Complete the following equations (Hint: Answers may include concepts learned from previous organic chapters):
c.

Question

Complete the following equations (Hint: Answers may include concepts learned from previous organic chapters):

c. Chemical equation showing the oxidation of 2 CH3CH2SH with an unknown product indicated by a question mark.

Accepted Answer

Hi, everybody. Welcome back. Our next problem says, predict the major product of the reaction. Below note, solution may involve concepts learned from different chapters. We have a pretty large alkene molecule here with several branches and it's reacting with hydrogen bromide. So we have a molecule with a carbon carbon double bond reacting with a strong acid that has a halogen atom in it. So this re involves a hydro halogen reaction. And when we look at how these reactions will occur, note that our question asks us to find the major product because in these reactions, we break the carbon carbon double bond or one of them, excuse me to leave it a single bond. We add hydrogen to one of those carbons and our halogen to the other. But we have to decide which are the preferred locations. And for that, we need to follow Markov Niko's rule with that when we have these hydro halogen reactions. So as a sample, we have a protein molecule plus HX, the major product is the one in which the halogen is added to the more substituted carbon. While the hydrogen excuse me is added to the less substituted carbon So for instance, in our reaction, our sample reaction shown here, the carbon. So we, we look at our double bond, we know we're going to break the one of the bonds and the two carbons involved the carbon I'm highlighting in blue in the middle is more substituted, it has a carbon on either side. Whereas the carbon I'm highlighting in red is less substituted. It's a terminal carbon. So when we draw the major expected product, let's draw our three carbon chain. We've broken the double bond. So we just have a single bond between those two carbons. And again, highlighting them in blue, the middle one and red, the terminal one, I know that the major product is one in which I will add the X to that blue carbon and the hydrogen to that red carbon. Now note this is just the major product, there will be some product produced in which it's swapped and the X is added to the red carbon. But that will be a minor product compared to the other way around because this provides a more stable product and a more stable intermediate. So with that in mind, let's turn to the problem. We've been given the molecule we've been given, looking at our cc double bond to see which is more substituted. So when we look at our structure, it's this um written as sort of a condensed structure, but with the branches indicated with lines rather than parentheses. So we have C three C with ach three branch ch, two C with ACH two C three branch, then the double bond and then Chch with ach two ch three branch ch two ch three. So I'll use blue highlighting for my more substituted carbon. So the one on the left of the double bond is more substituted. It has the chain, the main chain attached to it and then it has an Ayl chain attached to it as well. So it has bonds to three carbons in total. The one that has a double bond to and then two other carbons. Meanwhile, our less substitute carbon than I've highlighted in red has a bond to only one other carbon than one. It's in a double bond with. So our major product would be and we'll write out our structure here ch three ch with ach three branch CH two. And now we're at our carbon that was in the double bond. So we'll write c this is our blue carbon. So this is the one that gets the bromide. So we'll put br attached to it as well as the ethyl group down below CH two C three. It now just has a single bond with the red carbon. So I'll write that right next to it since we don't have single bonds written as lines on here. So there's the red carbon, it's had the hydrogen added to it from the hydrogen bromide. So instead of being ch it's now CH two. And then we have Ch with an ethyl branch, ch two, ch three up above. And then lastly CH two CH three. So we have our major product here. If we want to come up with its name, we have a 12345678 carbon chain. So we have an octane, we would number it in a way such that the bromine has the lowest substitute number. So we'll call the one on the left carbon number, one, the one on the right carbon number eight. So our bro means on carbon 1234. So we'll have four bromo. And then we have one methyl group and two ethyl groups. We go by alphabetical order. So we'll have and then let's number our carbons 234, 5678. So we have ethyl groups on carbons four and six. So we got four bromo, 46 di ethyl and we've got one methyl group on carbon two. So two metal octane. So four bromo, 46, diethyl, two methyl octane as the name of our final product here. So we'll go ahead and just highlight that in yellow. And again, it's, this is our major product. We could have a minor product where we add the bromine and the hydrogen on the other carbons, but that's not what our question is asking us about. So we've shown the product where the bromine is added to that more substituted carbon. See you in the next video.

Complete the following equations (Hint: Answers may include concepts learned from previous organic chapters):
c.

Verified video answer for a similar problem:

Key Concepts

Chemical Equations

Functional Groups

Reaction Mechanisms

Complete the following equations (Hint: Answers may include concepts learned from previous organic chapters):c.

Verified video answer for a similar problem:

Key Concepts

Chemical Equations

Functional Groups

Reaction Mechanisms

Complete the following equations (Hint: Answers may include concepts learned from previous organic chapters):
c.