# Organic chem mechanism [closed]

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?

• hydrolysis in step 1 produces acid, which condenses with -OH at C-5 to produce the lactone. – Aniruddha Deb Sep 22 '20 at 6:33
• @AniruddhaDeb not hydrolysis. Deprotonation of the -OH and intramolecular cyclisation by nucleophilic attack. – Waylander Sep 22 '20 at 6:44
• 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 – Yusuf Hasan Sep 22 '20 at 8:37
• I suggest you to change the title in order to reflect the body of question. – 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.