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Topic: Saponification help?  (Read 3273 times)

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Offline webassignbuddy

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Saponification help?
« on: October 31, 2013, 09:16:49 AM »


^ The answer to this isn't this:

O=C1CC=C([O-])O1

Is the answer this:

O=C=CCC(=O)[O-]

Offline discodermolide

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Re: Saponification help?
« Reply #1 on: October 31, 2013, 10:39:38 AM »
No it's not the ketene.
What else can happen to the anhydride upon base treatment?
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Offline webassignbuddy

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Re: Saponification help?
« Reply #2 on: October 31, 2013, 11:40:03 AM »
No it's not the ketene.
What else can happen to the anhydride upon base treatment?

I have absolutely no clue... :(
This reaction isn't in any of my teachers' slides
« Last Edit: October 31, 2013, 11:52:07 AM by webassignbuddy »

Offline discodermolide

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Re: Saponification help?
« Reply #3 on: October 31, 2013, 11:59:21 AM »
How about attack of the ethoxide at a carbonyl with ring opening?
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Offline webassignbuddy

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Re: Saponification help?
« Reply #4 on: October 31, 2013, 12:01:35 PM »
How about attack of the ethoxide at a carbonyl with ring opening?

O=C(O)CCC(=O)[O-]

Is that the answer?
At the bottom of the question there's a hint that says "There is NO acid workup"

Offline discodermolide

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Re: Saponification help?
« Reply #5 on: October 31, 2013, 12:13:11 PM »
Not quite, what happened to the ethoxide group and where is the Na+, if there is no acid work-up?
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Offline webassignbuddy

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Re: Saponification help?
« Reply #6 on: October 31, 2013, 12:20:05 PM »
Not quite, what happened to the ethoxide group and where is the Na+, if there is no acid work-up?

I have no idea...

1) The OH attacks the the left or right carbonyl.
2) 2 electrons are pushed onto the carbonyl's oxygen and then gets a negative charge
3) 2 electrons from that oxygen get pushed back down to kick out the ring's middle oxygen, which will now have a negative charge
4) Then that oxygen deprotonates the OH that was added, right? so I would get this....

O=C(O)CCC(=O)[O-]

And than the O with the negative charge attacks the Na+?

O=C(O)CCC(=O)O[Na]

But that doesn't look right, considering all of the other problems ive gotten right doesn't even have answers containing Na

Offline discodermolide

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Re: Saponification help?
« Reply #7 on: October 31, 2013, 12:22:34 PM »
what about this compound?
O=C(OCC)CCC([O-])=O.[Na+]
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Offline webassignbuddy

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Re: Saponification help?
« Reply #8 on: October 31, 2013, 12:24:43 PM »
what about this compound?
O=C(OCC)CCC([O-])=O.[Na+]

I...don't get it...how does the O get the ethyl group on it?
Is there another pushing mechanism for this?

Offline webassignbuddy

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Re: Saponification help?
« Reply #9 on: October 31, 2013, 12:27:53 PM »
Ohhhhhhhh

Na+ and OCH2CH3-

Gotcha....
thanks!

Offline orgopete

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Re: Saponification help?
« Reply #10 on: October 31, 2013, 06:17:30 PM »
Ohhhhhhhh

Na+ and OCH2CH3-

Gotcha....
thanks!

Of course, that is why we emphasis mechanisms. With a mechanism, it seems implausible the ethyl group wouldn't be connected.
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