# Synthesing tetrahydropyran from ethylene oxide

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?

• What if you used Grigniard reagent with chlorine present... – Mithoron Aug 1 '20 at 23:10
• Think about how many carbons you need to introduce. In addition, you need a some kind of functional group handle so that you have something to work with. – Zhe Aug 1 '20 at 23:18
• it would solve it, so basically what can my R chain contain in gringard, i thought only hydrocarbons, it can contain halogens then? – Katia Aug 1 '20 at 23:18
• Yes, halogen (fluorine in particular) may be present, but it may complicate preparation of the reagent. – Mithoron Aug 1 '20 at 23:57
• Alternatively, you can incorporate some unsaturation into your Grignard reagent, or a different functional group that's protected. – Zhe Aug 2 '20 at 0:59

1. Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.

2. 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.

3. Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.

4. Remove the THP group using cat. PTSA in MeOH

5. 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

$$\ce{HO-CH2-CH2-CH2-CH2-CH2-Br ->[OH-] C6H10O}$$

• – Waylander Aug 2 '20 at 12:32
• Actually, I'm not sure if you might actually just get spontaneous cyclisation during the deprotection. Would be a nice bonus. – orthocresol Aug 2 '20 at 13:38
• That should work – Waylander Aug 2 '20 at 15:08

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.

• I would suggest that instead of I, you might want to use Br. It is just a thought since Br would still satisfy the conditions and it seems to be more prevalently used. – Safdar Aug 2 '20 at 10:03
• How does the magnesium not react with chlorine? Can we use $\ce{SCH3}$ at the far end instead? – Oscar Lanzi Aug 2 '20 at 10:04
• Do you have a reference for the preparation and use of 3-halomagnesium halides? – Waylander Aug 2 '20 at 10:10
• @Waylander pubs.acs.org/doi/pdfplus/10.1021/acs.jchemed.5b00453?src=recsys Not exactly for the above compound, but preferential reaction is observed. – Aditya Roychowdhury Aug 2 '20 at 11:50
• I think the route falls over at that point. AFAIK you cannot form a Grignard of 3-halo propyl halides and get a standard addition. There is scope for all kinds of alternative reactions - cyclopropane formation possibly. – Waylander Aug 2 '20 at 12:13