How many stereoisomers are formed from the reaction of 3-methy...

Question

How many stereoisomers are formed from the reaction of 3-methylcyclohexene with NBS?

Accepted Answer

Hi, everybody. And welcome back. Our next question says, identify the number of products formed including stereo isomers from the reaction of three ethyl psycho butte one E with envious A one B two C four or D eight. Well, first, of course, we need to think about what type of reaction is going on. We have an alkene with a molecule of N BS. N BS is short for N bromo syin amide. It's what's known as a radical initiator because it has this weak bond that's easily broken. So it can have generate radicals when exposed to heat or light. In this case, it generates a bromine radical. So we have this bromine that reacts with our alkene and it adds on this bromine to the alkene with ali radical brominated, meaning it creates a radical, the aly position, which would be the carbon adjacent to the double bond. This means that the radical that's generated gene generated is stabilized by resonant structures. So let's take a look at our molecule and see what we expect to happen. So we have uh three apple cyclo butte one in. So we've got a cyclo butane. So four carbon ring it has a double bond at carbon one. So from carbon one to carbon two, we'll make that across the top and then an ethyl group at carbon three. So carbon three draw our ethyl group. Now we are going to have this aic brominated. What are our AIC carbons? There's two of them. So if we label our carbons 123 and four, our lily positions are we use red dot At carbon three, it is adjacent to carbon two and then a green dot At carbon four is adjacent to carbon one. However, when we look at this carbon three is a tertiary carbon, so it'll be a much more stable radical there. So rather than have it get brominated at carbon four, the other a lilic position, which is just a secondary carbon, it will prefer to react at carbon three. So we know that one of our products without steri chemistry indicated, we'll have this bromine added on the same carbon that carbon three as our ethyl group. However, we do need to indicate stereo isomers and that carbon is a chiral center, there's four different substituents. So we'll need to account for that. However, let's recall that the whole point of this Aic Broin nation is that we have our radicals stabilized by resonance structures. So what are our resonance structures here? Well, we've got the molecule we just drew and we're going to show the radical form with a radical here on carbon three. And we can have that electron come over single headed arrow. It's just one electron on to the bond between carbon two and three. And then we can have one of the electrons from the pi bond to come down and make that a double bond there between carbon two and three. And then that leaves the other electron over on carbon one. So this moves the radical around gives us this other resonance structure. So draw our cyclo beauty and ring ayl group. The double bond is now between carbons two and three and the radical is over here on carbon one. So again, we've got a radical stabilized or resident stabilized, excuse me, radical. So with the radical run on carbon one, then we know that our bromine can also add on to carbon one. Draw that leaving that double bond between carbon two and three and then our other group over on carbon three. But again, this is also the carbon with the bromine attached is a Carval center here as well. So we need to indicate or be ready for steri chemistry there. So what are our products including our stereo isomers? Well, let's draw our bromine added to the tertiary carbon. And we just need to indicate that we'll put the bromine on a solid wedge and then our ethel group on a dashed twitch for that first stereo isomer. And then we can reverse those. So put the bromine on dashed wedge and the ethel group on a solid wedge. So there's two and then we need to do the other thing with our other resonance form Dr Cyclo but double bond between two and three, we've got the ethel group down there. And then in this case, the bromine's added opposite to our ethyl group. So we'll add like one ST ice more that brought me not a solid wedge and then a second stereo isomer. See if we can fit this here, we need to scoot everything over just barely squeeze it in there with the bromine on a dashed wedge there. So there we have it, we've got four possible products there including their stereo isomers. So our answer choice is going to be choice C four, we know that if we have only two as in choice B that does not include stereo isomers and then if we'd chosen the answer one, that would ignore the fact that we have these different possible radicals based on resonance, that's why choices A and B one and two are incorrect. And then choice D eight most likely includes structures based on if you brominated at the other allelic position of carbon four. But again, that the tertiary carbon will generate just that much more stable radical. So that's why choice D eight is incorrect. So once again, number of products formed including stereo isomers from this reaction of our cyclen with N BS is choice C four. See you in the next video.

How many stereoisomers are formed from the reaction of 3-methylcyclohexene with NBS?

Key Concepts

Stereoisomerism

NBS (N-Bromosuccinimide)

Cycloalkene Reactivity

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