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Topic: Michael addition  (Read 3279 times)

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Offline thesurgeon

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Michael addition
« on: April 26, 2016, 10:41:29 AM »
Which statement about the Michael's addition is false? A) the nucleophile is generally the enolate ion of a diketone. B) the substrate is an a,b-unsaturated ketone. C) the enolate is weakly basic. D) the reaction is a conjugate addition. E) the reaction is a 1,4-addition.

I chose B) because it's not necessarily a ketone, but turns out it was E). How come Michael addition isn't a 1,4 addition?

Offline Dan

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Re: Michael addition
« Reply #1 on: April 27, 2016, 02:53:06 AM »
It's a terrible question if you ask me, a lot of ambiguity. I would not have chosen E either, I'd have gone for A or B. I'm pretty sure Michael addition is a 1,4-addition reaction by definition.

I think you have to ask the author of the question for an explanation.
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Offline orgopete

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Re: Michael addition
« Reply #2 on: April 29, 2016, 09:00:07 AM »
This looks like an error to me. I wasn't going to choose A or B though.
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Offline Babcock_Hall

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Re: Michael addition
« Reply #3 on: April 29, 2016, 10:27:24 AM »
I would have chosen C, but I suppose it depends on where one draws the line between weakly and strongly basic.

Offline Dan

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Re: Michael addition
« Reply #4 on: April 29, 2016, 10:52:23 AM »
There ate lots of different Michael acceptors and nucleophiles that participate in Michael reactions, so I'd say A and B are not necessarily true. Nor is C, since there is no definition of "weakly basic".

D and E are definitely true though.
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Offline orgopete

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Re: Michael addition
« Reply #5 on: April 30, 2016, 10:39:31 AM »
There ate lots of different Michael acceptors and nucleophiles that participate in Michael reactions, so I'd say A and B are not necessarily true. Nor is C, since there is no definition of "weakly basic".

D and E are definitely true though.

While I agree with not necessarily true, I'd also argue they are not false. I still think this problem is in error.
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Offline DeanC

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Re: Michael addition
« Reply #6 on: June 10, 2016, 08:51:06 AM »
I'd be drawn towards A to be honest, the nucleophile is an enolate ion but not always formed from a diketone. For example, in a Robinson annulation, cyclohexanone can be used as the Michael donor and the alpha carbon is deprotonated to give the enolate ion.

I'd like to know how some people think B is false tho? I was under the impression 1,4 additions can only occur in an α-β-unsaturated substrate? If it wasn't α-β-unsaturated, how could it be a Michael addition by definition and how could conjugate addition occur? Apologies of course if I am wrong but would love somebody with more experience than myself to clear this up.

D and E definitely look correct tho and, as others have pointed out, your 'weakly basic' may differ from my definition of it. You would need something to compare it to in order to give a definite answer for that one.

Offline orgopete

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Re: Michael addition
« Reply #7 on: June 10, 2016, 11:29:23 AM »
I'd be drawn towards A to be honest, the nucleophile is an enolate ion but not always formed from a diketone. For example, in a Robinson annulation, cyclohexanone can be used as the Michael donor and the alpha carbon is deprotonated to give the enolate ion.

I'd like to know how some people think B is false tho? I was under the impression 1,4 additions can only occur in an α-β-unsaturated substrate? If it wasn't α-β-unsaturated, how could it be a Michael addition by definition and how could conjugate addition occur? Apologies of course if I am wrong but would love somebody with more experience than myself to clear this up.

D and E definitely look correct tho and, as others have pointed out, your 'weakly basic' may differ from my definition of it. You would need something to compare it to in order to give a definite answer for that one.

I used an enolate of a diketone in my class for an example. I argued that if a 1,2-addition occurred, it would reverse and enable the 1,4-addition to take place. If 1,4-addition took place followed by protonation, the conditions are usually not sufficiently basic to reverse the addition. Hence I reasoned this is why weakly basic nucleophiles were more likely involved in 1,4-addition reactions. However, none of this matters. The poster was trying to understand why E should be the correct answer. The poster wasn't asking why the answer wasn't A. By my looking at all of the choices, this appears to be an error in the problem.
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Offline DeanC

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Re: Michael addition
« Reply #8 on: June 12, 2016, 02:52:53 PM »
I'd be drawn towards A to be honest, the nucleophile is an enolate ion but not always formed from a diketone. For example, in a Robinson annulation, cyclohexanone can be used as the Michael donor and the alpha carbon is deprotonated to give the enolate ion.

I'd like to know how some people think B is false tho? I was under the impression 1,4 additions can only occur in an α-β-unsaturated substrate? If it wasn't α-β-unsaturated, how could it be a Michael addition by definition and how could conjugate addition occur? Apologies of course if I am wrong but would love somebody with more experience than myself to clear this up.

D and E definitely look correct tho and, as others have pointed out, your 'weakly basic' may differ from my definition of it. You would need something to compare it to in order to give a definite answer for that one.

I used an enolate of a diketone in my class for an example. I argued that if a 1,2-addition occurred, it would reverse and enable the 1,4-addition to take place. If 1,4-addition took place followed by protonation, the conditions are usually not sufficiently basic to reverse the addition. Hence I reasoned this is why weakly basic nucleophiles were more likely involved in 1,4-addition reactions. However, none of this matters. The poster was trying to understand why E should be the correct answer. The poster wasn't asking why the answer wasn't A. By my looking at all of the choices, this appears to be an error in the problem.

I can see your logic there, fair points made. I think we can both agree that E is definitely true anyway, so there must be a mistake, as you have pointed out. Thank you for the input. 

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