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I don’t understand how the reaction produces this product.

In the starting material, there is an $\ce{-OH}$ group attached to carbon 5 but where did the $\ce{-OH}$ group go in the product?

  • $\begingroup$ hydrolysis in step 1 produces acid, which condenses with -OH at C-5 to produce the lactone. $\endgroup$ – Aniruddha Deb Sep 22 '20 at 6:33
  • 3
    $\begingroup$ @AniruddhaDeb not hydrolysis. Deprotonation of the -OH and intramolecular cyclisation by nucleophilic attack. $\endgroup$ – Waylander Sep 22 '20 at 6:44
  • $\begingroup$ Just for the sake of completeness: the second step using TBAF is basically a deprotection of the silyl protective group. A tentative mechanism is that the fluoride will first add to the silicon to form a penta-coordinated intermediate, then the alkoxide departs, in the process yielding a Bu-SiF as a side product. The alcohol will be fianlly generated by quenching, as can be seen here $\endgroup$ – Yusuf Hasan Sep 22 '20 at 8:37
  • $\begingroup$ I suggest you to change the title in order to reflect the body of question. $\endgroup$ – Nilay Ghosh Sep 22 '20 at 15:12

This is straightforward enough: The $\ce{NaOMe}$ deprotonates the $\ce{OH}$ group on $\ce{C}$5, the alkoxide then does an intramolecular nucleophilic attack on the ester to give the lactone. This is favoured because it is intramolecular and forming a [6]-ring. It is possible that they may be some ester exchange of the $\ce{tBuO}$ group with the $\ce{MeO}$ group forming methyl ester, but either is more readily attacked by the internal nucleophile so the end result is the same.


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