A formate ester, such as ethyl formate, reacts with an excess ...

Question

A formate ester, such as ethyl formate, reacts with an excess of a Grignard reagent to give (after protonation) secondary alcohols with two identical alkyl groups.

(a) Propose a mechanism to show how the reaction of ethyl formate with an excess of allylmagnesium bromide gives, after protonation, hepta-1,6-dien-4-ol.

Accepted Answer

Hey everyone and welcome back a reaction between an access of a green yard regent and a format Esther followed by a subsequent protein nation produce a secondary alcohol with c identical alcohol groups suggests a plausible mechanism for the reaction between an excess of allele magnesium bromide and metal format. Your mechanism should explain the formation of the product. We may start by drawing metal format. So that's an ester which has a company group with hydrogen on one end and meth oxy group on theater. That's metal format. According to the problem, we're going to treat it with an excess of a little magnesium bromide. We're going to draw in a little group bon appetit magnesium bromide. We're going to use access and there must be ether as our solvent. That's reaction number one. Reaction. Number two would be acidic World cup. Now let's think about the steps of this reaction. First of all, we are going to redraw hello magnesium bromide. And as you recall, the green region. First of all add tags the electra Philip position of the carbonell group. And we're breaking the pi bon transferring these five electrons onto oxygen in the form of a lone pair. So this would be our first stop which is the nuclear attack stop. We're forming our intermediate. And you will notice that the fact that we have broken that double bond. Now we have an elk oxide and iron on one end. We have hydrogen meth oxy group And now we are going to draw that add a little group. So that would be three carbon atoms. Were going to bond this carbon atom to the original carbon atom within the carbonell. That'd be 123. There's a double bond at this position. Now this intermediate is not really stable. So what we're going to do, we're going to restore the carbon bond using these electrons, there's an excess of electrons. We're going to use them into the formation of the pi bond and then we will have the loss of the living room because this is a good leaving group for us. That's why this reaction is possible here. So we're going to say that here we have the loss of the leaving group and we are forming an alga hydride. Because if you notice there is this carbon group now with hydrogen and on the other end we have our aloe group that we want to three carbon atoms now because the problem suggests there's an access. That makes complete sense. We once again have our carbon compound which can participate in a granular reaction. So we draw another equivalent of a low magnesium bromide. We have another nuclear attack. Stop in that nuclear attack. Stop just as we did before we're going to attack that carbonell and formed the elk oxide. And I am we can indicate that this is another nuclear attack. Stop and through the resultant Elcock side. So there will be oxygen with a negative charge Hydrogen on one end And now there will be two identical. A little groups. Monitor it just as the problem suggests, That's 1, 2, 3, that's our intermediate. Our elk oxide essentially were last step we want to perform here is the protein nation stop during the protein nation step. We're using hydro knee. Um so we can expand H O H. H with a former positive charge and oxygen and the negative charge and oxygen from elk oxide attacks one of the acidic protons followed by the loss of the water molecule. And this is how we are forming our found product. We can say that this is the protein ation stop and during that proto nation step we need to get our alcohol. So let's see how that looks like that would be. Ohh, we may not necessarily show that implicit Haijun. So let's just show the to allow groups of 1 to thursday double one over here. 123, they're say double bond at this position. So that's our found product. We got our alcohol. And let's just discuss the overall mechanism. So in the first step, we have the nuclear attack by the granite reagent which actually upon the formation of the elk oxide and I and restores the double bond. Just the fact that there is this good leaving group that comes from an ester. This is where important to understand asserts, can react with the granite region twice during the loss of the leaving group. Performing our AL to hide. And there's another equivalent of the granite region that comes in and forms the Elcock said. And on with these two identical AL kill chains in this case, those are a little substitutions that we have added. Right, just as the problem says, we may add two identical alcohol groups and this is what we did in the stop after the nuclear attack stop. We're performing the protein nation stop using acidic world cup were protein ating Alcock side informing our final alcohol. This is how the mechanism looks like, and we may simply conclude that the correct answer choice to the smoke. Bold choice question is C However, if you have any questions, don't hesitate to leave your comment down below in the comments section and thank you for watching

Key Concepts

Grignard Reagents

Ester Reactivity

Protonation and Alcohol Formation

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