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Topic: Synthesis, starting with cyclohex-2-enone  (Read 9752 times)

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Offline THC

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Synthesis, starting with cyclohex-2-enone
« on: March 04, 2008, 04:16:17 AM »
Q: Starting from cyclohex-2-enone (the first molecule in the attached fil) and any other reagents needed, how would you prepare the second molecule in the attached file? More than one step may be required.

A: I would prepare cyclohex-2-en-1-ol and chlorobenzene and let those react in a Friecel Craft alkylation.

Cyclohex-2-en-1-ol:
Reduce the C=O to C-OH with NaBH4.

Chlorobenzene:
This one is tricky.
Reduce the C=O to C-OH with NaBH4: cyclohex-2-en-1-ol
Eliminate -OH with KOH in CH3CH2OH: cyclohex-1,3-diene (is that possible? OH- is a bad leaving group. What if an acid is added, then the leaving group is H2O, which is better)
Eliminate 2 H from R2-CH-CH-R2: No idea how to do this.
Add Cl2 + FeCl3 to benzene: Chlorobenzene.

Then Friedel Craft alkylation?


I don't know... I don't think it'll work in the lab, but this is paper synthesis :) Still, I don't even think it works on paper. Anybody with a better suggestion?

Offline sjb

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #1 on: March 04, 2008, 04:41:03 AM »
I think you're making life very hard (but interestingly so) for yourself. By "any other reagents needed" I'd read that as chlorobenzene is available without making it.

As to your scheme for cyclohex-2-enone to chlorobenzene, looks fine - you can convert cyclohexa-1,3-diene to benzene via surrogate hydrogenation. Have a look at http://en.wikipedia.org/w/index.php?title=1%2C3-Cyclohexadiene&oldid=184791975

Alternatively, why not cyclohex-2-enone to the alpha-selenide at C6 (smelly) with (PhSe)2, oxidize to the selenoxide, eliminate to the diene then reduce the ketone to the alcohol, dehydrate (a bit like the use of hydroxylamine in modified Hantzch pyridine syntheses) to give benzene then chlorinate as before.

I'm not sure it's Friedel Crafts you want though at the end...

S

Offline macman104

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #2 on: March 04, 2008, 05:39:00 AM »
With the right nucleophile, couldn't you perform a 1,2 addition preferentially to the 1,4 addition?  Aren't grignard reagents sufficiently hard to prefer the 1,2 attack?

EDIT:  I did a quick check and organolithium regeants do preferentially attack the carbonyl instead of the conjugated alkene. 

1) Clemmensen reduction of cyclohex-2-enone -> cyclohexene
2) cyclohexene + HBr -> bromocyclohexane
3) bromocyclohexane + 2Li -> lithium+cyclohexane- + LiBr
4) lithium cyclohexane + cyclohex-2-enone -> product
« Last Edit: March 04, 2008, 05:25:44 PM by macman104 »

Offline sjb

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #3 on: March 05, 2008, 03:21:24 AM »
Good thinking....

But it's a phenyl ring on the righthand side, not a cyclohexane :)

Simply chlorobenzene to the lithio derivative or the Grignard and add to the enone were my initial thoughts.

S

Offline macman104

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #4 on: March 05, 2008, 03:48:11 AM »
Good thinking....

But it's a phenyl ring on the righthand side, not a cyclohexane :)
Oh!  So it is, lol!  And I did check my Organic Synthesis book, and grignards aren't usually preferential enough to selectively add to the carbonyl.

Offline sjb

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #5 on: March 05, 2008, 06:36:51 AM »
...grignards aren't usually preferential enough to selectively add to the carbonyl.

OK (it's always a good idea to think what might work before getting down to the nitty gritty of the gunk at the bottom of the rbf, or thumbing though the lit, IMO), at least we got the right disconnection and polarity :)


Offline adam

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #6 on: March 05, 2008, 07:05:36 AM »
You can use the Grignard reagent PhMgBr with Ce (III) as an additive to increase the regioselectivity of the reaction. Look this paper:

J. Am. Chem. Soc., Vol. 111, No. 12, 1989 4393

Offline THC

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Re: Synthesis, starting with cyclohex-2-enone
« Reply #7 on: March 05, 2008, 03:10:26 PM »
Thanks for your answers, I'll take them into consideration.

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