure is, I'm going to add ch three here at the, the first carbon in the chain. This branch sort of makes a y at the end of the molecule there, it essentially carbon two has two methyl groups attached to it. So adding a chlorine to either one will result in the same molecule since you can rotate these around and make it the same, make it an identical molecule. So that's why we're not going to add a chlorine to the methyl group. It will give us the same result as the molecule we just drew. So now let's move our chlorine along to carbon two. So it can extend down from carbon two. We've already added three hydrogens to our methyl groups. This carbon now has four bonds. So it doesn't get any hydrogens, then carbon three has two bonds. So it has two hydrogens and carbon four has three. So our second skeletal structure is CH three C with ach three and a chlorine on it, ch two ch three. So the structure number two, now we'll bump the chlorine along to carbon three. So again, draw four carmens and with bonds in between them and the methyl branch on carbon two. Now our chlorine's moved along to carbon three, let's film the hydrogens CH three on carbon one, see H on carbon two, which now has just three bonds, ch on carbon three, which has three bonds and CH three on carbon four. So this structure is CH three ch with ach three bonded to it, ch with AC L bonded to it and CH three. Now the last structure is one where we'll move the chlorine onto carbon four. This does not give the same molecule as having it on carbon one because that methyl groups makes this and uh makes this different. So carbon one and carbon four are different due to their distance from the methyl group. So again, our four carbon chain draw our four carbons with the bonds in between them methyl group on carbon two. And now chlorine on carbon four, fill in my hydrogens three hydrogens on carbon one, one hydrogen on carbon 22 hydrogens on carbon three and now two hydrogens on carbon four. Since it's got the chlorine on the other side, it's my fourth possible structure. Ch three ch with ach three bonded to it. Ch two ch two cl. So I have four possible halo alkane isomers where I have the chlorine on carbon one, I have a methyl group on carbon two. Since I have a chain of four carbons with five total carbons, then chlorine could be on carbon two. Chlorine could be on carbon three or chlorine could be on carbon four. So I have the condensed structural formulas for all of those possible arrangements. See you in the next video.

Draw the condensed structural formulas for all the possible haloalkane isomers that have four carbon atoms and a bromine.

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Key Concepts

Haloalkanes

Structural Isomers

Condensed Structural Formulas