DAT Destroyer OChem #6

[kxb4777]

Can anyone explain why this reaction would cut the lactone instead of just adding the methyl group to the carbonyl carbon?

Does the H3O acid form the hydroxyl group with the O on the ring (ester)?

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[gym.addict]

Look at it again, 2 methyl groups are added to the carbonyl carbon and the ester group is opened up and you get the two alcohol groups.

 

[whoaaitzkyle]

Can anyone explain why this reaction would cut the lactone instead of just adding the methyl group to the carbonyl carbon?

Does the H3O acid form the hydroxyl group with the O on the ring (ester)?

I'm not good with explaining ochem but I can explain to you what happens.

I'm not using the solutions so my reasoning might be a little off. You use 2 moles of CH3MgBr. The first mole is going to attack the carbonyl group creating an O- on it. Normally, you'd expect it to protonate and create an alcohol and you're done. However, the ester would be a good leaving group (as we have learned with other carbonyl groups such as acid halides) so the O- will go back and make a double bound (re-create the carbonyl group) and cleave the cyclic structure and now creating a new O- (the oxygen from the ester). That O- will be protonated to create your expected alcohol. Then you're going to have another mole of CH3MgBr attack the carbonyl group like usual and now you can protonate it to make your tertiary alcohol. So all in all, I believe you should be using 2 moles of CH3MgBr. Not sure if that's the "right" way to think of it but that's how I think of it and it's always worked for me. Hope that helps.

 

[gym.addict]

I'm not good with explaining ochem but I can explain to you what happens.

I'm not using the solutions so my reasoning might be a little off. You use 2 moles of CH3MgBr. The first mole is going to attack the carbonyl group creating an O- on it. Normally, you'd expect it to protonate and create an alcohol and you're done. However, the ester would be a good leaving group (as we have learned with other carbonyl groups such as acid halides) so the O- will go back and make a double bound (re-create the carbonyl group) and cleave the cyclic structure and now creating a new O- (the oxygen from the ester). That O- will be protonated to create your expected alcohol. Then you're going to have another mole of CH3MgBr attack the carbonyl group like usual and now you can protonate it to make your tertiary alcohol. So all in all, I believe you should be using 2 moles of CH3MgBr. Not sure if that's the "right" way to think of it but that's how I think of it and it's always worked for me. Hope that helps.

That's exactly how the mechanism works, but its a lot to type lol.

 

[kxb4777]

Okay, so I see how the ring is cleaved. I'm still uncertain as to why this reaction continues after the addition of the methyl group to the carbonyl. Why wouldn't the oxygen on the former carbonyl group just be protonated? The question itself doesn't mention 2 moles of CH3MgBr, btw.

You mentioned esters as good leaving groups...is that the reason the reaction continues on?

 

[kxb4777]

oh okay. i re-read it. lol answer's there. thanks.

 

[whoaaitzkyle]

Okay, so I see how the ring is cleaved. I'm still uncertain as to why this reaction continues after the addition of the methyl group to the carbonyl. Why wouldn't the oxygen on the former carbonyl group just be protonated? The question itself doesn't mention 2 moles of CH3MgBr, btw.

You mentioned esters as good leaving groups...is that the reason the reaction continues on?

Think of it like this. There's a lot of ring sterance. The most favored ring structure I think is a 6 carbon ring structure. Anything less than that has sterics. So when you add your grignard, it would be the most favorable to break the ring sterics. Once you've relieved the ring, you find you have your carbonyl again (refer to above explanation) so you will assume the grignard is in excess and you can do it again.

Also, if a reaction doesn't specifically state 1 partCH3MgBr, you always assume it's in excess.