What will be the major product obtained from the reaction of Br 2 with 1-butene if the reaction is carried out in a. dichloromethane? b. water?

Hi, everyone. Welcome back. Our next problem here says consider the reaction of cl two with 23 dimethyl but two in provide the major product if the reaction is carried out in the following. Number one, water and number two CH two cl two. So we have this reaction with a di Halide and an alkene. And we're doing this in two different solvents. So we have water and then we have this inert solvent of CH two cl two. So let's think about what type of reaction we have. It will obviously be a halogen nation. But when a halogen reaction takes place in water. So for number one, water, our product is a halo hydrant. So we have the addition of the first Halide to the double bond break, the double bond add on a Halide that forms an intermediate which is a cyclic halo hy excuse me, a cyclic hydro num ion. Let's try again say click halo ion. Sometimes these are tongue twisters and that ring is opened. But if the solvent is water, there's way more water molecules around than there are chlorine. So it's more likely that water is going to get in there and open that ring before a second chlorine atom. So that's why it'll end up with a halo hydrant in your ch two C two solvent. This is inert. So you'll just add two halogens two chlorine atoms to the cross that double bond. So the product it is a die hay ride. So let's think about each of these. So for number one, well, what's our reactant here? 23 dimethyl but two in. So we have four carbon change, we have a double bond between carbons two and three, put that right in the middle there and then carbons two and three each of a methyl group. So we add them on note right away. We have a symmetrical molecule here. Our double bond has a methyl group on either side. So we don't have to worry too much about stereo chemistry. Here is something Markov NK or not because our two carbons are identical, they're equally substituted and we won't have difference in stereo chemistry at the end. So makes things a little simpler for us. So there's a reactant molecule and now we're reacting with cl two in water. So we know we're gonna have a halo hydrant. So you know, we go through a stick lick polonium, I am as an intermediate. So if we had to worry about which carbon it would add to the ring would be broken in this case by a water molecule and it would add on to the most substituted carbon and then break open the ring. So the oh would be on the more substituted carbon. But again, our molecule is symmetrical. So we don't worry about that. We're just going to stick an oh group on one carbon and a chlorine on the other. So now we just have a single bond here and we end up with three groups on each side. So we have methyl, methyl and oh again, it doesn't matter which side I add it to because the symmetrical was a symmetrical molecule. Now it won't be, but it doesn't matter which side it adds on to. So, oh on the left side, in this case, and then chlorine on the other side. So if we want to name it just to describe it, we'd end up, let's number our carbons, the carbon close to the alcohol group is carbon one. So we'd end up with three chloro 23 dimethyl butane two, excuse me, butane two, all as our product. So that's in water. And then for number two, we have our reactant molecule there. It's reacting with cl two in the inner solvent, ch two cl two. And so we're going to have a die. He lied. So again, we'll end up with one chlorine on one side and one chlorine on the other. And we'll go through that. The click helenium ion is an intermediate with a second chlorine coming in and breaking open the ring. So one chlorine on either side, we don't worry about steri chemistry. We have no kwal centers here because of the two methyl groups. So we'd end up with 23 dielo, 23 dimethyl butane and just fix excellently out of the hyphen in there where we don't need one butane. So we've provided our major product. In each case. When we react with water, we end up with that halo hydran three chloro 23 dimethyl butane tool. When we react in an inert solvent, we end up with a di Halide 23 di Chloro 23 dimethyl butane. Hope to see you in the next video.

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10. Addition Reactions

Halohydrin

Organic Chemistry

10. Addition Reactions

Halohydrin

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6:00 minutes

Problem 25a,b

Textbook Question

What will be the major product obtained from the reaction of Br₂ with 1-butene if the reaction is carried out in
a. dichloromethane?
b. water?

Verified step by step guidance

Identify the type of reaction occurring in each case. For part (a), the reaction of Br2 with 1-butene in dichloromethane is an electrophilic addition reaction in a non-polar solvent. For part (b), the reaction of Br2 with 1-butene in water involves a bromohydrin formation due to the presence of a polar protic solvent.

For part (a), in the presence of dichloromethane, the Br2 molecule acts as an electrophile. The π-electrons of the double bond in 1-butene attack the Br2 molecule, forming a cyclic bromonium ion intermediate.

Still in part (a), the bromonium ion is then attacked by a bromide ion (Br⁻) from the back side, leading to the anti-addition of bromine across the double bond. This results in the formation of 2,3-dibromobutane as the major product.

For part (b), in the presence of water, the reaction mechanism begins similarly with the formation of a bromonium ion intermediate. However, instead of a bromide ion attacking the intermediate, a water molecule (H2O) acts as the nucleophile due to its abundance in the reaction medium.

In part (b), the water molecule attacks the more substituted carbon of the bromonium ion (Markovnikov addition), leading to the formation of a bromohydrin. After deprotonation of the attached water molecule, the major product is 2-bromo-1-butanol.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrophilic Addition

Electrophilic addition is a fundamental reaction mechanism in organic chemistry where an electrophile reacts with a nucleophile, typically involving alkenes. In the case of 1-butene reacting with Br2, the double bond acts as a nucleophile, attacking the bromine molecule, leading to the formation of a bromonium ion intermediate. This process is crucial for understanding how alkenes react with halogens.

Solvent Effects on Reaction Mechanisms

The choice of solvent can significantly influence the outcome of a chemical reaction. In non-polar solvents like dichloromethane, the reaction may favor the formation of a more stable product through a direct addition mechanism. Conversely, in polar solvents like water, the reaction can lead to different products due to the solvent's ability to stabilize charged intermediates, affecting the regioselectivity and stereochemistry of the final product.

Regioselectivity

Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others when multiple possibilities exist. In the reaction of Br2 with 1-butene, the regioselectivity will depend on the solvent and the stability of the intermediates formed. For example, in water, the formation of a bromohydrin may occur, while in dichloromethane, a dibromide may be the major product, illustrating how regioselectivity is influenced by reaction conditions.

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