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Topic: synthesis problem...hexanol  (Read 4151 times)

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Offline skido123

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synthesis problem...hexanol
« on: March 06, 2009, 01:24:49 AM »
Someone tried to make hexylisopropyl ether from hexanol and isopropanol(2-propanol) in the presence of a catalytic amount of H2SO4...
Would this be a good method pf synthesizing the desired product?
« Last Edit: March 06, 2009, 01:42:44 AM by skido123 »

Offline AWK

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Re: synthesis problem...hexanol
« Reply #1 on: March 06, 2009, 02:04:23 AM »
No, the main reaction will be elimination of  water from isopropanol
AWK

Offline aldoxime_amine

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Re: synthesis problem...hexanol
« Reply #2 on: March 06, 2009, 09:52:06 AM »
No, the main reaction will be elimination of  water from isopropanol

But can't temperature change the course of the reaction, at least theoretically...? :-\

Offline James Newby

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Re: synthesis problem...hexanol
« Reply #3 on: March 06, 2009, 03:54:35 PM »
A much easier way would be to convert one alcohol into a bromine using the Appel reaction (N-bromo succinamide, Triphenyl phosphine), purify then add the other alcohol under non nucleophilic basic conditions
4th year undergraduate at the University of Sheffield

Offline nj_bartel

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Re: synthesis problem...hexanol
« Reply #4 on: March 06, 2009, 05:21:16 PM »
Could also go PBr3 on the isopropanol and do a williamson synthesis using the sodium hexanolate

Offline macman104

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Re: synthesis problem...hexanol
« Reply #5 on: March 06, 2009, 05:56:17 PM »
nj, it would be much wiser to do it the other way.  In the presence of the base, your secondary bromide will more likely undergo E2 as you are going to form a more stable alkene (in comparison to having the bromide on the hexane chain).

If you reverse them hexyl bromide, and isopropanol, you avoid that issue more (still not full proof), and you also are now doing SN2 on a 1° carbon instead of a 2° one, which is also faster.

Offline nj_bartel

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Re: synthesis problem...hexanol
« Reply #6 on: March 06, 2009, 06:53:29 PM »
Sorry, for some reason I had stuck in my head that a cyclohexyl isopropyl ether was what was being formed, even though I wrote sodium hexanolate.  I don't understand myself.

At any rate, I agree with you.

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