How tBuLi Is Consumed During Exchange Instead of Accumulating | Visual AI Learning Guide

So why does tBuLi tolerate two equivalents? Because both equivalents have a job. The first one does the exchange: it takes the bromine off the aryl ring and leaves you with the aryllithium you wanted.

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When tBuLi performs Li–Br exchange with an aryl bromide, the reaction consumes two equivalents, and both are accounted for:

First equivalent — exchange: tBuLi attacks the bromine of ArBr, forming ArLi and tBuBr.
\[ \text{tBuLi} + \text{ArBr} \rightarrow \text{ArLi} + \text{tBuBr} \]
Second equivalent — dehydrohalogenation: the tBuBr formed is a tertiary alkyl bromide, which undergoes rapid E2 elimination with the remaining tBuLi to give isobutene and LiBr.
\[ \text{tBuLi} + \text{tBuBr} \rightarrow \text{isobutene} + \text{LiBr} + \text{isobutane} \]

This second step is fast even at low temperature because tertiary alkyl halides are excellent substrates for E2 elimination, and tBuLi is a strong base. The result is that no free tert-butyl nucleophile survives in the mixture. The by-product is isobutene, a neutral alkene that cannot attack the carbene carbon.

Contrast this with nBuLi: the nBuBr formed in the exchange is a primary alkyl bromide, which is a poor substrate for E2 elimination. It persists in solution as an intact electrophile that can react with the carbene.