Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: polymer man on June 12, 2012, 06:52:21 PM
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Hello,
I need your help with two mechanism
The first one is reduction:
http://imageshack.us/photo/my-images/694/78729026.jpg/
The procedure is:
A mixture of 2,6-dibromo-9,10-diphenyl-9,10-dihydroanthracene-9,10-diol (13.5 g, 15.20 mmol), KI (21.8 g, 131.48 mmol), Na2H2PO2.H2O (15.4g, 175.4 mmol) and acetic acid (152 mL) was heated at reflux for 40 min. The precipitate was collected and dissolved in dichloromethane (DCM). The organic solution was washed with brine, dried over anhydrous MgSO4.
The second one is reduction of nitro to amine:
http://imageshack.us/photo/my-images/72/50398641.jpg/
The procedure is:
To a stirred suspension of 2,6-dinitroanthracene-9,10-dione (9.4 g, 31.5 mmol) in ethanol (340 mL) was added a solution of sodium sulfide nonahydrate (34.1 g, 142 mmol) and sodium hydroxide (13.5 g, 338 mmol) in water (590 mL). The mixture was heated at reflux for 6 h and left to stand overnight. The ethanol was removed in vacuo and the residue cooled to 0-5 °C. The resulting precipitate was collected by filtration, repeatedly washed with water, and dried. Recrystallization from ethanol/ water afforded the product as an orange/red solid (7.35 g, 98%)
I appreciate your help.
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I got the second mechanism but the first one I could not. I think a radical needs to be formed in position 9 and 10 but how I do not know.
http://imageshack.us/photo/my-images/232/35173152.jpg/
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I got the second mechanism but the first one I could not. I think a radical needs to be formed in position 9 and 10 but how I do not know.
http://imageshack.us/photo/my-images/232/35173152.jpg/
I don't think it's a radical, it may be 2 benzylic carbonium ions formed by loss of 2 x water. The ions then re-aromatize.
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I didn't look up a reference to this, but it is an iodide reduction. The other product is I2.
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Thank you for replying.
What is making me to think it is a radical because if ions are formed in position 9 and 10, how double bond will be formed
http://img40.imageshack.us/img40/8558/28239906.jpg
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Thank you for replying.
What is making me to think it is a radical because if ions are formed in position 9 and 10, how double bond will be formed
http://img40.imageshack.us/img40/8558/28239906.jpg
You have the same problem with radicals, somewhere there will be two left over which have to pick up H radicals.
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unfortunately, I could not find clue about the first mechanism but it looks like iodine reduction.
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unfortunately, I could not find clue about the first mechanism but it looks like iodine reduction.
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Silly question. Is a hypervalent iodine species being formed here with one of the alcohols, this then oxidizes the other position?
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How about:
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Thank you for replying,,,,,,
It is really interesting, is it possible that iodine in one side and oxidation the other side and what is the role of NaH2PO2.xH2O
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Abstraction of I+ by I- to form I2 (which is what I drew) is well known.
I actually didn't read the conditions properly and assumed it was KH2PO4 and was simply acting as an acid catalyst for SN1 in the first step of the mechanism I drew. Realising now that it's potassium hypophosphite and not -phosphate, I doubt my suggestion is correct since there is no acid present. If the KI is catalytic, and not a stoichiometric reductant, it may simply be that it reduces the I2 back to iodide to turn over the catalytic cycle.
I'll have a think about it and try to get back to you, hopefully someone else can shed some light on it...
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Abstraction of I+ by I- to form I2 (which is what I drew) is well known.
I actually didn't read the conditions properly and assumed it was KH2PO4 and was simply acting as an acid catalyst for SN1 in the first step of the mechanism I drew. Realising now that it's potassium hypophosphite and not -phosphate, I doubt my suggestion is correct since there is no acid present. If the KI is catalytic, and not a stoichiometric reductant, it may simply be that it reduces the I2 back to iodide to turn over the catalytic cycle.
I'll have a think about it and try to get back to you, hopefully someone else can shed some light on it...
This is a potassium salt of phosphonic acid. I don't quite see how it can help but there is a reaction called the Atherton-Todd reaction in which if you react such phosphorus compounds with CCl4 or other halogen sources in the presence of an alcohol they are oxidized to the next highest in this case a phosphonate.
Perhaphs we are getting an intermediate I-P(O)OHO-K+ which reacts with the alcohol to give RO-P(O)OHO-K+ (R = the rest of the anthracene) which is the leaving group - a phosphate!
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This is a potassium salt of phosphonic acid.
I don't think it is, potassium phosphonate is KH2PO3
In this reaction it is potassium hypophosphite KH2PO2.
All I can find is vague references to it being a reducing agent. I am also trying to shoe-horn in some kind Atherton-Todd or Appel-type mechanism but not coming up with much. For these sort of things you need to start with an oxidising agent like iodine or a carbon tetrahalide to get the dehydration going. With only KI as the other reagent I don't understand what the hypophosphite is doing.
I can believe AcOH would protonate the OH of the starting material and promote SN1 substitution with I-. Lose water again and you can abstract I+ with I-. The only think I can think of is that the hypophosphite scavenges the I2 to initiate an Atherton-Todd or Appel-like Iodide substitution mechanism.
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This is a potassium salt of phosphonic acid.
I don't think it is, potassium phosphonate is KH2PO3
In this reaction it is potassium hypophosphite KH2PO2.
All I can find is vague references to it being a reducing agent. I am also trying to shoe-horn in some kind Atherton-Todd or Appel-type mechanism but not coming up with much. For these sort of things you need to start with an oxidising agent like iodine or a carbon tetrahalide to get the dehydration going. With only KI as the other reagent I don't understand what the hypophosphite is doing.
I can believe AcOH would protonate the OH of the starting material and promote SN1 substitution with I-. Lose water again and you can abstract I+ with I-. The only think I can think of is that the hypophosphite scavenges the I2 to initiate an Atherton-Todd or Appel-like Iodide substitution mechanism.
Do you think that you get something like this going on? Compound 1 then iodinates the OH like other P ylids?
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Yeah, exactly. What I was thinking of is attached below, but it doesn't balance the iodine so I don't think it's right (but maybe on the right track).
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In this reaction it is potassium hypophosphite KH2PO2.
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The only think I can think of is that the hypophosphite scavenges the I2 to initiate an Atherton-Todd or Appel-like Iodide substitution mechanism.
Honestly, I hadn't noticed that it was a hypophosphite. I agree that it probably scavenges iodine. However, the stoichiometric excess is such that it is difficult to be certain. If it were catalytic in iodine, then as an iodine scavenger would make better sense. I think we may also surmise the stoichiometry has not been optimized. It appears as though the inorganic reagents are inexpensive and if some is good, more is better.
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Yes Dan, I was thinking along these lines but couldn't put it down as eloquently as you did!
I'll keep thinking about this little problem. ::)
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Thank you for your help, I appreciate that,,,,
Many Thanks