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.

(a)

(b)

(c)

Accepted Answer

welcome back everyone in this problem. We wish to show the products of the d hydro elimination reactions of the following. L. Kill hey lights using potassium hydroxide and also label these products as major and minor. Now before we look at this problem, let's just recall the basic principles of the hydro hallucination reactions. And in such reactions we are removing a proton and a hey light and iron. And because we are moving a protein and a highlight and iron, we can simply state that such reactions could also be classified as elimination reactions. Because there are two types of elimination reactions. We first of all have to understand what kind of base we're dealing with. And because the problem suggests that we have potassium hydroxide was associates and potassium and hydroxide. The hydroxide and I and is classified as a strong base. It's basically charged And a strong base implies that such a reaction is E two. Well done. We are done with our analysis. We are ready to look at the actual problem and apply that knowledge. What we have to do now is simply recall the steps of an E two reaction. And in such a reaction, we're primarily interested in beta hydrants. So first of all we indicate the alpha carbon itself which is bonded to the leaving group chlorine. This is our alpha carbon and the neighboring carbons are beta carbons. We have three made of carbon to these physicians and these beta carbons contain beta hae jin's. So we can show at least one for each. So we have one over here here and here we may think about our products this way, how many different beta high genes do we have? And actually we have two different data hygiene is because two methyl groups are exactly the same. There's a free bond rotation making them equivalent. So these high regions are number one, they're equal and to each other and there's this another hydrogen number two which is different. We are ready to draw the mechanism and get our products. We already know that we will have two different products. So starting with a we will draw the original structure. This is our L kill Hayley and let's start with one of the two methyl groups. What do we simply remove one of the hygiene is bonded to that first metal group, The hydroxide and ion in an E two mechanism simply attacks Haijun. We have a transfer of these bonding electrons in say a double bond over here and we have the loss of the leaving group. So that's how we are forming our first product. We may now draw the resultant Elkin. This elkin well look as follows. So basically we still have a metal group bonnet but here we are forming a double bond. So that's our first product. Similarly, if we draw that structure once again and think about the other beta hygiene. The main issue is that if you look at this beta Haijun and recall that for an eater reaction. Your substitute parents should be anti per plane or to each other meaning they must be on opposite sides of that carbon carbon bond. Over here we first of all have to rotate our bond to make sure that we have the N. C. Per planer arrangement. And that said we're going to rotate that bond and read through our structure correctly for Annie's reaction to take place such that chlorine starts pointing downwards. So let's read through our structure in an anti per plane arrangement and this is the structure that we get. Now one of the muscle groups will start pointing up, another one will go down and chlorine will be down as well. Now we clearly see that this is the anti per plane arrangement meaning we're ready to eliminate that hygiene. And if you look at the steps of the elimination the hydroxide and iron, it's like the proton were forming a double bond and removing the leaving group. So the product in this case is again an elkin but it's a different out came We form a double bond between positions two and 3. Well done. We have our products, we can already label these as our products and the word last thing we want to do. We want to classify them as minor or major. Recall that according to the states of school, we try to form a more stable alki meaning more substituted Elkin. This is a dice of student alK in and this is a tricep student Elkin because respectively we have two Lkoh substitutions for the first sale king and three Lkoh substations for the 2nd 1, therefore a more stable alkaline would be the tri substituted products. We could say that this is our major product because it's more stable and the di substituted version would be the minor one because it's less stable. Great job. So let's move on to be First of all, we will just redraw our structure and we will do that by drawing the parent and then the muscle group. Our goal is exactly the same here. We wish to find the position of our alpha carbon. It's over here. So we noticed now that we only have two different beta high jeans and let's start with the bottom one of the metal group. To remove this beta hydrogen were again using our strong base D protein ating forming a double bond. And simultaneously we have the loss of the leaving group and we are forming our first product. So this is our first tell team. And similarly we have to think about the removable removal of the second proton, the second beta hydrants over here. So we will simply redraw our structure. We will show that beta hygiene that we're interested in and because again the hydrogen and chlorine are on the same side of the C C bond. We have to perform the bond rotation first to make sure that it's anti per planner. So we are just going to show that we are rotating the bond after the bond rotation. We get the structure that we are Actually are interested in because we want to perform an E two reaction. So the metal group will start pointing up and chlorine will be pointing down, making this an anti arrangement which is perfect for us. And now using the previous steps, the hydroxide, we'll attack the proton, we will form a double bond and we will have the loss of the leading group. And this is now our final product. We are forming a double bond between position two and and that's it. That's how Al came. Now let's think about the stability of these Elkins. Remember that a more stable Al King would be the major product. So this is a mono substituted and this is a di substituted. Notice that there's only one LK group bonded to the carbon that is within the double bond or sp two carbon. And here we have two LK groups bonded to sp two carbon atoms. That makes the the structure a decisive student, Hell King. Therefore a deist subsidy to the Elkin is more stable, meaning this is our major product while the previous one is our minor product. So, these are our answers. This is the minor product. This is the major product and we are ready to look at sea which is our final structure to be analyzed for C. Our al kill Hey line looks like it does. And notice that after careful announces, you may see that because there's an internal plan of symmetry, the two beta hydrants at these positions are exactly the same. So it doesn't really matter which one you remove, you will get exactly the same product. We are using our strong base D. Protein ating forming a double bond and simultaneously we have the loss of the living group to form our final product. So the final product in this case would be a single product, a di substituted elkin. So there's no minor or major version, there's just a single product form. So we don't have to worry about labeling, labeling it as a minor or a major product. And that's pretty much it. We have all of our products for A. B and C. The correct answer for this question is A and if you have any questions, feel free to leave them down below. 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.
(a)
(b)
(c)

Key Concepts

Dehydrohalogenation

Zaitsev's Rule

Regioselectivity

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