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Topic: iodine azide addition  (Read 8502 times)

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Offline beheada

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iodine azide addition
« on: November 02, 2006, 11:46:17 AM »
So I have this problem of adding IN3 to butene and it wants me to show a resonance structure of iodine azide, calculate the formal charges... THEN asks what is the polarity of the I-N bond?

So I made the first structure be I-N--N--N and calculated the formal net charge to be 0.
Then I made the resonance structure be I-N-N---N and calculated the same formal net charge @ 0.

The question also asks to draw a mechanism for the addition, and I assume it's the same as hydroboration... but the part that eludes me is what they mean by what IS the polarity of the I-N bond? I mean... in both instances the bond is non-polar, leaning toward the more electronegative... so what else could they possibly want and/or mean?


rayfe

Offline Dan

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Re: iodine azide addition
« Reply #1 on: November 02, 2006, 01:39:55 PM »
OK, I think you have he resonance structures right, but you didn't put any charges in them.
I'd write them like this:

[ I-N=N+=N- <> I-N--N+(triple)N ]

As for the polarity of I-N, I think they are just after direction of the polarity. This is clear fron the relative electronegativities of I and N, and also from the second resonance structure above.

As for the mechanism, I would be tempted to go via a 3 membered iodinium ion intermediate (the same intermediate as in I2 addition), but I'm not sure about that, it's just the first thing that came to my head.
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Offline beheada

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Re: iodine azide addition
« Reply #2 on: November 03, 2006, 11:03:58 AM »
About the mechanism... my book says that the iodine addition here is analogous to hydroboration/oxidation. During that mechanism, there is NO 3-membered intermediate because the H and the B attach from the SAME side at the SAME time. Therefore, the hint in the book suggests that the Iodine (which winds up on the LESS substituted carbon due to steric crowding) attaches at the same time as the N3 complex.

In that sense, why would the polarity effect the addition?

Offline movies

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Re: iodine azide addition
« Reply #3 on: November 03, 2006, 12:43:35 PM »
About the mechanism... my book says that the iodine addition here is analogous to hydroboration/oxidation. During that mechanism, there is NO 3-membered intermediate because the H and the B attach from the SAME side at the SAME time. Therefore, the hint in the book suggests that the Iodine (which winds up on the LESS substituted carbon due to steric crowding) attaches at the same time as the N3 complex.

The Encyclopedia of Reagents for Organic Synthesis says exactly the opposite.  The mechanism proceeds by making the iodonium and then the azide attacks at the more substituted end of the iodonium.  You can tell that from the stereochemistry of the products, as you said, but the stereochemistry observed is that the I and the N3 approach from opposite sides of the olefin.

What book are you looking at, by the way?  I would definitely trust what is the EROS, it's all based on the reports in the chemistry literature and has references even!

Offline beheada

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Re: iodine azide addition
« Reply #4 on: November 03, 2006, 01:49:14 PM »
My book is:

ORGANIC CHEMISTRY: 6th Edition by John McMurry

And the text from it states:
"Iodine azide, IN3, adds to alkenes by an electrophilic mechanism similar to that of boron. If a monosubstituted alkene such as 1-butene is used, only one product results:

CH3CH2CH=CH2 + I-N=N=N ---> CH3CH2CHCH2I2
                                                              l
                                                             N=N=N                             "

Maybe that should be BROMINE???? Then it would all make a lot more sense.




rayfe

Offline movies

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Re: iodine azide addition
« Reply #5 on: November 03, 2006, 05:38:58 PM »
Yeah, it must be that.  It wouldn't make sense for it to react like boranes, the orbitals don't work out right.

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