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Topic: periodic acid  (Read 7481 times)

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chanman1

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periodic acid
« on: June 09, 2005, 01:19:08 AM »
I had a sort of mechanism question about reacting a furanose sugar with periodic acid...
I'll attempt to draw the fisher projection of the sugar:
      CHO                1'
H----|-----OH        2'
H----|-----OCH3     3'
H----|-----OH        4'
      CH3                5'

On my test it was in cyclic form w/ 1' C -OH up (therefore beta sugar), 2' C -OH down, 3' C -OCH3 down, 4' C -OH up. The question was what the products would be if this sugar reacted with HIO4.
The answer is that you'd get formic acid, HCO2H, and this open chain sugar.
      CHO
H----|------OCH3
H----|------OH
      CH3

I thought that HIO4 cleaves in between 2 OH groups (such as R2-COH-COH-R'2 and only 2 OH groups (meaning there can't be another O on the other side). Apparently, HIO4 can also cleave between an aldehyde group and the -OH on it's alpha carbon. How does this work? and what other groups can HIO4 cleave aside from 2 -OH's, and an aldehyde and a -OH?

(lol i can't believe the fisher projection actually came out alright...)

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Re:periodic acid
« Reply #1 on: June 09, 2005, 01:37:58 AM »
This molecule most likely reacts from the cyclic form, right?  So you have a lactol with an OH adjacent to it, which is essentially like two adjacent hydroxyl groups.  Just draw out the ring-closed form of the sugar and it should make sense.

chanman1

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Re:periodic acid
« Reply #2 on: June 09, 2005, 02:03:39 AM »
Ok, so I have the cyclic form and see the -O-CHOH-CHOH- on the right side of the sugar. My next question is, why does formic acid come out, I thought that when you have 2 hydroxyl groups on a ring, they would open up the ring and create 2 -CO2H groups on the ends of where the -COH used to be.

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Re:periodic acid
« Reply #3 on: June 10, 2005, 03:09:50 AM »
Okay, so draw the product that you expect after the oxidation step is done.  How do you think water might react with that molecule?

chanman1

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Re:periodic acid
« Reply #4 on: June 10, 2005, 04:18:09 AM »
it would look like...
         CO2(-)
         O
H3C---|--------H
H------|--------OCH3
         CO2(-)

since the 2 carboxylic acids that were formed will be deprotonated since the surrounding water is at pH 7... dunno where to go from there.

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Re:periodic acid
« Reply #5 on: June 10, 2005, 12:37:52 PM »
Take a step back.

What would be the product if you reacted ethylene glycol (HOCH2CH2OH) with periodic acid?

chanman1

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Re:periodic acid
« Reply #6 on: June 16, 2005, 06:46:02 PM »
you would get 2 formaldehydes (H2C=O)

so, would you get 2 carbonyl groups in the ring between the 2 hydroxyl groups? This would react with water to give you the deprotonated carboxylic acid groups I think.

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Re:periodic acid
« Reply #7 on: June 16, 2005, 07:36:34 PM »
What happens to the C-C bond between the carbons bearing the two hydroxyl groups?  What happens to the ring?

Draw the product that you expect for me.

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