dome13
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1-pentylnaphthalene theory
Hi there!
My friend's organic chemistry teacher at university had a quieston in an exam like this: synthetise 1-pentylnaphthalene in 2 steps without producing
any isomer. My suggest was, first brominate the naphthalene to alpha position, then react with pentylmagnesium bromide. He replied that the idea is
good, but there would be also some product in beta position (I guessed similar, but maybe...). Anyway, last year no one can't solve this reaction.
Maybe this is already irrelevant, but I am still curious what way it can be produced to satisfy the condition. I hope you guys has something in your
mind. 
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mackolol
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If it doesn't have to be synthesised from naphthalene, you can take 1-naphthylamine and convert it via sandmeyer to 1-bromonaphthalene and then do the
grignard
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clearly_not_atara
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^agree, in two steps... from what?! here, i can synthesize 1-pentylnaphthalene in one step:
start with 1-naphthylbromopentane, dissolve in ether, add Mg, quench with water, reserve nonpolar phase. you're done!
or if i can get creative, start with tetralone, pentyl Grignard, H2SO4/SeO2 (I think this won't overoxidize).
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dome13
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Sorry for that! It must be started from naphthalene.
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DraconicAcid
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I was going to suggest acylating with pentanoyl chloride and then reducing with Zn/HCl, but apparently the acyalation gives a mixture of 1- and
2-acylnaphthalene.
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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myr
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My hunch is a radical halogenation can do so selectively. Maybe a photoredox bromination with NBS, some activator dye, and UV light to give 1-br
naphthalene, then do a cross-coupling.
A wilder idea is to use some enzyme, like an oxygenase, to make the a-naphthol selectively, then make the triflate ester, then cross-couple, but
that's three steps.
Edit:
Found something (https://onlinelibrary.wiley.com/doi/full/10.1002/anie.200906...): claims 95% yield of the 1-bromo isomer. Good enough I think.
[Edited on 10-3-2021 by myr]
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Triflic Acid
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Maybe first sulfonate the aromatic at high temp, to get 2 napthalene sulfonic acid, thats a meta directing group, then remove that group after doing a
fields craft with this molecule
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Texium
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Quote: Originally posted by Triflic Acid  | | Maybe first sulfonate the aromatic at high temp, to get 2 napthalene sulfonic acid, thats a meta directing group, then remove that group after doing a
fields craft with this molecule |
That’s three steps.
myr, that paper looks interesting. I’ll have to save that to my folder of modern reactions that might actually have some practical value.
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clearly_not_atara
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1. One-electron reduction of naphthalene with sodium to the radical sodium naphthalenide.
2. Reaction with pentyl bromide affords decane, 1,4-dipentylnaphthalene, and 1-pentylnaphthalene. See:
https://pubs.acs.org/doi/pdf/10.1021/ja01027a068?casa_token=...
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draculic acid69
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Friedel crafts rxn of benzene plus acyl chloride with alcl3 equals the corresponding ketone
but benzene+alkyl bromide with aluminium powder equals the corresponding alkyl benzene
So apply this to napthalene. Napthalene plus pentyl halide with aluminium should equal
Pentyl napthalene right. I'm not 100% certain but I think I'm right.i might be missing something
but I'm not thinking of anything that would prevent this
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DraconicAcid
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| Quote: Originally posted by draculic acid69 | Friedel crafts rxn of benzene plus acyl chloride with alcl3 equals the corresponding ketone
but benzene+alkyl bromide with aluminium powder equals the corresponding alkyl benzene
So apply this to napthalene. Napthalene plus pentyl halide with aluminium should equal
Pentyl napthalene right. I'm not 100% certain but I think I'm right.i might be missing something
but I'm not thinking of anything that would prevent this |
Naphthalene does FC-type additions at both 1 and 2 positions; the difficulty is in getting only the one isomer.
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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zed
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Ummm. Maybe there is an answer, maybe there isn't.
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clearly_not_atara
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I posted the answer. You react pentyl bromide with
sodium naphthalenide. They're simple reactions and it's been published multiple times. The yield is poor and there are side products, but no papers
mention any 2-isomer forming, because the structure of the radical anion breaks the symmetry.
Credits to myr and Triflic Acid for suggesting radical reactions and equilibrium respectively. Na naphthalenide
equilibrates to a 1-coordinated structure, apparently.
[Edited on 12-3-2021 by clearly_not_atara]
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zed
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I have no patience with professors that do not eventually provide answers to their test questions.
And, in some cases, there are no real answers to the questions posed.
At some Universities, exams are given wherein of 120 supposedly possible points, the top score is 35.
Anything might be asked.
It isn't my idea of a good system. My opinion won't change.
You want the "one true answer"? Ask your friends Organic Chem. teacher.
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myr
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I've had those kinds of professors before, and they are The Worst. It's an ego thing half the time: radix peccati superbia est and all that.
I think I am practically the most correct, with only one isomer made, but I bet atara's answer is the one the prof was looking for: from a didactic
sense, the professor was looking for a good understanding of radical chemistry, which is not really taught well (my experience, anyway)
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