Grignard formation and vicinal dihalides

Question

So let's say I have a symmetric 1,2-dihalide (the halides are identical) and I wanted to make a Grignard and I only add 1 eq. of magnesium metal ...

Will I get 1 eq. of asymmetric Grignards (i.e. only one side of the product has the C-MgBr bond) as my product or will:

a) Some of the product be "double" Grignards - i.e. both sides of what used to be the 1,2-dihalide now have C-MgBr bonds?

b) I get some E2 elimination product from Grignards robbing protons from unreacted 1,2-dihalides?

c) A combination of the above?

d) Something else entirely?

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Let's say the above reaction works the way I want it to and we get the 1 eq. of asymmetric Grignard reagent.

We then react it with a carbonyl containing compound such as acetone ... do we get an epoxide? Are there even any routes to an epoxide? This suggested as a possible product ...

But I don't see the epoxide.

Answer 1

Taken from the Virtual Textbook of Organic Chemistry, which is maintained by William Reusch.

If two halogen atoms are present in a given compound, reactions with reducing metals may take different paths depending on how close the carbon-halogen bonds are to each other. If they are separated by four or more carbons, as in the first example below, a bis-organometallic compound may be formed. However, if the halogens are bonded to adjacent (vicinal) carbons, an elimination takes place with formation of a double bond.

He does however not provide any explanation for this behavior. It could be that this result is largely empirically based, about some aspects of the Gringard reaction mechanism there is not much known about (or consensus among chemists). I looked at Clayden et al. and Vollhardt & Shore and neither give a satisfying answer. If I find any in an other perhaps more advanced textbook, I'll edit this answer.