Orgo Destroyer #143

[hope_to_match]

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This is from 2013 version. It says to complete the product and gives alkyne as a starting material that reacts with Ch3MgBr (first step) and then Acetone (second step followed by protonation).

My question is: i thought that alkynes react with grignard and hinder its formation? Is this why in this problem it reacts with Ch3MgBr and then alkyne with MgBr attacks acetone and the product is formed? Chad said that Ch4 would bubble out and no reaction would take place but in this problem we are making an acidic specie a nucleophile..Can someone help me? Im so confused 😱

 

[Jordwin]

This is from 2013 version. It says to complete the product and gives alkyne as a starting material that reacts with Ch3MgBr (first step) and then Acetone (second step followed by protonation).

My question is: i thought that alkynes react with grignard and hinder its formation? Is this why in this problem it reacts with Ch3MgBr and then alkyne with MgBr attacks acetone and the product is formed? Chad said that Ch4 would bubble out and no reaction would take place but in this problem we are making an acidic specie a nucleophile..Can someone help me? Im so confused 😱

Chad says that the grignard does react with the alkyne, but that all it does is take off the acidic hydrogen from the terminal alkyne. The alkyne then has a negative charge on the end. In chads videos, he adds the acid back, which would make it turn back to the alkyne with no charge and everything would happen the way you imagined it to.

In this case, though, before the negative charge can be taken off by an acid (h3o+ is added in step 2, AFTER acetone), acetone is added (ketone). This reaction would have the negatively charged alkyne (nucleophile) attacking the partial positive carbon of the carbonyl of acetone, kicks up the electrons of the double bond to oxygen to make it a negatively charged oxygen. Then the second step with the acid adds a proton to the oxygen with a negative charge, making an alcohol.

Does this make sense??

 

[Jordwin]

The main thing here is that the acid isn't used in the first step, so the alkyne keeps it's negative charge and can act as a nucleophile.

 

[hope_to_match]

The main thing here is that the acid isn't used in the first step, so the alkyne keeps it's negative charge and can act as a nucleophile.

i think im still confused. Im looking at Chads notes and it says 'Ch4 bubbles out to destroy grignard reagent'. Looking at the solution to this question, Ch4 bubbles out in the first step as well. Wouldnt it destroy the newly formed grignard?

I understand what you mean. Chads notes show acid and H3O+ in the first step, therefore it protonates it right back and nothing happens? As opposed to in this problem no H3O+ is used in this step and thats why no protonation and alkyne can act as a nucleophile? Then my next question is: if we don't add H3O+ in the first step in Chad's notes, would the rxn still happen? I thought the main thing there was the fact that we used an acidic specie to begin with (Acetic acid, alkyne etc...)

 

[Jordwin]

The grignard doesn't form in the way it normally does. It takes off the acidic hydrogen and does not attach to the alkyne with the negative charge, so the ch4 does bubble out still. This problem is still "destroying" the grignard by using an acidic hydrogen.

So to your second question. As you know we start with an acidic hydrogen, in this case the alkyne hydrogen. So once the grignard comes in, it will take off the hydrogen and give the alkyne a negative charge. This is the reaction that we want here. If you add acid before you add an electrophile such as the ketone, the alkyne will be protonated and the reaction will not proceed the way it was intended to. A rxn did happen, but the original reactant was reformed.

If the acid is added AFTER the electrophile, though, the nucleophilic alkyne with the negative charge can attack the electrophile and attach to the ketone in this case. The acid will then protonated the oxygen of the ketone, and will not be able to protonated the negatively charged alkyne because it has already reacted.

I'm not sure that this explains it much better, maybe someone else can give it a shot.

 

[hope_to_match]

The grignard doesn't form in the way it normally does. It takes off the acidic hydrogen and does not attach to the alkyne with the negative charge, so the ch4 does bubble out still. This problem is still "destroying" the grignard by using an acidic hydrogen.

So to your second question. As you know we start with an acidic hydrogen, in this case the alkyne hydrogen. So once the grignard comes in, it will take off the hydrogen and give the alkyne a negative charge. This is the reaction that we want here. If you add acid before you add an electrophile such as the ketone, the alkyne will be protonated and the reaction will not proceed the way it was intended to. A rxn did happen, but the original reactant was reformed.

If the acid is added AFTER the electrophile, though, the nucleophilic alkyne with the negative charge can attack the electrophile and attach to the ketone in this case. The acid will then protonated the oxygen of the ketone, and will not be able to protonated the negatively charged alkyne because it has already reacted.

I'm not sure that this explains it much better, maybe someone else can give it a shot.

yeah i get it. so the main point is adding acid after the electrophile. if we add it before no rxn will take place at all, correct?

 

[Jordwin]

yeah i get it. so the main point is adding acid after the electrophile. if we add it before no rxn will take place at all, correct?

The original alkyne will be the same, yes.

Although a grignard will be used and the ch4 will still bubble out, so in that regard a rxn took place. But this is most likely insignificant.