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Posted

Not sure if this is the right place to post this.

 

If you have an addition equation with a molecule containing a C=C bond and the COOH group is the C=C broken or is the COOH group removed?

 

e.g

 

...C=C...COOH + Br2

 

is the C=C bond broken and the Br2's added or does one Br stick to the end of the chain removing the COOH and producing CO2 + HBr

Posted

Not sure with the Br2 - Perhaps Hermantrude can help you - he's good at this stuff.

 

If it was HBr though - I would say you should look at the "Markovnikov rule". I think that would suggest that the H from the HBr would go to the Carbon from the double bond which origionally had the most Hydrogens already - and the Br would go to the Carbon nearest the COOH..... although I would wait for Hermantrude (or someone elase who knows what they are talking about) to reply before taking my word for it.

Posted

Neither. You will get addition of both halves of the Br2 across the double bond, which makes a vicinal-dibromide (it is an anti addition if you are working with a ring system). If you want to pluck the COOH group off, you want to look up the Hunsdiecker reaction, which requires silver carboxylate salts to work. If the C=C group is around, you'll need to convert it into a nonreactive group that can be returned to a C=C bond later unless you won't mind both reactions happening simultaneously.

 

By C=C bond broken, Im not sure what you mean. Usually this means, total scission of the carbon carbon bonds when I've heard it, but it's possible that you meant an addition of bromine across the double bond.

Posted

Thank you UC - now I've got home, thought about it some more and discussed it with the wife - I was about to say that the Br will add across both sides of the double bond. I think that's what dg means. As you stated - differing reaction conditions and addition of catalysts may give something else (like replacement of the acid group or OH)

Posted

sorry to confuse you yes i meant Br will add across both sides of the double bond.

 

basically a fatty acid of 14C with a C=C & a carboxylic group. it reacts with Bromine (addition reaction) with a catalyst.

 

So I guess the Bromine is just added & the product becomes: C14H26O2Br2. The full formula would include the - bonds right?

Posted

It won't need any catalyst at all :)

 

This is done on a massive industrial scale with intact fatty acid triglycerides (veggie oil) to make brominated vegetable oil. The bromine addition makes the oil more dense as more is added, and they stop when it is exactly the same density as the drink it's being put into. That way, the tiny droplets of oil don't float or sink and the drink is "cloudy" looking which is desirable in some cases.

 

If you can get any bromine, you can do this yourself by adding a drop or two to a little veggie oil and swirling it. The brown-red color vanishes as the Br2 adds across the double bonds in the oil. Eventually, you'll run out of double bonds though.

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