Free-radical bromination of the following compound introduces ...

Question

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

a. Propose a mechanism to show why free-radical halogenation occurs almost exclusively at the benzylic position.

b. Draw the two stereoisomers that result from monobromination at the benzylic position.

c. Assign R and S configurations to the asymmetric carbon atoms in the products.

Accepted Answer

Hey everyone. Let's solve this problem. It says. The following compound undergoes free radical domination at the Ben Zelic position. Give the mechanism for the mon abomination of the given compound and draw the two stereo I summer products also assign RNS configurations to the asymmetric card john Adams in the products. So the brahma nation of al canes in the presence of sunlight follows a free radical mechanism like the problem says. And the given rumination here occurs exclusively at the Benz Illich position. And why is that? Well, because there is the resonance stabilization of the Benz alec radical intermediate that is formed during this reaction. So since this is a free radical mechanism, the first step is a radical initiation and that is forming the bro mean radical through hemolytic cleavage of the bromine bromine bond in the presence of sunlight. And in the second step we have the radical propagation which is done by the abstraction of that Ben's Ilic hydrogen by the roman radical. And then the Benz Ilic radical formed is stabilized again by that resonance and undergoes rumination to finish the reaction. So let's draw first the initiation step of the mechanism. So we have the bro mean bro mean and we are using one headed arrows to move one electron onto each bromine atom, there's sunlight To form our two bromine radicals. Okay, now propagation, we are going to extract that Ben's Ilic hydrogen. So we have our bro mean radical hanging out over here and now we're going to have 31 headed arrows in the steps. So one coming from the bro mean and then one coming from this hydrogen bond that we're going to remove this hydrogen from. So those two before my bond. And then that extra electron will come onto the compound which is how we form that intermediate. That radical intermediate, just like that. And now since we have the radical next to the double bond, we can have the resonance structures. So let's draw all the resonance structures possible here. So now that double bond has moved to the outside of the ring, in the radical has moved to the carbon that is no longer in a pi bond. So we can have the same thing. Again pi bond moves over towards that radical and then our double bonds and the radical again has moved to the carbon that is no longer in the double bond pi bond moves towards the radical and the last one the radical or sorry, the double bond again moves to that radical and the radical will go back to its original position. And then all the double bonds are in the alternating positions from where they started. And we also have that HBR that has formed. So that is it for the intermediate. Now we will have the brahma nation of that intermediate Which will create our two stereo I summer products because that bro mean, can come from the front side or the back side. So again we're going to have one electron onto this bombing and then one electron reaching out towards that radical to form that bond. Or you can also have the termination step with that other br radical that was created in the initiation, simply forming the bond with that radical. So we have the bombing attaching to the front, in the booming, attaching to the back. So in order to assign the R. Or S. Configuration, we're going to have to use the con angled pre log nomenclature system. So let's quickly review that. So we do this on asymmetric carbons. Right. And asymmetric carbons have four different groups. So the first step is to assign priorities by decreasing atomic mass to these groups. So the highest atomic mass gets the first priority, lowest atomic mass gets the fourth priority. And if there is a tie let's say like two carbon atoms. Then we'll look at their adjacent atoms. To compare if we have a double bond, we count that as two single bonds and a triple bond. We count that as three single bonds. So say if we have a carbon double bonded to an oxygen, we count it as if that carbon has two single bonds to two oxygen atoms. Okay then once we have our priorities assigned, we have to assign our or S. And we look at the orientation of group number four. So if group number four is dashed, then if you connect groups 12 and three, then the pathway of those groups. If they are counterclockwise, that will be s orientation. If they are clockwise, that is our orientation. If it is wedge. If group number four is wedge rather. And you connect groups 12 and three and they go counterclockwise. Now that will be our orientation and clockwise will be S orientation. So the rubric changes depending on if group number four is dash or wedge. Okay, so let's give it a shot with our products here. So we have the four same four groups on these products so we can assign them the same priorities bro, money is going to be one, hydrogen will be four and then we have two carbon groups. This benzine ring is going to have a higher priority because it's bonded to carbon and a carbon and both of those carbons are in double bonds. So there's a lot more carbons. Whereas here this carbon is simply bonded to three hydrogen. So this will get to 32 and three priority. So now let's connect them while noting the orientation of Group four. So in the first product that hydrogen is dashed. And if you connect groups 12 and 312 and three, that is going in a counterclockwise position. So this will be S. Orientation. And the next product our hydrogen is wedge. We connect groups 12 and 3123. That is counterclockwise again. But now, since our hydrogen is wedge, this will be our orientation. Okay, so that is it for this problem. Making choice a the correct answer. That's it for this one. I hope this video was helpful, and thank you for watching.

Key Concepts

Free-Radical Halogenation

Benzylic Position

Stereoisomerism

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