epoxide to ether

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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  • $\begingroup$ What if you used Grigniard reagent with chlorine present... $\endgroup$ – Mithoron Aug 1 at 23:10
  • $\begingroup$ 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. $\endgroup$ – Zhe Aug 1 at 23:18
  • $\begingroup$ it would solve it, so basically what can my R chain contain in gringard, i thought only hydrocarbons, it can contain halogens then? $\endgroup$ – Katia Aug 1 at 23:18
  • $\begingroup$ Yes, halogen (fluorine in particular) may be present, but it may complicate preparation of the reagent. $\endgroup$ – Mithoron Aug 1 at 23:57
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    $\begingroup$ Alternatively, you can incorporate some unsaturation into your Grignard reagent, or a different functional group that's protected. $\endgroup$ – Zhe Aug 2 at 0:59

Synthetic scheme

  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.

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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

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

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Waylander Aug 2 at 12:32
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    $\begingroup$ Actually, I'm not sure if you might actually just get spontaneous cyclisation during the deprotection. Would be a nice bonus. $\endgroup$ – orthocresol Aug 2 at 13:38
  • $\begingroup$ That should work $\endgroup$ – Waylander Aug 2 at 15:08

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.

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New contributor
ADITYA ROYCHOWDHURY is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ 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. $\endgroup$ – Safdar Aug 2 at 10:03
  • $\begingroup$ How does the magnesium not react with chlorine? Can we use $\ce{SCH3}$ at the far end instead? $\endgroup$ – Oscar Lanzi Aug 2 at 10:04
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    $\begingroup$ Do you have a reference for the preparation and use of 3-halomagnesium halides? $\endgroup$ – Waylander Aug 2 at 10:10
  • $\begingroup$ @Waylander pubs.acs.org/doi/pdfplus/10.1021/acs.jchemed.5b00453?src=recsys Not exactly for the above compound, but preferential reaction is observed. $\endgroup$ – ADITYA ROYCHOWDHURY Aug 2 at 11:50
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    $\begingroup$ 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. $\endgroup$ – Waylander Aug 2 at 12:13

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