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Topic: BrCH2COBr  (Read 6213 times)

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Offline fran008

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BrCH2COBr
« on: January 15, 2006, 12:07:11 PM »
I have to produce (on paper!) CHBrCH2COBr from CH3CH2Br

The method I thought of was to convert to ethene then brominate and then turn to the aldehyde by using K2Cr2O7 in acid at a low temperature. But would that keep it from becoming BrOCCOBr?

Is my method wrong? Any help?

Offline FeLiXe

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Re:BrCH2COBr
« Reply #1 on: January 15, 2006, 01:29:32 PM »
you product has three Cs your educt two, so you have to a carbon atom in the process.

my suggestion:

Grignard for the acid
CH3CH2Br + Mg -> CH3CH2MgBr
+ CO2 -> CH3CH2COOMgBr
+ H20 -> CH3CH2COOH

Hell-Vollhardt-Zelinsky for the bromo-acid-bromide
+PBr + Br2 -> CH3-CH2Br-COBr

I just noticed that's not your substance. anyway that's what I can think about right now.

By the way: What substance are you actually producing? The second C has 5 bonds if it has a double bond to the left C
« Last Edit: January 15, 2006, 01:30:42 PM by FeLiXe »
Math and alcohol don't mix, so... please, don't drink and derive!

Offline fran008

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Re:BrCH2COBr
« Reply #2 on: January 16, 2006, 12:33:34 PM »
Oh I messed it all up! Typical! I cant copy well

It is from CH3CH2Br ------> BrCH2COBr!

sry :s

Offline fran008

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Re:BrCH2COBr
« Reply #3 on: January 19, 2006, 11:28:58 AM »
Hi! I managed to answer my own difficulty! Arent I good? So for anyone intrested, the bromoethane is converted to ethanol, which is then oxidised to a carboxyllic acid and then finally with PBr3 to give BrCH2COBr!

Regards

fran008

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