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I have managed to do the mechanism for the first step, but have no clue how 4 equivalents of MeLi are used for the second step of the reaction.

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The first eq. of MeLi deprotonates the alcohol. This does not cyclise as it would form a 7-membered ring and that is not strongly favoured.

The subsequent eqs of MeLi remove the methyl carbamate group as described by Sawyer, et al., by consecutive additions of Me to give first the N-acetamide then the amine anion plus acetone. The N anion cyclises onto the chloride to give the product. The final eq. of MeLi adds to the acetone to give Li t-butoxide which gives t-butanol C on workup.

J. S. Sawyer, B. A. Narayanan, Synth. Comm., 1983, 13, 135.

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