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I discussed the following today:

  1. If adding an iodine crystal to a Grignard to make the reaction go. will this tiny crystal force some of the metal or reagent to be used up. ergo, therefore, there is a lower yield (I know it is microscopic change, this is not the question)

It is more about if it takes some of the metal and uses it up, maybe forming Mg-Iodide, so it cannot be used in the Grignard reaction, and I am not sure if Mg-Iodide cannot be used or if it actually catalyzes the reaction.

It is in the first step of forming phenyl-Magnesium bromide (or other reagent of the same type) reagent for use in step 2 of Grignard. I write bromide on purpose, so you can see that it is another halide than iodide.

Could it be the iodine catalyzes the reaction, or actually, some of the phenylmagnesium bromide is turned to phenyl magnesium iodide, which can also be used in the reaction? Ergo there is no loss in yield at all?

Imagine for step 1, forming only the Grignard reagent, I used iodine equivalents to 10% of, for example, of bromobenzene, and used Mg (not in excess), but the exact amount of Mg equal to bromobenzene, as 1:1 ratio, and the reaction normally went to 100% yield (it does not) but imagine it. Would I lose 10% from the iodine crystals?

  1. can you tell me what the iodine crystals forms in the reaction and will it able to be used in the reaction or not?
  2. will it catalyze the reaction instead?

Will you show me, with drawings, the intermediates or products of the iodine crystal?

Will it interfere with the yield, as it takes some of the magnesium and forms most likely $\ce{Mg-I}$ that is not used in the reaction, or is it I am asking you? It doesn't give any hints regarding lowering of yield. Will it help catalyze or form Grignard reagent with iodide? Why?

A pointer to what I am actually asking is. what happens to the iodine that is not used to remove hydroxide or etch the surface then what comes out of that will it continually as catalytic etch give $\ce{Mg-I}$ then Grignard reagent with bromide and back or what happens with it? will it use op the Mg turnings lowering yield.

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    $\begingroup$ Note: 1) Caps Lock = SHOUTING. Need to avoid that 2) Spelling Mistakes: "Reagent", "Grignard". $\endgroup$ – Nilay Ghosh Feb 11 at 4:04
  • $\begingroup$ Since normal practice is to use excess magnesium, why would a small amount of iodine affect the yield? $\endgroup$ – matt_black Feb 11 at 12:11
  • $\begingroup$ because the question got hijacked and they added a thump upon Maurice's answer, and a thump down on my question, I did shout, it happens every time because someone it only answering for points, and this is a serious question for someone in Metal-coupling reactions. it still amazes me that the whole question got ruined by these since it was a good question I asked! now nobody wants to answer because of the negative 1 one my question and I have to ask it again somewhere else $\endgroup$ – user53617 Feb 11 at 16:19
  • $\begingroup$ if there is a spelling mistake and that is all you care about then you need to refocus yourself! there are also comma and period mistakes. $\endgroup$ – user53617 Feb 11 at 16:19
  • $\begingroup$ matt_black. this I already know. try to look at the example in the text and then what can you tell me, and are you saying then that it will lower yield but it doesn't happen in practice because there is still excess Magnesium turnings in the bottle? all I want to know is "does Iodine lower the yield if magnesium is in a 1:1 ratio with an organic halide and they normally react 100% with each other (they don't I know but imagine it)" $\endgroup$ – user53617 Feb 11 at 16:25
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No. Iodine is used to help magnesium react with the bromoderivate. If pure magnesium is added to a solution containing a bromoderivate, it should react quickly. But sometimes it does not react at all, because magnesium is covered by a thin and protective layer of magnesium oxide. Added iodine will etch this layer and the underlying metal. This reaction breaks the oxide layer, so that the metal can get in contact with the bromoderivate, and quickly react.

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  • $\begingroup$ Iodine does not diffuse through the passive film, it reacts. Magnesium oxide, or the hydroxide if it reacts with moisture, is basic and will thus [disproportionate the iodine to form relatively soluble iodide and iodate salts] (researchgate.net/figure/…). We can use iodine because with magnesia being a relatively strong base, we do not need a more aggressive halogen to react. $\endgroup$ – Oscar Lanzi Feb 10 at 21:12
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    $\begingroup$ PLSS UNDERSTANDS SINCE YOU ARE ANSWERING WHAT IT IS USED FOR, AND NOT MY QUESTION: WILL IT INTERFERE WITH THE YIELD, AS IT TAKES SOME OF THE Mg AND FORMS MOST LIKELY Mg-I THAT IS NOT USED IN THE REACTION, OR IS IT IAM ASKING YOU? because I already know what iT is used for. read the Whole thing the question at the bottom doesn't give a hint of anything else then will it lower the yield. will it or will it help BY CATALYSE OR FORM GRIGNARD REAGENT WITH IODIDE and why? $\endgroup$ – user53617 Feb 10 at 22:44
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    $\begingroup$ I cannot approve that as an answer like I did not know why we use it and how on earth did you guys misunderstand something very specific and complex, hard to know unless you worked with metal-coupling reactions to something anyone can find on the internet as to why we use Iodine hmm... point-hunters $\endgroup$ – user53617 Feb 10 at 22:49
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    $\begingroup$ @user53617 I repeat once again, don't use Caps lock. It is equivalent to shouting. The tone in your comment is very strong. Please be careful. $\endgroup$ – Nilay Ghosh Feb 11 at 4:40
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    $\begingroup$ @user53617 Maybe some magnesium is lost due to iodine having reacted with it. But this amount is small. And usually in Grignard synthesis metallic magnesium is added in excess, to compensate for this effect. $\endgroup$ – Maurice Feb 12 at 10:26

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