# Grignard formation and vicinal dihalides

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?

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.

• Definitely can't be an epoxide... I'm pretty sure you'd just get ethylene from this reaction. The anion (corresponding to the Grignard) will just kick out the other bromide. – jerepierre Feb 12 '15 at 21:20
• If a ring closure would occur after having added $\ce{-CH2-CH2-Br}$ the product wouldn't be an epoxide but an oxetane. – Klaus-Dieter Warzecha Feb 12 '15 at 21:33