# How can we synthesize this cycloester, starting with methyl 4-phenylbutanoate?

I was wondering how to realize following synthesis:

I am struggling with this one; for sure, this has something to do with the benzyl position which is particularly reactive, and something nucleophilic which can close the ring and form the cycloester. But how to do this? Thanks in advance.

Step 1 - benzylic bromination with NBS, Benzoyl peroxide

Step 2 - displace bromine with cyanide using KCN/DMSO

Step 3 - ester hydrolysis to acid with LiOH/THF/water

Step 4 - acid reduction to alcohol with BH3

Step 5 - nitrile hydrolysis with KOH

Step 6 - lactone formation with cat. PTSA/toluene reflux.

An alternative approach to the proposal of @Waylander for the synthesis of lactone 7 from ester 1 depends on the efficacy of the Barton nitrite photolysis, 2$$\rightarrow$$3. Examples of this reaction are typically conducted with secondary nitrite esters in rigid, steroidal frameworks with few degrees of freedom, an advantage that the transformation 2$$\rightarrow$$3 lacks. The generation of an alkoxy radical (8$$\rightarrow$$9) allows for 1,6-H transfer. In this case, the C-H bond has the advantage of being benzylic.

Hydrolysis of the oxime 3 affords the hydroxyketone 4 that may be in inconsequential equilibrium with its hemiketal 5. A Wittig reaction with (methoxymethylene)triphenylphosphane produces an enol ether that is a masked aldehyde group. Acid hydrolysis of the enol ether affords the hemiacetal 6, which can be readily oxidized by Jones reagent. In the event that a keto acid is produced upon oxidation, $$\ce{NaBH4}$$ reduction of the ketone and acid-catalyzed cyclization may be effected to form lactone 7.

• Neat! Silver Carbonate/Celite will do the lactol to lactone selectively if there is any problem with Jones reagent Sep 16 at 6:38
• @Waylander: I couldn't find 2-->3 in Chem. Absts. Who knows? Another Tet. Lett. for that ambitious young chemist? ;) Sep 16 at 13:16
• if only I still had a lab Sep 16 at 13:23
• @Waylander: Unfortunately, what we are doing currently is our default position. Sep 16 at 13:27