What is the major product of the reaction of 1 mol of propyne wit...

Question

What is the major product of the reaction of 1 mol of propyne with each of the following reagents?

j. the product of part i followed by 1-chloropropane

Accepted Answer

Hey, everyone. And welcome back to another video, determine the major product when one mole of 33 dimethyl butte, one ion reacts with sodium amide followed by two bromo butane. Let's start by drawing the the starting material. We have an alk at position number one, we have our triple bond as suggested by the name of the structure. So we are going to draw our triple bond and the parent is time. So we have a total of four carbon atoms 123 and we are going to add one more. And it also states that at position three, we have two me groups. So 123, we're going to add two methyl groups. Well done. So we got our starting material. First of all, we are using one mole of it, right. So let's keep in mind and we are using sodium amide. The problem doesn't state the amount that's fine. We can assume an excess. It doesn't really matter if we analyze that problem. We will notice that sodium amide essentially implies that we're using amide and ions, right? Because sodium is an iron spectator. So we are going to use A I which is a strong base. And we have to recall that it is a classical base whenever it comes to derotating terminal alkins. So the negative charge will attack our acidic hydrogen ripe, followed by the formation of a lone pair on carbon. And as a result, and a satellite ion will be formed. Now, we are, we can simply redraw the product that we got in this first step. And if we remove our hydrogen, we will simply notice that. Now we end up with a satellite. According to Stoch Mery, we are still theoretically forming one mole assuming 100% recovery. We're not really worried about the practical yield. But you may notice that according to Stoch Mery, assuming an excess of sodium amide, we are going to form one mole of a satellite. And afterwards, as the problem suggests, we are adding two bromo butane. So if we draw 1234 member chain, that's butane. Car number two has BROMO group, we have a basic essence reaction because we have a strong nuclear file. That's our a satellite because it has a negative charge on carbon. We have our secondary substrate. We have have our good leaving group, which is our BROMO group. And therefore we are observing an S N two reaction in which the negatively charged carbon. Well, let's tag the Electrophysics Center followed by the loss of the leaving group. So we are ready to draw the product. And because we are forming our carbon carbon bond. We can start with our nuclear file and we can draw that result on carbon carbon bond to begin with. And we also noticed that the carbon atom that we have just drawn will also have one methyl group. So we're going to add a methyl group, set it and also an ethyl group. So now the problem was asking us to determine the major product and the product that we have formed here is 225 trim meal helped 39. OK. So we can simply review everything that we did to get that product. First of all, we use sodium amide to dep proton, the terminal alkin, we have formed a satellite and I which is a strong nuclei in a, in an essence two reaction with an alkyl bromide. It produces a new internal alkin in this case, 225 trime hept three. I thank you for watching and I hope to see you in the next video.

What is the major product of the reaction of 1 mol of propyne with each of the following reagents? j. the product of part i followed by 1-chloropropane

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