Predict the dehydrohalogenation product(s) that result when th...

Question

Predict the dehydrohalogenation product(s) that result when the following alkyl halides are heated in alcoholic KOH. When more than one product is formed, predict the major and minor products.

(d)

(e)

Accepted Answer

Hi everyone and welcome back in today's video we are looking at the following problem which reads show the products of the E two D. Hydro hallucination reactions of the following al kill. Hey lights using potassium hydroxide also label the products as major and minor. Now we have two substrates. Let's redraw the first substrate to analyze it. Clearly we have an ethyl group pointing down bonded to the ring and the remaining substations are roaming pointing up deuterium which is an isotope of hydrogen. It has similar chemical properties to hydrogen and then high chin pointing down now because we are told that this is an E two D hydro hallucination using potassium hydroxide. So we may not worry about potassium hydroxide is a strong base here. We essentially want to first of all understand where the alpha carbon is is basically over here because it's the carbon atom that is directly bonded to the leaving group. The beta carbons were the surrounding carbons are these two carbon atoms and because this is an E two elimination, we want to identify beta hydrogen. So there's one beta hydrogen and disposition which is implicit implicit. And it's pointing up because the ethyl group is pointing down and similarly we have hygiene atom at this beta position and a deuterium atom pointing up. We have to recall that there's an anti co plane or requirement for E two reactions to occur. What that means is simply that when we find our leaving group, we want to make sure that the leaving group is pointing in opposite direction, then the hydrogen on our beta position, simply speaking. When we're looking at that ring. If bruning is pointing up, the only beta hygiene that that I can remove must be pointing down so I can immediately tell that this hydrogen is not something that I can remove because it's also pointing out there in sin arrangement. However, if I look at my remaining beta position, I can remove either the deuterium or hydrant because they have similar chemical properties. But the deuterium is in sin arrangement because it's also pointing up, hydrogen is pointing down which is perfect for us. It's anti co planner too bro mean because it's pointing down and bromine is pointing up. So this is the only atom that we could remove. We may now show that according to our mechanism the hydroxide would accept that hydrogen we would form a double bond because that's an elimination. We'll be forming an L keen and simultaneously we will have the loss of grooming and we will form our product for a. That's the only product that we can get. We are not showing any stereo chemistry here because that's an sp two carbon explainer. The only remaining part is this ethyl group. So that's our one and only product. There is no major minor product because as we previously saw there was only one beta hydrogen that we could remove. We're done well, great job. So now let's look at part B, part B is a bit more complex. Now we have two bromo substitue ints they are both good leaving groups. We want to identify beta hydrogen. So to do that. We also want to make sure that we understand what this group is. And this is a third be it cell group. It's very famous for being historically hindered, bulky group. That means we want to draw a chair confirmation first to understand where to place it. Now. Third Beatle group is a very hysterically hindered group and that means in order to make our compound stable, we want to keep it in an equatorial position. So we may choose this position to represent an equatorial arrangement of the third beetle group. We're going to place it in an equatorial position so that the remaining substitutions will be either excel or equatorial depending on on our structural arrangement of B. Now, if we count our carbon atoms for simplicity we can say that this is one, Then we have 2, 3, 4 and five. If we go clockwise, that will be position number two three, four and five position. Number three Hasbro main pointing up. In other words, recall that position. # three in particular has an axial position pointing down and an equatorial pointing up. So essentially bro. Mean a position. Number three will be on equatorial position. We may now draw that bromine in equatorial position essentially that bond should be parallel. So that would be bromine and hydrogen and that position is pointing down So it's axel position, number four blooming is pointing up position. Number four has an axial arrangement for an up position. So that's how blooming looks like it's an actual arrangement and the hydrogen atom and this position is pointing down which is an equatorial position. So we are going to join equatorial position for hydrogen in position number five has a metal group pointing up. We call the position number five has an axle going down. So that would be an equatorial position for a methyl group and also an axle for the remaining hydrogen. Now, if we evaluate everything that we have, we want to make sure that during the elimination we have that anti co planer requirements satisfied so that we want to have one comma two trans die axial arrangement. This pruning in particular position number three will not participate in an elimination reaction because it's in equatorial position. We want to make sure that both positions are in axial arrangement. So the only leaving group that works for us is this bro mean at position number four. So we can only eliminate this prominent position number four and we want to find the beta hydrogen which are in axial positions. So we see that the beta high jeans are these two positions, position number five, position number three we have two axial hydrants that we can remove and that said we can go to our original structure and we can say that We simply may remove these two beta high jeans during our elimination step. Now, because chair analysis might be confusing to some and we already have extracted all the information we needed. We may just redraw our original substrate. So we are going to throw our ring bromo group, another bromo a muscle group. And here we have a third beat cell group. We already know that the two hydrants that we can remove are these two hydrogen. So that means we will have two different products because these two hydrogen are antico planner to bromine. We're using potassium hydroxide is our strong base attacking this hydrogen, forming a double bond and simultaneously the leaving group leaves. So we may draw our first product for that reaction. Nothing changes the disposition. We did not affect stereo chemistry of the 3rd Beatle Group. Now, regarding this position. Same thing. We still have a metal group pointing up here, There's no more blooming. We already had it as a leaving group. However, we have a double bond at this position which makes this carbon planner and as a result there will be blooming attached on a solid line. On the other hand, what if we removed the remaining hydrogen? If we attack that hydrogen? We would form a double bond here and we would have the loss of the leaving group again and see co planner arrangements is satisfied hydrants pointing down roman is pointing out so our next product now could be tron, Nothing changes with respect to the 3rd Beatle Group. However, here we are forming a double bond meaning the muscle group is on a solid line. This is an sp two carbon L explainer. There's no more blooming but this blooming is still present. These are our two products for B. We just have to recall that the stability of Elkins depends on how substituted they are. And notice that this L king is more substituted, its tri substituted in this one regarding the number of carbons bonded. Said that would be di substituted now because this is a die substitute. Al keen, it's less stable and therefore that will be our minor product and the other one which is tri substituted is major because it's more stable based on the state of school. Alright, so because there was a lot of work here, let's just reveal what we did for part A. We got our product using a strong base. We said there must have been an anti co planer arrangement to remove that hydrogen and then form an L. Keen. All good. We got our product that's the L. Keen for reaction A. And have a look at reaction b. We got two products Minor and Major, which are these two products? As you may see the correct answer to this multiple choice question is B. However, if you feel that you have any questions, feel free to drop your comment down below in the comments section and thank you for watching this video

Predict the dehydrohalogenation product(s) that result when the following alkyl halides are heated in alcoholic KOH. When more than one product is formed, predict the major and minor products.
(d)
(e)

Key Concepts

Dehydrohalogenation

Zaitsev's Rule

Stereochemistry of Alkenes

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