Draw a mechanism for the reaction of propanoyl chloride with 2...

Question

Draw a mechanism for the reaction of propanoyl chloride with 2 moles of phenylmagnesium bromide.

Accepted Answer

All right. Hi, everyone. So for this question, let's propose a suitable mechanism to show how benzoyl chloride reacts with two equivalents of magnesium bromide. So before we begin, what sticks out to me is this is a beautiful magnesium bromide with extra emphasis on this magnesium bromide, right? Because recall that a magnesium bromide attached to an actual group like R S A bugle is indicative of a grand re agent, right? And if we have a green yard re agent, then of course, we have a green yard reaction. So for our green um reaction, we're going to see an addition of this is a beautiful group attached to our magnesium chloride magnesium bromide. Sorry to our benzo chlorine. That's what I meant to say. So now, in order to illustrate that, let's start off by drawing our benzoyl chloride, which is a benzene ring and an ASO group attached chloride to be exact. And for the sake of this video, I'm going to abbreviate our benzene ring as ph or a fennel group. So this is our benzoyl chloride and our isil beauty of magnesium bromide. I'm going to go ahead and I'm going to draw out in blue. So here's our magnesium bromide and here's the rest of our little group. So recall that in a Green Yard reaction are magnesium bromide has a partial positive charge, which means that the carbon its attached to has a partial negative charge. And this partial negative charge on carbon makes it nuclear filic. So your first step in a green Yard reaction is actually going to be the nuclear attack of this partial negative carbon to the carbon, no carbon in our benzoyl chloride, right. So once that happens, the pi electrons of your carbonnel are going to shift towards oxygen. And that means that in our next step, we're going to see an intermediate in which we have an O minus charge. So this is going to be our little group now attached to benzoyl chloride, right? And now here is gonna be our benzoyl chloride R P H CEO and oh minus like I mentioned before. So this oh minus is going to require a counter ion, which in this case is going to be magnesium bromide, which has since detached itself from our green yard re agent. So now in our next step, this all minus that we have here is actually going to become a carbonnel once again, after it donates a pair of electrons between this oxygen and the carbon is attached to. And that process is actually going to kick out chloride, so to speak. So we're going to end up with a ketone in our next step. This is our isil beetle group once again and now attached to it, we're going to have our final group and a car bottle indicating that this is a ketone now, but this is not exactly the end of the road, right? Because we have two equivalent of isil Butto magnesium bromide reacting with benzoyl chloride. So what that means is that I need to give myself a bit of space because the same process is going to repeat itself once again. So I'm going to draw out another equivalent of isil Butto magnesium bromide, but in a different color. So I'll pick green. And so the same process is going to happen one more time, right? Because our magnesium bromide group still has a partial positive, which means that our carbon here is partially negative and therefore nuclear filic. So for our next step here, we're going to see an additional nuclear filic attack from this second equivalent of the Bureau magnesium bromide, which is going to displace the pi electrons of our card bottle towards oxygen. So now We are going to end up with two is a beauty groups attached. Alright. This is gonna be our 1st 1. Here's our carbon in the center with our O minus in our final group. And here is going to be our second, so beautiful. So now we're almost done. Now, we're almost done and this reaction is actually going to end up with our final step requiring a proto nation, right? Because recall that a common product of the green, your creation is generally an alcohol. And in order to generate an alcohol, here, we are going to need acid. So we're gonna need protons floating around in solution. But before I do that, let's first point out that magnesium bromide is going to function as a counter ion once again. So here's our proton, which oxygen is gonna go ahead and take. And so now we're going to end up with our proto native alcohol. At the end of this process, he's our first isil beautiful group. Here's our ohh now with our fennel group And our 2nd is a beautiful group. So now this tertiary alcohol here is going to be your final product and this corresponds to option d within the multiple choice. So just to recap our grand re agent functions as a nuclear file in particular, the carbon. So that nuclear filic carbon in our grand re agent attacked the carbonnel of our starting material to generate aqui tone after the first reaction. But because there were two equivalents of green, you'd re agent, the process repeated itself. This nuclear feel like attack repeated itself until we ended up with an alcohol after acidic worker. So with that being said, thank you very much for watching. And I hope you found this helpful

Draw a mechanism for the reaction of propanoyl chloride with 2 moles of phenylmagnesium bromide.

Key Concepts

Grignard Reagents

Nucleophilic Addition Mechanism

Acyl Chlorides

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