Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: combatwombat on February 18, 2009, 01:37:45 PM
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All carbons must come from benzene or acetaldehyde
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All carbons must come from benzene or acetaldehyde
I count 3 carbon's that do not come from benzene
Acetaldehyde has only 2....
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how efficient does the synthesis need to be? Is this homework? Any reagent restrictons (besides sources of carbon)?
I can do it in 8 steps.
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I count 3 carbon's that do not come from benzene
Acetaldehyde has only 2....
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi529.photobucket.com%2Falbums%2Fdd336%2Ffr_slingsandarrows%2Fcaptain_obvious_rock_gift.jpg&hash=83f51bae41fecefc2fc12c86bd393fe192b53f89)
No reagent restrictions, though we are doing aldol reactions right now so that's probably involved.
I started by Friedel-Crafts'ing acetyl chloride to benzene, then hitting it with KMnO4 to make benzaldehyde.
Then, I made acetaldehyde into a Grignard reagent by reducing it with NaBH4 then changing the OH into a Br with PBr3, then treating it with magnesium in /\o/\
Finally, I then reacted that with the benzaldehyde, remade the ketone with PCC, and then eliminated a beta-hydrogen with NaOEt. If I am correct, you'd have this:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimages.yuku.com%2Fimage%2Fjpeg%2F37836c554cb1583250b72aa2449101b56c3f6249.jpg&hash=af59e41bc66485eab5282b4715e04df444d43a04)
Assuming the benzene ring was an appropriate nucleophile to react with the beta carbon, it would react with itself and you'd get the desired product.
Does this sound OK? If so, could I have used H2CrO4 instead of PCC? (my professor always wants us to use the cheaper reagents, and I guess PCC is pretty costly - but I am not sure if it cleaves carbon chains like KMnO4 does).
How did you do it?
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hitting it with KMnO4 to make benzaldehyde.
Aldehydes are way more reactive than just about anything. KMnO4 will oxidize it all the way to the acid.
http://www.cem.msu.edu/~reusch/VirtTxtJml/benzrx2.htm#benz9a
eliminated a beta-hydrogen with NaOEt
NaOEt will not eliminate a beta-hydrogen atom. You'd need a better leaving group than hydride to get to the unsaturation.
we are doing aldol reactions right now
yup, mine involves an aldol condensation.
Assuming the benzene ring was an appropriate nucleophile
Benzene's a pretty poor nucleophile. The only reason it works in EAS is because you generate such a hot electrophile in situ (R+, acylium ion, etc). The final step will not spontaneously react.
could I have used H2CrO4 instead of PCC
In theory, yes. But I don't think you have a workable synthesis at this point.
I started with F.C. acylation, too, though. My 2nd step was also conversion to benzoic acid, but I used the iodoform reaction instead of KMnO4 (http://www.cem.msu.edu/~reusch/VirtTxtJml/suppmnt3.htm#halfm). From there we differ. you're off to a good start, imho.
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I'm very interested in this style of chemistry so i thought id have a go! Some of the reactions will give a poor yield but i think it should work in principle:
1) Benzene to toluene via freidel-crafts alkylation with CH3Cl, AlCl3.
2) Toluene brominated on methyl group using Br2 in UV
3) Acetaldeyde to acetic acid using Cr2O7 then to acyl chloride using SOCl2.
4) Friedel Crafts acylation on brominated toluene should substitute in the ortho and para positions
5) SiMe3Cl in a suitable base to form the enol equivalent. Removal of the SiMe3 group from the oxygen can produce the 5 membered ring via the alkene and SN2 elimination on the bromine.
So...
Benzene to toluene.
Toluene to phenyl bromide
Acetaldeyde to carboxylic acid then to acyl chloride
friedel crafts acylation of acyl chloride onto ortho position of benzene ring
enol formation using SiMe3Cl or another suitable reagent.
Lone pair on oxygen reforms the ketone, double bond attacks the methyl group eliminating the bromine via SN2.
Potential Problems:
Toluene to phenyl bromide could over substitute
friedel crafts acylation could affect the bromine-carbon bond and give the para product.
The base required in forming the enol equivilent could substitute with the bromine
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Interesting route. Some notes:
freidel-crafts alkylation with CH3Cl
We are restricted to only using Acetaldehyde as our carbon source. Can't use methyl chloride.
Friedel Crafts acylation on brominated toluene
We also have an alkyl halide (benzyl bromide). That can also be converted to an electrophile upon treatment with AlCl3. Benzyl bromide will likely react with itself.
SiMe3Cl in a suitable base to form the enol equivalent.
I see where you're going with this, and there's nothing academically wrong with the step. However, there's no need to trap the enolate as the silyl enol ether. Deprotonation with base is really all that would be needed.
The synthesis looks ok. The only problem step I see is the F.C. acylation in the presence of a potent F.C. alkylation electrophile.
As to your potential problems, the oversubstitution is likely. and as long as you use a non-nucleophilic base (LDA), you'll probably be ok in the final step.
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Quote: We also have an alkyl halide (benzyl bromide). That can also be converted to an electrophile upon treatment with AlCl3. Benzyl bromide will likely react with itself.
This is the step i struggled with. I also thought about a Palladium cross coupling reactioon by having a benzene-halogen bond and adding it to an alkene but couldnt get the number of carbons right. Could a hydroxyl group be used in place of the benzene, do the friedel crafts acylation, then convert the hydroxyl group to a bromine group via the Appel reaction (PPh3, Br2 or NBS)? This could then self condense with LDA forming the 5 membered ring.
As for making toluene from benzene and acetaldehyde, im stumped! Azmanam could you PM your route to me please?
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Ugh, I can't believe I tried to use H- as a leaving group. That's what I get for trying to do syntheses while watching the simpsons.
If I make this, could I cleave off the last carbon with KMnO4 (leaving the 3rd carbon as RCOOH)? Or would that only work if it were closer to the aromatic ring?
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Pretty sure that'll just take you to ketone.
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Chewing off an alkyl group with KMnO4 and hot alkaline solutions to leave a carboxylic acid only works with groups that have a benzylic hydrogen.
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So, going back to this compound:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimages.yuku.com%2Fimage%2Fjpeg%2F37836c554cb1583250b72aa2449101b56c3f6249.jpg&hash=af59e41bc66485eab5282b4715e04df444d43a04)
Is there no way this could react to form the 3-membered ring? This is the one step I'm not sure about
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So, going back to this compound:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimages.yuku.com%2Fimage%2Fjpeg%2F37836c554cb1583250b72aa2449101b56c3f6249.jpg&hash=af59e41bc66485eab5282b4715e04df444d43a04)
Is there no way this could react to form the 3-membered ring? This is the one step I'm not sure about
Adding catalytic amount of acid may give it in low yield. Most probably (always?) the resonance stabilised carbocation will form, so formation of a highly strained cyclobutane ring may take place.
My synthesis uses 8 steps, the carbon reduction step being the haloform reaction.
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What about chlorinating, then acylating benzene with that substituent you have in your picture, then nitrating, then hydroboration/oxidation, then PBr3, then Mg, which would hopefully lead to NAS of the chloride.
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convert the hydroxyl group to a bromine group via the Appel reaction (PPh3, Br2 or NBS)?
probably not on an aryl alcohol (a phenol). The Appel is really only useful for alkyl alcohols.
As for making toluene from benzene and acetaldehyde, im stumped!
My route (not that it's the best or the only way to do it) does not go through toluene. Hint: I have benzaldehyde as an intermediate in my route.
then PBr3, then Mg, which would hopefully lead to NAS of the chloride.
if I'm following correctly, you add Mg to the alkyl bromide in the presence of the aryl chloride? You'll likely get insertion without selectivity - that is, you'll get the aryl Grignard as well as the alkyl Grignard.
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Yeah, that was the plan. Completely slipped my mind I'd form another grignard.
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My 8 steps: ( i am posting because others have also posted theirs..)
1. Do aldol condensation on the aldehyde. (donot dehydrate)
2. haloform reaction, acidify.
3. Protect aldehyde by making the acetal
4. Chlorinate using SOCl2
5. AlCl3 + benzene + compound in 6
6. Wolff kisnher (or Red P +HI; not a nice lab procedure; any alternatives can anyone suggest?)
7. Cleave the acetal and subsequent oxidation simulataneously using hot aqeous acidified KMnO4
8. SOCl2; AlCl3
(My last step is actually 2 steps lol)
Doubtful step: second beacuse aldol product may again form
@azmanam Can you post your scheme?
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2. haloform reaction, acidify.
The product of the aldol add'n is the beta-hydroxy aldehyde. You'll have to oxidize the alcohol to get the methyl ketone. But then the methylene protons in the middle are the most acidic. I'd acetal protect the aldehyde, then oxidize the alcohol, then iodoform.
7. Cleave the acetal and subsequent oxidation simulataneously using hot aqeous acidified KMnO4
I'd also be afraid you'd cleave the benzylic carbon to get benzoic acid and clip that alkyl group right off.
Here's what I did... again, not that it's the best or most efficient (in fact, I think it's rather inefficient). In looking back through it, the problem step is the 2nd F.C. acylation... I'd be afraid it would cleave the acetal p.g. and kill the regioselectivity of the F.C. acylation.
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To get around the 2nd F.C. problem, I guess you could not oxizide benzyl alcohol (product of LAH reduction) to benzaldehyde, but instead protect as TBS ether. Then F.C., deprotect, oxidize to aldehyde, aldol, reduce.
I think a better 2nd generation synthesis, though is a hybrid of mine and aldoxime_amine... I think tethering one long chain to form the 5-membered ring is probably better than tethering 2 short chains off the benzene ring.
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Wow, this problem was even tougher than I thought. How's this pathway:
1. Make enolate of acetaldehyde, react with benzaldehyde and NaOH to get the Claisen-Schmidt product
2. Treat with H2, Pd to get rid of C-C double bond
3. Use KMnO4 to turn aldehyde into carboxylic acid
4. SOCl2 to change the OH into a Cl
5. AlCl3 to attach it to the benzene ring
Is this workable? The only sketchy part seems to be making the enolate of acealdehyde - it might react with other acetaldehyde molecules to make unwanted products...
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yeah, that's essentially the 2nd half of the hybrid route. again, KMnO4 will probably clip off the entire alkyl group. The crossed aldol will need to be acidic, not basic, to get the elimination product.
You're right that acetaldehyde will probably self-condense if you add base (or acid) to a mixture of benzaldehyde and acetaldehyde. To overcome that, you change the order of addition. Add acetaldehyde SLOWLY to a mixture of benzaldehyde and acid. This will ensure that as soon as the enol forms, it reacts with benzaldehyde - present in MUCH greater excess) and not with itself. this only works if 1 aldehyde does not have alpha protons (if it is not enolizable).
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But the methyl carbinol is oxidized to the ketone in situ. Here is a reference.
http://en.wikipedia.org/wiki/Haloform_reaction (http://en.wikipedia.org/wiki/Haloform_reaction)
An alternative to the KMnO4 step in my synthesis could be probably Fehlings' solution.
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But the methyl carbinol is oxidized to the ketone in situ
really? i didn't know that. thanks for the link
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Regarding your first synthesis (azmanam) I think the second F.C. is slightly doubtful because of steric hindrance..?
But your second synthesis is very nice. That strange reaction what is it called (the one using tosyl hydrazine)? any links?
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tosyl hydrazine, yes. very nice, easy, reaction I've done several times on a,b-unsaturated esters. Generates diazene in situ. Ene reaction to hydrogenate C=C and liberate N2.
http://en.wikipedia.org/wiki/Diazene
http://www.jce.divched.org/Journal/Issues/1965/May/jceSubscriber/JCE1965p0254.pdf (subscription might be required).
For more fun, (but more dangerous conditions), try this paper. I've used to to reduce a,b-unsat. ketones in the presence of terminal alkenes no problem.
http://dx.doi.org/10.1021/jo00354a027