March 28, 2024, 05:23:01 AM
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Topic: Urgent help needed ASAP! Major product produced from the reaction below is what?  (Read 6647 times)

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Offline theanonymous

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The structure (below) I drew at the bottom is wrong :(
Here's the SMILES string...
CC1=CCC=CC1

In the first electron transfer, the carbon to which the sodium electron adds creates a carbanion and the carbon "para" to the carbanion is a carbon radical which becomes a carbanion upon electron transfer from the second sodium atom. Since the carbanion is negatively charged, it would not be low in energy if it formed on a carbon bonded to an alkyl group (an electron pump). This results in the alkyl group being bonded to a double bond carbon (sp2 carbon) and not to an sp3 carbon.

But what's the answer?

Offline Calicum

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Birch Reduction

Offline theanonymous

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Offline discodermolide

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Development Chemists do it on Scale, Research Chemists just do it!
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Offline theanonymous

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Look here: http://en.wikipedia.org/wiki/Birch_reduction

I looked at that like 4 min ago.
That's what I did and I still got it wrong...

Offline discodermolide

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How about this product?
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Offline Dan

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That's what I did and I still got it wrong...

You replaced the isopropyl group with a methyl group.
My research: Google Scholar and Researchgate

Offline Borek

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Here's the SMILES string...
CC1=CCC=CC1

If you know SMILES, why don't you use [smiles]CC1=CCC=CC1[/smiles] tags?

CC1=CCC=CC1
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline curiouscat

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Here's the SMILES string...
CC1=CCC=CC1

If you know SMILES, why don't you use [smiles]CC1=CCC=CC1[/smiles] tags?

CC1=CCC=CC1

Adding frills

CC(C)C1=CCC=CC1>CO>CC1=CCC=CC1

Offline theanonymous

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Offline theanonymous

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That's what I did and I still got it wrong...

You replaced the isopropyl group with a methyl group.

Ohhhh... for some reason I thought the 2 extra lines on the isopropyl group were implied Hydrogens...
Thanks a bunch!!

Offline theanonymous

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How about this product?

That makes more sense!
Thanks!!

Offline theanonymous

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What about this one though?
Do I do the same thing with the Birch Reduction?



Like this?

CCC1C=CC(C)C=C1

Because this is what I had in my notes...

In the first electron transfer, the carbon to which the sodium electron adds creates a carbanion and the carbon "para" to the carbanion is a carbon radical which becomes a carbanion upon electron transfer from the second sodium atom. Since the carbanion is negatively charged, it would not be low in energy if it formed on a carbon bonded to an alkyl group (an electron pump). This results in the alkyl group being bonded to a double bond carbon (sp2 carbon) and not to an sp3 carbon.

Offline theanonymous

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Either that or it's this, no?

CCC1=CCC(C)=CC1

Offline theanonymous

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I think I got the one above right...

I'm not sure about this one though...



I got:
O=C(O)c1ccc(C(=O)O)cc1

Does that look right?
Because if the alkyl group possesses a benzylic hydrogen then that carbon will be oxidized to a carboxylic acid.

So if THIS one is right, then what's the product for this one? o.O



I tried doing Oc1ccccc1O but Webassign said I was wrong...
« Last Edit: January 17, 2013, 08:40:19 AM by theanonymous »

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