
I thought that the first step must be using a Grignard reagent to add carbons, but I need a hydroxyl group on the carbon on the other side to make it a diol and produce the product given, but nothing seems to be working.
How do I proceed?
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Sign up to join this communityI thought that the first step must be using a Grignard reagent to add carbons, but I need a hydroxyl group on the carbon on the other side to make it a diol and produce the product given, but nothing seems to be working.
How do I proceed?
Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.
Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol.
Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.
Remove the THP group using cat. PTSA in MeOH
Form the alkoxide using LiHMDS in Et2O and cyclisation will occur to give the pyran. Isolate by careful distillation.
For preparation of the Grignard reagent, use allylmagnesium bromide.
$$\ce{CH2=CH-CH2-Br ->[Mg][ether] CH2=CH-CH2-Mg+Br-}$$
Now we react the Grignard reagent formed with ethylene oxide, our starting compound.
$$\ce{C2H4O ->[CH2=CH-CH2-Mg+Br-][Et2O] HO-CH2-CH2-CH2-CH=CH2}$$
This gives us pent-4-en-1-ol. Now we protect the alcoholic group by esterification, using trifluoroacetic anhydride ($\ce{(CF3CO)2O }$).
$$\ce{HO-CH2-CH2-CH2-CH=CH2 ->[(CF3CO)2O][Pyridine] CF3-COO-CH2-CH2-CH2-CH=CH2}$$
Now, we use $\ce{HBr/R2O2}$ to form 5-bromopentyl 2,2,2-trifluoroacetate.
$$\ce{CF3COO-CH2-CH2-CH2-CH=CH2 ->[HBr/R2O2] CF3COO-CH2-CH2-CH2-CH2-CH2-Br}$$
Now we remove the trifluoroacetate protecting group to get 5-bromopentanol.
$$\ce{CF3COO-CH2-CH2-CH2-CH2-CH2-Br ->[K2CO3/MeOH] HO-CH2-CH2-CH2-CH2-CH2-Br}$$
Now, under basic conditions, the compound undergoes intramolecular SN2 to form the six member cyclic ether
$$\ce{HO-CH2-CH2-CH2-CH2-CH2-Br ->[OH-] C6H10O}$$
Just summarizing the steps already written in the comments,
One equivalent magnesium will react with iodine preferentially due to it being a better leaving group than chlorine. The bond also has more covalent character.
Subsequent step of reaction will open the oxirane ring easily.
Then hydrolysis produces 5-chloropentan-1-ol.
This is followed by intramolecular SN2 attack by lone pair on oxygen of hydroxide at the chlorine carbon causing cyclized product formation.