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What does $\ce{Cu}$ do? And why do we need it? Can we not do $\mathrm{S_N}2$ without it? And is there any other metal that can help in this? Maybe something like $\ce{Ag}$? I'm guessing this because the special thing about $\ce{Cu}$ is that it is worse reducing agent than $\ce{H}$.

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Copper is used to avoid lithium-halogen exchange. Additional information on that reaction can be found in this discussion. Organolithiums will often, instead of acting as a nucleophile, form a new organolithium: $$\ce{RLi + R'X <=> RX + R'Li}$$ Using $\ce{Cu(I)}$ allows the formation of an organocuprate, which acts as a nucleophile, allowing alkylation rather than metallation: $$\ce{2RLi + CuX -> LiR2Cu + LiX}$$ $$\ce{LiR2Cu + R'X -> R'R + RCu + LiX}$$ I'm not sure if silver could be used instead; I'm not seeing any reference for it, but I may just not be searching the right terms. The Wikipedia page on organosilver chemistry isn't terribly extensive, but I'm guessing stoichiometric use of silver, while not terribly expensive as transition metals go, would certainly be more so than analogous use of copper.


Update:

It seems that analogous chemistry cannot be performed with silver or gold. Lithium diorganoargentates ($\ce{LiR2Ag}$) are apparently too unstable. Lithium diorganoaurate ($\ce{LiR2Au}$) complexes are apparently either not good nucleophiles, or the triorganogold(III) ($\ce{R2AuR'}$) intermediate complexes are too stable to perform similar chemistry. This is per this 2005 paper. A fair amount of gold chemistry has come out since then, so I'm not sure how much of this has changed since.


Second update (though I should have answered this to begin with):

Also, I'd like to add that this is not an $\text{S}_\text{N}2$ reaction. The organocuprate undergoes oxidative addition into the $\ce{R'X}$ bond. After that, it loses the $\ce{X^-}$ ligand to form $\ce{LiX}$ and $\ce{R2CuR'}$, then undergoes reductive elimination to form $\ce{R'R + CuR}$.

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  • $\begingroup$ I see, that helps, but why does RLi undergo that reaction? Isn't R-Li a strongly polarized bond? What kind of bond is it actually? And what reaction is this undergoing? $\endgroup$ – Shodai Mar 7 '16 at 8:40
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    $\begingroup$ The Li-X exchange mechanism isn't really known for sure yet. This presentation from the MacMillan group talks about three proposed mechanisms: radical exchange, nucleophilic attack on the halogen, and a four-centered intermediate. I would guess it doesn't undergo normal nucleophilic attack due to hard-soft acid-base chemistry, i.e. poor orbital overlap. I'm not totally sure, though. $\endgroup$ – SendersReagent Mar 7 '16 at 16:25
  • $\begingroup$ I've updated my answer to include a link to another Chemistry.SE answer about lithium-halogen exchange. $\endgroup$ – SendersReagent Mar 7 '16 at 19:41

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