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Author Topic: Grignard Problem  (Read 7538 times)

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gojase

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Grignard Problem
« on: May 31, 2011, 11:40:02 PM »

I've been working on some multiple choice problems in this study guide of I bought, and there's a predict-the-product(s) question that I don't understand.  Hopefully some of you can help me out.  Here it goes:

3-hydroxybutanal reacting with 1) methylmagnesium bromide [1 equiv.], then 2) H3O+.

According to the book, the correct answer is 3-hydroxybutanal + methane, but I don't see how the answer I choose, pentane-2,4-diol, is incorrect.  What is the reasoning for this?


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Åke

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Re: Grignard Problem
« Reply #1 on: June 01, 2011, 11:58:49 AM »

Is the Grignard reagent beaing an acid or a base in this reaction? What would happen if two equivalents of the Grignard reagent were used?
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gojase

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Re: Grignard Problem
« Reply #2 on: June 01, 2011, 08:13:08 PM »

Hello, Ake.

It's definitely not acting as an acid, but I'm unfamiliar with Grignards reacting as bases with carbonyl compounds, except during failed reactions--like the quenching that occurs when reacting with carboxylic acid.  I tried to work on this problem earlier today, but the more I worked on it, the more confused I got.  The only way I can see 3-hydroxybutanal and methane resulting as the final product, is if the Grignard were to deprotonate the alcohol, which would get re-protonated when the acidic water is introduced.  Another failed reaction.

But this doesn't seem likely.  If the Grignard were acting as a base, I'd think it'd abstract the alpha-hydrogen and undergo a dehydration reaction before it deprotonates the alcohol.  This would be the first alpha-carbon that I've encountered that gets deprotonated by a Grignard, but I was banking on the chance that the alcohol on the beta-carbon increased the alpha-carbon's acidity through induction.  But still, this results in but-2-enal, which is still incorrect.  My guess is that the activation energy too high for this to occur without heat.

Had the book not provided me with the answer, I would've never considered any of the above and been absolutely certain that the Grignard would act as a nucleophile and attack the carbonyl carbon.  The intermediate, I would've thought, would've been an alkoxide that gets protonated by the acidic water during the second step.

If two equivalent moles of the Grignard reagent were used, I think the final product would be pentane-2,4-diol.  That's assuming that my first scenario--where the alcohol is deprotonated--is the correct scenario.  The first Grignard would be quenched by the alcohol.  The second Grignard would come in unhindered and act as a naked nucleophile.  The intermediate would have two alkoxide ions, one on carbon 2 and one on carbon 4.

That's all I could think of, and I really thought it out.  It's frustrating. There's probably a really simple explanation... I just can't find it.


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Åke

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Re: Grignard Problem
« Reply #3 on: June 01, 2011, 10:21:20 PM »

Hi gojase!

If the intention is to attain precisely that transformation, why would you consider it a “failed reaction”? There are situations when Grignard reagents act as bases rather than nucleophiles, usually when the electrophiles contain relatively acidic protons, like for example those of alcohols. The elimination into crotonaldehyde can't happen under the basic conditions, unless the hydroxyl group is first made into a leaving group.
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gojase

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Re: Grignard Problem
« Reply #4 on: June 02, 2011, 04:38:07 AM »

Hey, Ake.

Thank you for your response.  I called it a failed reaction because the the net reaction results in no transformation on the substrate.  The only thing undergoing transformation is the reagent.  If someone had acidic water, methylmagnesium bromide, and 3-hydroxybutanal, I don't see why they'd bother with the first step of the reaction, when they could just introduce the water to the Grignard without the presence of the adol.  It's just an extra step with an extra chemical.

And assuming that there were two equivalent moles of the Grignard reagent being used, wouldn't that be considered basic enough conditions for the dehydration to occur?   Because the addition of the acidic water is a separate step.

What is going on with this reaction, by the way?  Is it the alcohol that gets deprotonated?

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gojase

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Re: Grignard Problem
« Reply #5 on: June 02, 2011, 04:50:59 AM »

Oops! Ignore my second paragraph. I was getting my scenarios confused.  But still, even with just one equivalent mole of the Grignard, wouldn't that be basic-enough conditions for the dehydration to occur?
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gojase

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Re: Grignard Problem
« Reply #6 on: June 02, 2011, 12:53:14 PM »

I got it! I asked my friend about it.  He pointed out to me that the same alkyl haldes that can't form into Grignard Reagents because of their acidic H's (R-COOH, R-NH, R-SH, R-OH), are the same groups that quench the Grignard Reagents when they're on the substrate.  It's what I was looking for: a generalization.  Something finally makes sense to me!  Since it was in the Nucleophilic Addition for Carbonyl Compounds, I guess it was a trick question.  But I still don't see how the alcohol is more acidic than the alpha hydrogen.  I'm guessing that they're at least close.  Alpha hydrogens typically have a pKa of around 16-20, and alcohols around 16-18.  I'm under the impression that the alcohol's inductive effects would cause the a-carbon to be deprotonated first.  At this point, I guess it really doesn't matter--especially since there isn't an option for such a product.

If I said anything incorrect, please correct me!  Take care.
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BluePill

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Re: Grignard Problem
« Reply #7 on: June 02, 2011, 04:10:10 PM »

Either way, even if you deprotonate the alcohol or the alpha hydrogen, it would still form the same compound. If you deprotonate  the alpha hydrogen, it form an enol intermediate, which could tautomerize into a carbonyl again.
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TMS lover

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Re: Grignard Problem
« Reply #8 on: June 18, 2011, 05:32:24 PM »

I believe all that happens is that the 3-hydroxybutanal being an acid reacts with the grignard reagent which can act as a verry strong base. The result of this neutralization gives you methane and and an alkoxide ion. The alkoxide is then neutralized by the acid work up in the next step. This process is also known as "killing" the Grignard reagent.
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BluePill

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Re: Grignard Problem
« Reply #9 on: June 18, 2011, 07:01:06 PM »

It's not an acid. There are acidic protons, however.
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orgopete

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Re: Grignard Problem
« Reply #10 on: June 19, 2011, 09:09:11 AM »

I think the first problem with this problem is that it probably isn't a real reaction. No chemist would try this reaction as the OH would quench the Grignard. If that were the end of the problem, then it could work as indicated. However, I rather strongly believe that if you treated 3-hydroxybutanal with a base, such as t-butoxide, it would effect an elimination of hydroxide. In this case, the alpha hydrogens are close in acidity to the OH, therefore I would expect some enolization to occur and this would convert the RO(-) into an ROH and loss of OH(-) could occur. It could also cause some aldol condensation reactions to generate some water which can protonate more of the alkoxides.

Since it wasn't specified and people often prepare Grignard reagents first and then add their reactant to the Grignard. In the midst of the addition, deprotonation would make the anion of the hydoxybutanal and until 0.5 equivalents of alcohol were added, there would still be an excess of Grignard present. It would add to the aldehyde.

I think the best way to interpret this problem is from what was intended, not from what actually might occur. I think the book intended that an alcohol should quench the Grignard, so even though an aldehyde is present, the Grignard wouldn't add.
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napoleon79

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Re: Grignard Problem
« Reply #11 on: July 03, 2011, 06:58:35 PM »

I think that :
Proton of HO- group has harder electrophile effective than proton of alpha carbon. So CH3- of CH3MgBr will attact to proton of HO- group that is the first with 1:1 equiv. Then RO- group was protonation by acid.

If you want methyl group attacted to carbonyl group of aldehyde, you need to protect hydroxyl group such as THP.

thanks  :)
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orgopete

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Re: Grignard Problem
« Reply #12 on: July 04, 2011, 08:57:36 AM »

If I had a 3-hydroxyaldehyde and wanted to convert it into a 1,3-diol, the very first thing I would try would be to add it to 2 eq of Grignard. I would skip the protection/deprotection. Oh, I concede that I waste one equivalent of Grignard.
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