[Rutherford]

What is the mechanism of the transformation from the upper compound in the presence of MeLi to the propellane?

I thought that the two -CH₂-Cl carbons lose a proton and attack the -CBr₂ carbon. But how are the two chlorines removed?

[Dan]

Look up lithium-halogen exchange reactions for some inspiration!

[Rutherford]

So lithium exchanges with the two chlorines, and then the product is hydrolyzed to produce the propelanne, right?

[Dan]

Can you explain where you think a hydrolysis happens?

Note the rate of Li-X exchange X = I>Br>Cl

[Rutherford]

The hydrolysis happens at the end, after the two bromines are removed through nucleophilic attacks of carboanions.

[Enthalpy]

Hey, propellane is one way to synthesize tricyclo[1,1,1]pentane, which is one the the best possible rocket fuels!
As soon as you've filled your test tube, could you produce a few 100t please? 

[Enthalpy]

For my personal information:

Why are bromo-chloro- reactants so common, rather than all-chloro or all-bromo? Are they (a1) easier to synthesize? Or do they (a2) undergo better controlled reactions?

When LiMe reacts with a thing-Cl, (b1) why isn't thing-Me produced? Does it (b2) need LiMe to be much more abundent than thing-Cl? Or (b3) being a tetramer, LiMe brings at once all the necessary lithium, so that high concentration isn't essential?

With the drawn reactant, once an intermediate dibromo-chloro-lithium-thing is obtained, (c1) why does it eliminate LiBr instead of LiCl? Or (c2) is the spiropentane produced as well?

What would you think of using low-pressure gaseous (d1) K, Cs, Rb or (d2) Zn, Cd instead of LiMe for this propellane synthesis?

Thank you!