michalJenco
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Unsuccessful Grignard with PhMgBr - what did I make?
Hello chemists,
I wanted to prepare 2-phenyl-2-propanol (a-cumyl alcohol, b.p. 202 °C) by reacting acetone with phenylmagnesium bromide. However, at the final distillation, all of
my product came over at 170-176 °C.
I have done about 15 successful grignard reactions before, all of them with aliphatic reagents; this was my first attempt of an aromatic alcohol.
I am baffled as to:
- why I did not get the desired product
- what my actual product is
My only lead right now is 2-phenyl-1-propene - has a b.p. of 165-169°C and it is the dehydration product of 2-phenyl-2-propanol. I don't know how it could form without me
deliberately trying to dehydrate the cumyl alcohol, though.
My grignard procedure is as follows:
- create the grignard reagent in diethyl ether
- drip in carbonyl in ~10% excess
- aqueous and acidic (40% H2SO4) workup until both layers clear
- separation of layers and extraction of aqueous layer with more ether
- drying organic layer over K2CO3 to remove water (4 days in this particular reaction)
- decant from K2CO3, wash K2CO3 with a bit of ether
- distill off ether
- change setup for a high temperature distillation to get the desired product
I welcome your ideas!
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DraconicAcid
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40% sulphuric acid? I'm not surprised that would dehydrate a benzyl alcohol, if the product you got was indeed the dehydration one.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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michalJenco
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So even at the b.p. of ether 40% sulfuric acid dehydrates benzyl alcohols?
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DraconicAcid
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Since you're making a conjugated alkene, I wouldn't be surprised at all to see it happen at room temperature. I could be wrong, though- I'm not an
organic chemist.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Tsjerk
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0.5 molar equivalent of sulfuric acid is enough, it would probably already be enough to just use it as a catalyst (or even do it with plain water) but
with 0.5 molar equivalent you at least keep the magnesium in solution instead of having it precipitate as the hydroxide.
If you use exactly 0.5 equivalent, you will have your alcohol, one MgBr2 and one MgSO4. A small excess of sulfuric acid doesn't hurt, but I would
personally go for 5% or so sulfuric acid. That won't dehydrate but will get the job done. And I would aim for a volume of dilute acid that would be
1.2 times the required amount to keep the magnesium in solution.
[Edited on 8-2-2020 by Tsjerk]
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DavidJR
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Dehydration is fast for tertiary alcohols in general. Why don't you test the product with bromine water to see if it is indeed an alkene?
Also I generally prefer to use saturated aqueous ammonium chloride for the workup of Grignard reactions.
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Amos
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And even if you don't have bromine you can generate a small amount by adding bleach or acidic hydrogen peroxide to a solution of a bromide salt.
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Tsjerk
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Quote: Originally posted by Amos | And even if you don't have bromine you can generate a small amount by adding bleach or acidic hydrogen peroxide to a solution of a bromide salt.
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Correct topic?
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DraconicAcid
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Yes- the use of bromine was mentioned as a way to test for the presence of an alkene.
And I second the use of ammonium chloride to neutralize the alkoxide. Concentrated acid will dehydrate the alcohol, and concentrated base may also do
that.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Tsjerk
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Quote: Originally posted by DraconicAcid |
Yes- the use of bromine was mentioned as a way to test for the presence of an alkene.
And I second the use of ammonium chloride to neutralize the alkoxide. Concentrated acid will dehydrate the alcohol, and concentrated base may also do
that. |
Got it! Ammonium chloride seems the way to go.
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DavidJR
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It is also possible to just use water alone without any mineral acid. However this tends to result in insoluble magnesium junk.
Saturated NH4Cl is just the best all round option in my opinion.
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michalJenco
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Thank you all for suggestions!
I knew NH4Cl is the official way to go but I just .. hate working with anything ammonia. I want to make alcohols even more, though, so I will repeat
the experiment with saturated NH4Cl next time and I will report the results here.
Thanks for suggesting testing with in-situ bromine water, I forgot about that! I will try that when I get back.
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draculic acid69
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U wanted p2pol which is a secondary alcohol right but grignard+acetone equal tertiary alcohols right and if I'm not mistaken the alcohol will form in
the one position on the benzene ring (tert OH) not along the chain (sec OH) and then using h2so4 will dehydrate the tert alcohol to a alkene even at
low temperatures.i think this can happen at about 40'c
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Texium
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Quote: Originally posted by draculic acid69 | U wanted p2pol which is a secondary alcohol right but grignard+acetone equal tertiary alcohols right and if I'm not mistaken the alcohol will form in
the one position on the benzene ring (tert OH) not along the chain (sec OH) and then using h2so4 will dehydrate the tert alcohol to a alkene even at
low temperatures.i think this can happen at about 40'c | No- he stated in the OP
2-phenyl-2-propanol and even linked the structure. Read carefully before jumping to assumptions.
[Edited on 2-9-2020 by Texium (zts16)]
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draculic acid69
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Quote: Originally posted by Texium (zts16) | Quote: Originally posted by draculic acid69 | U wanted p2pol which is a secondary alcohol right but grignard+acetone equal tertiary alcohols right and if I'm not mistaken the alcohol will form in
the one position on the benzene ring (tert OH) not along the chain (sec OH) and then using h2so4 will dehydrate the tert alcohol to a alkene even at
low temperatures.i think this can happen at about 40'c | No- he stated in the OP
2-phenyl-2-propanol and even linked the structure. Read carefully before jumping to assumptions.
[Edited on 2-9-2020 by Texium (zts16)] |
I didn't see the first 2.my bad.disregard the above post
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michalJenco
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So I did the test with bromine water (NaBr with a bit of H2SO4 and H2O2) and the color disappeared immediately on shaking. Looks like I made myself
the aromatic propene after all.
Also, I failed to mention a fact which didn't occur connected to me then - in the final destilation of the product a LOT of water (~15mL) came along
with organics before the final fraction of 170-177°C (and that fraction was also cloudy and cleared up with molecular sieves). This was after 3 days
of drying the ether solution over anhydrous K2CO3. I distilled 3 other alcohols from other grignards (3-heptanol, 2-nonanol and diisoamyl methanol) in
the same session and none of them presented any water. At that time I just assumed the drying wasn't enough. Now I am convinced the alcohol dehydrated
during the distillation, as it seems impossible to me that so much water could be dissolved in the ~60mL of organics I distilled.
However; by drying over base all the acid from workup should have been long neutralised, hence I still don't quite understand the dehydration. As
sources list the b.p. at atmospheric pressure, I suppose that means the molecule is able to survive such distillation.
What are your opinions on this?
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Tsjerk
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Did you filter of the K2CO3? Or was it present in the bioling flask?
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michalJenco
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I did, however I am not a 100% sure a small amount did not make it through. If catalytic amounts are enough to dehydrate everything at the
distillation temperature, that must be the culprit.
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Tsjerk
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I don't know, but I think it would definitely not help. The boiling temperature is close to the temperature K2CO3 starts to lose water, it is probably
around an equilibrium. So the anhydrous stuff is able to both pull water from your molecule, and to lose the water again.
The dehydrated propene is lost together with the water from the boiling flask, and at the same time that evaporation keeps the temperature down so you
won't collect any alcohol.
This might be what is going on.
[Edited on 13-2-2020 by Tsjerk]
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michalJenco
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That is an excellent explanation, must be what happened.
Thank you all for the input.
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