September 23, 2019, 05:28:31 AM
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Topic: Alkene formation: excercise doubt  (Read 313 times)

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Offline xshadow

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Alkene formation: excercise doubt
« on: September 09, 2019, 09:42:52 AM »
Hi,I don't understand   the last step

First one is a "normal halogenation"

Which reactant should I use to get that alkene and reforming C=C double  bond?

Thanks!

Offline chenbeier

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Re: Alkene formation: excercise doubt
« Reply #1 on: September 09, 2019, 09:45:24 AM »
Which reaction, no pictures visible.

Offline xshadow

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Re: Alkene formation: excercise doubt
« Reply #2 on: September 09, 2019, 11:18:37 AM »

Offline chenbeier

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Re: Alkene formation: excercise doubt
« Reply #3 on: September 09, 2019, 11:32:37 AM »
Reduction with Lithium, Magnesium or Calcium.

Better is to add HI instead of I2 and do E-Reaction with Hydroxide. Side product is the alcohol.

Offline Babcock_Hall

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Re: Alkene formation: excercise doubt
« Reply #4 on: September 09, 2019, 12:02:58 PM »
xshadow,

Please show us your attempt or give us your thoughts before we can help you.  That is a Forum Rule (see red link above).

Offline xshadow

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Re: Alkene formation: excercise doubt
« Reply #5 on: September 09, 2019, 03:50:42 PM »
I ' ve asked you  all because nothing come to mind thinking at the reactions seen in org chem1


I  have a  "" I-C-C-I" structure and I need to take out the two Iodine  in order to create a double  bond C=C"..

But in which  way?

A  sn2  reaction on a  alkyl halide with an H- (witch whic species? NaBH4 or LiAlH4 ?? I've senn these species in reduction of  C=O ,not too much in Sn2 with alkyl halide)as nucleophile can take away the iodium as I-...
And IF only one of the two halide go out I could think in the
 Next  step in an elimination reaction E1/E2 on the second C-Halide bond




BUT this Sn2 reaction occours on both the C-I bond and I'll get an alkane (so no elimination reaction on the second C-I)



Some help? Thanks all!!!


Offline xshadow

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Re: Alkene formation: excercise doubt
« Reply #6 on: September 09, 2019, 04:04:16 PM »
Reduction with Lithium, Magnesium or Calcium.

Better is to add HI instead of I2 and do E-Reaction with Hydroxide. Side product is the alcohol.

Reduction using a source of H-  nucleophile,like LiAlH4 or NaBH4??

A pseudo Sn2 reaction  on boh the C-halide bond?
But then I need to convert a single bond CH-CH  in a C=C

Thanks

Offline Babcock_Hall

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Re: Alkene formation: excercise doubt
« Reply #7 on: September 09, 2019, 04:26:33 PM »
With respect to cis- versus trans-alkenes, where does the equilibrium lie?

Offline xshadow

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Re: Alkene formation: excercise doubt
« Reply #8 on: September 10, 2019, 03:03:35 AM »
With respect to cis- versus trans-alkenes, where does the equilibrium lie?

Obviously trans!!

But as I said before my doubt is about the reactant to use in order to convert I--C-C-I in a C=C bond


Offline rolnor

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Re: Alkene formation: excercise doubt
« Reply #9 on: September 10, 2019, 09:01:09 AM »
I dont you need any reagent, this type of diiodides are not stable, they collapse into the trans-alkene.

Offline Babcock_Hall

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Re: Alkene formation: excercise doubt
« Reply #10 on: September 10, 2019, 10:19:28 AM »
@OP, When something speeds up a reaction but is not a reactant or product, what is it?

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