Write the structures of all singly chlorinated products that form when 2,4-dimethylpentane is reacted with Cl₂.

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Hi, everyone. Welcome back. Let's look at our next question. What are the singly chlorinated products that result from the reaction of three methyl pentane with CL two? And we're given a structure of three methyl pentane. It's a five carbon chain with a methyl group, one carbon three. And all of our carbons and hydrogens are specifically written out here reacting with one molecule of CL two. So we want to look at the structure of our reactant here thinking about where we could substitute a chlorine for a hydrogen. And we also need to think about which products would be unique and which substitutions might generate a product that would be identical to another one. Well, what determines whether products would be unique the symmetry of the molecule? So I've drawn a skeletal version of this structure just because I find that makes it a little easier to evaluate the geometry at times. Sometimes all those hydrogens written on there makes it difficult to really see the arrangement. And first I want to evaluate is this a symmetrical molecule? Well, it is, it has a plane of symmetry going through carbon three and our methyl group. So just note that adding, say one chlorine on one side can result in an identical product. When we add it on the other side, they'll be super imposable because they're on either side of this plane of symmetry. So we need to keep that in mind. For instance, if we start with carbon number one and add a chlorine there, well, that's a site we could add a chlorine to, but that would generate the exact same molecule as adding a chlorine to carbon five. So we only need to consider one of those and I'm going to raise the arrow to carbon five. Since that's not a unique product, do we need to think about whether we generate a different product by replacing different hydrogens? We have three hydrogens on that terminal carbon. Well, no, we don't because of the tetrahedral geometry of are carbon with four single bonds. If we imagine it has its one bond to the other carbon and then three hydrogens, we imagine that sort of tripod shape, replacing any one of these hydrogens would result in the same molecule. So we have just one product resulting from adding a chlorine to the terminal carbons. Now we just march along our molecule here and contemplate each carbon in turn. So carbon two is also a location where chlorine could go, it can replace one of those two hydrogens. But note that that would give us the same result as a chlorine on carbon four. Again, due to that plane of symmetry. So I will erase that arrow. I just wrote in on carbon four. It's not a unique product. So we've got carbon 12. Now we're on to carbon three. Carbon three has that methyl branch. It's got three bonds, but it's still got one hydrogen left that could be replaced with the chlorine. So carbon three is another possible site. It is on the plane of symmetry. So that's a unique product. There won't be a, an identical molecule generated elsewhere. And then finally, our methyl branch is a carbon as well. Chlorine could be added to the carbon of the methyl branch. So there's another product possibility. All that's left are carbons four and five which again, we know generate those identical products. So our question asked for those four products, we should therefore generate them in the same type of structure drawing as the problem gave us. So I will write out all my carbons in the structure. Our first product will be adding that chlorine to carbon. Number one. So I've done that, it replaces one of its three hydrogens. So it's now just ch two. And now we'll just write in the rest of the hydrogens on this structure. And we have our first product chlorine on carbon. Number one, just going to highlight that chlorine on carbon. Number one, lets move on to structure number two or possibility of number two, this one is generated by adding a chlorine to carbon. Number two, it originally had two hydrogens. So it's just got one left and now we just go through and fill in the remaining hydrogens on our other carbons. So there we go. Product number two, chlorine on carbon two. Now we look at product number three. So we're talking about adding chlorine to carbon three. So we put that there. Carbon three only had one hydrogen. So now it has no hydrogens, so no hydrogens there, but we fill the hydrogens on the rest of the molecule and we have product three chlorine on carbon three. Finally, our possible product number four, this was generated by adding chlorine to the methyl group at the bottom. It originally had three carbons. Now it just says two and then we just fill in the hydrogens on the rest of the molecule for our final fourth product chlorine added to our methyl group. So there we have it, we have four unique singly chlorinated products that are generated from this reaction. See you in the next video.

Write the structures of all singly chlorinated products that form when 2,4-dimethylpentane is reacted with Cl₂.

Verified video answer for a similar problem:

Key Concepts

Chlorination Reaction

Stereochemistry

Product Distribution

Write the structures of all singly chlorinated products that form when 2,4-dimethylpentane is reacted with Cl2.

Verified video answer for a similar problem:

Key Concepts

Chlorination Reaction

Stereochemistry

Product Distribution

Write the structures of all singly chlorinated products that form when 2,4-dimethylpentane is reacted with Cl₂.