In the addition of just 1 mole of bromine to 1 mole of hex-1-y...

Question

In the addition of just 1 mole of bromine to 1 mole of hex-1-yne, should the hex-1-yne be added to a bromine solution or should the bromine be added to the hex-1-yne? Explain your answer.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem for the addition of one mole of roaming to one mole of TB time, which of the following is a better approach to the reaction. A blooming solution is slowly added to the to be time solution and B two B time solution is slowly added to the booming solution. Now let's visualize A and B. Essentially we can just think of two different flasks in particular. We are going to draw two different flasks where we have our reactions. So first of all, let's vote that we have this one and then that one for B. Now, for a the scenario is the following. It says the booming solution is added to the to be time solution. So that means here we have T B time, that's our solution and we're adding blooming to it. So that's our first scenario just so that we can visualize it more easily. And for B we have grooming solution and the region that we're adding is TB time solution. Now, in each case we have an equivalent reaction. It's it's just that we are switching our solution to a different one and then changing our region. But you may notice that we have exactly the same to react. It's it's just that the order in which we are adding them or mixing them is different. So we're going to think about this reaction in general and why this is important, why there is a better approach. Why don't they really give the same product or not necessarily, maybe just something is different about it. So let's understand it. Now, if we start by drawing our al kind, we have to be times. So that means we have four carbon atoms and then there must be triple bond between carbons two and three. So that's to be time now, what kind of reactions do we get when it reacts with blooming? We call that first of all, if we draw our blooming and if we expand that blooming blooming bond when booming approaches the pi bond, it gets slightly polarized. So we have a partially positive and and partially negative and because electrons shift towards this booming gT the electron repulsion between the pi bond and this blooming that is incoming. So now essentially we may notice that first of all we have an attack by the pi bond on this bromine, bromine has a lone pair that attacks back. Right. So we can essentially show that it attacks back and then that's followed by the loss of the bromide and ion. And in this stuff, we are forming our intermediate right? We are going to form the following structure. You may 1st of all notice that you are breaking one of your pi bonds. So you end up with a double bond than there are two metal substitute prints and then we are getting our blooming with a positive charge because now it has two bonds. Right? So now when we formed this intermediate, we are going to perform the backside attack the bromide. And I am well essentially attack any of the two carbons right there equally substituted. And the structure is symmetrical. So we are going to let's attack this one. This would be followed by the loss of the leaving group. And you may notice how we would end up with the anti or basically trans arrangement of the two bromo substitutes, right? Because we have our backside attack. So in this step we are forming our product and this is what we get when we add one mole of roaming. So let's draw it clearly notice how we can start by drawing our double bond first. And as we said, during the loss of the leaving group, this bromo substitution will be pointing up. We had a backside attack. So this promo substation will be pointing down with respect to the double bond. And then we are going to add our two methyl groups. So we have our trans structure Right now, think about this reaction. What if it was more than one more of roaming added essentially. You will end up with the following structure. You can have one more addition. So that would be your another equivalent of grooming, right? Because we still have one more pi bond. So you will form your tetra substituted elkin that be your four carbon atoms. So let's throw our four member chain 1234. And then you will notice how you're adding one more broom in here. So you'll have tea total right? You will have t total here and t more here. So there's a chance to form a tetra substituted structure. Right? This was our first equivalent. We want to stop here, right? We want to get that product, we don't want to get the other one. So now thinking about these structures, this is really important. Which pathway will we choose? Going back to our representation of the two scenarios. We may see that in b we're using bromine as our starting solution and we are slowly adding to be time. Now this implies that bro mean is an access right? Because roominess and access and you're slowly adding to be time. There's a high chance to form a Tetra substituted product, the one that we discussed. Right? So essentially you might expect to form that Tetra substituted product here because you have the solution of grooming and that implies blooming is an access because you're slowly adding your limiting reagent. So this would essentially give you a Tetra substituted product but for this one TB time is an access because that's your starting solution and you're slowly adding your limiting reagent. Right? Slowly adding booming and that means there's a higher chance or basically that would be the main product right? Because you're slowly adding blooming until you get that first product. So you would basically form your di substituted product. Just as we said before you're going to get to como substitutions added. So basically that would be our analysis. We understand that if we compare A and B, which one of the two would be a better approach. It would be a right. So we can say that A. With correspond to a better approach. We can label this is our final answer. Blooming solution is slowly added to the T. B. Team solution would be a better approach for our reaction because the starting solution contains TB time and access and then grooming is consumed first and TB teen remains in excess. So basically this allows us to perform a better reaction because we expect to have this is our major product and for B we expect to have our Tetra substituted product as the major product that would be it based on our answer. We can conclude that A is the right choice. However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching.

In the addition of just 1 mole of bromine to 1 mole of hex-1-yne, should the hex-1-yne be added to a bromine solution or should the bromine be added to the hex-1-yne? Explain your answer.

Key Concepts

Electrophilic Addition

Reactivity of Alkynes

Solubility and Reaction Medium

Watch next