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Topic: Curly Arrows Question  (Read 4384 times)

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Offline sammmyx_

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Curly Arrows Question
« on: September 28, 2014, 08:38:25 AM »
What is the product of this curly arrow process?





I think it's the first one. Can anyone confirm?

Thanks!

Offline Hunter2

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Re: Curly Arrows Question
« Reply #1 on: September 28, 2014, 09:14:36 AM »
No the charge must be still there, so which one can it only be?

Offline sammmyx_

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Re: Curly Arrows Question
« Reply #2 on: September 28, 2014, 09:18:17 AM »
Well, the 4th one still has a charge on the hydrogen, would that be it?

Offline Hunter2

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Re: Curly Arrows Question
« Reply #3 on: September 28, 2014, 09:19:53 AM »
No think logical. You add a positive charged Ion to a neutral molecule, what can only be the result.

Offline Arkcon

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Re: Curly Arrows Question
« Reply #4 on: September 28, 2014, 09:22:07 AM »
Or to look at it another way, what was "pushed" by the curly arrow?  How did that change the H+ ion?  And what would happen next? To both molecules/atoms?  Try to show us which ones of the multiple guess options we can exclude.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline sammmyx_

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Re: Curly Arrows Question
« Reply #5 on: September 28, 2014, 09:23:53 AM »
Oh! The third one!

Offline Hunter2

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Re: Curly Arrows Question
« Reply #6 on: September 28, 2014, 09:25:10 AM »
Yes correct. But this molecule is very unstable. What will happen to it next. Where will be the positive charge will go?

Offline sammmyx_

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Re: Curly Arrows Question
« Reply #7 on: September 28, 2014, 09:26:32 AM »
Well it wouldn't separate from the hydrogen, would it?

Offline Hunter2

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Re: Curly Arrows Question
« Reply #8 on: September 28, 2014, 09:29:08 AM »
The positive charge is now at the Oxygen. There are two possibilties to make it more stable. Think about charge and double bond. Is  this stable? What try to do the bonds do in the molecule.

Offline Babcock_Hall

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Re: Curly Arrows Question
« Reply #9 on: September 29, 2014, 07:17:12 PM »
I realize that I am coming to this problem late, but another way to approach this problem is to calculate the formal charge of the oxygen atom.

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